Solution Manual for Engineering Mechanics Statics and Dynamics 2nd Edition by Plesha

Solutions Manual

30

Chapter 2 Solutions Problem 2.1

E

jE j

If vavg is the average velocity of a point P over a given time interval, is vavg , the magnitude of the average velocity, equal to the average speed of P over the time interval in question? Solution

jE j

In general, vavg is not equal to v avg. To see this, consider a car that drives along a loop of length  L over a time interval  t such that the departure and arrival points coincide. Since the departure and arrival positions coincide, vavg is equal to zero. This, implies that vavg is also equal to zero. By contrast, the average speed will be different from zero becaus e it is equal to the ratio  L=t .

E

jE j

This solution s manual, in any print or electronic form, remain s the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplicat ion or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

June 25, 2012

Full file at http://testbankwizard.eu/Solution-Manual-for-Engineering-Mechanics-Statics-and-Dynamics-2nd-Edition-by-Plesha

Dynamics 2e

31

Problem 2.2 A car is seen parked in a given parking space at 8:00 A . M . on a Monday morning and is then seen parked in the same spot the next morning at the same time. What is the displacement of the car between the two observations? What is the distance traveled by the car during the two observations?

Solution

The displacement is equal to zero because the difference in position over the time interval considered is equal to zero. As far as the distance traveled is concerned, we cannot determine it from the information given. To determine the distance traveled we would need to know the position of the car at every time instant during the time interval considered instead of just at the beginning and end of the time interval in question.


June 25, 2012


Solutions Manual

32

Problem 2.3

E

Is it possible for the vector v shown to represent the velocity of the point P ?

Solution

E

No, because the vector v shown is not tangent to the path at point P , which it must.


June 25, 2012


Dynamics 2e

33

Problem 2.4

E

Is it possible for the vector a shown to be the accelerati on of the point P ?

Solution

E

No, because a does not point toward the concave side of the trajectory of P , which it must.


June 25, 2012


Solutions Manual

34

Problem 2.5 Two points P and Q happen to go by the same location in space (though at different times). (a) What must the paths of P and Q have in common if, at the location in question, P and Q have identical speeds? (b) What must the paths of P and Q have in common if, at the location in question, P and Q have identical velocities?

Solution

Part (a) In the first case, what we can expect the paths to shar e is that point in space which, at dif ferent instants, is occupied by P and Q . Part (b) In the second case, the pat hs in question wil l not only share a point, lik e in the previous case, but will also have the same tangent line at that point, since the velocity vector is always tangent to the path.


June 25, 2012


Dynamics 2e

35

Problem 2.6

D

r The position of a car traveling between two stop signs along a straight city block is given by Œ9t  .45=2/ sin.2t=5/ ç m, where t denotes time (in seconds) , and where the argument of the sine function is measured in radians. Compute the displacemen t of the car between 2:1 and 3:7 s, as well as between 11:1 and 12:7 s. For each of these time intervals compute the average velocity. STOP

STOP

Solution

We denote the quantities computed between 2:1 and 3:7 s by subscript 1, and between 11:1 and 12:7 s by subscript 2 . Using the definition of displacement, we have

E D Œr.3:7 s/

r1



O

r.2:1 s/ç ur

and

E D Œr.12:7 s/

r2



O

r.11:1 s/ç ur :

(1)

Applying the definition of average velocity we have

E

.vavg/1

D r.3:73:7ss/

 

r.2:1 s/ ur 2:1 s

O

and

E

.vavg/2

s/ D r.12:7 12:7 s

 

r.11:1 s/ ur : 11:1 s

O

(2)

Using the expression for r.t/ in the problem statement, the expressions in Eqs. (1) and (2) can be evaluated to obtain

E D 8:747 uO r m

and

r2

D 5:467 uO r m=s

and

.vavg/2

r1

E D 13:73 uO r m;

and

E

.vavg/1

E

D 8:579 uO r m=s:


June 25, 2012


Solutions Manual

36

Problem 2.7 A city bus covers a 15 km route in 45 min. If the initial departure and final arrival points coincide, determine the average velocity and the average speed of the bus over the entire duration of the ride. Express the answers in m =s.

Solution

Since the departure and arrival points coincide, the displacement vector over the duration of the ride is equal to zero. This implies that the average velo city of the bus over the duration of the ride is equal to zero:

E D 0E:

vavg

Letting d denote the total distance traveled by the bus and letting  t denote the time to travel the distance d , the average speed over the duration of the ride is

vavg Since d

D 15 km D 15

 103 m and

t

D dt :

(1)

D 45 min D 2700 s, we can evaluate the above expression to obtain v D 5:556 m=s: avg


June 25, 2012





Solutions Manual

40

Problem 2.11 The position of a car as a function of time t , with t > 0 and expressed in seconds, is

E D Œ.5:98t 2 C 0:139t 3

O C .0:523t 2 C 0:0122t 3 Determine the velocity, speed, and acceleration of the car for t D 15 s. r.t/



0:0149t 4 / {



0:00131t 4 / | ç ft:

O

Solution

The velocity is obtained by taking the derivative of the position with respect to time. This gives

E D Œ.11:96t C 0:4170t 2

v



0:05960t 3 / {

O C .1:046t C 0:03660t 2



0:005240t 3 / | ç ft=s:

(1)

0:005240t 3 /2 ft=s;

(2)

O

The speed is the magnitude of the velocity . Using Eq. (1), we have

v

D

q

.11:96t

C 0:4170t 2



0:05960t 3 /2

C .1:046t C 0:03660t 2



which can be simplified to

v

Dt

p

144:1

C 10:05t



1:261t 2  0:05009t 3

C 0:003580t 4 ft=s:

(3)

The acceleration is computed by taking the derivative of the velocity with respect to time. Using Eq. (1), we have a Œ.11:96 0:8340t  0:1788t 2 / { .1:046 0:07320t  0:01572t 2 / | ç ft=s2 : (4)

ED

C

Evaluating Eqs. (1), (3), and (4) for t

E

v.15 s/

OC

D 15 s, we have

D .72:08 {O C 6:240 |O/ ft=s;

v.15 s/

C

D 72:34 ft=s; aE.15 s/ D

O


2

O C 1:393 |O/ ft=s :

.15:76 {

June 25, 2012


Dynamics 2e

41


E D Œ12:3.t C 1:54e 0:65t / {O C 2:17.t C 1:54e 0:65t / |Oç m:

r.t/

Find the difference between the average velocity over the time interval0  t  2 s and the true velocity 1 s. Repeat the calcula tion for the time interval computed at the midpoint of the interval, i.e., at t 8 s  t  10 s. Explain why the difference between the average velocity and the true velocity over the time interval 0  t  2 s is not equal to that over 8 s  t  10 s.

D

Solution

The velocity is obtained by taking the derivative of the position with respect to time. This gives

E D Œ12:30.1 1:001e0:6500t / {O C 2:170.1 1:001e0:6500t / |Oç m=s: Using Eq. (1), for t D 1 s we have vE.1 s/ D .5:872 {O C 1:036 |O / m=s: The average velocity over the time interval 0 t 2 s, which we will denote by .vE /1 , is .vE /1 D rE .2 s/ rE.0/ D .5:410 {O C 0:9545 |O / m=s: 2s Letting  vE1 D . vE /1 vE.1 s/, using the results in Eqs. (2) and (3), we have vE1 D .0:4623 {O C 0:08155 |O / m=s: Using Eq. (1), for t D 9 s we have vE.9 s/ D .12:26 {O C 2:164 |O / m=s: The average velocity over the time interval 8 s t 10 s, which we will denote by .vE /2 , is rE.10 s/ rE.8 s/ .vE /2 D D .12:26 {O C 2:163 |O/ m=s: v







(2)

avg



avg

avg

(1)



(3)









(4)

avg



avg

E D .vE

Letting  v2

avg/2 

(5)

2s

E

v.9 s/, using the results in Eqs. (4) and (5), we have v2

E D

.0:002550 {

O C 0:0004499 |O/ m=s:

E ¤ E

We observe that v1  v2 . This is due to the fact that, in general, the approxima tion of the true velocity by the average velocity over a given time interval is a function of the time interval in question, i.e.,  v changes depending on the interval on which it is computed. This solution s manual, in any print or electronic form, remain s the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplicat ion or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

E

June 25, 2012


Solutions Manual

42


E D Œ.66t

r.t/



O C .1:2 C 31:7t

120/ {



8:71t 2 / | ç ft:

O

If the speed limit is 55 mph, determine the time at which the car will exceed this limit.

Solution

In order to solve the problem we need to determine the speed of the car. So, we first determine the velocity of the car and then we compute its magnitude. The velocity is found by taking the time deriv ative of the position. This yields,

E D Œ66 {O C .31:7

v



O

17:42t/ | ç ft=s:

(1)

The speed is the magnitude of the velocity . Using Eq. (1), we have

v

D

v

D


q

662

C .31:7

5361  1104t

17:42t/ 2 ft=s;

(2)

C 303:5t 2 ft=s: 55 mph D 80:67 ft=s, and solving for t we have ) 5361 1104t C 303:5t 2 D .80:67 ft=s/2:

5361  1104t

p

Setting the speed in Eq. (3) equal to the speed limit

p



C 303:5t 2 ft=s D 80:67 ft=s



(3)

(4)

The second of Eqs. (4) is a second order algeb raic equation in t with the following two roots:

t

D

0:8427 s

and

t

D 4:482 s:

(5)

Since t > 0 we can only accept the second of the two roots in Eq. (5). Therefore, we conclude that the car will exceed the given speed limit at

t

D 4:482 s:


June 25, 2012


Dynamics 2e

43


O C .1:2 C 31:7t 8:71t 2/ |Oç ft: Determine the slope  of the trajectory of the car for t 1 D 1 s and t 2 D 3 s. In addition, find the angle  between velocity and acceleration for t 1 D 1 s and t 2 D 3 s. Based on the values of  at t 1 and t 2 , argue E D Œ.66t

r.t/



120/ {



whether the speed of the car is increasing or decreasin g at t 1 and t 2 .

Solution

Since the velocity is always tangent to the path, the angle  can be computed by finding the velocity and then determining the orientation of the velocity relative to the horizontal direction. The velocity is the time derivative of the position. Differentiating the given expression for the position with respect to time, we have

E D Œ66 {O C .31:7

v



O

17:42t/ | ç ft=s:

(1)

The orientation  of the velocity vector can be computed as:



D tan1

v  y

vx

;

(2)

where, referring to Eq. (1),

vx

D 66 ft=s

vy

and

D .31:7



17:42t/ ft=s:

Substituting Eqs. (3) into Eq. (2) and evaluating the corresponding expression for t t 2 3 s, we have

D D

1

D 12:21 ı

and

2

D

ı

17:30

(3)

t

D t1 D 1 s and

;

(4)

where  1 and  2 are the values of  at t 1 and t 2 , respectively. To determine the angle  , we first determine the acceleration as the time derivative of the velocity. Differentiating Eq. (1) with respect to time gives

ED

a

2

O ft=s :

17:42 |

(5)

The angle  is obtained as



    31:7/ D cos1 vEvE aaEE D cos1 17:4217:42.17:42t ; 662 C .31:7 17:42t/ 2 



ˇˇ ˇ

p

(6)



where we have used Eqs. (1) and (5). Denoting by  1 and  2 the values of  at times t 1 and t 2 , respectively, Eq. (6) gives

1

D 102:2 ı

and

2

D 72:70 ı:

(7)

90ı , at

t 1 the acceleration has a component opposite to the velocity . This indicates that the speed Since  1 > of the car is decreasing at t 1 . Since  2 < 90ı , at t 2 the acceleration has a component pointing in the same direction as the velocity. This indicates that the speed of the car is increasing at t2 . This solution s manual, in any print or electronic form, remain s the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplicat ion or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

June 25, 2012


Solutions Manual

44

Problem 2.15

Œt { .2 3t 2t 2 / | ç m describe the motion of the point P relative to the Cartesian frame Let r y.x/ for the trajectory of P for of reference shown. Determine an analytic expression of the type y 0  t  5 s.

ED OC C C

O

D

Solution

The position of P is given as

E D x.t/ {O C y.t/ |O;

r.t/ where

x.t/

D tm

and

y.t/

(1)

D 2 C 3t C 2t 2 m:

(2)

Solving the first of Eqs. (2) with respect to time, we have

t

D x:

(3)

Substituting Eq. (3) into the second of Eqs. (2), we obtain:

y.x/

D 2 C 3x C 2x 2 m:

(4)

Now we observe that the coordinate x is an increasing funct ion of time. Therefore, the range of x covered for  5 s is determined by computing the value of x corresponding to t 0 and t 5 s. Using the first of Eqs. (2), we have x.0/ 0 and x.5 s/ 5 m: (5)

D

0t

D

So, the trajectory of P for 0



t



y.x/

D

D

5 s is given by

D 2 C 3x C 2x 2 m

for

0x



5 m:


June 25, 2012


Dynamics 2e

45

Problem 2.16

Œt { .2 3t 2t 2 / | ç ft describe the motion of a point P relative to the Cartesian frame of Let r reference shown. Recalling that for any two vectors p and q we have that p q p q cos ˇ , where ˇ is the angle formed by p and q , and recalling that the velocity vector is always tangent to the trajectory, determine the function  .x/ describing the angle between the acceleration vector and the tangent to the path of P .

ED OC C C O E E

E

E

E E D j EjjEj 

Solution

The velocity vector is the time derivative of the position vector:

E D Œ1 {O C .3 C 4t/ |Oç ft=s:

v

(1)

The acceleration vector is the time derivative of the velocity vector. Therefore, differentiating both sides of Eq. (1) with respect to time, we have a 4 | ft =s2 : (2)

ED O

As stated in the problem,  is the angle between the acceleration vector and the tangent to the path of P . Since v is always tangent to the path,  can be computed as the angle formed by the vectors a and v , which is

E

E



 vE aE 

D cos1 vE aE 

ˇˇ ˇ

From Eqs. (1) and (2) we have that

jEvj D

q

12

C .3 C 4t/2



ft=s

D

:

(3)

p10 C 24t C 16t  2

ˇ ˇ ˇ Eˇ

EE E

E

ft=s and

2

jEaj D 4 ft=s :

(4)

Substituting the expressions for v , a, v , and a into Eq. (3), we have

 Since we have that r x

C 4t D cos1 p10 C3 24t C 16t 2





:

(5)

D x D t ft, we can replace t with x in the last of Eqs. (5) to obtain:  .x/

D cos1 p10 C324x C 4xC 16x 2 :


June 25, 2012


Solutions Manual

46

Problem 2.17

E D 2pt {O C D 3 s.

The motion of a point P with respect to a Cartesian coordinate system is described by r 4 ln.t 1/ 2t 2 | ft, where t denotes time, t > 0, and is expressed in seconds. Determine the angle  formed by the tangent to the path and the horizontal direct ion at t



C C

 O

Solution

Since the velocity is always tangent to the path, we can find the angle  by determining the angle formed by the velocity vector and the horizontal direction. The velocity is the time derivative of the position. Hence, differentiating the position with respect to time, we have

E D p1t {O C 4 t C1 1 C t |O

v Evaluating Eq. (1) at t







ft=s:

(1)

D 3 s, we have vE.3 s/ D .0:5774 {O C 13:00 |O / ft=s:

Since both components of the velocity are positive, at t



D tan1

(2)

3 s, the angle  can be computed as follows

 vD.3 /  s

y

vx .3 s/

;

(3)

where, referring to Eq. (2),

vx .3 s/

D 0:5774 ft=s

and

vy .3 s/

D 13:00 ft=s:

(4)

Substituting Eqs. (4) into Eq. (3), we have



D 87:46ı:


June 25, 2012


Dynamics 2e

47

Problem 2.18

ED  p OC

2 t{ The motion of a point P with respect to a Cartesian coordinate system is described by r 4 ln.t 1/ 2t 2 | ft, where t denotes time, t > 0, and is expressed in seconds. Determine the average acceleration of P between times t1 4 s and t 2 6 s and find the difference between it and the true acceleration of P at t 5 s.



C C

 O

D

D

D

Solution

By definition, the average acceleration over a time interval t 1

E D t2 1 t1

aavg



Z



t

 t2

is

t2

E

adt: t1

(1)

E D d vE=dt , we can write aE dt D d vE. Substituting this expressio n into Eq. (1), gives v E2 vE1 ; aE D t2 t1

Recalling that a



avg



(2)

where v1 and v2 are the values of v at times t 1 and t 2 , respectively.

E

E

E  1  1 vE D p {O C 4 (3) C t |O ft=s: t C1 t Evaluating the expression in Eq. (3) at t D t1 D 4 s and t D t2 D 6 s, and using the results to evaluate We now proceed to determine the velocity as the time derivative of the position. This gives

Eq. (2), we have

E D.

aavg

2

O C 3:886 |O/ ft=s :

0:04588 {

(4)

We now determine the (true) acceleration as the time derivative of the velocity. Using Eq. (3), this gives

ED

a At t

D 5 s, Eq. (5) gives



E



1 2t 3=2

a.5 s/

D.

O C4

{



1

1 .t

C

1/2



| ft=s2 :

(5)

2

(6)

O

O C 3:889 |O/ ft=s :

0:04472 {

Subtracting Eq. (6) from Eq. (4) side by side, we have

E

E

aavg  a.5 s/

D.

O

0:001154 { 

0:003175 | / ft=s2 :

O

The above results allow s one to measure the error made in approximating the true acceleration at t aavg.

E


D s s with

June 25, 2012



Dynamics 2e

49

Problem 2.20 As part of a mechanism, a peg P is made to slide within a rectilinear guide with the follow ing prescribed motion: r.t/ x 0 sin.2 !t /  3 sin. !t/ {;

O E D  where t denotes time in seconds, x 0 D 1:2 in:, and ! D 0:5 rad=s. Determine the displacement and the

distance traveled over the time interval 0  t  4 s. In addition, determi ne the corresponding average velocity and average speed. Express displacement and distance traveled in ft , and express velocity and 2 cos2 ˇ  1. speed in ft =s. You may find useful the following trigonometric identity: cos .2ˇ/

D

Solution

E

The function that describes r.t/ is the sum of two periodic funct ions. The period of the function sin .2 !t / is half the period of the function sin . !t /. Hence, the overall period p of r.t/ coincides with the period of sin. !t/. We determine p as follows:

E

 !p

D 2 )

D 2=! D 4 s; D

p

D

(1)

D D

0:5 rad=s. Therefore, letting t i 0 and tf 4 s p , we see that t i where we have used the fact that ! and tf are exactly one period apart. This implies that the position of the peg at times t i and tf is identical. These considerations tell us that the displacement over the given time interval is equal to zero:

E D 0E;

r

E D rE.tf / rE.ti /. This implies that the correspondin g average velocity is also equal to zero: vE D 0E:

where  r



avg

To determine the distance traveled, we begin by observing that the peg starts its motion at the srcin, i.e.,

E D 0E. Next we determine the veloci ty of the peg as the time deriva tive of the position:   vE D  !x 0 2 cos.2 !t/ 3 cos. !t/ {O: (2) Using the expression in Eq. (2), at time t D 0 , we have vE.0/ D  !x 0 {O. This result implies that, at t D 0 ,

r.0/





the peg is moving to the left. In order to come back to its initial position, the peg must revers e the direction of motion. This observation is important in that it leads us to a strategy to determine the distance traveled. Let the total time interval starting at t i and ending at tf be subdivided as follows:

0

t i < t1 < t2 <



< tn < tf

D

4 s;

(3)

D

where t 1 , t 2 , .. . , and t n , are the times at which the peg changes direc tion of motion. If we can determine the n times t 1 , t2 , .. . tn , then the total distance trave led is the sum of the length of each segment trav eled between the time instants in Eq. (3), that is,

j

(4)


June 25, 2012

d

D jx.ti /



x.t1 /

j C jx.t1/



x.t2 /

j C C jx.t n/ 



x.tf / ;


Solutions Manual

50 where, referring to the problem statement,

x.t/

D x0sin.2 !t/





3 sin. !t / :

(5)

To determine the times at which the peg changes the direction of motion, we need to determine the times at which the x component of velocity changes sign, which correspond s to the times at which the velocity equals zero. Referring to Eq. (2), this requires that we solve the equation

2 cos.2 !t /  3 cos. !t/

D 0:

(6)

Using the trigonometric identity provided in the problem statement, Eq. (6) can be rewritten as

4 cos2 . !t/  3 cos. !t/  2

D 0;

(7)

which is a quadratic equation in cos . !t/ whose solution is cos. !t /

p D 3 ˙8 41 )

D 1:176

cos. !t /

and

cos . !t /

D

0:4254:

(8)

The first root is not acceptable because the cosine function cannot take on values larger than one. Hence, the times at which the x component of the velocity is equal to zero are given by the following sequence of time values:

D 1! cos1



3

p41

!

D 0 ;1;2;:: : (9) Since the time values we are interested in must be between t i D 0 and tf D 4 s, then the only acceptable solutions are t1 D 1:280 s and t2 D 2:720 s: (10) t

8

˙ 2 n;

n

Now that the times at which the peg changes direction of motion are known, referring to Eq. (4), we can then apply the formula giving the distance traveled:

d

D jx.0/



x.t1 /

j C jx.t1/



x.t2 /

j C jx.t2/



j

x.4 s/ :

(11)

Using Eqs.(5) and (10), we have

x.0/

D0;

x.t 1 /

D

0:3485 ft;

x.t 2 /

D 0:3485 ft;

x.4 s/

D 0:

(12)

Using the (full precision value s of the) results in Eq. (12), we can evalua te Eq. (11) to obtain

d

D 1:394 ft:

(13)

Now that the distance traveled is known, the corresponding average speed is obtained by dividing the distance traveled by the length of the time interval considered:

vavg

D 4ds :

(14)

Using the (full precision value of the) result for d in Eq. (13) we have

vavg

D 0:3485 ft=s:


June 25, 2012


Dynamics 2e

51

Problem 2.21 The position of point P as a function of time t , t in seconds, is



0 and expressed

E D 2:0 Œ0:5 C sin.!t/ ç {OC9:5 C 10:5 sin.!t/ C 4:0 sin2.!t/  |O; where ! D 1:3 rad=s and the position is measured in meters. Find the trajectory of P in Cartesian components and then, using the x component of rE.t/ , find the maximum and minimum value s of x r.t/

reached by P . The equation for the trajectory is valid for all values of x , yet the maximum and minimum values of x as given by the x component of r.t/ are finite. What is the srcin of this discrepanc y?

E

Solution

We begin by writing the position of P as

E D x.t/ {O C y.t/ |O;

r

(1)

where

x.t/

D Œ2:0.0:5 C sin !t/ç m

and

y.t/

D .9:5 C 10:5 sin !t C 4:0 sin2 !t / m:

(2)

Next, we solve the first of Eqs. (2) for sin !t as a function of x :

D x 2 1: 

sin !t

(3)

Substituting Eq. (3) into the second of Eqs. (2) we have

y.x/

D



9:5

C 10:5

x 1 

2

C 4:000

2

x 1  

2

m

)

y.x/

D 5:250 C 3:250x C 1:000x2 m:

(4)

We now need to determine the range of x covered by the motion of P . To do so, referring to the first of Eqs. (2), we observe that the minimum and maximum values of x are achieved when the function sin !t achieves its minimum and maximum values, respectively, which are the values 1 and 1, respectively. Substituting the values 1 and 1 for sin !t in the first of Eqs. (2), we have, respectively,

C

C

xmin

D

1 m

and

xmax

D 3 m:

The maximum and minimum values of the x coordinate occur due to the presence of a periodic function in the equation for this component. Therefore, the trajectory, which is valid for all times t , is constrained to oscillate between 1 m and 3 m on the x axis.


June 25, 2012


Solutions Manual

52

Problem 2.22 The position of point P as a function of time t , t in seconds, is



0 and expressed

E D 2:0 Œ0:5 C sin.!t/ ç {OC9:5 C 10:5 sin.!t/ C 4:0 sin2.!t/  |O; where ! D 1:3 rad=s and the position is measured in meters. r.t/

(a) Plot the trajectory of P for 0 2:3 s, and 0  t  5 s.



t



(b) Plot the y.x/ trajectory for 10 m

0:6 s, 0



x





t



1:4 s, 0



t



10 m.

(c) You will notice that the trajectory found in (b) does not agree with any of those found in (a). Explain this discrepa ncy by analytically determining the minimum and maximum values of x reached by P . As you look at this sequence of plots, why does the trajectory change between some times and not others?

Solution

D 1:3 rad=s, we write the x and y coordinates of point P as x.t/ D f2:0Œ0:5 C sin.1:3t/ çg m and y.t/ D Œ9:5 C 10:5 sin.1:3t/ C 4:0 sin2 .1:3t/ ç m:

Part (a).

Since !

(1)

One strategy to plot the trajectory of point P is to plot the line that connects the points of coordinates

Œx.t/;y.t/ ç as time t varies within a given time interval. This way of plotting the trajectory does not involve finding y as a function of x . Rather, it consists of generating a list of .x ; y/ values, each of which is computed by first assigning a specific value of time. This procedure is called a parametric plot, where the parameter used to generate the plotted poin ts is time and does not appear directly on the plot (i.e., the plot uses x and y axes, but it does not show the time values corresponding to the points on the plot). Parametric plots can be generated using any appropriate numerical software such as MATLAB or Mathematica. The parametric plots of the trajectory of P shown below were generated in Mathematica with the following code:


June 25, 2012


Dynamics 2e

53

Part (b). In this part of th e problem we first need to write the tra jectory in the form y we start with solving the first of Eq. (1) for sin .1:3t/ as a function of x . This gives

D y.x/ . To do so,

1:000 m ; D x 2:000 m 

sin.1:3t/

(2)

which can then be substituted into the second of Eqs. (1) to obtain

y.x/

D



9:500

C 10:50

x

1:000 2:000





C 4:000

x

1:000 2:000



)

2



y.x/

m

D 5:250 C 3:250x C 1:000x2 m:

(3)

Now that we have the trajectory in the form y.x/ , we can plot it over the given interval 10 m  x  10 m as shown on the right. This plot was generated using Mathematica with the following code:

Part (c). 3 s: 2.1:3/

Referring to Eq. (1), for x max; sin 1:3t



D C1 ) t D 2.1:3/ s and for x

min ;

Thus, the minimum and maximum values are, respectively,

xmin

D

1:000 m

and

xmax

sin 1:3t

D 1)t D 

D 3:000 m:

(4)

The minimum and maximum values of the x coordinate occur due to the presence of a periodic functio n in the equation for this component. Therefore, the trajectory, which is valid for all times t , is constrained to oscillate between 1:000 m and 3:000 m on the x axis. This fact explains why in the plot sequence in Part (a) the trajectory seems not to change after a while: point P keeps tracing the same curve segmen t again and again. The periodicity of the motion of P also explains the discrepancy between the trajectory obtained in Part (a) and that obtained in Part (b). In fact, since the plot generated in Part (a) was based on a direct application of

Eqs. we contrast, see that the in question correctly reflects periodic time the coordinates of point(1), P . By the plot trajectory computed in Part (b) nothe longer carries anydependence direct relationship with time. The trajectory plotted in Part (b) does contain the trajectory plotted in Part (a) as a sub-curve. The problem with the trajectory in Part (b) is that we have no direct way of knowing what part of the entire curve actuall y pertains to the motion of point P .


June 25, 2012


Solutions Manual

54

Problem 2.23

D

20 mph on a A bicycle is moving to the right at a speed v0 horizontal and straight road. The radius of the bicycle’s wheels is R 1:15 ft. Let P be a point on the periphery of the front wheel. One can show that the x and y coordinates of P are described by the following functions of time:

D

x.t/

D v 0t C R sin.v0t=R/

and

y.t/

D R 1 C cos.v0t=R/ :

Determine the expressions for the velocity, speed, and acceleration of P as functions of time. Solution

The velocity of P is the time derivative of P ’s position, which, in the coordinate system shown, is

E D v0t C R sin.v0t=R/  {O C R1 C cos.v0t=R/  |O:

r

(1)

Differentiating the above expression with respect to time, we have



E D v 0 1 C cos

v Since v 0

 v t  0

R

O

{  v0 sin

v t  0

R

O

|:

(2)

D 20 mph D 20 5280 3600 ft =s and R D 1:15 ft, we have E D .29:33 ft=s/Œ1 C cosŒ.25:51 rad=s/t ç {O

v



O

.29:33 ft=s/ sinŒ.25:51 rad=s/t ç | :

The speed is the magnitude of the velocity vector:

v

q D

vx2 .t /

C

vy2 .t/

s

D v 0 2 C 2 cos

v t  0

R

)

v

p

D .29:33 ft=s/ 2 C 2 cosŒ.25:51 rad=s/t ç: (3)

The acceleration of P is the time derivative of P ’s velocity. From Eq. (2), we have

ED

a Since v 0



v02 v0 t sin R R

 

O

{

v02 v0 t cos R R

 

O

|:

(4)

5280 ft =s and R D 1:15 ft, we have D 20 mph D 20 3600

ED

a



ft s2 sinŒ.25:51 rad=s/t ç {

748:2 = 

O



ft s2 cosŒ.25:51 rad=s/t ç | :

748:2 = 


O

June 25, 2012


Dynamics 2e

55

Problem 2.24

D


D

x.t/


y.t/

and


Determine the maximum and minimum speed achieved by P , as well as the y coordinate of P when the maximum and minimum speeds are achieved. Finally, compute the acceleration of P when P achieves its maximum and minimum speeds. Solution

The speed of P is the magnitude of the velocity of P . Hence, we first compute the veloc ity of P , which is the time derivative of P ’s position. In the coordinate system shown , the position of P is


r

(1)

Differentiating the above expression with respect to time, we have



E D v 0 1 C cos

v

 v t  0

R

O

{  v0 sin

v t  0

R

O

|:

(2)

Hence, the speed is (3) C vy2.t / D v0 2 C 2 cos vR0t ; which implies that v is maximum when cos .v0 t=R/ D 1 and minimum when cos .v0 t=R/ D 1, i.e., v D 2v 0 D 58:67 ft=s and v D 0 ft=s; (4) 5280 where we have used the fact that v 0 D 20 mph D 20 3600 ft =s. Since v D v when cos.v0 t=R/ D 1 and v D v when cos .v0 t=R/ D 1 , using the expression for

v

max

 

vx2 .t/

D

q

s



min



min

max

the y component of the position of P in Eq. (1), we have

D 0 ft and where we have used the fact that R D 1:15 ft. yvmin

yvmax

D 2R D 2:300 ft;

To determine the acceleration corresponding to vmin and v max, we first determine the acceleration of P by differentiating with respect to time the expression in Eq. (2). This gives

v02 v0 t sin R R

2

O vR0 cos vR0t |O: (5) Now, recall that for v D v we have cos .v0 t=R/ D 1 and for v D v we have cos .v0 t=R/ D 1. In 0 both cases, we have sin .v t=R/ D 0 . Using these consider ations along with Eq. (5), we have v2 v02 aEv D 0 |O D .748:2 ft=s / |O and aEv D |O D . 748:2 ft=s / |O ; R R where we have used the following numerical data: v 0 D 20 mph D 20.5280=3600/ ft=s and R D 1:15 ft. ED

a



 

{



min

max

2

min

 

max






2

June 25, 2012


Solutions Manual

56

Problem 2.25

D


D

x.t/


y.t/

and


Plot the trajectory of P for 0  t  1 s. For the same time inter val, plot the speed as a function of time, as well as the components of the velocity and acceleration of P . Solution

The velocity of P is the time derivative of P ’s position, which, in the coordinate system shown, is


r

(1)

Differentiating Eq. (1) with respect to time, we have



E D v 0 1 C cos

v

 v t  0

R

O

{  v0 sin

v t  0

R

O

|:

(2)

The speed is now found by taking the magnitude of the velocity vector. Hence, we have 2

v

D

q

v0 t

2

vx .t /

C vy .t/ D v 0 2 C 2 cos  R :

s

(3)

The acceleration of P is the time derivative of P ’s velocity. Hence, from Eq. (2), we have

ED

a



v02 v0 t sin R R

 

O

{

v02 v0 t cos R R

 

O

|:

(4)

Trajectory of P . We can plot the trajectory of P for 0  t  1 s by plotting the line connecting the x and y coordinates of P computed as functions of the parameter t . The x and y coordinates of P are the component of r (in Eq. (1)) in the x and y direction, respectively. This can be done with a variety of pieces 20 mph 29:33 ft=s and R 1:15 ft, the plot presented below has been of numerical software. Since v 0 obtained using Mathematica with the following code

E

D

D

D


June 25, 2012


Dynamics 2e

57

D

D

20 mph 29:33 ft=s Speed of P . The speed of P can be plotted for 0  t  1 s using Eq. (3), with v 0 and R 1:15 ft. The plot shown below was generate d using Mathematica with the following code:

D

Velocity Components The components of the velocity of P can be plotted for 0  t  1 s using the expressions in Eq. (2) with v0 20 mph 29:33 ft=s and R 1:15 ft. The plot shown below was generated using Mathematica with the following code:

D

D

D


June 25, 2012


Solutions Manual

58

Acceleration Components The components of the acceleration of P can be plotted for 0  t  1 s using 20 mph 29:33 ft=s and R 1:15 ft. The plot shown below was the expressions in Eq. (4) with v 0 generated using Mathematica with the following code:

D

D

D


June 25, 2012


Dynamics 2e

59

Problem 2.26 Find the x and y components of the acceleration in Example 2.3 (except for the plots) by simply differenti ating Eqs. (4) and (5) with respect to time. Verify that you get the results given in Example 2.3. Solution

Referring to Eqs. (4) and (5) of Example 2.3 on p. 38 of the textbook, we recall that the x and y components of the velocity are, respectively,

v0 y x

2v0 a

P D y 2 C 4a 2 and yP D y 2 C 4a 2 : To determine xR , we differentiate xP with respect to time with the help of the chain rule: d xP xR D D ddyxP dy D yP ddyxP : dt dt

p

p

(1)

(2)

Differentiating the first of Eqs. (1) with respect to y and substituting the result along with the second of Eqs. (1) into Eq. (2) we then have

RD

x

R



2

y vC y4ayP 0

2 3=2

2

P

C

P ) y 2 C 4a 2 v0 y

p

RD

x

8v02 a3 y2

C 4a 22

:

(3)

To determine y we differentiate y with respect to time with the help of the chain rule. This gives

y

P

dy

P

d y dy

y

P:

dy

(4)

R D dt D dy dt D P dy

Substituting the second of Eqs. (1) into Eq. (4) and simplifying, we have

RD

y

2 a2 y 4v0

y2

C 4a 22

:

Our results match those in Example 2.3.


June 25, 2012


Solutions Manual

60

Problem 2.27 Find the x and y components of the acceleration in Example 2.3 (except for the plots) by differentiating the first of Eqs. (3) and the last of Eqs. (1) with respect to time and then solving the result ing two equations for x and y . Verify that you get the results given in Example 2.3.

R

R

Solution

We recall that the first of Eqs. (3) and the last of Eqs. (1) in Example 2.3 on p. 38 are, respecti vely,

v2 0

x2

y2

and

2y y

DP CP

4ax:

(1)

PD P

Recalling that v 0 and a are constants, differentiating Eqs. (1) with respect to time gives

0

D 2xP xR C 2yP yR

and

R D yP 2 C y y:R

2a x

We now view Eqs. (1) as a system of two equations in the two unknowns question gives

PD

x

yv0

p

y2

C 4a 2

and

PD

y

(2)

P

P

x and y . Solving the system in

2av0

p

y2

C 4a 2 ;

(3)

where, similarly to what was done in Example 2.3 on p. 38 of the textbook, we have enforced the condition that y > 0. Next, we view Eqs. (2) as a system of two equations in the two unknowns x and y whose solution is

P

3 R D y xP CyP 2a yP

x

and

R

R

2 R D y xP CxP yP2a yP : 

y

(4)

Substituting Eqs. (3) into Eqs. (4) and simplifying, we have

RD

x

8v02 a3

4a 2

C

2 y2



and

RD

y

2 a2 y 4v0

4a 2

C y 22

:

Our results match those in Example 2.3.


June 25, 2012


Dynamics 2e

61

Problem 2.28 Airplane A is performing a loop with constant radius  follows:

.x  xC /2

D D

D

C .y



D 1000 ft. The equation describing the loop is as y C /2 D  2 ;

where x C 0 and y C 1500 ft are the coordinates of the center of the loop. If the plane were capable of 160 mph, determine the velocity and acceleration of the maintaining its speed constant and equal to v 0 plane for  30 ı .

D

Solution

We begin by expressing the position as follows:

E D x {O C y |O :

r

(1)

D 30ı, the plane is in the lower right quadrant of the loop and the expression for the path of the airplane can be written as y D yC 2 x 2 ; (2) where we have accounted for the fact that x C D 0 . We observe that x D  sin  : (3) For 



q



Equations (1) and (2) combined indicate that we can regard position as being a known function ofx . With this in mind, we can write an expression for the velocity of the airplane by differentia ting Eq. (1) with respect to time with the help of the chain rule. This gives

dy xP |O ) vE D xP {O C E D xP {O C dx

v

P

P

xx

p

2  x 2

O

|:

(4)

The quantity x in the last of Eqs. (4) is unknown, but it can be found by enforcing the condition that the magnitude of the velocity is equal to v 0 . Doing so gives

v2

x2

DP C

x 2x 2

P

2 

where we have accounted for the fact that, at Eqs. (4), we have

x

v0

2  x 2 ;

(5)

D ) PD   D 30 ı , xP > 0. Substituting the last of Eqs. (5) into the last of

E D v0

v

v 02

x2

q

q

O C v 0x |O:

2  x 2 {


(6) June 25, 2012


Solutions Manual

62 Since v 0

5280 D 160 mph D 160 3600 ft =s,  D 1000 ft, and using Eq.(3), we can evaluate Eq. (6) to obtain

E D .203:2 {O C 117:3 |O/ ft=s:

v

To determine the acceleration we differentiate the expression for the velocity in Eq. (6) with respect to time, which gives

ED

a

P {O C v0 xP |O : 

v0 x x





p

(7)

2  x 2

Substituting the last of Eqs. (5) into Eq. (7) and simplifying, we have

ED

a Recalling that v 0



v02 x { 2

O C v022

q

2  x 2 | :

O

(8)

5280 ft =s, and using Eq.(3), we can evalua te Eq. (8) to obtain D 160 mph D 160 3600

E D.

a

2

O C 47:69 |O/ ft=s :

27:53 {


June 25, 2012


Dynamics 2e

63

Problem 2.29 An airplane A takes off as shown with a constant speed equal to v0 160 km=h. The path of the airplane is described by the equa6  104 m1 . Using the comp onent tion y  x 2 , where  system shown, provide the expression for the velocity and accelera400 m. Express the velocity in m =s tion of the airplane when x and the accelerat ion in m =s2 .

D

D

D

D

Solution

The position of the airplane can be described as

E D x {O C y |O :

(1)

E D x {O C  x2 |O:

(2)

E D xP {O C 2 xxP |O :

(3)

r Since y

D

 x 2 , Eq.

(1) becomes

r

Using the chain rule to differentiate Eq. (2) with respect to time, the velocity of the airplane is

v

P

The term x in Eq. (3) can be determined by enforcing the conditio n that the magnitude of the velocity is equal to v 0 . This gives

P C 4  2x2xP 2 D v02 ) xP D p1 Cv40  2x2 ;

x2

P

(4)

where we have chosen x > 0 because the airplane is moving in the positive x direction. Substituting the last of Eqs. (4) into Eq. (3) gives v0 2 v0 x v { : (5) 2 2 1 4 x 1 4  2x 2 1000 For v 0 160 km=h 160 3600 m =s,  6  104 m1 , and x 400 m, we can evaluate Eq. (5) to obtain

D

E D Dp C

O C p CD vE D .40:07 {O C 19:23 |O / m=s:

D

The acceleration is the time derivative of the velocity. Using Eq. (3) and the chain rule, we have

  E D ddxxP xP {O C 2 xP 2 C 2 x ddxxP xP |O:

a

(6)

Differentiating the last of Eqs. (4) with respect to x , we have

PD

dx dx



2

1 C44 vxx 0

2

2 3=2

:

(7)

Substituting the last of Eqs. (4) and Eq. (7) into Eq. (6), after simplification, gives

ED

a For v 0



4  2 v02 x

1 C

2 4  2x 2



OC

{

2 v02 2

2 2

1 C 4 x 

O

|:

(8)

4  D 160 km=h D 160 1000 m , and x D 400 m, Eq. (8) gives 3600 m =s,  D 6 10 aE D . 0:7516 {O C 1:566 |O / m=s : 



1

2


June 25, 2012


Solutions Manual

64

Problem 2.30 A test track for automobi les has a portion with a specific profile describe d by:

y

D

D

D h 1





sin.x=w/ ;

8 ft, and where the argument of the sine function is understood to be in radians. where h 0:5 ft and w A car travels in the positive x direction such that the horizontal component of velocity remains constant and equal to 55 mph. Modeling the car as a point moving along the given profile, determine the maximum speed of the car. Express your answer in ft =s.

Solution

Letting x and y represent the coordinat es of the car, the position of the car is

E D x {O C h1

r



O

sin.x=w/ | :

(1)

The velocity is the time deriva tive of the position. Using Eq. (1) and the chain rule, we have

E D xP {O

v



h x x |: cos w w

 

PO

(2)

To determine the maximum speed, we first determine the speed, which is the magnitude of the velocity. From Eq. (2), we have

v

ˇD P ˇs C x

1

h2 x : cos 2 w2 w

 

(3)

From Eq. (3) we see that the speed varies because of the presence of the cosine function, whose maximum possible value is equal to one. We conclude that the maximum value of the speed is

vmax

ˇD P ˇs C x

1

h2 : w2

(4)

P D 55 mph D 55 5280 3600 ft =s, h D 0:5 ft, and w D 8 ft, Eq. (4) gives

For x

vmax

D 80:82 ft=s:


June 25, 2012


Dynamics 2e

65


y

D

D

D h 1



cos.x=w/ ;



0:20 m and w 2 m, and where the argument of the cosine function is understood to be where h in radia ns. A car travels in the positive x direction with a constant x component of velocity equal to 100 km=h. Modeling the car as a point moving alon g the given profile, deter mine the velocit y and 24 m. acceleration (expressed in m=s and m =s2 , respectively) of the car for x

D

Solution

E D x {O C y |O , i.e.,

Letting x and y represent the coordinat es of the car, the position of the car is r

 rE D x {O C h 1



 .x=w/ |O :

cos

(1)


E D xP {O C wh sin

v

x w

PO

x |:

(2)

The acceleration is the time derivative of the velocity. Recalling that x is constant, differentiating Eq. (2) with respect to time and using the chain rule, we have

ED

a For x 100 km=h 100 1000 3600 m =s, h in Eqs. (2) and (3) to obtain

PD

D

E D .27:78 {O

v

h x cos x2 | : w2 w

 

P

P O

(3)

D 0:20 m, w D 2 m, and x D 24 m, we can evaluate the expressions 

O

1:490 | / m=s and

2

E D 32:56 |O m=s :

a


June 25, 2012


Solutions Manual

66


y

D

D

D h 1





cos.x=w/ ;

0:75 ft and w 10 ft, and where the argument of the cosine function is understood to be in where h radians. A car drives at a constant speed v0 35 mph. Modeling the car as a point moving along the given profile, find the velocity and accelera tion of the car for x 97 ft. Express velocity in ft =s and acceleration in ft =s2 .

D

D

Solution

Letting x and y represent the coordinat es of the car, the position of the car is bvr

 rE D x {O C h 1



D x {O C y |O , i.e.,

 .x=w/ |O :

cos

(1)


E D xP {O C wh sin

v

x w

PO

x |:

(2)

The quantity x is currently unknown but it can be determined by enforcing the conditio n that the speed is equal to v 0 . Recalling that v

D xP 2 C yP 2, from Eq. (2), setting the speed equal to v0 gives x 2 2 wv 0 xP 2 C 2 sin2 xP D v 0 ) xP D ; w w 2 w C h2 sin2 .x=w/ where we have taken xP > 0 because the car moves in the positive x direction. Substituting the last of P

h2

p

q

(3)

Eqs. (3)

into Eq. (2), we have

ED

wv 0

q

{

2

w2

E D .51:32 {O

v

OC

hv 0 sin.x=w/

|O: w 2 C h2 sin2 .x=w/ C h2 sin .x=w/ 5280 For v 0 D 35 mph D 35 3600 ft =s, h D 0:75 ft, w D 10 ft, and x D 97 ft, Eq. (4) gives v



(4)

q

O

1:046 | / ft=s:

The acceleration is the time derivative of the velocity. Differentiating Eq. (2) with respect to time and using the chain rule, we have

2      E D ddxxP xP {O C wh xwP cos wx C sin wx ddxxP xP |O:

a


(5)

June 25, 2012


Dynamics 2e

67

Differentiating the last of Eqs. (3) with respect to x we have

PD

dx dx

2



 hv 2 w Ch

0 sin .2x=w/

2

2 sin2 .x=w/ 3=2

where we have used the trigonometric identity 2 sin ˛ cos ˛ Eq. (6) into Eq. (5) and simplifying, we have

ED

a



h2 wv 02 sin .2x=w/ 2 w2

C h2 sin2.x=w/



2



OC

{



D

D

ED

a

(6)

D sin.2˛/. Substituting the last of Eqs. (3) and hw 2 v02 cos.x=w/

O

|;

C h2 sin2.x=w/ 2   2 sin ˛ cos ˛ D sin.2˛/ . w2

where, again, we have used the trigonometric identity 35 5280 0:75 ft, w 10 ft, and x 97 ft, Eq. (7) gives 3600 ft =s, h

D

;

For v0

(7)

D 35 mph D

2

O C 19:00 |O/ ft=s :

.0:3873 {


June 25, 2012


Solutions Manual

68

Problem 2.33 The orbit of a satellite A around planet B is the ellipse shown and is described by the equation .x=a/ 2 .y=b/ 2 1, where a and b are the semimajor and semiminor axes of the ellipse, respectively. When x a=2 and y > 0, the satellite is moving with a speed v 0 as shown. Determine the expression for the a=2 and y > 0. satellite’s velocity v in terms of v 0 , a , and b for x

D

C

D

E

D

Solution

D

a=2 and the We begin by identifying the value of the y coordinate of the satellite corresponding to x condition y > 0. Setting x a=2 in the equation describ ing the path of the satellite, we have

D

1 4

2

C yb2 D 1 )

y

D

p3 2

b;

where we have selected the positive root to satisfy the requirement that relation describing the path of the satellite with respect to time, we have

P

Substituting x

P

xx

yy

2

2

(1)

y > 0. Next, differentiating the

0:

(2)

a b D D a=2 and the last of Eqs. (1) in Eq.C(2), after simplifying, we have p a 3yP xP C D 0:

(3)

b

We also know that

x2

P C yP 2 D v02;

(4)

where we treat the speed v 0 as a known quantity. Equations (3) and (4) form a system of two second order algebraic equations in the two unknowns x and y . This system has two solutions. However, as shown in the figure, at the instant considered A is moving upward and to the left. Hence, we have that x < 0, which is sufficient to allow us to determine the following unique solution:

P

P

p 0 P D p3a23av C b2

x



Recalling that the velocity is given by v

x{

P

and

P D p3av20bC b 2 :

y

(5)

y | , using Eq. (5), we can express the velocity as

EDPOCP O p  v0 vE D p 3a {O C b |O : 2 2 3a C b 


June 25, 2012


Dynamics 2e

69

Problem 2.34 In the mechanism shown, block B is fixed and has a profile described by the following relation:

y

"

D h 1 C 12

 x  1 x  # 2

4



d

4 d

:

The follower moves with the shuttle A , and the tip C of the follower remain s in contact with B . 0:25 in:, d 1 in:, and the horizontal position of C is x d sin .!t/ , where Assume that h ! 2 rad=s, and t is time in seconds. Determine an analytical expre ssion for the speed of C as a 0 , x 0:5 in:, and function of x and the parameters d , h , and ! . Then, evaluate the speed of C for x x 1 in. Express your answers in ft =s.

D

D D

D

D

D

D

Solution

The position of C is

E D x {O C y |O :

r C

(1)

B

C

y

x

Since remains in contact with , then the vertical of , namely , is given as a function of the problem statemen t. Hence, we can rewrite Eq. (1) position as follows:

"

E D x {O C h 1 C 21

r

 x  1 x  # 2

in

4



d

4 d

O

|:

(2)

The velocity of C is obtained by differe ntiating Eq. (2) with respect to time. Using the chain rule, this gives

E D xP {O C h

v

x

d2



x3 x |: d4



PO

(3)

Recalling that the speed is the magnitude of the velocity, using Eq. (3) we have

v Since x

ˇD P ˇs C x

1

h2

x

d2



x3 d4



2

:

(4)

D d sin .!t/ , differentiating this expression with respect to time, we have xP D d ! cos.!t/:

(5)

Equation (5) implies that

ˇ Pˇ D x

j

d ! cos.!t/

ˇˇ q

j ) xP D d !

1  sin2 .!t/

ˇˇ q

) xP D d !


1  .x=d/ 2 ;

(6)

June 25, 2012


Solutions Manual

70

D

d sin .!t/ . Substituting the last of where, in obtaining the last of Eqs. (6), we have used the fact that x Eqs. (6) into Eq. (4), we have the expression for the speed requested by the problem statement:

v

D!

p

d2 

x2

s

1

C h2

x

d2



x3 d4



2

:

(7)

0:25 0:25 in: Recalling that h 12 ft, d 0:5 x 0 , x 0:5 in: ft, and x 1 in: 12

D

D

D D

D

v

D

D 1 in: D 121 ft, and ! D 2 rad=s, we can evaluate Eq. (7) for D 121 ft to obtain, respectively,

D 0:5236 ft=s;

v

D 0:4554 ft=s;

and

v

D 0:


June 25, 2012


Dynamics 2e

71

Problem 2.35 In the mechanism shown, block B is fixed and has a profile described by the following relation:

y

"

D h 1 C 12

 x  1 x  # 2

4



d

4 d

:

The follower moves with the shuttle A , and the tip C of the follower remain s in contact with B . 20 mm, and A is made to move from x d to x d with a constant Assume that h 2 mm, d speed v 0 0:1 m=s. Determine the acceleration of C for x 15 mm. Express your answer in m =s2 .

D

D

D

D

D

D

Solution

The position of C is

E D x {O C y |O :

r

(1)

Since C remains in contact with B , then the vertical position of C , namely y , is given as a function of x in the problem statemen t. Hence, we can rewrite Eq. (1) as follows:

"

E D x {O C h 1 C 21

r

2

4

 x  1 x  # 

d

4 d

O

|:

(2)

The velocity of C is obtained by differe ntiating Eq. (2) with respect to time. Using the chain rule, this gives

E D xP {O C h

v

x

d2



x3 x |: d4



PO

(3)

Since the follower moves with A , and since A moves to the right with the constant speed v 0 , we then have

P D v 0:

x

(4)

Substituting Eq. (4) into Eq. (3) we then have that the velocity of C is

x

E D v 0 {O C v0h

v

d2



x3 d4



O

|:

(5)

The acceleration of C is obtained by differentiating the expression of the velocity in Eq. (5) with respect to time. Using the chain rule, this gives 2

E D hvd 20

a



13

x2 d2



O

|;


(6)

June 25, 2012


Solutions Manual

72

D D

2 mm where we have accounted for the fact that v 0 is constant. Recalling that h 20 0:1 m=s, we can evaluate the expressi on in Eq. (6) for x 15 mm 1000 m, v 0

D

ED

a

2 D 1000 m, d D 20 mm D 15 D 1000 m to obtain

2

O m =s :

0:03438 |


June 25, 2012


Dynamics 2e

73

Problem 2.36 The Center for Gravitational Biology Research at NASA’s Ames Research Center runs a large centrifuge capable of 20g of acceleration, where g is the acceleration due to gravity ( 12:5g is the maximum for 25 ft. The human subjects). The distance from the axis of rotati on to the cab at either A or B is R moves at the constant speed vA and yA > 0.

D

q

D

q

R 2  xA2 for yA  0 and by yA  R 2  xA2 for yA < 0. If A 120 ft=s, determine the velocity and acceleration of A when xA 20 ft

trajectory of A is described by yA

D

D

D

Photo credit: NASA

Solution

Starting with the equation of the trajectory for yA > 0, and differentiating it with respect to time, we have

P D

yA

P

xA xA

q

R2  xA2

:

(1)

2

We now recall that the speed of A can be computed as vA

2

D xPA C yPA, which implies 2 2 ) xPA2 C Rx2AxPAx2 D vA2 ;

xA2

P C yPA2 D vA2

q

(2)

 A

P

where the last of Eqs. (2) was obtained by using Eq. (1). The last of Eqs. (2) can be solved for xA to obtain

P D

xA

vA  R

q

R2  xA2 ;

(3)

P

where, referring to the figure in the problem statement, we have chosen the root with xA < 0 since, for xA 20 ft, A is moving down and to the left. Substituting Eq. (3) into Eq. (1), we have

D

P D vARxA :

yA

(4)

E D xPA {O C yPA |O , using Eqs. (3) and (4), we can now express the velocity of

Since vA

v



vA

R2  x 2 {

vA xA

A as

|:

(5)

AOC R EA D R O For vA D 120 ft=s, R D 25 ft, and xA D 20 ft, we can evaluate Eq. (5) to obtain vEA D .72:00 {O C 96:00 |O / ft=s:

q






June 25, 2012


Solutions Manual

74

Since the acceleration is the time derivative of the velocity, we can determine the acceleration of differentiating Eq. (5) with respect to time. Using the chain rule, this gives

E D vRA

aA

P

xA xA

q

R 2  xA2

A by

O C vRA xPA |O :

(6)

q

(7)

{

Substituting Eq. (3) into Eq. (6) and simplifying, we have

E D

aA For vA



vA2 xA R2

O

{

vA2 R2

R2  xA2 | :

O

D 120 ft=s, R D 25 ft, and xA D 20 ft, we can evaluate Eq. (7) to obtain aEA D .460:8 {O 345:6 |O / ft=s : 



2


June 25, 2012


Dynamics 2e

75

Problem 2.37 sliderPoint C is a point on the connecting rod of a mechanism called a crank. The x and y coordinates of C can be expressed as follows:

D R cos  C 12

p

L2  R 2 sin2  and y C .R=2/ sin  , where  describes !t, the position of the crank. The crank rotates at a constant rate such that  where t is time. Find expressions for the velocity, speed, and acceleration of C as functions of the angle  and the parameters, R , L , and ! . xC

D

D

Solution

Using the coordinate system and expression s given in the problem statement, the position of point C can be expressed as as a function of  as follows:

R

E D xC {O C yC |O D

rC

cos 

C 21

p

L2  R 2 sin2 

 {O C  R  |O: 1 2

sin 

(1)

The velocity is the time derivative of the position. Hence, differentiating Eq. (1) with respect to time and using the chain rule, we have

P

 R

vC

2

E D

Since 

2

R sin  cos 

sin 

Cp

L2 

R 2 sin2 

D ! t and that therefore P D ! , we can rewrite Eq. (2) as E D

vC

!R

2



2 sin 

C

R sin  cos 

p

L2  R 2 sin2 

 {O C 

P

R

cos  | :

2

(2)

O

O C !R cos  |O: 2

{

(3)

The speed is now found by taking the magnitude of the velocity vector. Using Eq. (3), this gives

vC

D !R 2

s

4 sin2 

4R sin2  cos 

C

p

2

L2  R2 sin 

2

2

2

 C RL2 sinR2 cos C cos2  : sin2 

(4)



The acceleration is found by taking the derivative of the velocity. Hence, differentiating Eq. (3) with respect to time, using the chain rule and recalling that  ! , we have

PD

E D

aC

! 2 R

2



2 cos 

2 

C R.cos2

p

L



sin2  /

R 2 sin2 

C .LR2

3 cos2 

sin 2 

2  R2 sin  /3=2



O

{

! 2R sin  | : 2


O

(5)

June 25, 2012


Solutions Manual

76

Problem 2.38 Point C is a point on the connecting rod of a mechanism called a slidercrank. The x and y coordinates of C can be expressed as follows:

D R cos  C 12

p

L2  R 2 sin2  and y C .R=2/ sin  , where  describes !t, the position of the crank. The crank rotates at a constant rate such that  where t is time. Let t be expressed in seconds, R 0:1 m, L 0:25 m, and ! 250 rad=s. Plot the trajectory of point C for 0  t  0:025 s. For the same interval of time, plot the speed as a function of time, as well as the components of the velocity xC

D

D

D

D

D

and acceleration of C .

Solution

The velocity of point C is the time derivative of the position of C . Using the coordinate system shown and since  ! t , can be written as

D

R

E D xC {O C yC |O D

rC

cos !t

C 12

p

L2  R 2 sin2 !t

 {O C  R 1 2

O

sin !t | :

(1)

Hence, differentiating the above expression with respect to time and simplifying, we have

E D

vC

!R

2



2 sin !t

R sin !t cos !t

C

L2 

R 2 sin2 !t

p



O C !R cos !t |O: 2

{

(2)

The speed is now found by taking the magnitude of the velocity vector:

vC

D

!R 2

s

4 sin2 !t

C

4R sin2 !t cos !t

p

L2 

R 2 sin2 !t

2

2

2

C RL2sin R!t2 sincos2 !t!t C cos2 !t :

(3)



The acceleration is found by taking the derivative of the velocity. Hence, differentiating Eq. (2) with respect to time, we have

E D

aC

! 2 R

2



2

2 cos !t

2

C R.cos2 !t 2 sin2 !t / C .LR2

p

L





R sin !t

3 cos2 

sin 2 !t

2  R2 sin !t /3=2



O

{

! 2R sin !t | : 2

O

(4)

Plot of the trajectory and speed of C . Plots of the trajectory and speed of C for 0 < t < 0:025 s can be generated with appropriate numerical software. The plots presented below were generated using Mathematica with the following code:


June 25, 2012


Dynamics 2e

77

E

E

Plots of the components of vC and aC . The components of point C ’s velocity and acceleration are shown below in the plots to the left and right, respectively. These plots were obtained using the following code:


June 25, 2012


78

Solutions Manual


June 25, 2012


Dynamics 2e

79

Problem 2.39 The following four problems refer to a car traveling between two stop signs, in which the car’s velocity is assumed to be given by v.t/ Œ9  9 cos.2t=5/ ç m=s for 0  t  5 s. Determine vmax , the maximum velocity reached by the car. Furthermore, determine the position svmax and the time tvmax at which v max occurs.

STOP

STOP

D

Solution

The expression for v.t/ consists of the difference between the constant 9 m=s and a cosine function multiplied by the value 9 m=s. The cosine function can only range between 1 and 1. The corresponding range of v is 18 m=s, when cos .2t=5/ 1, and v 0 , when cos .2t=5/ 1 . Hence, the maximum value therefore v of v.t/ is

D

D

D

vmax Since v

Dv

max

D 18:00 m=s:

D

)

1

(1)

1, we have

when the cosine function is equal to cos.2t vmax =5/

D

2t vmax =5

D s )

where the last of Eqs. (2) is the only admissible solution with four significant figures, we have

0



t



tvmax

D .5=2/ s;

5 s. Evaluating the last of Eqs. (2) to

D 7:854 s:

tvmax

To find the position s vmax at which v max is achieved, we first determine s.t/ . Recalling that v can write

ds

D v. t/ dt )

where the limits of integration reflect the fact that s

s

D



9t



Z

s

ds 0

Z D

v.t/dt;

(3)

0

D 0 for t D 0. Integrating v.t/ , we have

45 2t sin 2 5

 

D

svmax

D ds=dt , we

t

m:

Given that we have already determin ed t vmax , we can find s vmax by letting t .45  =2/ m, which, to four significant figures, gives

svmax

(2)

(4)

D tv

max

in Eq. (4). This gives

D 70:69 m:


(5)

June 25, 2012


Solutions Manual

80

Problem 2.40 The following four problems refer to a car traveling between two stop signs, in which the car’s velocity is assumed to be given by v.t/ Œ9  9 cos.2t=5/ ç m=s for 0  t  5 s. Determine the time at which the brakes are applied and the car starts to slow down.

STOP

STOP

D

Solution

D

dv=dt to go from a positive value to a negative value. Applying the brakes causes the acceleration a tbraking at which the brakes are applied corresponds to the time instant at which a 0. The instant t Differentiating the given expression for v.t/ with respect to time and setting the result equal to zero, we have

D

 18  2t 5

sin

braking

5



D

D0 )

where we have selected the only solution in the range 0 siginificant figures, we have

tbraking



tbraking t



D 52 s;

(1)

5 s . Expressing the above result to four

D 7:854 s:


(2)

June 25, 2012


Dynamics 2e

81

Problem 2.41 The following four problems refer to a car traveling between two stop signs, in which the car’s velocity is assumed to be given by v.t/ Œ9  9 cos.2t=5/ ç m=s for 0  t  5 s. Determine the average velocity of the car between the two stop signs.

STOP

STOP

D

Solution

By definition, the average velocity is the change in position divided by the time it takes for the change in position to occur. We denote by t i and tf the times at which the car starts moving and comes to a stop, respectively (the subscripts i and f stand for initial and final, respectively). The car starts moving at time ti 0 . For t tf , v 0 , so that we can find tf by solving the equat ion v.tf / 0 , i.e.,

D

D

D

D

Œ9  9 cos.2tf =5/ç m=s

D0 )

cos.2tf =5/

D1 )

To compute the change in position, we first determine s.t/ . Recalling that v s

D 5  s:

(1)

D ds=dt , we can write

t

D 0 v.t/dt; where the limits of integration reflect the fact that s D 0 for t D 0 . Integrating v.t/ , we have ds

D v. t/ dt )

tf

s

D



9t



ds

(2)

0

Z

45 2t sin 2 5

Z

 

m:

(3)

Evaluating Eq. (3) with the value of the last of Eqs. (1), we have

s.tf / As stated earlier,

vavg Recalling that t i

D 45 m:

D s.tftf/



s.t i /

 ti

(4)

:

D 0, s.ti / D 0, and using the results in Eqs. (1) and (4), we can evaluate v D 9:000 m=s:

(5)

v avg to obtain

avg


June 25, 2012


Solutions Manual

82

Problem 2.42 The following four problems refer to a car traveling between two stop signs, in which the car’s velocity is assumed to be given by v.t/ Œ9  9 cos.2t=5/ ç m=s for 0  t  5 s. Determine a max , the maximum of the magnitude of the acceleration reached by the car, and determine the position(s) at which a max occurs.

D

STOP

STOP

jj

jj

Solution

The acceleration is the time derivative of the velocity:

ˇˇ

 ˇˇ

D dv D 185 sin 2t5 m=s ) jaj D 185 sin 2t5 m=s : (1) dt The maximum value of jaj corresponds to the maximum value of j sin.2t=5/ j, which is equal to one. Therefore jaj D .18=5/ m=s , which, expressed to four significant figures, is

 

a

max

2

2

2

ˇ

2

jaj D 3:600 m=s : max

Letting t jajmax denote the time at which a

max

is achieved, we have that sin.2t jajmax =5/

t  D j.5j =4/ t  t  t  D 3:927 

jamax j 1

t 

s and

jamax j 2

where we have considered the only solutions in the time interval 0 figures, times jamaxj 1 and jamax j 2 are jamax j 1

jj

To find the positions at which a write

max

ds

D .15=4/ s; 

t 

s and

jamax j 2

t



1, i.e.,

D˙

5 s. Expressed to four significant

D 11:78 s:

is achieved, we first determine s.t/ . Recalling that v

Z

D v. t/ dt )

where the limits of integration reflect the fact that s

s

D



9t



s

ds 0

Z D

(2)

D ds=dt , we can

t

v.t/dt;

(3)

0

D 0 for t D 0. Integrating v.t/ , we have

 

45 2t sin 2 5

m:

(4)

Evaluating s in Eq. (4) at the times in Eqs. (2), we have

s 

jamax j 1

D 12:84 m

and

s 

jamax j 2

D 128:5 m:


June 25, 2012


Dynamics 2e

83

Problem 2.43

D p

The acceleration of a sled is prescribed to have the following form: a ˇ t , where t is time expressed 0 . Determine ˇ in such a way that the in seconds, and ˇ is a constant. The sled starts from rest at t distance traveled after 1 s is 25 ft.

D

Solution

The acceleration is a variable as follows:

D dv=dt . Since a is given as a function of time, we can separate the velocity and time dv dt

D a.t/ ) dv D a. t/ dt: (1) p Recalling that a.t/ D ˇ t and that the velocity is equal to zero for t D 0 , we integrate the last of Eqs. (1) as follows:

Z

v

D

dv 0

Z

t

p

ˇ t dt 0

)

v.t/

D 23 ˇt 3=2:

(2)

We recall that the velocity is ds=dt , so that we can write

ds dt Letting s 0 denote the value of s for t integrate the last of Eqs. (3):

s0

ds

D v. t/ dt:

(3)

D 0 and using the expression for v.t/ in the last of Eqs. (2), we can

s

Z

D v.t/ ) t

ds

D

Z

0

2 ˇt 3=2 dt 3

s  s0

)

D 154 ˇt 5=2:

(4)

D 1 s, we can rewrite the last of Eqs. (4) as follows: 4 d D 15 ˇt 5=2;

Letting d denote the distance traveled during  t

(5)

which can be solved for ˇ to obtain

ˇ Recalling that d

D 415dt 5=2 :

(6)

D 25 ft and that t D 1 s, we can evaluate Eq. (6) to obtain ˇ

D 93:75 ft=s = : 5 2


June 25, 2012


Solutions Manual

84

Problem 2.44

D p D D

ˇ1 t , a ˇ2 t , The acceleration of a sled can be prescribed to have one of the following forms: a ˇ3 t 2 , where t is time expressed in seconds, ˇ 1 1 m=s5=2 , ˇ 2 1 m=s3 , and ˇ 3 1 m=s4 . and a The sled starts from rest at t 0 . Determine which of the three cases allows the sled to cover the largest distance in 1 s. In addition, determ ine the distance cover ed for the case in question.

D

D

D

D

Solution

We will determine how position depends on time for each of the three cases. We observe that the acceleration a , we can separate velocity and is given as a function of time in each of three cases. Recalling that dv=dt time by writing

D

dv dt

D a.t/ ) dv D a. t/ dt: Recalling that the velocity is equal to zero for t D 0 , we can integrate the last of Eqs. (1) as follows:

Z

v

dv 0

D

Z

(1)

t

a.t/dt:

(2)

0

Using the expression for a.t/ for the three given cases, we have, respectively,

v

D 23 ˇ1t 3=2;

v

D 12 ˇ2t 2;

and

v

D 13 ˇ3t 3:

(3)

We recall that the velocity is ds=dt . Since the velocity is now known as a function of time, we can separate the position and time by writing

ds dt

D v.t/ ) ds D v. t/ dt: Letting s 0 denote the value of s for t D 0 and using the expression for Eqs. (4) as follows:

Z

s

ds s0

D

Z

t

)

v.t/dt 0

s  s0

D

Z

(4)

v.t/ , we can integrate the last of

t

v.t/dt:

(5)

0

Substituting into the last of Eqs. (5) the expression s for v.t/ given in Eqs. (3), we have

d

D 154 ˇ1t 5=2;

d

D 61 ˇ2t 3;

and

d

D 121 ˇ3t 4;

(6)

where we have denoted by d the distance s  s0 traveled by the sled as a function of time. Recalling that ˇ1 1 m=s5=2 , ˇ 2 1 m=s3 , ˇ 3 1 m=s4 , we can evaluate Eqs. (6) for t 1 s to obtain, respectively,

D

D

D

d

D 0:2667 m;

D

d

D 0:1667 m;

and

d

D 0:08333 m:

(7)

Comparing the three values of d we conclude that Largest distance traveled in 1 s is d

D 0:2667 m, corresponding to a D ˇ 1pt :


June 25, 2012


Dynamics 2e

85

Problem 2.45 A peg is constrained to move in a rectilinear guide and is given the following acceleration: a 20 ft=s2 , ! 250 rad=s, and t is time expressed in seconds. where a 0 0 and v 0 for t 0 , determine the position of the peg at t 4 s. If x

D D

D

D

D

D a0 sin !t ,

D

Solution

D dv=dt . Since a is given as a function of time, we can separate the variables dv D a. t/ dt ) dv D a 0 sin !t dt: Since v D 0 for t D 0 , we can integrate the last of Eqs. (1) as follows:

Recall that a writing

Z

v

D

dv 0

Z

t

a0 sin !t dt 0

)

v

D a!0 .1



v and t by (1)

cos !t/:

(2)

D

v . Since the velocity is now a known function of time, applying separation of Next recall that dx=dt variables for this case, we have dx

D v. t/ dt )

dx

D a!0 .1

where we have made use of the last of Eqs. (2). Recalling that x Eqs. (3) as follows:

Z Recalling that a 0

x

D

dx 0

2

D 20 ft=s

Z

t 0

and !

a0 .1 !



cos !t/dt

)

x



cos !t/dt;

(3)

D 0 for t D 0, we can integrate the last of D a!0



t



1 sin !t : !



(4)

D 250 rad=s, at t D 4 s, we have x.4 s/ D 0:3197 ft:


June 25, 2012


Solutions Manual

86

Problem 2.46 A peg is constrained to move in a rectilinear guide and is given the following acceleration: a 20 ft=s2 , ! 250 rad=s, and t is time expressed in seconds. where a 0 0 so that x.t/ is periodic. Determine the value of the velocity of the peg at t

D

D

D a0 sin !t ,

D

Solution

Recall that the acceleration is dv=dt . Hence, applying separ ation of variables for the case in which acceleration is given as a function of time, we can write

D a. t/ dt ) dv D a0 sin !t dt: Letting v 0 be the value of the velocity for t D 0 , we can integrate the last of Eqs. (1) as follows: dv

Z

v

dv v0

D

Z

t

)

a0 sin !t dt 0

D v0 C a!0 .1

v



(1)

cos !t/:

(2)

D

v . Since the velocity is now a known function of time, applying separation of Next recall that dx=dt variables for this case, we have dx

D v. t/ dt )

dx

D

C a!0 .1



v0





cos !t/ dt;

where we have made use of the last of Eqs. (2). Letting x 0 be the value of x for t last of Eqs. (3) as follows:

Z

x

dx 0

D



v0

C a!0 .1





)

cos !t/ dt

x

D x0 C v0t C a!0



t

(3)

D 0, we can integrate the 

1 sin !t : !



(4)

To answer the question in the problem statement, we rewrite x as follows:

x

D x0 C



v0

C a!0



t



a0 sin !t: !2

(5)

We now observe that the first term on the right-hand side of Eq. (5) is a constant and therefore is a special case of a periodic function. The second term on the right-hand side of Eq. (5) is linear in t and therefore is not periodic. Finally, the third term term on the right-hand side of Eq. (5), is a periodic function of time with period p 2  rad=! . We therefore concl ude that for the motion to be periodic, we must require that

D

v0 Recalling that v 0

C a!0 D 0 )

v0

D



a0 : !

(6)

D v.0/, a0 D 20 ft=s , and ! D 250 rad=s, we can evaluate the last of Eqs. (6) to obtain v.0/ D 0:08000 ft=s: 2




June 25, 2012


Dynamics 2e

87

Problem 2.47

D

2:5 m off the ground A ring is thrown straight upward from a height h 3:45 m=s. Gravity causes the rin g to and with an initial velocity v 0 9:81 m=s2 . Determine h max , have a constant downward acceleration g the maximum height reached by the ring.

D

D

Solution

We note that the motion of the ring occurs with constant acceleration equal to g and directed downward. Denoting the release height of the ring by s 0 , and using the constant acceleration equation relating position and velocity, we have v 2 v 02  2g.s  s0 /; (1)

D

where v 0 is the value of the speed corresponding to the position value s 0 . The maximum height is the value of s corresponding to v 0 . Hence, setting s h max and v 0 in the above equation, and solving for h max , we have

D

D

D

2

hmax Recalling that v

0

D

3:45 m=s, s

0

D

2:5 m, and g hmax

v0 D 2g C s0 :

(2)

9:81 m=s2 , we can evaluate Eq. (2) to obtain

D D 3:107 m:


June 25, 2012


Solutions Manual

88

Problem 2.48

D

A ring is thrown straight upward from a height h 2:5 m off the ground. Gravity causes the ring to have a constant downward acceleration g 9:81 m=s2 . Letting d 5:2 m, if the person at the window is to receive the ring in the gentlest possible manner, determine the initial velocity v 0 the ring must be given when first released.

D

D

Solution

We note that the motion of the ring occurs with constant acceleration equal to g and directed downward. Denoting the release height of the ring by s 0 , and using the constant acceleration equation relating position to velocity, we have v 2 v 02  2g.s  s0 /; (1)

D

where v 0 is the value of the speed corresponding to the position value s 0 . The person receives the ring in the gentlest possible manner when the ring reaches the height h d with velocity equal zero. Hence, substituting s 0 h , s .h d /, and v 0 into Eq. (1) and setting the outcome equal to zero, gives 0 v 02  2gd; (2)

D

D C D

D

C

which is an equation that can be solved for v 0 to obtain

v0

D

p

2gd;

(3)

where we have selected the positive root since the ring is thrown in the positive s direction. Recalling that g 9:81 m=s2 and d 5:2 m, we can evaluate Eq. (3) to obtain

D

D

v0

D 10:10 m=s:


June 25, 2012


Dynamics 2e

89

Problem 2.49

7 m=s when A hot air balloon is climbing with a velocity of a sandbag (used as ballast) is released at an altitude of 305 m. Assuming that the sandbag is subject only to gravity and that therefore its acceleration is given by y g , g being the acceleration due to gravity, determine how long the sandbag takes to hit the ground and its impact velocity.

RD

Solution g ,

Since the acceleration of the sandbag is constant and equal to following equation

y

P

D y0 C yP0t

where y 0 and y0 are the position and velocity at time t sandbag hits the ground, at t t impact we must have

D

0

D y0 C yP0t

position and time are related by the

1 2  2 gt ;

D 0. Letting t 1

(1) impact denote the time at which the

2

impact  2 gt impact:

(2)

This is a second order algeb raic equation for t impact whose only physically admissi ble solution is

timpact Recalling that g

D g1 yP0 C yP02 C 2gy0:

q

2

(3)

D 9:81 m=s , yP 0 D 7 m=s, and y0 D 305 m, we can evaluate Eq. (3) to obtain t D 8:631 s: impact

The expression of the velocity can be obtained by differentiating Eq. (1) with respect to time. This gives

P D yP0

y



gt :

(4)

The impact velocity can be found by substituti ng t impact from Eq. (3) into the (constant acceleration) Eq. (4), which, after simplification, gives

q

P D yP02 C 2gy0: Recalling again that g D 9:81 m=s , yP 0 D 7 m=s, and y 0 D 305 m, we can evaluate Eq. (5) to obtain yP D 77:67 m=s: yimpact

2

impact



(5)




June 25, 2012


Solutions Manual

90

Problem 2.50 Approximately 1 h 15 min into the movie King Kong (the one directed by Peter Jackson), there is a scene in which Kong is holding Ann Darrow (played by the actress Naomi Watts) in his hand while swinging his arm in anger. A quick analysis of the movie indicates that at a particular moment Kong displaces Ann from rest by roughly 10 ft in a span of four frames. Knowing that the DVD plays at 24 frames per second and assuming that Kong subjects Ann to a constant acceleration, determine the acceleration Ann experiences in the scene into question. your on answer in terms of the acceleration due gravity Express g . Comment what would happen to a person really subjected to this acceleration. Solution

Let  t denote the time it takes to play four frames at 24 frames per second. Hence we have

t

frames D 244frames D 16 s: =s

(1)

D

t is also the time it takes for King Kong to displace Ann by a distance  s 10 ft. Since we are assuming that the acceleration is constant, the motion is rectilinear and we can apply the constant acceleration equatio n relating position to time: 1 s s 0 v0 .t  t0 / ac .t  t0 /2 ; (2) 2 where s is the position at time t , and where s0 and v0 are the position and velocity at time t t 0 , respectively.

D C

C

D

Letting  t

Dt

 t0

and  s

Ds

 s0 , solving Eq. (2) for

ac Since  t

D2

ac

gives s  v0 t

t 2

:

(3)

D 16 s, s D 10 ft, and v0 D 0, we can evaluate ac =g, with g D 32:2 ft=s , to obtain ac =g D 22:36: 2

(4)

Finally, in terms of g we can then say that the acceleration to which Ann is subject is

ac

D 22:36 g:

Since the human body cannot withstand much more that 10–15g of acceleration, an acceleration of more than 22g would likely kill Ann.


June 25, 2012


Dynamics 2e

91

Problem 2.51

D

D D

A car travels on a rectilinear stretch of road at a constant speed v 0 65 mph. At s 0 the driver applies the brakes hard enough to cause the car to skid. Assume that the car keeps sliding until it stops, and assume 0:76 is the kinetic that throughout this process the car’s acceleration is given by s k g , where  k friction coefficient and g is the acceleration of gravity. Compute the car’s stopping distance and time.

RD

Solution

To determine the stopping distance, we recall that

R D ddtsP ) sR D ddssP ds ) sR D sP ddssP ; dt

s

(1)

where we have used the chain rule and the definition of velocity to obtain the second and the third of Eqs. (1), respectively. Recalling that s k g , using the last of Eqs. (1), we can write

RD

D sP d sP )

k g ds

Z

s 0

k g ds

D

D

Z

sP

P P

s d s; sP0

(2)

P

0 is the position at which the brakes are applied, and where s0 is the velocity of the car for s where s Carrying out the integration in the last of Eqs. (2), we obtain k gs

1 2 .s  s02 /: 2

(3)

D P P denote the stopping distance, we have that sP D 0 for s D s

Letting s stop Eq. (3) and solving for s stop , we have

D 0.

stop .

Enforcing this condition in

2

D 2sP0k g : (4) 5280 Recalling that sP0 D v 0 D 65 mph D 65 3600 ft =s,  k D 0:76, and g D 32:2 ft=s , we can evaluate Eq. (4) to obtain s D 185:7 ft: To determine the stopping time, we recall that sR D d sP=dt . Also, recalling that sR D k g , we can then write k g dt D d sP and integrate as follows: sstop

2

stop





Z

tstop 0

k g dt

D

Z

0

P )

ds sP0

k gt stop

D sP0 ) 

tstop

D sP0=.k g/;

(5)

D 0 is the time at5280 which the brakes are applied and t denotes the stopping time. Recalling that 3600 k P D D 65 mph D 65 ft=s,  D 0:76, and g D 32:2 ft=s , we can evaluate the last of Eqs. (5) to obtain t D 3:896 s:

where t

s0

stop

2

v0

stop


June 25, 2012


Solutions Manual

92

Problem 2.52 If the truck brakes and the crate slides to the right relative to the truck, the horizontal acceleration of the 0:87 is the kinetic friction crate is given by s gk , where g is the acceleration of gravity,  k coefficient, and s is the position of the crate relative to a coordinate system attached to the ground (rather than the truck). Assuming that the crate slides without hitting the right end of the truck bed, determine the time it takes to stop if its velocity at the start of the sliding motion is v 0 55 mph.

RD

D

D

Solution

R D d sP=dt . Since sR D

To determine the stopping time, we recall that s d s and integrate as follows:

P

Z

tstop 0

k g dt

D

Z

k g , we can then write k g dt

D

D sPk0g ;

(1)

0

P )

ds sP0

k gt stop

D sP0 ) 

tstop

D 0 is the time at which the crate starts sliding and t denotes the stopping time. Recalling that P D v 0 D 55 mph D 55 5280 3600 ft =s,  k D 0:87, and g D 32:2 ft=s , we can evaluate the last of Eqs. (1) to

where t

stop

2

s0

obtain

tstop

2:880 s:

D


June 25, 2012


Dynamics 2e

93

Problem 2.53 If the truck brakes and the crate slides to the right relative to the truck, the horizontal acceleration of the 0:87 is the kinetic friction crate is given by s gk , where g is the acceleration of gravity,  k coefficient, and s is the position of the crate relative to a coordinate system attached to the ground (rather than the truck). Assuming that the crate slides without hitting the right end of the truck bed, determine the distance it takes to stop if its velocity at the start of the sliding motion is v 0 75 km=h.

RD

D

D

Solution

To determine the stopping distance, we recall that

R D ddtsP ) sR D ddssP ds ) sR D sP ddssP ; dt

s

(1)

where we have used the chain rule and the definition of velocity to obtain the second and the third of Eqs. (1), respectively. Recalling that s k g , using the last of Eqs. (1), we can write

RD

k g ds

D sP d sP )

Z

s 0

k g ds

D

Z

sP

P P

s d s; sP0

(2)

where s 0 is the position at which the brakes are applied and s0 is the velocity of the crate at s Carrying out the integration in the last of Eqs. (2), we obtain

D

P

D 21 .sP2 sP02/: Letting s denote the stopping distance, we have that sP D 0 for s D s Eq. (3) and solving for s , we have 2 s D sP0 : k gs

D 0. (3)



stop .

stop

Enforcing this condition in

stop

stop

(4)

2k g

P D v 0 D 75 km=h D 75 1000 3600 m =s,  k D 0:87, and g D 9:81 m=s , we can evaluate Eq. (4) s D 25:43 m: 2

Recalling that s0 to obtain

stop


June 25, 2012



Dynamics 2e

95

Problem 2.55 A sphere is dropped from rest at the free surf ace of a thick polym er fluid. The g  v , where g is the acceleration due acceleration of the sphere has the form a to gravity,  is a constant, and v is the sphere’s velocity. 50 s1 determine the velocity of the sphere after 0:02 s. Express the If  result in feet per second.

D

D

Solution

D

dv=dt . Here the acceleration is given as a function of velocity, i.e., a We recall that a we have a.v/ dv=dt so that we can separate the v and t variables by writing

D

dt

D

dv D a.v/ )

dt

D g dvv :

(1)



D

0:02 s and vf v.tf /. Since the sphere is released from rest at time t Let tf last of Eqs. (1) as follows:

Z

tf

dt 0

D

Z

vf 0

dt g  v

)

tf

D





D a.v/. Therefore

1 ln .g   v/ 

D 0, we can integrate the

vf



0

)

tf

D



g   vf 1 ln  g





;

(2)

D

ln .a=b/ . The last of Eqs. (2) where we have used the logarithm property according to which ln a  ln b can be solved for vf . To do so, we first isolate the logarithmic term and then take the exponential of both sides of the resulting equation:  tf

Recalling that vf Eqs. (3) to obtain

D ln

g

vf g





)

e  tf

Dg

 vf g



)

vf

D g 1



e  tf :



(3)

D v.0:02 s/, g D 32:2 ft=s ,  D 50 s , and tf D 0:02 s, we can evaluate the last of 2

v.0:02 s/

1

D 0:4071 ft=s:


June 25, 2012


Solutions Manual

96

Problem 2.56 The motion of a peg sliding within a rectilinear guide is controlled by an actuator in such a way that a 0 .2 cos 2! t  ˇ sin !t/, where t is time, a 0 3:5 m=s2 , the peg’s acceleration takes on the form x ! 0:5 rad=s, and ˇ 1:5 . Determine the expressions for the velocity and the position of the peg as functions of time if x.0/ 0 m=s and x.0/ 0 m.

D

RD

D

D

P D

D

Solution

R D d x=dt P D a0.2 cos 2! t

Since x



ˇ sin !t/, we can write



ˇ sin !t/dt

D d xP )

Z

Z

t

xP

D 0 d x;P (1) 0 where, in choosing the limits of integration, we accounted for the fact that xP D 0 for t D 0 . Carrying out the integration in the second of Eqs. (1) and solving for xP , we have a0 xP D . sin 2! t C ˇ cos !t ˇ/: (2) ! Recalling that a 0 D 3:5 m=s , ! D 0:5 rad=s, and ˇ D 1:5 , we can express Eq. (2) as a0 .2 cos 2! t

a0 .2 cos 2!t



ˇ sin !t/dt



2

x

7:000 sinŒ.1:000 rad=s/t ç

PD where t is in seconds. Recalling that xP D dx=dt

˚

10:50 cosŒ.0:5000 rad=s/t ç  10:50 m=s;

C so that we have dx

write

a0 . sin 2!t !

C ˇ cos !t

D xP d t and using the expression of xP in Eq. (2), we can

Z

t

a0 . sin 2! t !

Z

x

C ˇ cos !t ˇ/dt D 0 dx ; (3) 0 where, in choosing the limits of integration, we accounted for the fact that x D 0 for t D 0 . Carrying out the 

ˇ/dt

D dx )





integration in the second of Eqs. (3) and solving for x , we have

D 2!a02 .1 cos 2! t C 2ˇ sin !t 2ˇ!t/: Recalling that a 0 D 3:5 m=s , ! D 0:5 rad=s, and ˇ D 1:5 , we can express Eq. (4) as x



(4)



2

x

D

7:000  7:000 cosŒ.1:000 rad=s/t ç

C 21:00 sinŒ.0:5000 rad=s/t ç



.10:50 s1 /t m=s2 ;

˚

where, again, t is in seconds.


 June 25, 2012


Dynamics 2e

97

Problem 2.57 The motion of a peg sliding within a rectilinear guide is controlled by an actuator in such a way that a 0 .2 cos 2! t  ˇ sin !t/, where t is time, a 0 3:5 m=s2 , the peg’s acceleration takes on the form x ! 0:5 rad=s, and ˇ 1:5 . Determine the total distance traveled by the peg during the time interval 0 s  t  5 s if x.0/ a 0 ˇ=! .

D

RD

D

D P D

Solution

R D d x=dt P D a0.2 cos 2! t

Since x

a0 .2 cos 2!t





ˇ sin !t/, we can write

Z

D d xP )

ˇ sin !t/dt

t

a0 .2 cos 2! t



D

ˇ sin !t/dt

0

Z

xP a0 ˇ !

P

d x;

(1)

P D a0ˇ=! at t D 0. Carrying

where, in choosing the limits of integrati on, we accounted for the fact that x out the integration in the second of Eqs. (1) and solving for x , we have

P

P D a!0 .sin 2! t C ˇ cos !t/: (2) To find the total distance traveled we must first determine when xP .t/ changes sign during the specified time x

interval. To do so, we first rewrite Eq. (2) as follows:

(3) P D a!0 cos !t 2 sin !t C ˇ ; where we have used the trigonometric identity sin 2! t D 2 sin !t cos !t and then factored the cos !t term. 5 s, Then, referring to Eq. (3), and keeping in mind that we are only interested in the peg’s motion for 0 t

x.t/



we see that cos !t > 0

  rad 0
)

for 0 < !t <

where we have used the fact that !

D



D  s < 5 s;

(4)

C ˇ > 0 for all 0 < t < 5 s: (5) 5 s, xP changes sign when t D  s , the distance traveled must be computed by integratin g 2 sin !t

Since, for 0  t  the velocity as follows:

d

Z

s

D 0 a!0  cos !t.2 sin !t C ˇ/ dt D 2!a02 cos 2! t C 2ˇ sin !t 0



s

a0 2! 2

h n



2 cosŒ2!.  s /ç

i

s



s

Z

5s s

a0 2! 2

4ˇ sinŒ!. s /ç

h

a0 cos !t.2 sin !t !





1

cos 2! t

C ˇ/  dt

C 2ˇ sin !t

i

5s

s

cosŒ2!.5 s/ç  2ˇ sinŒ!.5 s/ç :

(6)

D C C C Recalling a0 D 3:5 m=s , ! D 0:5 rad=s, and ˇ D 1:5 , we can evaluate the result in Eq. (6) to obtain d D 52:42 m: 2


o

June 25, 2012


Solutions Manual

98

Problem 2.58

D

0 with an initial speed v 0 . The incline is coated with a thin A package is pushed up an incline at x viscous layer so that the acceleration of the package is given by a .g sin   v/ , where g is the acceleration due to gravity,  is a constant, and v is the velocity of the package. If  30 ı , v 0 10 ft=s, and  8 s1 , determine the time it takes for the package to come to a stop.

D

D

D

C

D

Solution

D

We recall that a dv=dt and since the acceleration is given as a function of velocity, we can separate the variables v and t by writing

dv (1) D a.v/ ) dt D g sindv C v : be the time at which v D 0 and given that v D v0 at t D 0, we can integrate the second of dt



Letting t stop Eqs. (1) as follows: tstop

0

dt 0

Z

D



v0

dv g sin   v

C

Z

)

tstop

D



1 ln .g sin  

D

ˇ

C v/ v0   ) t D 1 ln g sing sinCv0 ; 0

stop

where we have used the logarithm property ln a  ln b ln.a=b/ . Given that  30 ı , and v 0 10 ft=s, we can evaluate t stop to obtain 

D

D

tstop

(2)

D 8 s , g D 32:2 ft=s , 1

2

D 0:2233 s:


June 25, 2012


Dynamics 2e

99

Problem 2.59

D

0 with an initial speed v 0 . The incline is coated with a thin A package is pushed up an incline at x viscous layer so that the acceleration of the package is given by a .g sin   v/ , where g is the acceleration due to gravity,  is a constant, and v is the velocity of the package. 25 ı , v 0 7 m=s, and  8 s1 , determine the distance d traveled by the package before it If  comes to a stop.

D

D

D

C

D

Solution

D

We recall that a dv=dt and that, since we need to relate a change in position to a corresponding change in vdv=dx . Then, since the accelera tion is given as a function of velocity, the chain rule allows us to write a velocity, we can separate the x and v variables as follows:

D

dx Since v

vdv D a.v/ )

dx

D



v g sin 

C v d v:

(1)

D v0 for x D 0 and v D 0, for x D d , we can integrate the second of Eqs. (1) as follows: d

0

dx

Z

0



DZ

v0

0

v g sin 

dv

d

C v

)

C D

C



DZ

v0

v g sin 

d v:

(2)

C v

To facilitate the integration of the right-hand side of the last of Eqs. (2), we observe that the integrand can be rewritten as follows: v 1 g sin    : (3) g sin   v  .g sin   v/

C

Substituting Eq. (3) into the last of Eqs. (2) and carrying out the integratio n, we have

d

 0 D v0 C g sin ln.g sin  C  v/ v ) 2

ˇ

0

d

 D v0 C g sin ln 2



g sin  ; g sin   v0

D

ln.a=b/ . Given that v0 where we have used the logarithm property ln a  ln b g 9:81 m=s2 ,  25 ı , we can evaluate the last of Eqs. (4) to obtain

D

D

d

C



(4)

D 7 m=s,  D 8 s , 1

D 0:7017 m:


June 25, 2012


Solutions Manual

100

Problem 2.60 Referring to Example 2.8 on p. 56, and defining terminal velocity as the velocity at which a falling object stops accelerating, determine the skydiver’s terminal velocity without performing any integrations.

Solution

From Example 2.8 on p. 56, we have that the acceleration of the skydiver is

a

Dg



Cd 2 v : m

D r D

Denoting the terminal velocity by v term, we have that a and solving for v term, we have

vterm From Example 2.8, we have that C d Eq. (2) to obtain

43:2 kg=m, m

D

vterm

0 for v

(1)

Dv

term .

mg : Cd

(2)

9:81 m=s2 , so that we can evaluate

110 kg, and g

D

Enforcing this condition in Eq. (1)

D 4:998 m=s:

D


June 25, 2012


Dynamics 2e

101

Problem 2.61 Referring to Example 2.8 on p. 56, determine the distance d traveled by the skydiver from the instant the parachute is deployed until the difference between the velocity and the terminal velocity is 10% of the terminal velocity.

Solution

The acceleration can be related to the velocity and position as follows:

ds : (1) D dv D dv D v dv dt ds dt ds From Example 2.8 on p. 56 we have that a D g Cd v 2 =m . Hence, substituting this expression into Eq. (1), a



we can separate the variables s and v as follows:

v

ds

d v:

2

Dg



(2)

Cd v =m

D

v0 From Example 2.8, we know that when the parachute is deployed, the velocity of the skydiver is 44:5 m=s. In addit ion, we know that the termi nal velocity is vterm mg=C d , where m 110 kg, g 9:81 m=s2 , and C d 43:2 kg=m. Letting v qt 1:1vterm denote the value of the velocity that is 10% away from that of the terminal velocit y (where the subscript ‘qt’ stands for quasi-terminal), and letting d be the distance traveled to achieve v qt starting from v 0 , we can integrate Eq. (2) as follows:

D

D

D

D

Z

d

ds 0

D

Z

vqt v0

p

D

v d v: g  .Cd =m/v 2

(3)

Carrying out the above integrations, we have

d

D



2 g  .Cd =m/v qt m ln 2Cd g  .Cd =m/v 02





D



m g.1  1:12 / ln ; 2Cd g  .Cd =m/v 02





(4)

D

where we have used the expression of v qt to obtain this last expres sion. So, recalling that m 110 kg, v0 44:5 m=s, C d 43:2 kg=m, and g 9:81 m=s2 , we can evaluate the above expression to obtain

D

D

D

d

D 7:538 m:


June 25, 2012


Solutions Manual

102

Problem 2.62 In a physics experiment, a sphere with a given electric charge is constrained to move along a rectilinear a 0 sin.2 s= /, where a 0 8 m=s2 ,  is measured in radians, guide with the following acceleration: a s is the position of the sphere measured in meters,   s  , and  0:25 m. 0 and then gently nudged away from this position, what is the If the sphere is placed at rest at s maximum speed that the sphere could achieve, and where would this maximum occur?

D

D D

D

Solution

D

Recall that a dv=dt . In this problem, the accelerati on is given as a function of position. Hence, to relate a change in velocity to a correspondent change in position, we start by rewriting the acceleratio n via the chain rule: a vdv=ds . This allows us to write

D

v dv

D a. s/ ds )

v dv

D a0 sin.2 s=/ds;

(1)

where we have separated the variables v and s and used the given expressio n for the acceleration. Recalling that v 0 for s 0 , we can now integra te the last of Eqs. (1) as follows:

D

D

Z

Z

v

s

D a0 Œ1 cos.2 s= /ç: (2) p The speed is the magnitude of the velocity, namely, jv j. Since jv j D v 2 , we can solve the second of Eqs. (2) for jv j to obtain v dv

0

D

)

a0 sin.2 s= / ds

0

v2



a0 

jv j D

(3)  Œ1  cos.2 s= /ç: To determine the maximum possible value of the speed, we observe that the cosine function under the square root in Eq. (3) can vary only between the values 1 and 1 . Hence, the maximum possib le value of the speed is achieved where the cosine function takes on the value 1, which gives

r

jv j D max

Recalling that a 0

D 8 m=s

2

and 

r

2a 0  : 

(4)

D 0:25 m, we can evaluate Eq. (4) to obtain jvj D 1:128 m=s: max

jj

We have already argued that v max occurs where the cosine function under the square root in Eq. (3) achieves the value 1. In turn, this implies that

2 sjvjmax 

D  C 2 n;

n

D 0; ˙1; ˙2;::: )

sjvjmax

D 12 .1 C2n/;

n

D 0; ˙1; ˙2;:::;

(5)

v max. Recalling that  0:15 m and that   s  , where s jv jmax denotes the value of s for which v we can evaluate the second of Eqs. (5) to obtain followi ng two values for s jvjmax that are within the admissible range for s :

j jDj j

D 0:1250 m and which correspond to n D 0 and n D 1, respectively. sjvjmax

D

sjvjmax

D

0:1250 m;




June 25, 2012


Dynamics 2e

103

Problem 2.63 In a physics experiment, a sphere with a given electric charge is constrained to move along a rectilinear a 0 sin.2 s= /, where a 0 8 m=s2 ,  is measured in radians, guide with the following acceleration: a s is the position of the sphere measured in meters,   s  , and  0:25 m. Suppose that the velocity of the sphere is equal to zero for s  =4. Determine the range of motion of the sphere, that is, the interval along the s axis within which the sphere moves. Hint: Determine the speed of the sphere and the interval along the s axis within which the speed has admissible values.

D

D D

D

Solution

D

dv=dt . Since the acceler ation is given as a function of position, to relate a change in velocit y Recall that a dv=dt via the chain rule: a vdv=ds . to a correspondent change in position, we start by rewriting a This allows us to write v dv a. s/ ds v dv a 0 sin.2 s= /ds; (1)

D

D

)

D

D

where we have separated the variables v and s and used the given expressio n for the acceleration. Recalling 0 for s  =4, we can now integra te the last of Eqs. (1) as follows: that v

D

D

Z

v

v dv 0

D

Z

s  4

)

a0 sin.2 s= / ds

v2

D



a0  cos.2 s= /: 

(2)

j j D pv2, the second of Eqs. (2) implies

jj

The speed is the magnitude of the velocity, namely, v . Since v

jv j D

r



The result in Eq. (3) is acceptable if cos .2 s= / 1 2

C 2 n



2 s 

3  2

C 2 n;

n

a0  cos.2 s= /: 



(3)

0. From a mathematical viewpoint, this implies that

D 0; ˙1; ˙2;::: ) 41 .1 C 4n/



s

3  4 .1

C4n/

n

D 0; ˙1; ˙2;:::

(4)

However, from a physical viewpoint, we do not expect the sphere to “jump” from an admissible range of a0 > 0 for s =4, the sphere will move to the right when it is motion to another. Observing that a released from rest at  =4. Therefore, referring to the last of Eqs. (4), the only acceptable range of motion is 1 3 s n 0 . Since  0:25 m, we have that the range of motion is 4    4 , corresponding to

D

D

D

D

0:06250 m  s



0:1875 m:


June 25, 2012


Solutions Manual

104

Problem 2.64 The acceleration of an object in rectilinear free fall while immersed in a linear viscous g  Cd v= m, where g is the acceleration of gravity, C d is a constant fluid is a drag coefficient, v is the object’s velocity, and m is the object’s mass. 0 and v 0 0 , find the velocity as a function of time and find the Letting t 0 terminal velocity.

D

D

D

Solution

Acceleration is given as a function of velocity, so we first find time as a function of velocity and invert that result to determine the velocit y as a function of time. Recalling that a dv=dt , we can write

D

dt

D dva ) )

Z

t

dt 0

t

D

Z D 

v 0

dv g  .Cd =m/v

m ln g  .Cd =m/v Cd



ˇˇ

v 0

D

D

m ln g  .Cd =m/v Cd m g  .Cd =m/v  ln Cd g

D



˚









ln g



D Cmd ln 



1

Cd v ; mg



D

where we have used t 0 0 and v 0 0 to obtain the lower limits on the two definite integrals. Solving this result for t as a function of v to find v.t/ , we obtain

mg v.t/

Cd t=m

D C d 1



e

: !1

To find the terminal velocity, we can either take the limit as t of v.t/ or we can determine the velocit y at which a 0 in the given expression for the acceleration. Doing the latter, we obtain

D

0

Dg



Cd vterm m

)

vterm

D mg . Cd


June 25, 2012


Dynamics 2e

105

Problem 2.65 The acceleration of an object in rectilinear free fall while immersed in a linear viscous g  Cd v= m, where g is the acceleration of gravity, C d is a constant fluid is a drag coefficient, v is the object’s velocity, and m is the object’s mass. Letting s 0 0 and v 0 0 , find the position as a function of velocity.

D

D

D

Solution

Recall that acceleration, velocity, and position can be related as follows:

a

D v dv ) ds

ds

: D vdv a

(1)

Since the accelerat ion is given as a function of the velocity, we can determine the position as a function of the velocity as follows:

s.v/

D

Z

v 0

v dv g  .Cd =m/v

D

Z

v 0

C

.Cd =m/v g  g dv .Cd =m/Œg  .Cd =m/v ç

Z v

D

0

mg=C d g  .Cd =m/v



m dv ; (2) Cd

C mg ln Cd

Cd v mg



which can be evaluated to obtain

s.v/

D

mg

Cd

m g  .Cd =m/v ln Cd g







m v Cd

)

s.v/

D



m v Cd






1



:

(3)

June 25, 2012


Solutions Manual

106

Problem 2.66 A 1:5 kg rock is released from rest at the surface of a calm lake. If the resistance offered by the water as the rock falls is directly proportional to the rock’s velocity, g  Cd v= m, where g is the acceleration of gravity, the rock’s acceleration is a Cd is a constant drag coefficient, v is the rock’s velocity, and m is the rock’s mass. Letting C d 4:1 kg=s, determine the rock’s velocity after 1:8 s.

D

D

Solution

We recall that a

D dv=dt . Using the given expre ssion for a we can write Cd dv g vD ) dt D g .Cdvd =m/v ; m dt 

(1)



D

0 at where, in writing the second of Eqs. (1), we have separated the variables v and t . Observing that v t 0 , and letting vf (f stands for final) denote the value of v at t tf 1:8 s, we can integrate the last of Eqs. (1) as follows:

D

Z

D D

tf

dt 0

D

Z

vf 0

dv g  .Cd =m/v

)

tf

D



g  .Cd =m/vf m ln Cd g



)

 tf

D



m Cd ln 1  vf : (2) Cd mg





Solving the last of Eqs. (2) for vf , we have

vf Given that m

D mg Cd

1





e Cd tf =m :

(3)

2

D 1:5 kg, g D 9:81 m=s , Cd D 4:1 kg=s, and tf D 1:8 s, we can evaluate vf to obtain vf

D 3:563 m=s:


June 25, 2012


Dynamics 2e

107

Problem 2.67 A 3:1 lb rock is released from rest at the surface of a calm lake, and its acceleration g  Cd v=m, where g is the acceleration of gravity, C d 0:27 lb  s=ft is a is a constant drag coefficient, v is the rock’s velocity, and m is the rock’s mass. Determine the depth to which the rock will have sunk when the rock achieves 99% of its terminal velocity.

D

D

Solution

We begin by determining the expression of the terminal velocity, which we denote by vterm . This is the velocity at which the accelerati on is equal to zero. Using the given expres sion for the acceleration, we have

0

Dg



Cd vterm m

)

vterm

D mg : Cd

(1)

We denote by v qt (qt stads for quasi-terminal) the value of v corresponding to 99% of v term, i.e.,

vqt

99 mg D 100 : Cd

(2)

D dv=dt . To relate the acceleration to position, we can use the chain rule and write D .dv=ds/.ds=dt/ D vdv=ds . Using this expressio n and the given expression for a , we can then write

Next, we recall that a

a

g

Cd v m

v

D

dv ds

v d v; g  .C =m/v

ds

)

D D

(3)

d

D

where, in writing the second of Eqs. (3) we have separated the variables v and s . We now observe that v 0 for s 0 and letting d be the value of s corresponding to v v qt , we can integrate the second of Eqs. (3) as follows:

D

Z

d

ds 0

D

Z

vqt 0

v dv g  .Cd =m/v

)

d

D

Z

vqt 0

C

.Cd =m/v g  g dv .Cd =m/Œg  .Cd =m/v ç vqt mg=C d d g  .Cd =m/v 0

)

D

Z





m dv; (4) Cd




d

D



m2 g



Cd

ln 1  vqt mg Cd2





m vqt Cd

2

)

d

D mC 2g d



where, in writing the last of Eqs. (5), we have used Eq. (2). Recalling that m Cd 0:27 lb  s=ft, we can evaluate the last of Eqs. (5) to obtain

D

d

ln.100/ 

99 ; 100



(5) 2

D 3:1 lb=g, g D 32:2 ft=s , and

D 14:80 ft:


June 25, 2012


Solutions Manual

108

Problem 2.68 A 3:1 lb rock is released from rest at the surface of a calm lake, and its acceleration g  Cd v=m, where g is the acceleration of gravity, C d 0:27 lb  s=ft is a is a constant drag coefficient, v is the rock’s velocity, and m is the rock’s mass. Determine the rock’s velocity after it drops 5 ft.

D

D

Solution

D dv=dt . To relate the acceleration to position, we can use the chain rule and write D .dv=ds/.ds=dt/ D vdv=ds . Using this expressio n and the given expression for a , we can then write Cd dv g v Dv ) ds D g .Cvd =m/v d v; (1) m ds where, in writing the second of Eqs. (1) we have separated the variables v and s . We now observe that v D 0 for s D 0 so that we can integrate the second of Eqs. (1) as follows: s v v v .Cd =m/v C g g ds D dv ) s D dv

We recall that a

a



Z

0

Z

0



Z

g  .Cd =m/v



0

.Cd =m/Œg



.Cd =m/v ç

Z v

)

s

D

0

mg=C d g  .Cd =m/v



m dv; (2) Cd




s

D



m2 g Cd2



ln 1 

Cd v mg





D

m v: Cd

(3)

5 ft. Since this cannot be done analytically, We now need to solve the above equation for v after setting s 3:1 lb=g , g 32:2 ft=s2 , and C d 0:27 lb  s=ft, the we will need to do it numerically. Given that m solution presented below was obtained using Mathematica via the following code:

D

D

D

where we note that, as required by most root finding algorithms, one needs to specify an initial guess for the 1 ft=s). The execution of the above code, gives the followi ng result (expressed to solution (we have used v 4 significant figures)

D

v.5 ft/

D 10:07 ft=s:


June 25, 2012


Dynamics 2e

109

Problem 2.69

D

g  Cd v= m, where the Suppose that the acceleration of an object of mass m along a straight line is a constants g and C d are given and v is the object’s velocity. If v.t/ is unknown and v.0/ is given, can you determine the object’s velocity with the following integral?

Z t

v.t/

D v.0/ C

0

g

Cd v dt: m



Solution

No, because the integrand is not an explicit function of the variable of integration, which is t . Clearly, if v.t/ is provided as an explicit function of time then one could integrate . However, if v.t/ were given it would seem superfluous to perform the integration to obtain it a second time.


June 25, 2012


Solutions Manual

110

Problem 2.70 Heavy rains cause a particular stretch of road to have a coefficient of friction that changes as a function of location. Specifically, measurements indicate that the friction coefficien t has a 3% decrease per meter. Under these conditions the acceleration of a car skidding while trying to stop can be approximated by s .k  cs/g (the 3% decrease in friction was used in deriving this equation for acceleration), where  k is the friction coefficient under dry conditions, g is the acceleration of gravity, and c , with units of m 1 , describes

RD

1

m Deterthe rate friction decrement. , and 0:5 , ty c of 0:015  k veloci v 45ofkm =h, where v 0 is theLet initial the car. 0

D

D

D

mine the distance it will take the car to stop and the percentage of increase in stopping distance with respect to dry conditions, i.e., when c 0.

D

Solution

D

dv=dt . The acceleration can be related to the position and velocity We recall that the acceleration is a .dv=ds/.ds=dt/ vdv=ds , where we have also used the fact that ds=dt v. using the chain rule: a Hence, observing that a s .k  cs/g , we can write

D DRD

D

D

D v dv ) ds

(1) D v dv; where we have separated the variables s and v . We now observe that v D v 0 for s D 0 . In addition, letting s denote the stopping distance over wet ground, we have that v D 0 for s D s . Therefore, the last of .k 

cs/g

. k 

cs/ g ds

wet

wet

Eqs. (1) can be integrated as follows:

Z

swet 0

. k 

cs/ g ds

D

Z

0

v dv

v0



)



k swet

C 12 cs 2

g D

wet

1 2  2 v0 :

(2)

Dividing the last of Eqs. (2) by g , multiplying by 2, and rearranging terms, we have

q

C v02=g D 0 ) s D 1c k ˙ 2k .cv 02/=g : (3) Given that c D 0:015 m ,  k D 0:5 , v 0 D 45 km=h D 45.1000=3600/ m=s, and g D 9:81 m=s , we can 2 cs wet  2k swet



wet

1





2

evaluate the last of Eqs. (3) to obtain the followin g two values of swet :

.swet /1

D c1



k 

q

2k



.cv 02 /=g



D 26:31 m

and

.swet/2

D c1



k

C

q

2k



.cv 02 /=g



D 40:35 m:

(4) Only the solution swet .s wet/1 is meaningful because for the is not assumed to keep moving after it comes .s wet/1 . Hence, we have to a stop at s wet

D

D

swet

D 26:31 m:

For dry conditions, c 0 so that s k g , which implies that the acceleration is constant and we can use v 02 2a c .s  s0 / to determine the stopping distance. Denoting by s dry the position s at the equation v 2 which v 0 under dry conditions, we have

D

D D C

0

D v02

RD

2



2k g.s dry  s0 /

)

sdry

D 2v0k g : )

sdry

D 15:93 m;


(5) June 25, 2012


Dynamics 2e

111

D

D

D

0, and where we have used the following numerical values: k 0:5, v0 where we have set s0 12:50 m=s, and g 9:81 m=s2 . Using the values of swet and sdry in the first of Eqs. (4) and the last of Eqs. (5), respectively, the percentage increase in stopping distance is calculated as follows:

D

.swet  sdry/ .100%/ sdry

D 2v2k g 0

1

0:015 m1 ,  k 0:5, v 0 Recalling that c can evaluate the above expression to obtain

D

D

c

k



q

2k



.cv 02 /=g





v02 / .100%/: 2 k g



(6) 2

D 45 km=h D 45.1000=3600/ m=s, and g D 9:81 m=s , we

.swet  sdry/ .100%/ sdry

D 65:21%:


June 25, 2012


Solutions Manual

112

Problem 2.71 A car stops 4 s after the application of the brakes while covering a rectilinear stretch 337 ft long. If the motion occurred with a constant acceleration a c , determine the initial speed v 0 of the car and the acceleration ac . Express v 0 in mph and a c in terms of g , the acceleration of gravity.

Solution

D

D

dv=dt . Since a ac is constant, we can separate the v and t variables by writing Recall that a ac d t . Letting t 0 be the time at which the brakes are applied and for which v v0 , we can integrate the expression dv a c d t as follows

dv

D

D

Z Recalling that v

D

D

v

D

dv v0

Z

t

ac d t 0

)

v

D v 0 C ac t:

(1)

D ds=dt , we can write ds D v dt and use the expressi on in the last of Eqs. (1) to write

Z

s

ds 0

D

Z

t

.v0 0

C ac t/dt )

s

D v 0t C 12 ac t 2;

(2)

D

0 to be the value of s when the brakes are applied. Letting t s and d s be stopping where we have chosen s time and distance, respec tively, and using the last of Eqs. (1) and of Eqs. (2), we must have 0

D v 0 C ac ts

and

ds

D v 0ts C 21 ac ts2:

(3)

This is a system of two equations in the two unknowns v 0 and ac whose solution is

v0 Recalling that t s

D 2ds =t s

and

ac

D

2 2d s =t s :

(4)

D 4 s and ds D 337 ft we can evaluate the above results to obtain v0 D 114:9 mph and ac D 1:308g; 

where we have expressed the acceleration in term of g

2

D 32:2 ft=s , the acceleration of gravity.


June 25, 2012


Dynamics 2e

113

Problem 2.72 As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given by  .g=L/ sin  , where g is the acceleratio n of gravity and L is the length of the pendulum cord. Derive the expression of the angular velocity  as a function of the angular coordinate  . The initial conditi ons are  .0/  0 and  .0/ 0 .

RD

P

D

P DP

Solution

R D d P=dt . Applying the chain rule, we write d P d P d  d P R D P (1) D D ) P d P D R d  : dt d  dt d Substituting the given expression for R into the last of Eqs. (1), we can then write P d P D .g=L/ sin  d  ; (2) where we have separated the variables P and  . Since P D P 0 for  D  0 , we can integrate Eq. (2) as follows: Recall that 



P

P

Z

0

P P



 d

.g=L/ sin 

DZ P

)

0

P

P

 2  0

1 2

d



Solving the last of Eqs. (3) for  as a function of  , we have

r

P D ˙ P02 C 2 g  cos  L

 . /



g

 D L.

cos 



cos 0 /:

(3)



cos 0 :


June 25, 2012


Solutions Manual

114

Problem 2.73 As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given by  .g=L/ sin  , where g is the acceleratio n of gravity and L is the length of the pendulum cord. 1:5 m. If  3:7 rad=s when Let the length of the pendulum cord be L  14 ı , determine the maximum value of  achieved by the pendulum.

RD

PD

D

D

Solution

PD RD P P P d P R D D dd ddt D P dd ) P d P D R d  : (1) dt Substituting the given expression for R into the last of Eqs. (1), we can then write P d P D .g=L/ sin  d  ; (2) where we have separated the variables P and  . Letting Pi D 3:7 rad=s and  i D 14 ı be the initial angular 0 . To find  max we need to relate Let  max be the maximum value of  . This value is achieved when  velocity with position. To do so, we recall that  d  =dt and, applying the chain rule, we write



velocity and the initial angle, respectively, we can then integrate Eq. (2) as follows:

Z

0 Pi

P PD

d

Z

max

.g=L/ sin 

)

d

i

P D g .cos  L

1 2  2 i

max 

cos i /:

(3)

The last of Eqs. (3) can be solved for  max to obtain

max Recalling that L

D cos1



cos i



P

Li2 : 2g



(4)

D 1:5 m, i D 14ı, Pi D 3:7 rad=s, and g D 9:81 m=s  D 94:38 ı:

2

max


June 25, 2012


Dynamics 2e

115

Problem 2.74 As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given by  .g=L/ sin  , where g is the acceleratio n of gravity and L is the length of the pendulum cord. The given angular acceleration remains valid even if the pendulum cord is re5:3 ft and assume that the placed by a massless rigi d bar. For this case, let L 0 ı . What is the minimum angular velocity at pendulum is placed in motion at  this position for the pendulum to swing through a full circle?

RD

D

D

Solution

P

D

0 for which the pendulum swings through a full circle is the value that The minimum value of  at   rad with  0 . This means that we need to relate velocity allows the pendulum to reach the angle  d  =dt and, applying the chain rule, we write with position. To do so, we recall that 

PD D RD P P P d P R D D dd ddt D P dd ) P d P D R d  : (1) dt Substituting the given expression for R into the last of Eqs. (1), we can then write P d P D .g=L/ sin  d  ; (2) where we have separated the variables P and  . Then, using the above observation concerning the value of P for   rad, we can integrate Eq. (2) as follows: D 0 1 P2 .g=L/ sin  d  ) P d P D D 2 Lg : (3) 2 P 0 

Z

Z

min

 rad





min



P

The last of Eqs. (3) can be solved for min to obtain

P D ˙2

min Recalling that g

2

D 32:2 ft=s

and L

r

g : L

(4)

D 5:3 ft, the result in Eq. (4) can be evaluated to obtain P D 4:930 rad=s; min

P

where we have considered only the positive value of min because in going from 0 to  rad the pendulum bob moves counterclockwise.


June 25, 2012


Solutions Manual

116

Problem 2.75 As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given .g=L/ sin  , where g is the acceleratio n of gravity and L is the length of by  the pendulum cord. Let L 3:5 ft and suppose that at t 0 s the pendulum’s position is  .0/ 32 ı with  .0/ 0 rad=s. Determine the pendulum’s period of oscillation, i.e., from its initial position back to this position.

RD

D P D

D

D

Solution

To determine the period of the pendulum using the given initial conditions, we need to establish a relationship between the angle  and time. To do so, we begin by establis hing a relation between the angular velocity  and and swing angle  , and then we integrate that result to determine  .t/. To find  . /, we begin by applying the chain rule as follows

P

P

R D d P D d P d D P d P ) P d P D R d  ) P d P D R d  : dt d  dt d Substituting the given expression for R into the last of Eqs. (1), we can then write P d P D .g=L/ sin  d  : 

(1)

(2)



ı

P D 0 when  DP 0 D 32 , we can integrate Eq. (2) as follows: g P d P D .g=L/ sin  d  ) 12 P 2 D .cos 

Since 

Z



0

Z





L

0

P

Solving for  , we have

r



cos 0 /:

(3)

2g P D ˙ . cos  cos 0 /: (4) L ı ı Once the pendulum is released from  0 D 32 , the angle  will decrease until it becomes 32 . Then the ı ı 



pendulum will then swing back to the srcinal angle. The time taken to go from 32 to 32 is equal to the time taken to swing back from 32ı to 32 ı . Hence, the period of oscillation, which we will denote by p , is twice the time that the pendulum takes to go from 32 ı to 32ı . With this in mind, referring to Eq. (4), and because  will initially decrease after release, the expression for  to use when  goes from 32 ı to 32ı is

P

r PD





2g . cos  L



cos 0 /;

(5)

P D d =dt , we can write d t D d =P , which, because of Eq. (5), we can write

Now, recalling that 

dt

D



s

L 2g

pcos d cos  

:

(6)

0


June 25, 2012


Dynamics 2e

117

We now integrate over one half of a complete swing to obtain

Z

ı

dt 0

ZD s 32

p=2



32

ı

L 2g

pcos 



ZD s 32

ı

d cos 32ı

)

p

The above integral can be evaluated numerically. Recalling that L Mathematica with the following code:



32

ı

2L g

pcos  d cos 32ı : 

2

D 3:5 ft and g D 32:2 ft=s , we have used

which yields

p

D 2:113 s:


June 25, 2012


Solutions Manual

118

Problem 2.76 As we will see in Chapter 3, the acceleration of a particle of mass m suspended by a linear spring with spring constant k and unstretched length L0 (when the spring x length is equal to L0 , the spring exerts no force on the particle) is given by g  .k=m/.x  L0 /. Derive the expression for the particle’s velocity x as a function of position x . Assume that at t 0 , the particle’s velocity is v 0 and its position is x 0 .

RD

P

D

Solution

RDP P

xd x=dx , which can then be The acceleration can be related to the position and the velocity as follows: x rewritten as xd x xd x . This latter expression can be integrated as follows:

P PDR

Z

xP

Z x

k 



(1)

where, as indicated in the problem statement, v 0 is the value of x for x us the velocity as a function of x .

D x0. Evaluating the integral gives

P PD

xdx v0

g

x0

m

.x  L0 / dx ;

P

1 2 x  12 v02 2

P

D g.x



x0 / 

k x 2  x02 2m



 C kL .x 0

m



x0 /:

(2)

P

Solving for x , we have

x

P D˙

s

v02

C 2g C kLm0 .x



x0 /  k x 2  x02 : m






(3)

June 25, 2012


Dynamics 2e

119

Problem 2.77 As we will see in Chapter 3, the acceleration of a particle of mass m suspended by a linear spring with spring constant k and unstretched length L0 (when the spring x length is equal to L0 , the spring exerts no force on the particle) is given by g  .k=m/.x  L0 /. 100 N=m, m 0:7 kg, and L 0 0:75 m. If the particle is released from Let k rest at x 0 m, determine the maximum length achieved by the spring.

RD

D D

D

D

Solution

RDP P


P PDR

Z

xP v0

Z x

P PD

xdx

g

x0

k  m



.x  L0 / dx ;

P

(1)

D

where as indicated in the problem statement, v 0 is the value of x for x x 0 . Now, in this particular problem, since the particle starts from rest at x 0 , we set v 0 0 and x 0 0 . We now observe that the coordinate x (when positive) measures the length of the spring. In addition, denoting the maximum length by x max , we observe that x max is achieved when x 0 , i.e., when the spring has stretched to the point that its velocity is equal to zero (before recoi ling back). Using these considerat ions, we can rewrite Eq. (1) as

D PD

Z

0

P PD

xdx 0

Z

xmax

 k  g

0

m

D



.x  L0 / dx

)

D

0

D gx

max 

k 2 x 2m max

C kLm0 x

max :

(2)

Solving the last of the above equati ons for x max, we obtain

D 2.mg Ck kL0/ : (3) Recalling that m D 0:7 kg, g D 9:81 m=s , k D 100 N=m, and L 0 D 0:75 m, we can evaluate Eq. (3) to obtain x D 1:637 m: xmax 2

max


June 25, 2012


Solutions Manual

120

Problem 2.78 As we will see in Chapter 3, the acceleration of a particle of mass m suspended by a linear spring with spring constant k and unstretched length L0 (when the spring x length is equal to L0 , the spring exerts no force on the particle) is given by g  .k=m/.x  L0 /. 8 lb=ft, m 0:048 slug, and L 0 2:5 ft. If the particle is released from Let k 0 ft, determine how long it takes for the spring to achieve its maximum rest at x length. Hint: A good table of integrals will come in handy.

RD

D D

D

D

Solution

RDP P


P PDR

Z

xP

P PD

xdx v0

Z

x

 k  g

x0

m



.x  L0 / dx ;

P

where as indicated in the problem statement, v 0 is the value of x for x the velocity as a function of x . 1 2 x  12 v02 2

P

D g.x



x0 / 

k x 2  x02 2m



(1)

D x0. Evaluating the integral gives us

 C kL .x 0

m

x0 /:



(2)

P

Then, keeping in mind that we are interested in the motion of the spring for x  0 (i.e., we are not interested in the recoiling motion of the spring after it has stretched to its maximum length), and solving for x , we have

PD

x

s

C 2 g C kLm0

v02





D r PD

0 , we have v 0

Since the particle is released from rest at x simplified to obtain

2.mg

x

.x  x0 / 

k 2 x  x02 : m





(3)

D 0. Consequently, the above equation can be

C kL0/ x

m

P



k 2 x : m

(4)

P D 0. Hence, setting xP D 0

Next, we determine the maximum length of the spring, which is achieved when x and x x max in Eq. (4) and solving for x max , we have

D

2.mg

C kL0/ x

m

k 2 x m max

max 

D0 )

xmax

D 2.mg Ck kL0/ )

2.mg

C kL0/ D x

maxk:

(5)

Substituting the last of Eqs. (5) into Eq. (4), we have

k k xx max  x 2 m m

PD

x

r Z

k m

) xP D

x xmax  x :

r q  rZ p D

(6)

P D dx=dt . Therefore, we can rearrange the terms in the last of Eqs. (4) to integrate as

Now we recall that x follows:

P D dx ) dt

x

txmax

0

dt

D

Z

xmax 0

dx x

P )

txmax

m k

xmax

0

dx x .x max  x/


:

(7)

June 25, 2012


Dynamics 2e

121

The integral on the right-hand side of the last of Eqs. (7) can be carried out by substitution, or by consulting a table of integrals, or by using a symbolic mathematical software. Regardles s of the method, we have

Zp

dx x .x max 

D 2 sin1 x/

r x  xmax

C C;

(8)

where C is a constant of integration . Then using the above resul t we have that the last of Eqs. (7) gives

txmax

D

r

m 2Œ sin1 .1/  sin1 .0/ç k

D

where we have used the following numerical data: m

r

m k

)

txmax

D 0:2433 s;

(9)

D 0:048 slug and k D 8 lb=ft.


June 25, 2012


Solutions Manual

122

Problem 2.79

D

18 kg is attached to the free end of a nonlinear spri ng such A weight A with mass m g  . =m/.y  L0 /3 , where g is the acceleration that the acceleration of A is a 0:5 m. Determine  such that A does not fall due to gravity,  is a constant, and L 0 below y 1 m when released from rest at y L 0 .

D

D

D

D

Solution

D

Recall that a dv=dt . In this problem, the accelerat ion is given as a function of position. Hence, to relate, a change in velocity to a correspondent change in position, we start by rewriting the acceleratio n via the chain vdv=dy . This allows us to write rule: a

D

v dv

D a.y/ dy )

v dv

D



g

 .y m





L0 /3 dy;

(1)

where we have separated the variables v and s and used the given expressio n for the acceleration. Recalling that v 0 for y L 0 , we can now integra te the last of Eqs. (1) as follows:

D

D

v

y

v dv

QD

Z

0

g

DZ  L0



.y  L0 /3 dy

m



v2

)

2g.y

D

Q



L0 / 



.y  L0 /4 :

1 m. In order for the weight not to fall below y , the speed must become equal to zero at y Let y Hence, letting v 0 for y y , the last of Eqs. (2) gives

D

DQ

0

D 2g.yQ

(2)

2m

D yQ .

 .y  L 0 /4 ; 2m

Q

(3)

D .yQ 4mgL0/3 :

(4)



L0 / 

which is an equation in  whose solution is

 Recalling that m



2

D 18 kg, g D 9:81 m=s , L0 D 0:5 m, and yQ D 1 m, we can evaluate Eq. (4) to obtain 

D 5651 kg=.m

2

2

s

/:


June 25, 2012


Dynamics 2e

123

Problem 2.80 Two masses mA and mB are placed at a distance r 0 from one another. Because of their mutual gravitational attraction, the acceleration of sphere B as seen from sphere A is given by

RD

r

G

m

A

C mB  ;

r2

where G 6:674  1011 m3 =.kg  s2 / 3:439  108 ft3 =.slug  s2 / is the universal gravitational constant. If the spheres are released from rest, determine

D

D

(a) The velocity of B (as seen by A ) as a function of the distance r . (b) The velocity of B (as seen by A ) at impact if r 0 0:7 lb, and

D 7 ft, the weight of A is 2:1 lb, the weight of B is

D 1:5 ft and dB D 1:2 ft, respectively.

(i) The diameters of A and B are dA

(ii) The diameters of A and B are infinitesimally small.

Solution

R D rP d rP=dt , we can

Part (a). Recalling that we can relate the accelerat ion to the veloci ty and position as r then write rP

R D rP ddrrP )

rdr 0

Z

which can be evaluated to obtain 1 2 r 2

P D G mA C mB 

r

P D

r

1 1  r



G .mA

)

r0

PD

r



C mB /

p

r

Z

1 dr; r2

0

2G .mA

C mB /

r

r0  r ; rr0

(1)

where we have chosen the negative root because the masses are moving toward each other. Part (b). Now that we have the velocity as a function of positio n, we can proceed to answer the ques tions posed in Part (b) of the problem. For question (i), when the masses touch r rA rB .dA dB /=2, so that

D C D



p

2G .mA



3:439   Recalling that G 1:2 ft, and r 0 7 ft, we can evaluate the above expression to obtain

dB

D

D

C

C mB / dA C2 dB r10 : (2) 108 ft3 =slug s2 , mA D 2:1 lb=g , mB D 0:7 lb=g , g D 32:2 ft=s , dA D 1:5 ft, PD

r

D

s

2

5

r

PD

5:980 

For part (ii), we take the limit of Eq. (2) as r

10

ft=s:

(3)

! 0 to obtain rP ! 1: 


June 25, 2012


Solutions Manual

124

Problem 2.81 Two masses mA and mB are placed at a distance r 0 from one another. Because of their mutual gravitational attraction, the acceleration of sphere B as seen from sphere A is given by

RD

r

G

m

A

C mB  ;

r2

where G 6:674  1011 m3 =.kg  s2 / 3:439  108 ft3 =.slug  s2 / is the universal gravitational constant. Assume that the particles are released from rest at r r0.

D

D

D

(a) Determine the expressio n relating their relativ e position r and time. Hint:

Zp

x=.1  x/ dx

D sin1px 



p

x.1  x/:

(b) Determine the time it takes for the objects to come into contact if r 0 1.1 and 2:3 kg, respectively, and (i) The diameters of A and B are dA

D 3 m, A and B have masses of

D 22 cm and dB D 15 cm, respectively.

(ii) The diameters of A and B are infinitesimally small.

Solution

Part (a). To find the relation between position and velocity, we observe that we can relate the acceleration to the velocity and position as r r d r=dt . Hence, we can then write

RDP P d rP rR D rP )

Z

dr

rP

P D

rdr 0

G .mA

C mB /

where we have used the fact that the spheres are released from rest so separation distance be r 0 . These integrals can be evaluated to obtain 1 2 2r

P D G mA C mB 

1 1  r



) rP D

r0



p

Z

r r0

2G .mA

1 dr; r2

P D 0 and have let their initial

r

C mB /

r

r0  r ; rr0

(1)

where we have chosen the negative root because the masses are moving toward each other and so r is decreasing. dr=dt and we write d t d r= r . Using this expression and the expression for Next, we observe that r r in the last of Eqs. (1), we can then write

P

PD

Z

t

dt 0

D

D

P

C mB /

Zr r

1



2G .mA

r0

r d r; 1  r= r0

where we have divided both the numerator and the denominator of the fraction under the square root by r 0 0 . Making the substitution x r=r 0 so that d r r 0 dx , and evaluating and we have used the fact that t 0 the integral on the left-hand side, we obtain

p

D

D

3=2

t

D



p

r0

2G .mA

C mB /

Z r r=r 0

1

D

x dx : 1x


June 25, 2012


Dynamics 2e

125

Using the hint given in the problem statemen t, this becomes becomes 3=2

t

D



p

r0

2G .mA

C mB /

h

sin1

px

) Part (b). We are given that r 0 11

3

 2

m =kg s . 22 cm (i) Using dA Eq. (2), we obtain

t

r=r 0

p iˇˇ " r s Dp C



x .1  x/

1

;

3=2 r 0

2G .mA

mB /

sin1

r r0



r r0



1

r r0





#

 : 2

(2)

D 3 m, mA D 1:1 kg, and mB D 2:3 kg, and we know that G D 6:674



10

D

D 0:2200 m, dB D 15 cm D 0:1500 m, and r D .dA C dB /=2 D 0:1850 m in

D 380;600 s: (ii) If the diameters are infinitesimally small, r ! 0 . Hence, from Eq. (2) we obtain t

3=2

t

D

p

.=2/r 0 2G.mA

C mB / )

t

D 383;100 s.


June 25, 2012


Solutions Manual

126

Problem 2.82

R

Suppose that the acceleration r of an object moving alon g a straight line takes on the form

RD

r

G

m

A

C mB  ;

r2

P

where the constants G , mA , and mB are known. If r.0/ is given, under what conditions can you determine r.t/ via the following integral?

P

t

r.t/

r.0/

P DP

G



Z

0

mA

C mB dt:

r2

Solution

P

¤ 0 during


June 25, 2012

r.t/ can be determined if the position r of the object is known as a function of time t and if r.t/ the time interval of interest.


Dynamics 2e

127

Problem 2.83 If the truck brakes hard enough that the crate slides to the right relative to the truck, the distance d between the crate and the front of the trailer changes according to the relation

RD

d

(

k g k g

C aT

for t < ts ; for t > ts ;

where ts is the time it takes the truck to stop, aT is the acceleration of the truck, g is the acceleration of gravity , and  is the kinetic friction coefficient between the truck and the crate.k Suppose that the truck and the crate are initially traveling to the right at v 0 60 mph and the brakes are applied so that aT 10:0 ft=s2 . Determine the minimum value of  k so that the crate does not hit the right end of the truck bed if the initial distance d is 12 ft. Hint: The truck stops before the crate stops.

D

D

Solution

Referring to figure on the right, the acceleration of the truck relative to the crate is given by

(

R D xR 1 D  k g C aT xR 2 D  k g

d

for t < ts ;

(1)

for t > ts ;

where ts is the time at which the truck comes to a stop, and where the subscripts 1 and 2 are used to distinguish expressions corresponding to t < ts from those for t > ts , respectively. Using the constant accelerat ion equation of the type v v0 ac .t  t0 /, the time ts at which the truck stops is

C

0

D v 0 C aT ts )

ts

D



v0 ; aT

D (2)

where v 0 is common the initial speed of the truck and crate. Letting x 1 be the position of the truck relative to the crate at the time the truck comes to a stop, using the constant acceleration equation of the type 1 2 s s 0 v0 t 2 ac t , we have

D C

C

x1

D x0 C xP 0ts C 21 xR 1ts2 D d C 21 .k g C aT / ts2;

D

(3)

P D

where x 0 d is the initial position of the truck relative to the crate and x0 0 is the initial velocity of the truck relative to the crate. After the truck comes to a stop, the truck continues to slide relative to the crate with the acceleration x2

R

in Eq. (1). Using this accelera tion, the distance the truck moves relati ve to the crate after the truck comes to a stop can be found using xf2 x12 2x2 xf  x1 ; (4)

P D

P DP C

R



P

where xf 0 is the final velocity of the truck relative to the crate, x1 is the velocity of the truck relative to 0 is final position of the truck relative to the crate, and x 1 is the position of the the crate at time t s , xf

D


June 25, 2012


Solutions Manual

128

P

truck relative to the crate at time t s and is given by Eq. (3). We now need to find x1 , which can be done using:

P D xP 0 C xR 1ts ) xP 1 D .k g C aT / ts ; where we have again used the fact that xP 0 D 0 . Substituting Eqs. (1), (3) and (5) into Eq.   0 D Œ. k g C aT / ts ç2 C 2. k g/ d 21 . k g C aT / ts2 : x1



(5) (4), we obtain (6)



Finally, substituting in t s from Eq. (2), we get the final equation for  k :

0

. D

k g

aT /

C

2

 av  C 2. 0



T

which, after simplification, becomes

k g/

d 

1

. k g

C

2

"

aT /

2

 av  # ; 

0

(7)

T

2

0 Solving for  k we get

D avT0 .k g C aT /



(8)

2

D g 2dva0 aT v2 ) T5280 0ft=s, aT D where we have used v 0 D 60 mph D 60 3600 k

obtain the numerical result.

2d k g:



D 0:3012, (9) 10 ft=s , g D 32:2 ft=s , and d D 12 ft to

k 

2


2

June 25, 2012


Dynamics 2e

129

Problem 2.84

D

D

72 mph and vB 67 mph, Cars A and B are traveling at vA respectively, when the driver of car B applies the brakes abruptly, causing the car to slide to a stop. The driver of car A takes 1:5 s to react to the situation and applies the brakes in turn, causing car A to slide as well. If A and B slide with equal accelerations, i.e., sA sB k g , where  k 0:83 is the kinetic friction coefficient and g is the acceleration of gravity, compute the minimum distance d between A and B at the time B starts sliding to avoid a collision.

R D

R D

D

Solution

We will denote by dA and dB the stopping distances of cars A and B , respectively. The stopping distance of car B is completely determined by the application of the brakes. The stopping distance of car A is determined by both the reaction time and the subsequent applica tion of the brakes, that is,

dA

D dA C dA ; r

(1)

b

where dAr is the distance trave led by A during the reaction time and dAb is the distance trave led by A during the application of the brakes. During the reaction time, the speed of A remains constant. Therefore,

dAr

D

D vAtr ;

(2)

where t r 1:5 s is the reaction time of the driver of car A . After both cars apply the brakes, for both cars we v 02 2ac .s  s0 / to relate the speeds of the cars to their can use the constant acceleratio n equation v 2 acceleration and stopping distances. Setting to zero the final velocities of both cars, we have

D C

0 0

D vA2 D vB2



2k gdAb



2k gdB

)

dAb

D 2vA2k g ;

(3)

2 vB

)

dB

D 2 k g :

and

dB

D 2vBk g :

(4)

From Eqs. (1)–(4), we have 2

dA

D vAtr C 2vAk g

2

(5)

We observe that, in order to avoid a collision, the separation d between A and B at the moment that B applies dB d . Then, using Eqs. (5), we have the brakes must be such that dA

D C d D dA

Recalling that  k

tr

0:83, g



dB

D 21k g vA2

32:2 ft=s2 , vA

D 1:5 s, we can Devaluate d Dto obtain

2  vB

72 mph

D d

D D 186:4 ft:

Cv t :

(6)

A r

72 5280 3600 ft =s, vB

67 5280 3600 ft =s, and

67 mph

D


D

June 25, 2012


Solutions Manual

130

Problem 2.85 The spool of paper used in a printing process is unrolled with velocity vp and acceleration ap . The thickness of the paper is h , and the outer radius of the spool at any instant is r . If the velocity at which the paper is unrolled is constant, determine the angular accelerati on ˛ s of the 0:0048 in:, for vp 1000 ft=min, and spool as a function of r , h , and vp . Evaluate your answer for h two values of r , that is, r 1 25 in: and r 2 10 in:

D

D

D

D

Photo credit: © David Lees/CORBIS

Solution

The radius decreases by the paper thickness h for every one revolution. Hence, letting  be the angle measuring the angular position of a fixed radial line on the spool (  increases when the spool turns clockwise), then we have r h : (1)  2 Assuming that the decrease in radius can be viewed as occurring continuously, we can change the above relation into a relation in terms of differentials. That is we can write

D

dr d

D 2h : (2) Next, observing that the angular velocity of the spool is ! s D d  =dt , we can then use the above equation to 

relate the time rate of change of r to ! s by applying the chain rule as follows:

P D ddtr D ddr ddt ) rP D 2h !s ) 

r

!s

D

 2

h

P

r:

Recalling that the linear velocity of the paper is related to the angular velocity of the spool as can use Eq. (3) to relate vp to r as follows:

P vp D r !s )

vp

D

 2

h

P

) rP D

rr

hvp

2 r

(3)

vp

D r !s , we

:

(4)

Taking the time derivative of vp in the second of Eqs. (4), accounting for the fact that vp is constant (i.e., ap 0 ), and using the last of Eqs. (4), we have

D

a

p

0

D D

 2

h

rP

2

C

rr

R

)

0

h2 vp2

 2

D

h



4 2 r 2

rr :

C R

(5)

!

Taking the time derivati ve of the last of Eqs. (3) and solving the last of Eqs. (5) to find expressions for ˛ s and r , respectively, we have 2 h2 vp  2 ˛s r and r ; (6) 2 3 h 4 r

R

D

R

RD


June 25, 2012


Dynamics 2e

131

which, when combined, imply that 2

D 2hv rp3 : Evaluating the expression above for h D 0:0048 in: D .0:0048=12/ ft, vp D 1000 ft=min D .1000=60/ ft=s, r1 D 25 in: D .25=12/ ft, and r 2 D 10 in: D .10=12/ ft, we have ˛s

˛s

ˇ

2

r Dr1

D 0:001956 rad=s

and

˛s

ˇ

2

r Dr2

D 0:03056 rad=s :


June 25, 2012


Solutions Manual

132

Problem 2.86 The spool of paper used in a printing process is unrolled with velocity vp and acceleration ap . The thickness of the paper is h , and the outer radius of the spool at any instant is r . If the velocity at which the paper is unrolled is not constant, determine the angular acceleratio n ˛ s of 0:0048 in:, vp 1000 ft=min, the spool as a function of r , h , vp , and ap . Evaluate your answer for h ap 3 ft=s2 , and two values of r , that is, r 1 25 in: and r 2 10 in:

D

D

D

D

D


Solution

The radius decreases by the paper thickness h for every one revolution. Hence, letting  be the angle (in radians) measuring the angular position of a fixed radial line on the spool (  increases when the spool turns clockwise), then we have r h : (1)  2 Assuming that the decrease in radius can be viewed as occurring continuously, we can change the above

D

relation into a relation in terms of differentials. That is we can write

dr d




r

!s

D

 2

h

P

r:


P

vp

D r !s )

vp

D

 2

h

P

rr

) rP D

hvp

2 r

(3)

vp

D r !s , we

:

(4)

Taking the time derivative of the second of Eqs. (4) and then using the last of Eqs. (4) to substitute for the term r , we have h2 v 2 2  p (5) vp ap h 4 2 r 2 r r :

P

P D D

C R To find an expression for the term rR , we take the time derivative of rP in Eq. (3) and obtain h rR D ˛s : 2



!




(6) June 25, 2012


Dynamics 2e

133

Substituting the result from Eq. (6) into Eq. (5), we have

ap

D

 2

h



h2 vp2 4 2 r 2



!

hr ˛s : 2

(7)

Then, solving for ˛ s and simplifying, we obtain 2

˛s

D arp C 2hv rp3 :

Evaluating the expression above for h 0:0048 in: .0:0048=12/ ft, vp 1000 ft=min 3 ft=s2 , r 1 25 in: .25=12/ ft, and r 2 10 in: .10=12/ ft, we have

ap

D

D

D

D

˛s

ˇ

D DD 2

r Dr1

D 1:442 rad=s

and

˛s

D

ˇ

D .1000=60/ ft=s,

2

r Dr2

D 3:631 rad=s :


June 25, 2012


Solutions Manual

134

Problem 2.87 The spool of paper used in a printing process is unrolled with velocity vp and acceleration ap . The thickness of the paper is h , and the outer radius of the spool at any instant is r . If the velocity at which the paper is unrolled is constant, determine the angular acceleration ˛ s of 0:0048 in: and vp 1000 ft=min as a the spool as a function of r , h , and vp . Plot your answer for h function of r for 1 in:  r  25 in. Over what range does ˛ s vary?

D

D


Solution

The radius decreases by the paper thickness h for every one revolution. Hence, letting  be the angle measuring the angular position of a fixed radial line on the spool (  increases when the spool turns clockwise), then we have r h : (1)  2 Assuming that the decrease in radius can be viewed as occurring continuously, we can change the above relation into a relation in terms of differentials. That is we can write

D

dr d




r

!s

D

 2

h

P

r:

(3)


P vp D r !s )

vp

D

 2

h

P

) rP D

rr

hvp

2 r

vp

D r !s , we

:

(4)

Taking the time derivative of vp in the second of Eqs. (4), accounting for the fact that vp is constant (i.e., ap 0 ), and using the last of Eqs. (4), we have

D

 2

ap

D0 D

h

h2 vp2

 2

2

rP

C r rR 

)

0

D

h



4 2 r 2

C r rR

!

:

(5)

R

Taking the time derivative of the last of Eqs. (3) and the last of Eqs. (4) to find expressions for ˛ s and r , we can write 2 h2 vp  2 ˛s r and r ; (6) 2 3 h 4 r

D

R

RD


June 25, 2012


Dynamics 2e

135

from which we have 2

˛s

D 2hv rp3 :

To plot the above function, we first substitute the values of the known coefficients, i.e., we recall that h 0:0048 in:, and vp 1000 ft=min 200:0 in:=s and rewrite ˛ s as

D

D

D

3

˛s

2

D 30:56rin3 : =s :

(7)

The above function can now be plotted. The plot below was generated using Mathematica with the following code:

The quantity ˛ s appears to vary from 30:5 rad=s2 to close to zero as r varies from 1 in: to 4 in:


June 25, 2012


Solutions Manual

136

Problem 2.88 Derive the constant acceleration relation in Eq. (2.32), starting from Eq. (2.24). State what assumption you need to make about the acceleration a to complete the derivation. Finally, use Eq. (2.27), along with the result of your derivat ion, to derive Eq. (2.33). Be careful to do the integral in Eq. (2.27) before substituti ng your result for v.t/ (try it without doing so, to see what happens). After completing this problem, notice that Eqs. (2.32) and (2.33) are not subject to the same assumption you needed to make to solve both parts of this problem. Solution

Assuming that the acceleration is not equal to 0 and integrating Eq. (2.24), we have

t.v/

D t0 C a1c

Z

v

)

dv v0

t.v/

D t 0 C a1c .v



v0 /

)

v

D v0 C ac .t

 t0 /:

(1)

Integrating Eq. (2.27) we have

s

D s0 C a1c

Z

v

)

v dv v0

s

D s0 C 2a1 c

v

2





v02 :

(2)

Substituting for v from Eq. (1), we obtain

s

D s0 C 2a1 c

which can be simplified to obtain

h

ac2 .t

 t0 /

2

C 2v0ac .t

 t0 /

D s0 C v0.t

 t0 /

C 2 ac .t

;

(3)

2

1

s

i

 t0 /

:


(4)

June 25, 2012


Dynamics 2e

137

Problem 2.89 The discussion in Example 2.12 revealed that the angle  had to be greater than  min analytical expression for  min in terms of h , w , and d .

D 0:716ı. Find an

Solution

The smallest possible angle (with respect to the horizontal) corresponds to the straight-line trajectory that going from the point at which the ball is hit to the top of the center field wall.

Using elementary trigonometry, we have that the analytical expression of the slope of the straight-line trajectory is



D tan1

w h 

d

:

To achieve this trajectory the ball would need to be imparted an infinite speed. That is, the straight-line trajectory cannot be achieved in practice.


June 25, 2012


Solutions Manual

138

Problem 2.90 A stomp rocket is a toy consisting of a hose connected to a “blast pad” (i.e., an air bladder) at one end and to a short pipe mou nted on a tripod at the other end. A rocket wit h a hollow body is mounted onto the pipe and is propelled into the air by “stomping” on the blast pad. Some manufacturers claim that one can shoot a rocket over 200 ft in the air. Neglecting air resistance, determine the rocket’s minimum initial speed such that it reaches a maximum flight height of 200 ft. Solution

The maximum height depends on the vertical component of the launch velocity. The higher this component the higher the height. Therefore, the minimum value of the speed needed to reach the desired height is found by launching the rocket purely in the vertical direction. Referring to the figure at the right, we consider the case in which the motion is completely in the y direction. Since the positive y direction is opposite to gravity, we have that the acceleration of the rocket is y g constant. We can relate velocity to position using the following constant acceleration equation: y 2 y02  2g.y  y0 /; (1)

RD

D

P DP where yP 0 is the velocity of the rocket for y D y 0 , and where we choose y 0 to denote the launch position of the rocket. Setting y 0 D 0 and recalling that the maximum height is achieved when yP D 0 , for y D h Eq. (1) becomes 0 D yP02 2gh ) yP0 D 2gh ; (2) max



max

p

max

where we have chosen the positive root since the rocke t is initi ally launched upward. Recalling that hmax 200 ft and g 32:2 ft=s2 , and observing that the initial speed coincides with y0 , we can evaluate the last of Eqs. (2) to obtain

D

D

P

vmin

D 113:5 ft=s:


June 25, 2012


Dynamics 2e

139

Problem 2.91

D

D

150 ft and at a constant speed v 0 80 mph drops a An airplane flying horizontally at elevation h package P when passing over point O . Determine the horizontal distance d between the drop point and point B at which the package hits the ground.

Solution

We model the motion of the package as projectile motion. Observing that the positive direction of the y axis shown is opposite to that of gravity, we have that the components of the constant acceleration of the package P are x 0 and y g: (1)

RD

D

RD

0 denote the instant at which the package is released, Eqs. (1), along with Letting t denote time and t Eq. (2.33) on p. 49, tell us that the x and y coordinates of the package as functions of time are x.t/

D x.0/ C xP .0/t

and

y.t/

D y.0/ C y.0/t P

1 2  2 gt ;

(2)

where x.t/ and y.t/ are the x and y components of the package’s velocity. The package is released at a height h over point O . Also, the velocity of the package at the instant of release is equal to the velocity of the plane. Hence, at the instant of release, we have

P

P

x.0/

D0;

y.0/

D h; xP .0/ D v0;

and

P D 0:

y.0/

(3)

Substituting Eqs. (3) into Eqs. (2), x.t/ and y.t/ become:

x.t/

D v 0t

and

y.t/

Dh

1 2  2 gt :

Let t i denote the time at which the package impacts the ground. Since y.ti / Eqs. (4), we have

h  21 gt i2 We now observe that d

D0 )

D

s

2h : g

D x .ti /. Substituting the second of Eqs. (5) into the d

Recalling that v 0 to obtain

ti

(4)

D 0, referring to the second of (5) first of Eqs. (4), we have

2h : g

D v0

sD

(6) 2

D 80 mph D 80.5280=3600/ ft=s, h 150 ft, and g D 32:2 ft=s , we can evaluate Eq. (6) d D 358:1 ft:


June 25, 2012


Solutions Manual

140

Problem 2.92

D

D

60 m and at a constant speed v 0 120 km=h drops a An airplane flying horizontally at elevation h package P when passing over point O . Determine the time it takes for the package to hit the ground at point B . In addition, determ ine the velocity of the package at B .

Solution

We model the motion of the package as projectile motion. Observing that the positive direction of the y axis shown is opposite to that of gravity, the components of the constant acceleration of the package P are

R D0

x

and

RD

y

g:

(1)

D

0 denote the instant of release, Eqs. (1), along with Eq. (2.32) on p. 49 and Letting t denote time and t Eq. (2.33) on p. 49, tell us that the x and y components of the velocity of the package, along with the x and y coordinates of the package as functions of time are 1 2 (2) P D xP .0/; y.t/ P D y.0/ P gt; x.t/ D x.0/ C xP .0/t; and y.t/ D y.0/ C y.0/t P 2 gt ; where xP .t/ and yP .t/ are the x and y components of the package’s velocity. Since the package is released at a

x.t/





height h over point O while traveling with the airplane, at the instant of release we have

x.0/

D0;

y.0/

D h; xP .0/ D v0;

and

P D 0:

y.0/

(3)

Substituting Eqs. (3) into Eqs. (2), we have

P D v0; y.t/ P D

x.t/

gt;

D v0t;

x.t/

and

y.t/

Dh

1 2  2 gt :

(4)

D 0, referring to the last of Eqs. (4), we have (5) D 2h=g:

Let t i be the time at which P impacts the ground. Since y .ti /

h  12 gt i2 2

D0 )

ti

p

D 60 m and g D 9:81 m=s , we can evaluate the last of Eqs. (5) to obtain ti D 3:497 s: Denoting the velocity of P at B by vEB , we now observe that the velocity at B is vEB D xP .t i / {O C y.t P i / |O. Recalling that h

Hence, substituting the second of Eqs. (5) into the first two of Eqs. (4), we have

vEB D v 0 {O g 2h=g |O : D 120 km=h D 120.1000=3600/ m=s, h D 60 m, and g D 9:81 m=s , Eq. (6) gives vEB D .33:33 {O 34:31 |O / m=s: 

Recalling that v 0

p

(6)

2




June 25, 2012



Solutions Manual

142

Problem 2.94

D

6 m=s and angle The jaguar A leaps from O at speed v0 ˇ 35ı relative to the incline to try to intercept the panther B at C . Determine the distance R that the jaguar jumps from O to C (i.e., R is the distance between the two points of the trajectory that intersect the incline), given that the angle of the incline is  25 ı .

D

D

Solution

The acceleratio n of A is completely in the vertical directi on. Hence, referring to the figure at the right, the components of the acceleration g sin  and y g cos  . of A in the x and y directions are x v 0 cos ˇ and The components of the initial velocity of A are v x0 vy0 v0 sin ˇ . Observing that the x and y components of acceleration of A are both constants, we can use constant acceleration equations to write

RD

RD D

D

D v0.cos ˇ/t C 21 g.sin  /t 2 D tv0 cos ˇ C 21 g.sin  /t; (1) y D v 0 .sin ˇ/t 12 g. cos  /t 2 D t v0 sin ˇ 12 g. cos  /t ; (2) where we have accounted for the fact that, at time t D 0 , A is at the srcin of the chosen coordinate system. x





Denoting by t C the time at which A reaches C , we observe that for t

tC , y

0 . Therefore, from Eq. (2),

D D ) tC D 2vg 0cossinˇ :

D v0 sin ˇ 12 g.cos  /tC Observing that, for t D t C , x .tC / D R , substituting the last of Eqs. (3) into Eq. (1), we have 0



R

D 2vg0cossinˇ




v0 cos ˇ

C 12 g.sin  / 2vg0cossinˇ

(3)



;

(4)

2

D 2vg0cossinˇ cos ˇ C tan  sin ˇ: (5) Recalling that v 0 D 6 m=s, ˇ D 35ı , g D 9:81 m=s , and  D 25ı , we can evaluate the above result to obtain R D 5:047 m: R

2


June 25, 2012




Dynamics 2e

145

Problem 2.97 A golfer chips the ball into the hole on the fly from the rough at the 4ı and d 2:4 m, verify that the edge of the green. Letting ˛ golfer will place the ball within 10 mm of the center of the hole if 5:03 m=s and an angle the ball leaves the rough with a speed v 0 ˇ 41 ı .

D

D D

D

Solution

Referring to the figure at the right, we will use the coordinate system with axes x 1 and y 1 , which are horizontal and vertical, respectively. The acceleration of the ball in this coordinate system has components

xR 1 D 0 and yR1 D g: (1) D 0 be the initial time, and using constant accelera tion equations, 

Letting t we have

P D v 0 cos.˛ C ˇ/

x1

P D v 0 sin.˛ C ˇ/

y1

and



gt;

(2)

E D v0.cos.˛ C ˇ/ {O C sin.˛ C ˇ/ |O/. D 0 and y1 D 0 for t D 0, we have y1 D v 0 sin.˛ C ˇ/t 12 gt 2 : (3)

where the have used the fact that the initial velocity of the ball is v.0/ Integrating Eqs. (2) with respect to time, and enforcing the fact that x 1

x1

D v 0 cos.˛ C ˇ/t

From the first of Eqs. (3) we have t

and



x 1 =Œv0 cos.˛

ˇ/ ç. Substituting this resul t into the second of Eqs. (3),

D C  g sec2.˛ C ˇ/  2 y1 D tan .˛ C ˇ/x 1 x1 : (4) 2v02 Recalling that ˛ D 4 ı and ˇ D 41 ı , so that ˛ C ˇ D 45 ı , tan .˛ C ˇ/ D 1 and sec 2 .˛ C ˇ/ D 2 , so that Eq. (4) simplifies to: g 2 y1 D x 1 x : (5) v02 1 The x 1 and y 1 coordinates of the point at which the ball lands must satisfy the condition x 1 tan ˛ D y1 .

we have





Combining this requirement with Eq. (5) we have

2

(6) D vg0 1 tan ˛ D 2:399 m; where we have used the following numerical values: v 0 D 5:03 m, g D 9:81 m=s , and ˛ D 4 ı . The value

x1 tan ˛

D x1



g

x12 v02

)

x1



2

of x 1 in Eq. (6) is the x 1 coordinate of the ball’s landing spot. With this information , letting dL denote the distance between the ball’s landing spot and the the golfer, we can determine dL using trigonometry as follows:

x1

dL

D cos ˛ D 2:405 m ) d dL D 0:005 m < 10 mm; where we have used the fact that ˛ D 4 ı and d D 2:4 m. We can then conclude that

(7)



The golfer’s chip shot is successful.


June 25, 2012


Solutions Manual

146

Problem 2.98 In a movie scene invol ving a car chase, a car goes over the top of a ramp at A and lands at B below. 20 ı and ˇ 23 ı , determine the distance d covered by the If ˛ car if the car’s speed at A is 45 km=h. Neglect aerodynamic effects.

D

D

Solution

The acceleration of the car in the xy coordinate system shown is a g sin ˇ {  g cos ˇ | x g sin ˇ and y g cos ˇ:

ED

O

O ) RD

D D

RD

(1)

0 to be the time at which the car jumps off at the srcin A . Hence, We set t y 0 at t 0 . Also, at t 0 , the velocity components of the cars are

D

vx .0/

D

D v 0 cos.˛ C ˇ/

and

vy .0/

D v0 sin.˛ C ˇ/:

(2)

Hence, using the above considerations along with constant acceleration equations, the y coordinate of the car as a function of time is given by

y

D v 0 sin.˛ C ˇ/t

2 1  2 g. cos ˇ/t :

(3)

D 0, from Eq. (3) we have ) tB D 2v0 gsincos.˛ ˇC ˇ/ :

Letting tB denote the time at which the car lands at B , since yB

v0 sin.˛

C ˇ/tB

1 2  2 .g cos ˇ/tB

D0

(4)

Next observing that the x component of the acceleration in the second of Eqs. (1) is also constant, using constant acceleration equations, the x coordinate of the car as a function of time is given by

x

D v 0 cos.˛ C ˇ/t C 12 .g sin ˇ/t 2:

(5)


xB

D

2 D d D 2v0 gsincos.˛ ˇC ˇ/ Œcos.˛ C ˇ/ C tan ˇ sin.˛ C ˇ/ç:

D

45 km=h 45.1000=3600/ m=s, ˛ Recalling that v0 evaluate the above result to obtain d

(6)

D 20ı, ˇ D 23ı, and g D 9:81 m=s , we can 2

D 24:09 m:


(7)

June 25, 2012


Dynamics 2e

147

Problem 2.99 In a movie scene invol ving a car chase, a car goes over the top of a ramp at A and lands at B below. Determine the speed of the car at A if the car is to cover distance d 150 ft for ˛ 20 ı and ˇ 27 ı . Neglect aerodynamic effects.

D

D

D

Solution

The acceleration vector of the car in the xy coordinate system shown is a g sin ˇ {  g cos ˇ | x g sin ˇ and y g cos ˇ:

ED

O

O ) RD

D D

RD

(1)

0 to be the time at which the car jumps off at the srcin A . Hence, We set t y 0 at t 0 . Also, at t 0 , the velocity components of the cars are

D

vx .0/

D

D v 0 cos.˛ C ˇ/

and

vy .0/

D v0 sin.˛ C ˇ/:

(2)

Hence, using the above considerations along with constant acceleration equations, the y coordinate of the car as a function of time is given by

y

D v 0 sin.˛ C ˇ/t

2 1  2 g. cos ˇ/t :

(3)

D 0, from Eq. (3) we have ) tB D 2v0 gsincos.˛ ˇC ˇ/ :

Letting tB denote the time at which the car lands at B , since yB

v0 sin.˛

C ˇ/tB

1 2  2 .g cos ˇ/tB

D0

(4)

Next observing that the x component of the acceleration in the second of Eqs. (1) is also constant, using constant acceleration equations, the x coordinate of the car as a function of time is given by

x

D v 0 cos.˛ C ˇ/t C 12 .g sin ˇ/t 2:

(5)


xB

2 D d D 2v0 gsincos.˛ ˇC ˇ/ Œcos.˛ C ˇ/ C tan ˇ sin.˛ C ˇ/ç;

(6)

which can be solved for v 0 to obtain

v0

D

s

dg cos ˇ

C ˇ/Œ cos.˛ C ˇ/ C tan ˇ sin.˛ C ˇ/ ç : Recalling that d D 150 ft, ˛ D 20 ı , ˇ D 27 ı , and g D 32:2 ft=s , Eq. (7) can be evaluated to obtain v0 D 52:82 ft=s: 2 sin.˛

(7)

2


June 25, 2012


Solutions Manual

148

Problem 2.100 The M777 lightweigh t 155 mm howitzer is a piece of artillery whose rounds are ejected from the gun with a speed of 829 m=s. Assuming that the gun is fired over a flat battlefield and ignoring aerodynamic effects, determine (a ) the elevation angle needed to achieve the maximum range, (b ) the maximum possible range of the gun, and ( c ) the time it would take a projectile to cover the maximum range. Express the result for the range as a percentage of the actual maximum range of this weapon, which is 30 km for unassisted ammunition. Photo credit: U.S. Army Photo

Solution

We will solve the problem using the xy coordinate system show n in the figure at the right. The x axis is assumed to lie on the ground. We assume that the projectile is launched at t 0 from the srcin. The acceleration of the projectile is constant and has horizontal and 0 and y g , respectively. Then, letting  be the elevation angle vertical components x and v 0 the intial speed of the projectile, and using constant acceleration equations, the x and y coordinates of the projectile are given by

D

RD

RD

D .v0 cos  /t and y D .v0 sin  /t 21 gt 2: (1) Denoting the time of flight by tf , then y D 0 for t D tf . Hence, from the second of Eqs. (1), we have tf D 2v 0 sin  =g: (2) The range R is given by the value of x for t D tf . Substituting Eq. (2) into the first of Eqs. (1), and using the trigonometric identity 2 sin  cos  D sin 2 , we have R D .v 02 =g/ sin 2 : (3) Part (a). From Eq. (3), R occurs when sin 2 D 1 . This equation has infinitely many solutions for  , x



max

but the only meaningful solution is

2Rmax Part (b).

D 2 rad )

Rmax

D 4 rad D 45ı:

(4)

Substituting the last of Eqs . (4) into Eq. (3), we ha ve that R max is

Rmax

D

D

2

D v02=g:

(5)

Recalling that v 0 829 m=s and g 9:81 m=s , and expressing R max as a percentage of the actual range of 30 km, we can evaluate the above expression to obtain

Rmax

D 233:5% of the actual maximum rang e :

Part (c). Denoting the time the projectile takes to cover R max by t Rmax , we have that t Rmax is equal to tf for   Rmax . Hence, substituting the last of Eqs. (4) into Eq. (2), we have

D

tR D p2v0 =g: D 829 m=s and g D 9:81 m=s , we can evaluate the above result to obtain tR D 119:5 s:

(6)

max

Recalling that v 0

2

max


June 25, 2012


Dynamics 2e

149

Problem 2.101 You want to throw a rock from point O to hit the vertical advertising 30 ft away. You can throw a rock at the speed sign AB , which is R v0 45 ft=s. The bottom of the sign is 8 ft off the ground and the sign is 14 ft tall. Determine the range of angles at which the projectile can be thrown in order to hit the target, and compare this with the angle subtended by the target as seen from an observer at point O . Compare your results with those found in Example 2.11.

D

D

Solution

This problem can be solved as illustrated in Example 2.11 on p. 70 of the textbook. We recall Eq. (7) in Example 2.11 on p. 70 of the textbook:

tan 

D

v02

˙

q

v04  g.gR2 gR

C 2y v02/

:

(1)

D

45 ft=s, Observe that we are given all of the data needed to use the above equation. Namely, we havev 0 32:2 ft=s2 , R 30 ft, so that substituti ng in the above equati on yA 8 ft and yB 22 ft we have

g

D

D

D

D

( (

D 30:43ı; 2 D 74:50 ı ; 56:84ı ; ) 1 DD 69:41 ı: 2

y

D yA D 8 ft )

y

D yB D 22 ft

1

(2)

(3)

Following the same logic as in Example 2.11, we obtain the two ranges of firing angles as

30:43ı







56:84ı

and

69:41ı

and

2







74:50 ı :

The sizes of these intervals are, respectively,

1

D 26:41ı

D 5:087ı:

The angle subtended by the target as seen from an observer at point O is

ˇ

D tan1

 22  30



tan1

8 30

)

ˇ

D 21:32ı:

(4)

Unlike Example 2.11, the difference between the angle subtended by the target and  1 or  2 is significant. In addition, we see that the value of  1 is much closer to ˇ than 2 .


June 25, 2012


Solutions Manual

150

Problem 2.102 Suppose that you can throw a projectile at a large enough v 0 so that it can hit a target a distance R downrange. Given that you know v 0 and R , determine the general expressions for the two distinct launch angles  1 and  2 that will allow the projectile to 30 m=s and R 70 m, determine numerical hit D . For v 0 values for  1 and  2 .

D

D

Solution

RD

RD

0 and y g , Using the axes in the figure, the components of the acceleration of the projectile are x where g is the acceleration due to gravity. Therefore, the acceleration of the projectile is constant and applying the constant accele ration relation in Eq. (2.33) on p. 49 of the textbook, we have

D y0 C .v0 sin  /t 12 gt 2; (1) where we have accounted for the fact that the projectile is at O for t D 0 , and where we have denoted by  the orientation of the initial velocit y of the projectile. For x D R we have that y D 0. Enforcing this D x0 C .v0 cos  /t;

and

y

D x0 C .v0 cos  /tD ;

and

0

x



condition, Eqs. (1) give

R

D y 0 C .v0 sin  /tD

1 2  2 gtD ;

(2)

where tD is the time the projectile takes to go from O to D . Eliminating tD from Eqs. (2), with x 0 y0 0 , we have gR  2v02 sin  cos  0 sin.2 / gR=v02 ;

D

D 0 and (3)

D ) D where, in writing the second of Eqs. (3), we have used the trigonometric identity sin .2 / D 2 sin  cos  . As long as the values of R and v 0 are such that gR=v02  1, and observing that the physically acceptable values of  lie in the range 0    90ı , we have that the last of Eqs. (3) admits the following two solutions:

1 For the given values of R and  2 to obtain

D 21 sin1.gR=v02/

and

2

D 90ı

1  2

sin 1 .gR=v02 /: 2

D 70 m and v0 D 30 m=s, and recalling that g D 9:81 m=s , we can evaluate 1 1

D 24:86 ı

and

2

D 65:14ı:


June 25, 2012


Dynamics 2e

151

Problem 2.103 An alpine ski jumper can fly distances in excess of 100 m by using his or her body and skis as a “wing” and therefore, taking advantage of aerodynamic effects. With this in mind and assuming that a ski jumper could survive the jump, determine the distance the jumper could “fly” without aerodynamic effects, i.e., if the jumper were in free fall after clearing the ramp. For the purpose of your 11ı (slope of calculation, use the following typical data: ˛ 36ı (average slope of the hill), ramp at takeoff point A), ˇ

D D 0 D 86 km=h (speed at A ), h D 3 m (height of takeoff point with v respect to the hill). Finally, for simplicity, let the jump distance be the distance between the takeoff point A and the landing point B .

Solution

We will solve the problem using a Cartesian coordinate system with srcin at A and axes x and y oriented such that the x axis is parallel to the hill (see figure at the right). In the chosen coordinate system, the veloci ty of the jumper at A is

E D v0 cos.ˇ

vA



O C v0 sin.ˇ

˛/ {



O

˛/ | :

(1)

Once airborne, the acceleration of the jumper is

E D g sin ˇ {O

a



O

g cos ˇ | :

(2)

Using constant acceleration equations, we then have that

x

v 0 cos.ˇ  ˛/t

1 2 2 .g sin ˇ/t

and

y

v 0 sin.ˇ  ˛/t

1 2  2 .g cos ˇ/t ;

(3)

D t D 0 to be theCtime at which the jumperDtakes off at A, and where we have accounted where we have set for the fact that, at t D 0 , the velocity of the jumper is that in Eq. (1). Letting tB denote the time at which the jumper lands at B , we can replace t with tB in the second of Eqs. (3) and enforce the condition that yB D h cos ˇ . This gives h cos ˇ D v 0 sin.ˇ ˛/tB 21 .g cos ˇ/tB2 v sin.ˇ ˛/ ˙ v02 sin 2 .ˇ ˛/ C 2hg cos2 ˇ ) tB D 0 : (4) g cos ˇ 









q



The solution for tB corresponding to the minus sign in front of the square root is negativ e. Hence, the 86 km=h 86.1000=3600/ m=s, only acceptable value for tB is that with the sign. Recalling that v 0 ˇ 36 ı , ˛ 11 ı , h 3 m, and g 9:81 m=s2 , we then have

D

D

D

C

D

tB

D

D

D 2:765 s:

(5)

Using the data listed right above Eq. (5) and substituting tB into the first of Eqs. (3), we have xB 

yB



D 81:92 m.

Then, recalling that between points A and B can be h cos ˇ 2:427 m, we have that 2 the 2distance calculated using the Pythagorean theorem, i.e., dAB xB yB which gives

D

D

dAB

q D

C

D 81:96 m:


(6)

June 25, 2012


Solutions Manual

152

Problem 2.104 A soccer player practices kicking a ball from A directly into the goal (i.e., the ball does not bounce first) while clearing a 6 ft tall fixed barrier. Determine the minimum speed that the player needs to give the ball to accomplish the task. Hint: Consider the equation for the projectile’s trajectory of the form y C 0 C1 x C2 x 2 , with the y axis parallel to the direction of gravity, for the case in which the ball reaches the goal at its base. Solve this equation for the initial speed v 0 as a function of the initial angle  , and finally find .v 0 /min as you learned in calculus. Don’t forget to check whether or not the ball clears the barrier.

D C

C

Solution

The coordinate system shown at the right has srcin at A , the position of the ball 0 . The trajectory of the ball has the form y C 0 C1 x C2 x 2 . To find at t C0 , C 1 and C 2 we proceed as follows. First, we observe that y 0 for x 0, 0 . Second, recall that the velocity is always tangen t to the which implies that C 0 0, the trajectory. Therefore, given that ˇ is the orientation of the velocity at t 0 must be equal to tan ˇ , i.e., C 1 .dy=dx/ xD0 tan ˇ . We know that the slope of the trajectory at x . tan ˇ/x C2 x 2 . To find C 2 , we now recall that y g. Using the chain rule trajectory has the form y to differentiate the trajectory with respect to time, we have

D

D C

C D

D

D

D

D

D

D

C

D RD

P D .tan ˇ/xP C 2C2xxP ) yR D .tan ˇ/xR C 2C2xP 2 C 2C2xxR : (1) Since xR D 0, xP is constant and therefore equal to its initial value, i.e., xP D v0 cos ˇ . Substituting this condition into the last of Eqs. (1) along with yR D g , we have g D 2C 2 .v0 cos ˇ/ 2 ) C2 D g sec2 ˇ=.2v 02 /: (2) y







In summary, the trajectory of the ball is given by

y

D

D .tan ˇ/x



Œg sec2 ˇ=.2v 02 /çx 2 :

(3)

D 0 be the coordinates of the base of the goal. For the ball to land at the base of the 2 0 D . tan ˇ/x G Œg sec2 ˇ=.2v 02 /çxG ) v0 D gxG = sin 2ˇ: (4)

80 ft and y G Let x G goal, we have

p



Minimizing v 0 with respect to ˇ requires making the denominator of the fraction under the square root of the last of Eqs. (4) as large as possible. The maximum value of the sine function is 1, which is achieved when 2ˇ .  =2/ rad. Hence, we have

D

ˇ

D .=4/ rad )

.v0 /min

D pgxG )

.v0 /min

D 50:75 ft=s,

(5)

where we have used the fact that g 32:2 ft=s2 and x G 80 ft. Substituting ˇ . =4/ rad and v 0 .v0 /min into Eq. (3), and computing the value of y corresponding to x 58 ft (which is the x coordinate of the barrier), we have y.58 ft/ 15:94 ft > 6 ft;

D

D

D

D

D

D

that is, the ball clears the obstacle in front of the goal. This solution s manual, in any print or electronic form, remain s the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplicat ion or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

June 25, 2012


Dynamics 2e

153

Problem 2.105 A soccer player practices kicking a ball from A directly into the goal (i.e., the ball does not bounce first) while clearing a 6 ft tall fixed barrier. Find the initial speed and angle that allow the ball to barely clear the barrier while barely reaching C1 x  C2 x 2 , where the the goal at its base. Hint: A projectile’s trajectory can be given the form y coefficients C 1 and C 2 can be found by forcing the parabola to go through two given points.

D

Solution

As explained in the hint, the trajectory can be given the form

y

D C 1x

C2 x 2 ;



(1)

where, referring to the figure at the right, it is understood that the srcin of the 58 ft and yB 6 ft be the coordinates of the top of the coordinate system used is at point A . Let xB 80 ft and y G 0 be the coordinates of the base of the goal. The ball must barely barrier. Also, let x G clear the barrier and then it must barely reach the goal. Hence, we have

D

D

D

yB

D C1xB



2 C2 xB

D

0

and

D C 1 xG



2 C 2 xG :

(2)

The above equations form a system of two equations in the two unknowns C 1 and C 2 whose solution is

xG yB

C1

D xB .xG

and

yB

C2

:

(3)

xB / xB .xG  xB / We now need to relate the coefficients C 1 and C 2 to the initial speed and angle of the ball. To do so, we begin

D



with noticing that since the velocity is tangent to the trajectory, and since the initial orientation of the velocity 0 must be equal to tan ˇ . That is is the angle ˇ , the slope of the trajectory at x

D

D C1 D tan ˇ ) ˇ D tan1 xB .xxGG yB xB / ) ˇ D 20:62ı, where we have used the fact that x G D 80 ft, yB D 6 ft, and xB D 58 ft. Next we recall that yR D



.dy=dx/ xD0





(4)

g . Using

the chain rule to differentiate Eq. (1) with respect to time, we have

P D C1xP 2C2xxP ) yR D C1xR 2C2xP 2 2C2xxR: (5) We now observe that we have xR D 0 . This also implies that xP is constant and therefore equal to its initial value, i.e., xP D v 0 cos ˇ . Enforcing these conditions, along with yR D g , in the last of Eqs. (5), we have g D 2C2 .v0 cos ˇ/ 2 ) v0 D g=.2C 2 / sec ˇ: (6) y













80 ft, yB Now that ˇ is known, recalling again that x G Eqs. (3) to evaluate C 2 , we can evaluate v 0 to obtain

D D v0 D 62:52 ft=s;

where we have also used the fact that g

p

6 ft, and xB

58 ft and using the second of

D

2

D 32:2 ft=s .


June 25, 2012


Solutions Manual

154

Problem 2.106 In a circus act a tiger is required to jump from point A to point C so that it goes through the ring of fire at B . Hint: A projectile’s trajectory can be given the form y C 1 x  C2 x 2 , where the coefficients C 1 and C2 can be found by forcing the parabola to go through two given points. 5:5 m from A. Determine the tiger’s initial velocity if the ring of fire is placed at a distance d

D

D

Furthermore, determine the slope of the tiger’s trajectory as the tiger goes through the ring of fire.

Solution

Referring to the figure at the right, we adopt a Cartesian coordinate system with srcin at A . The trajectory of the tiger is of the form: y C 1 x  C2 x 2 : (1)

D

We find C 1 and C 2 by observing that the tiger passes through .5:5; 3/ m, and lands at C point B of coordinates .xB ; yB / .9; 0:5/ m. Using Eq. (1) to enforce of coordinates .x C ; yC / these conditions, we have

D

D

yB

D C 1xB



2 C2 xB

and

yC

D C 1 xC



2; C 2 xC

(2)

which is a system of two equations in the two unknowns C 1 and C 2 whose solution is 2

C1

2

D xxCCxyBB.xCxB yxCB / 



and

C2

D xxCCxyBB.xCxB yxCB / : 

(3)



We now need to relate the C 1 and C 2 to the initial speed and angle of the tiger. To do so, we notice that since the velocity is tangent to the trajectory, and since the initial orientation of the velocity is the angle ˇ , the slope of the trajectory at x 0 must be equal to tan ˇ . That is

D

.dy=dx/ xD0

D C1 D tan ˇ )

ˇ

D tan1

2 2 C B  xB yC

x y

xC xB .xC

D



xB /



D 52:75 ı;

(4)

D

.5:5; 3/ m and .x C ; yC / .9; 0:5/ m. Next we recall th at where we have used the fact that .x B ; yB / y g . Using the chain rule to differentiate Eq. (1) with respect to time, we have

RD

2

y

(5) P D C1xP 2C2xxP ) yR D C1xR 2C2xP 2C2xxR: We now observe that we have xR D 0 . This also implies that xP is constant and therefore equal to its initial value, i.e., xP D v 0 cos ˇ . Enforcing these conditions, along with yR D g , the last of Eqs. (5) gives g D 2C2 .v0 cos ˇ/ 2 ) v0 D g=.2C 2 / sec ˇ ) v0 D 9:781 m=s; (6) 











p


June 25, 2012


Dynamics 2e

155

D

D E

.9; 0:5/ m, we have evaluated v 0 by first where, again recalling that .xB ; yB / .5: 5; 3/ m and .x C ; yC / evaluating C 2 in the last of Eqs. (3) and then the angle ˇ in the last of Eqs. (4). Now that v 0 and ˇ are known, observing that vinitial v 0 cos ˇ { v 0 sin ˇ | , we can evaluate vinitial to obtain

E

D

OC E

vinitial

O

D .5:920 {O C 7:786 |O/ m=s:

The slope of the trajectory is obtained by differentiating Eq. (1) with respect to x :

dy=dx

D C1

Recalling that .xB ; yB / .5:5; 3/ m and .x C ; yC / can evaluate to slope at x xB to obtain

DD

dy dx

ˇˇ

x DxB



2C2 x:

(7)

D .9; 0:5/ m, and evaluating C1 and C2 in Eqs. (3), we D

0:2244:


June 25, 2012


Solutions Manual

156

Problem 2.107 In a circus act a tiger is required to jump from point A to point C so that it goes through the ring of fire at B . Hint: A projectile’s trajectory can be given the form y C 1 x  C2 x 2 , where the coefficients C 1 and C2 can be found by forcing the parabola to go through two given points. Determine the tiger’s initial velocity, as well as the distance d so that the slope of the tiger’s trajectory

D

as the tiger goes through the ring of fire is completely horizon tal.

Solution

Referring to the figure at the right, we will be using a Cartesian coordinate system with srcin at A . The trajectory of the tiger is

y

D C 1x



C2 x 2 ;

(1)

where C 1 and C 2 are constants to be determined by making sure that the tiger passes through point B of coordinates .xB ; yB / .d;3 m/ with zero slope, and then lands on point C of coordi.9 m;0:5 m/. Using Eq. (1) to enforce these nates .x C ; yC /

D

D

conditions, we have

D C1xB D0 x Dd y C D C 1 xC yB

dy dx

ˇˇ



2 C2 xB

D C 1d C2d 2; 0 D C 1 2d C2 0:5 m D .9 m/C1 .81:00 m /C2 :

) ) )

2  C2 xC

3m

(2)



(3)





2

(4)

The last of Eqs. (2)–(4) form a system of three equations in the three unknowns d , C 1 , and C 2 , which can be solved numerically. For example we have used Mathematica with the following code

which yields the following two solutions:

D C11:275 D

C1 0:05809

D 0:0002812 m  2 D 0:1355 m

C

2

1

D 103:3 m; d D 4:705 m: d

1

C

(5) (6)

Because the first solution implies that d > xC , the solution in question is not acceptable and therefore we have that the only acceptable solution is

C1

D1:275;

C

2

D 0:1355 m ; 1

d

D 4:705 m.


(7)

June 25, 2012


Dynamics 2e

157

Now we turn to the determination of the initial velocity of the tiger. To do so, we notice that since the velocity is tangent to the trajectory, and since the initial orientation of the velocity is the angle ˇ , the slope of the trajectory at x 0 must be equal to tan ˇ . That is

D

D tan1.C1/ D 51:89ı; (8) where we have used the numerical solution for C 1 . Next we recall that yR D g . Differentiating Eq. (1) with D C1 D tan ˇ )

.dy=dx/ xD0

ˇ



respect to time, we have

P D C1xP

y



P ) yR D C1xR

2C2 x x



2C2 x 2  2C2 x x:

P

R

(9)

We now observe that we have x 0 . This also implies that x is constant and therefore equal to its initial value, i.e., x v 0 cos ˇ . Enforcing these conditions, along with y g , the last of Eqs. (9) gives

RD

PD

g

D

2C2 .v0 cos ˇ/

P RD

2

)

v0

D

p

g=.2C 2 / sec ˇ

)

v0

D 9:749 m=s;

(10)

9:81 m=s2 , and, again, we have used the numerical solution for C 2 and where we have used the fact that g ˇ . Now that v 0 and ˇ are known, observing that vinitial v 0 cos ˇ { v0 sin ˇ | , we can evaluate vinitial to obtain

D

E

vinitial

E

D

OC

O

E

D .6:017 {O C 7:671 |O/ m=s:


June 25, 2012


Solutions Manual

158

Problem 2.108 A jaguar A leaps from O at speed v0 and angle ˇ relative to the incline to attack a panther B at C . Determine an expression for the maximum perpendicular height h max above the incline achieved by the leaping jaguar, given that the angle of the incline is  .

Solution

We will use a Cartesian coordinate system aligned with the incline as shown at the right. The acceleration vector is then given by

E D g sin  {O

a



O )

g cos  |

ay

D

g cos  :

Applying the constant acceleratio n equation in the y direction, we have 2 vy2 v 0y 2ay .y  y0 /: (1)

D

C

At the h max position the y component of velocity must be equal to zero. Enforcing this condition, we have

0

D .v 0 sin ˇ/2



2g cos  .hmax  0/:

(2)

Solving Eq. (2) for h max , we obtain 2

hmax

2

D v2g0 sincos ˇ :


June 25, 2012


Dynamics 2e

159

Problem 2.109 The jaguar A leaps from O at speed v 0 and angle ˇ relative to the incline to intercept the panther B at C . The distance along the incline from O to C is R , and the angle of the incline with respect to the horizontal is  . Determine an expression for v 0 as a function of ˇ for A to be able to get from O to C .

Solution

Using the xy coordinate system shown at the right, we write the following two constant acceleration equations describing the x and y coordinates of jaguar A as a function of time:

x y

D x0 C v0x t; D y0 C v0yt

(1)

1 2  2 gt ;

(2)

D

t 0, and where it is understood that the jaguar leaps at time where .x0 ; y0 / are the coordinates of the point from which the jaguar leaps. Since the jaguar leaps from the srcin of the chosen coordinate system, letting t C denote the time at which jaguar A arrives at C , we have R cos  R sin 

D Œv0 cos.ˇ D Œv0 sin.ˇ



 /ç tC ;

(3)



 /ç tC  12 gt C2 ;

(4)

where R is distance from O to C . Eliminating t C from Eqs. (3) and (4) gives  sin 

D cos  tan.ˇ



Multiplying all terms in the above equation by cos.ˇ 

sin  cos .ˇ   /

)

D cos  sin.ˇ

sin  cos .ˇ   /



/ 

C cos  sin.ˇ

where we have used the identity sin .A we have



gR

/  

gR

cos2 

2v02

cos2 .ˇ   /

:

(5)

 /, we have

cos2 

2v02 cos.ˇ   / /

2

D 2vgR2 coscos .ˇ 0



/

)

sin ˇ

2

D 2vgR2 coscos .ˇ 0



/

;

C B/ D sin A cos B C cos A sin B . Solving the last of Eqs. (6) for v0

D

gR 2

r p

(6)

v0 ,

cos  : sin ˇ cos.ˇ   /


June 25, 2012


Solutions Manual

160

Problem 2.110 The jaguar A leaps from O at speed v 0 and angle ˇ relative to the incline to intercept the panther B at C . The distance along the incline from O to C is R , and the angle of the incline with respect to the horizontal is  . Derive v0 as a function of ˇ to leap a given distance R along with the optimal value of launch angle ˇ , i.e., the value of ˇ necessary to leap a given distance R with the minimum v0 . Then plot v0 as a function of ˇ for g 9:81 m=s2 ,

D ı 7 m, and  D 25 , and find a numerical value of the D optimal ˇ and the corresponding value of v 0 for the given set

R

of parameters. Solution

Using the xy coordinate system shown at the right, we write the following two constant acceleration equations describing the x and y coordinates of jaguar A as a function of time:

D x0 C v0x t; y D y 0 C v0y t x

(1)

1 2  2 gt ;

(2)

D

t 0, and where it is understood that the jaguar leaps at time where .x 0 ; y0 / are the coordinates of the point from which of the jaguar leaps. Since the jaguar leaps from the srcin of the chosen coordinate system, letting t C denote the time at which jaguar A arrives at C , we have

D Œv0 cos.ˇ R sin  D Œv 0 sin.ˇ

R cos  



 /ç tC ;

(3)

1 2   /ç tC  2 gt C ;

(4)

where R is distance from O to C . Eliminating t C from Eqs. (3) and (4) gives  sin 

D cos  tan.ˇ



Multiplying all terms in the above equation by cos.ˇ 

sin  cos .ˇ   /

D cos  sin.ˇ



/ 

gR

/  

gR

cos2 

2v02

cos2 .ˇ   /

:

(5)

 /, we have

cos2 

2v02 cos.ˇ   / 2

2

  C cos  sin.ˇ  / D 2vgR2 coscos ) sin ˇ D 2vgR2 coscos ; .ˇ  / .ˇ  / 0 0 where we have used the identity sin .A C B/ D sin A cos B C cos A sin B . Solving the last of Eqs. (6) for

)

sin  cos .ˇ   /







(6)

v0 ,

we have

v0

D

r p gR 2

cos  sin ˇ cos.ˇ   /

:

(7)

Recalling that g 9:81 m=s2 , R 7 m, and  25 ı , we can plot the above function with any appropriate mathematical software. The plot shown below was obtained using Mathematica with the following code:

D

D

D


June 25, 2012


Dynamics 2e

161

We can find the optimal value of ˇ to reach a distance of R and setting it equal to zero. Recalling that  25 ı , this gives

D

dv 0 dˇ

D



1 2

r

gR cos ˇ cos.ˇ   /  sin ˇ sin.ˇ   / cos  2 Œsin ˇ cos.ˇ   /ç3=2

)

D 7 m by differentiating v0 with respect to ˇ D0

cos ˇ cos.ˇ  25ı /  sin ˇ sin.ˇ  25ı /

D 0:

(8)

The above equation is a transcendental equation that we will solve numerically. Again, this can be done with any appropriate mathematical software. We have used Mathematica with the following code

Note that the use of root finding algorit hms generally requir es the user to provide a guess of the value of the solution. As can be seen in the above code (see information provided at the end of the code line), we have provided a guess of 25 ı . The outcome of this calculation giv es

ˇoptimal Then, using the above value of ˇ along with g that the corresponding value of v 0 is

D 57:52ı:

D 9:81 m=s , R D 7 m, and  D 25ı, from Eq. (7) we have

.v0 /optimal

2

D 6:297 m=s:


June 25, 2012


Solutions Manual

162

Problem 2.111 A stomp rocket is a toy consisting of a hose connected to a blast pad (i.e., an air bladder) at one end and to a short pipe mounted on a tripod at the other end. A rocket with a hollow body is mounted onto the pipe and is propelled into the air by stomping on the blast pad. 120 ft=s, and if the rocket’s landing spot at B is If the rocket can be imparted an initial speed v 0 at the same elevation as the launch point, i.e., h 0 ft, neglect air resistance and determine the rocket’s launch angle  such that the rocket achieves the maximum possible range. In addition, compute R , the rocket’s maximum range, and tf , the corresponding flight time.

D

D

Solution

Referring to the figure at the right, we will use an xy coordinate system with srcin at the launch point of the rocket. Let  be the elevation 120 ft=s be the intial speed. The acceleration of the angle and v 0 rocket is equal to  g in the y direction and zero in the x direction. Hence, we can use the constant acceleration equation s v 0 t 12 ac t 2 , to express the x and y coordinates of the rocket as a function of time. This gives

D

D C

x y

D .v 0 cos  /t; .v 0 sin  /t

(1)

1 2  2 gt :

(2)

D D 0, the time taken by the rocket to arrive at

Since in this problem point B lies on the line y obtained by equating Eq. (2) to zero. This gives

D 2v0gsin  :

tB

(3)

Substituting Eq. (3) in place of t into Eq. (1), using the trigonometr ic identity 2 sin  cos  observing that xB R , we get v02 sin 2

D

R

D

Substituting  Rmax

D 2 rad )

Rmax

2

Observing that the time of flight is the same as

2v0 sin Rmax =g , which can be evaluated to obtain tf

(4)

D 1, i.e.,

D 4 rad D 45:00ı:

D 45:00ı into Eq. (4) with g D 32:2 ft=s Rmax

D sin 2 , and

;

g

where R is the range of the rocket. The maximum valu e of R occurs when sin 2

2

B can be

(5)

gives

447:2 ft:

DtB given in Eq. (3), for



D

Rmax , we have tf

D

D 5:270 s:


June 25, 2012


Dynamics 2e

163

Problem 2.112 A stomp rocket is a toy consisting of a hose connected to a blast pad (i.e., an air bladder) at one end and to a short pipe mounted on a tripod at the other end. A rocket with a hollow body is mounted onto the pipe and is propelled into the air by stomping on the blast pad. Assuming the rocket can be given an initial speed v 0 120 ft=s, the rocket’s landing spot at B is 10 ft higher than the launch point, i.e., h 10 ft, and neglecting air resistance, find the rocket’s launch angle  such that the rocket achiev es the maximum possible range. In addition, as part of the solution, compute the corresponding maximum range and flight time. To do this:

D

D

(a) Determine the range R as a function of time. (b) Take the expression for R found in (a), square it, and then differentiate it with respect to time to find the flight time that corresponds to the maximum range, and then find that maximum range. (c) Use the time found in (b) to then find the angle required to achie ve the maximum rang e.

Solution

xy coordinate Referring to the figure on the right, we will use an system with src in at the laun ch point of the rock et. Let  be the elevation angle and v0 120 ft=s be the intial speed. The acceleration of the rocket is equal to g in the y direction and zero in the x

D

direction. Hence, we can use the constant acceleration equation s 1 2 2 ac t , to express the x and y coordinates of the rocket as a function of time. This gives

v0 t

D

C

x y

D .v 0 cos  /t; D .v 0 sin  /t

(1)

1 2  2 gt :

(2)

D tf the rocket is at B , so that we must have R D v 0 .cos  /tf ) cos  D v0Rtf : 2h C gt 2 1 2 h D v 0 .sin  /tf gt ) sin  D 2v0tf f ; 2 f

Letting tf denote the time of fligh t, for t

(3)

(4)



Next, recalling that sin2 

C cos2  D 1, using the last of Eqs. (3) and (4), we have 2h



2

gt 2

2

C f C  v0Rtf  D 1; 2v0 tf

!

(5)

which, for convenience, we view as an equation for R 2 whose solution is

R2

D v02tf2



1 2

2 2 f

h C gt  :


(6) June 25, 2012


Solutions Manual

164

Now we maximize R with respect to tf . Since the value of tf for which R is maximum coincides with the value of tf for which R 2 is maximum, we can find the value of tf in question by differentiating Eq. (6) with respect to tf and then setting the res ults to 0. This gives

dR 2 dtf Recalling that g

D 0 D 2v 0tf

Rmax



 2 hC

1 2 gt 2 fRmax

.gt

fRmax /

)

tfRmax

D

s

2

D 32:2 ft=s , v0 D 120 ft=s, and h D 10 ft, we can evaluate tf

Rmax

tfRmax

2 2 .v  hg/: g2 0

(7)

to obtain

5:211 s:

D

To find R max , we substitute the expression of tfRmax from the last of Eqs. (7) into Eq. (6) and then we take a square root. To find the corresponding value of  we substitute the value of tfRmax from the last of Eqs. (7) 32:2 ft=s2 , v 0 120 ft=s, h 10 ft, into the last of Eqs. (4) and solve for  . Recalling that we have g these operations yield the following results:

D

Rmax

D 437:1 ft

and

Rmax

D

D

D 45:66ı:


June 25, 2012


Dynamics 2e

165

Problem 2.113

D E D OC O

A trebuchet releases a rock with mass m 50 kg at the point O . The .45 { 30 | / m=s. If one initial velocity of the projectile is v0 were to model the effects of air resistance via a drag force directly proportional to the projectile’s velocity, the resulting accelerations . =m/ x and y g  in the x and y directions would be x .=m/ y , respectively, where g is the acceleration of gravity and  0:64 kg=s is a viscous drag coefficient. Find an expression for the trajectory of the projectile.

RD

P

D

P

RD

Solution

We can integrate the x and the y components of acceleration to obtain the x and y displacement as a function d xP of time. The problem state s that x .=m/ x . Then, recalling that x , we can write dt

RD

P D ddtxP )

 x m

 dt m

P

D dxPxP )

RD

Z

xP

P P D

dx x

Z

t

 dt m

 m

) xP D .v 0/x e t ; (1) where .v 0 /x is the x component of the velocity of the projectile at t D 0 . Next, we recall that xP D dx=dt . 



.v0 /x



0

So, using the last of Eqs. (1) we have

dx dt

D .v0/x e

 mt

)

dx

D .v 0/x e

Z

)

 mt

dt

x

dx 0

D .v0/x

Z

t



e  m t dt

)

0

D m.v0/x

x

1





e  m t : (2)



We can now repeat these steps starting with the acceleration in the y direction. Doing so, we have 

Z

yP .v0 /y

g

C

P

dy . =m/ y

PD

Z

t

dt 0

) yP D mg 

e

 m t

where .v 0 /y is the y component of the velocity of the projectile at t respect to time, we obtain 2

y

D

m g 2



mgt 

C m .v0/y

2

 m g 



1

 C .v / e 0 y

 m t

(3)

D 0. Integrating Eq. (3) again wit h

C m .v0/y





e m t :

(4)

m x ln 1  :  m.v0 /x

(5)

2

From Eq. (2) we find 

e m t

D



1

x m.v0 /x



)

t

D



Substituting the last of Eqs. (5) into Eq. (4) and recalling that m 0:64 kg=s, we obtain and 

D

y

D 59:88

 10

3



D 50 kg, .v0/x D 45 m=s, .v0/y D 30 m=s

ln 1  2:844  104 x





 C 17:70x

m:


(6)

June 25, 2012


Solutions Manual

166

Problem 2.114

D

0:64 kg=s, deterContinue Prob. 2.113 and, for the case where  mine the maximum height from the ground reached by the projectile and the time it takes to achieve it. Compare the result with what you would obtain in the absence of air resistance.

Solution

When the projectile is at its maximum height, the y component of its velocity must be zero. Therefore, we first need to find y as a function of time. Using y d y=dt and y g  .=m/ y , we obtain

P

RD P RD P yP t d yP d yP m    yP Œg C . =m/ yP ç ) dt ) ln g C yP D D Dt dt g C .=m/ yP  m v 0 v     ) ln g C m yP ln g C m v0y D  t=m ) ln ggCC mvy0yP D  t=m m 



Z

Z

0y













g C m v  e mPD

 t=m

0y

D





mg



C v0y



y

PD

v0y e  t=m

)

C



mg



P

PD

 mg



g

0y

; (1)   where we have used t0 0 and have let the y component of the initial velocity be v 0y . Now that we have an expression for y.t/ , we again observe that the maximum height is reached when y 0 . Letting H max denote the maximum height achieved by the projectile and t max the time at which this height is achiev ed, setting the last of Eqs. (1) to zero, we obtain

)

y



ˇˇ !

e  tmax =m 

mg 

D0 ) )

mg mg  v0y m mg  v0y ln  mg

 tmax =m

tmax

D

D ln





C

C

 )

tmax

D 3:000 s,

(2)

where we have used m 50 kg, g 9:81 m=s2 , v 0y 30 m=s, and  0:64 kg=s to obtain the numerical result. The maximum height in the presence of air resistance can be calculated as H max y max h, where h is t max . Therefore, we now need to find y.t/ by integrating given as 4:5 m and y max is the value of y when t Eq. (1) one more time using

D

D

D

D

D

D

P D dy D dt

y

 mg 

C v0y



e  t=m



mg 

) )

Z

y

t

D

D

m  

0

y

Z 

dy

0

 mg 

mg 

C v0y

C v0y





C

mg dt  mg 1  t; (3) 

e  t=m

e  t=m








June 25, 2012


Dynamics 2e

167

D

D 0. Substituting t D t D 3:000 s from Eq. (2), m D 50 kg, D 0:64 kg=s into Eq. (3), we obtain y D 44:71 m. Therefore, H H D y C h D 49:21 m ) H D 49:21 m, D 4:5 m. In the absenc e of air resist ance, the maxi mum height is given by .H / D C h, where .y / can be calculated by letting yP D 0 in the projectile motion equatio ns with

0 and t0 where we have used y0 g 9:81 m=s2 , v 0y 30 m=s, and  is

D

D

max

where h

max

max

max

max

max

max no air

.ymax /no air

max no air

constant gravity. Therefore,

y2

P D yP02

2



2g.y max /no air

D0 )

.ymax/no air

D P D

0 , y0 30 m=s, and g where we have used the fact that y 0 the absence of air resistance .H max /no air is given by .Hmax /no air where h

D .y

max/no air

D y2gP0 D 45:87 m;

(4)

2

D 9:81 m=s . Hence, the maximum height in

C h D 50:37 m;

(5)

D 4:5 m. Therefore, the percent incre ase in height with no air resistance is

percent increase in height with no air resistance

)

D 50:3749:2149:21 



100%

percent increase in height with no air resistance


D 2:363 %.

June 25, 2012


Solutions Manual

168

Problem 2.115

D

0:64 kg=s, deterContinue Prob. 2.113 and, for the case where  mine tI and xI , the value of t , and the x position corresponding to the projectile’s impact with the ground.

Solution

We begin by working part of the solution to Prob. 2.113. Specifically, we can integrate the x and the y components of acceleration to get the x and y displacement as a function of time. The problem states that d xP x . =m/ x . Then, recalling that x , we can write dt

RD



P

P D ddtxP )

 x m



RD  d xP dt D ) m xP

Z

xP .v0 /x

PD

dx x

P



Z

t 0

 dt m

) xP D .v0/x e

 t m

(1)

where .v 0 /x is the velocity component of the projectile. Integrating Eq. (1) again with respect to time, we obtain x

t

dx

Z

.v /

D

0 x

Z



e  m t dt

mvx0

x

)

D

1





e m t :

 Proceeding similarly to obtain the expression of y as a function of time, we have 

Z

yP

.v0 /y

0

g

C

P

dy . =m/ y

PD

0

Z

t

dt 0

) yP D mg 

e

 m t



1

(2)



 C .v / e 0 y

 m t

:

(3)

Integrating Eq. (3) again with respect to time, we obtain 2

y

D

m g 2

mgt 



C m .v0/y

2

 m g 

C m .v0/y





e m t :

(4)

m x : ln 1  m.v0 /x 

(5)

2

Next, from Eq. (2) we find 

e m t

D



1

x m.v0 /x



)

t

D







Substituting the last of Eqs. (5) into Eq. (4), we have 2

y

2

2

D  m2g C m2g ln 1



x m.v0 /x

 C m .v /   m g C m .v / 1 

0 y



2



0 y



x m.v0 /x

:

(6)

To find the time of impact tI and the location xI of the impact we observe that the impact is characterized h 4:5 m. Hence we can use a numerical root finding method to find the value by the condition y of t in Eq. (4) for which the condition y h is satisfied. Similarly, we can use a numerical root finding

D

D

D


June 25, 2012


Dynamics 2e

169

D

method to find the the value of x in Eq. (6) for which the condition y h is satisfied. Because the majority of root finding methods require us to provide a guess of the solution, before using any such methods, we proceed to plot y.t/ as given in Eq. (4) and y.x/ as given in Eq. (6). Recalling that we are given m 50 kg, .v 0 /x 45 m=s, .v 0 /y 30 m=s and  0:64 kg=s, the plots shown below were obtained using Mathematica with the following code:

D

D

D

D

which gives

From the above two plots, we see that tI is close to 6 s and xI is close to 250 m. Hence, we will use the values just listed as guesses in an appropriate root finding numerical method to find more accurate values to the quantities tI and xI . For example, this can be done using Mathematica with the following code:

Evaluating the outcome of the above code, we have

tI

D 6:189 s

and

xI

D 267:7 m:


June 25, 2012


Solutions Manual

170

Problem 2.116 With reference to Probs. 2.113 and 2.115, assume that an experiment is conducted so that the measured value of xI is 10% smaller than what is predicted in the absence of viscous drag. Find the value of  that would be required for the theory in Prob. 2.113 to match the experiment.

Solution

For the case of no air drag, the time of impact can be calculated by equating the constant acceleration equation for the y coordinate of the projectile to h. Letting tI denote the time of impact, we would have

y

D .v 0/y t

1 2  2 gt

D h  ) gtI2 2.v0/y tI  2h D 0 ) tI D .1=g/ .v0/y ˙ v0y2 C 2gh ) tI D 6:263 s;

)

.v0 /y tI

1 2  2 gtI







q

(1)

where we have discarded the solution with the negative square root because it yields a negative time value, 30 m=s, g 9:81 m=s2 , and h 4:5 m. and where we have used the following numerical data: .v0 /y The impact distance xI for the case of no air drag can be calculated again by using constant acceleration

D

equations (with ax

D 0). This gives x D .v 0 /x t )

xI

D

D

D

D .v 0/x tI D 281:8 m;

(2)

where .v 0 /x 45 m=s and we used the expressi on for tI in Eq. (1). The problem state ment indicates that the x position of the rock in the presence of air drag is: .x I /air 0:9xI , i.e.,

.xI /air

D D 253:6 m:

(3)

To be able to use these results, we first determine the trajectory of the projectile in the presence of air resistance. We begin by working part of the solution to Prob. 2.113. Specifically, we can integrate the x and the y components of acceleration to get the x and y displacements as a function of time. Starting with d xP the given acceleration components, using x , we can integrate the expression for the x component of dt acceleration to get x . xP t  dx   dt x .v 0 /x e  m t (4)

RD

P

Z

.v0 /x

P

Z

PD

x

0

m

) PD

where .v 0 /x is the velocity component of the projectile. Integrati ng Eq. (4) with respect to time, we obtain x

RD

Using y

d yP , dt 

Z

0

t

dx

D

.v 0 /x

Z

0

 m t

e

dt

)

x

D

 mvx0 m t 1e : 





(5)

P

we can integrate the expression for the y component of acceleration to get y .

Z

yP .v0 /y

g

C

P

dy . =m/ y

P

Z D

t

dt 0

) yP D

mg   t e m 





 1 C .v / e 0 y


 m t

:

(6) June 25, 2012


Dynamics 2e

171

Integrating Eq. (6) with respect to time, we obtain 2

y

D

m g

mgt 



2

C m .v0/y

2

 m g 

C m .v0/y





e m t :

(7)

m x : ln 1  m.v0 /x 

(8)

2

Next, from Eq. (5) we find 

e m t

D



1

x m.v0 /x



)

t

D








y

2

D  m2g C m2g ln 1

2 

x m.v0 /x

 C m .v /   m g C m .v / 1 

0 y



2



0 y



x m.v0 /x

:

(9)

We now observe that at impact the x and y coordinates of the rock are .xI /air and h. By enforcing this condition in Eq. (9) we obtain an equation in  that can be solved numerically. Because most root finding algorithms require the user to supply a guess of the solution, we begin by plotting the value of y for x .xI /air 253:6 m (see Eq. (3)) as a function of . The plot presented below was obtained in Mathematica using the following code:

D

D

D

D

0 for x .xI /air when  is a bit greater than 1 kg=s. Hence, we will From the plot above, we see that y 1 kg=s as the guess for a root finding algorithm in order to find a more accurate value of use the value  the value of  . When using Mathema tica, this can be done with the following code

D

which yields the following solution



D 1:345 kg=s:


June 25, 2012


Solutions Manual

172

Problem 2.117

E D 2 {O C 1 |O C 7 kO and bE D 1 {O C 2 |O C 3 kO . Compute the following quantities.

Consider the vectors a

E (b) bE (c) a E (d) a E (e) .a E (f) a E

E aE bE C bE aE aE aE/ bE .aE bE/

(a) a  b   











Parts (a)–(d) of this problem are meant to be a reminder that the cross product is an anticommutative operation, while Parts (e) and (f) are meant to be a reminder that the cross product is an operation that is not associative. Solution

Using the vectors given in the problem statement, various properties of the cross-product are illustrated through a few simple exercises. Part (a)

The commutative relationship for the cross-product is demonstrated by first evaluating

ˇ ˇ

ˇ ˇ

{O |O kO aE bE D det 2 1 7 ) 

Part (b)

1 2

3

E ED

O C 1 |O C 3 kO

ab

11 {



:

(1)

The cross-product is again evaluated, but this time in the opposite order such that

E ED

ba

ˇˇO ˇ

O O

{ | k det 1 2 3 2 1 7

ˇˇ ˇ

E E D 11 {O

)

ba



O

1|



O

3k :

(2)

Thus the cross-product is anti-commutative because the results are equal in magnitude, but opposite in direction (sign). Part (c) The fact that the cross-product relation between two vectors is anti-commutative is also demonstrated through the equation below, where

ˇˇ ˇ

ˇˇ ˇ

ˇˇ )ˇ

ˇˇ EˇC E

{O |O kO {O |O kO  aE bE C bE aE D det 2 1 7 C det 1 2 3 D 



1 2

3

2 1

E

  O C 1 |O C 3 kO C 11 {O

11 {

7

ab



O

1|



O

3k ;

E D 0E:

ba


(3)

(4)

June 25, 2012


Dynamics 2e Part (d)

173

The cross-product of a vector wit h itself, such as

E ED

aa

ˇˇO ˇ

O O

{ | k det 2 1 7 2 1 7

will always yield the zero vector.

ˇˇ ˇD E

0;

(5)

Part (e) Demonstrating the non-associative nature of the cross-product, the example shows one possible way to calculate the product of three vectors, where

aE aE 



E D 0E bE D 0E:

b

(6)



Part (f) The associative property does not hold for cross-products since the result of part (e) is not equal to the result of

E

a

aE bE D aE 

ˇˇˇO

O O

{ | k  det 2 1 7 1 2 3

   ) aE aE bE D 4 {O 





ˇˇˇ D



ˇˇˇ O

O O

ˇˇˇ

{ | k det 2 1 7 ; 11 1 3

O C 13 kO

83 |

:


(7)

June 25, 2012


Solutions Manual

174

Problem 2.118

E D 1 {O C 2 |O C 3 kO and bE D 6 {O C 3 |O. (a) Verify that aE and bE are perpendicular to one another. (b) Compute the vector trip le product aE .a E bE/. (c) Compare the result from ca lculating aE .a E bE/ with the vector jEaj2 bE. The purpose of this exercise is to show that as long as aE and bE are perpendicular to one another, you can always write aE .aE bE/ D jE aj2 bE. This identity turns out to be very useful in the study of the planar motion of rigid bodies. Consider two vectors a



















Solution

Two vectors are perpendicular if their dot product is zero. Thus,

Part (a)

E ED1

a b 

 6

C 2 3 C 3 0 D 0: 

(1)



E

E

ˇˇ O ˇ

| k 2 3 ; 18 15

Part (b) The triple product is evaluated by first calculating the cross-product of a and b and then taking the cross-product of a with the cross-product of a and b . The calculation proceeds by

E

E

a

E

E

aE bE D aE 

ˇˇ O )ˇ

ˇˇ  DE O  Eˇ E D  O

O O

{ | k  det 1 2 3 6 3 0

E

a

a ab

9 { 

 18 |O C 15 kO D

{ det 1 9

O C 0 kO  :

84 {  42 |

Part (c) The vector expression is evaluated as

jEaj2bE D 12 C 22 C 32 2   D 84 {O 42 |O C 0 kO ; aE bE. which is the same as aE 



p



O

O

ˇˇ ˇ

 O C 3 |O C 0 kO ;

6 {

(3)

(4) (5)





(2)




June 25, 2012


Dynamics 2e

175

Problem 2.119

E E D O E E E E

Let r be the position vector of a point P with respect to a Cartesian coordinate system with axes x , y , r x { ry | (i.e., r k 0). Also, and ´ . Let the motion of P be confined to the xy plane, so that r !r k be the angular velocity vector of the vector r . Compute the outc ome of the products let !r !r  .!r  r / and !r  .r  !r /.

E

E E

ED OC O

E OD 

Solution

Use the property verifie d in part (c) of the solution to Problem 2.118 :

E

!r



!E rE D r 



jE!r j2rE D

2 ! r

r {O C r |O  : x

(1)

y

The cross-product is anti-commutative, as verified in part (b) of the solution to Problem 2.117. Therefore the triple cross-product !r  r  !r is Eq. (1) multiplied by 1.

E

E E  ! E rE !E  D !E r 



r

r  Œ

2 r

!E rE D ! r {O C r |O  : r 

ç

x

(2)

y


June 25, 2012


Solutions Manual

176

Problem 2.120 The three propellers shown are all rotating with the same speed of 1000 rpm about different coordinate axes.

angular

(a) Provide the proper vector expressions for the angular velocity of each of the three propellers. (b) Suppose that an identical propeller rotates at 1000 rpm about the axis ` oriented by the unit vector u` . Let any point P on ` yP ´P . Find the vector have coordinates such that xP

D

O

D

representation of the angular velocity of this fourth propeller. Express the answers using units of radians per second.

Solution

Part (a)

D .100=3/ rad=s D 104:7 rad=s. The angular veloc ity vectors can be written as     !E1 D 104:7 kO rad=s; !E 2 D 104:7 {O rad=s and ! E3 D  104:7 |O rad=s:

1000 rpm



Part (b)

(1)

The unit vector in the ` direction is

O D p13 {O C |O C kO



u`

:

(2)

Therefore, the angular velocity is expressed as

E D 100 p {O C |O C kO 3 3

!`





rad=s



D 60:46 {O C |O C kO



rad=s:


(3)

June 25, 2012


Dynamics 2e

177

Problem 2.121 Point P is constrained to move along a straight line ` whose positive orientation is described by the unit vector u` . Point A is a fixed reference point on ` . Let the vector rP =A denote the position of P relative to A and let uP =A be a unit vector pointing from A to P . Use the concept of time derivat ive of a vector to describe the velocity and acceleration of P . In addition, comment on what happens to the description of the velocity and acceleration when P happens to coincide with the fixed point A .

O

E

O

Solution

O E D rP =A uO P =A;

The motion of P is rectilinear. Using the unit vector uP =A, we write the position vector of P as

rP =A

(1)

where rP =A is the distance between P and A . We can calculate the velocity of P using Eq. (2.48) on p. 81 of the textbook: vP rP =A uP =A !r  rP =A; (2)

E DP O CE E E uOP P =A D !E r uO P =A:

E

O

where !r is the angular velocity vecto r of rP =A, which is the same as that of the unit vector uP =A, and by Eq. (2.46) on p. 81 of the textbook, we can write (3)



Since the motion of P is rectilinear, uP =A is essentially a constant (see discussion at the end of the solution). Referring to Eq. (3) (and since uP =A can never be zero), this means that, for rectilinear motions,

O O

E D 0E:

(4)

E D rPP =A uO P =A:

(5)

!r Hence the expression for the velocity of P is

vP

To derive the acceleration vector of P , we can proceed in the same manner as for the velocity to obtain (6) E D vEPP D rRP =A uO P =A C !E v vEP ; where ! E v is the angular velocity of the vector vEP . As with rEP =A, the angular velocity of vEP is the same as that of u O P =A, which, by Eq. (4), is zero. Therefore , the acceleration of P is aEP D rRP =A uO P =A: (7) We stated that u O P =A is essentially a constant. Referring to the problem’s figure, we have

aP

uP =A.t/



(8) P follows A, O D uuOO `` when when P precedes A . When P coincides with A , then uO P =A is undefined. Hence we can say that uO P =A is constant except when P coincides with A , at which case uO P =A is undefined and therefore cannot be used to describe the position, and

(



consequently velocity and acceleration, of P relative to A .


June 25, 2012


Solutions Manual

178

Problem 2.122

E

Starting with Eq. (2.48), show that the second derivative with respect to time of an arbitrary vector A is given by

ER D AR uOA C 2!EA AP uOA C !EP A AE C !EA !EA AE :

A









Keep the answer in pure vector form, and do not resort to using components in any component system. Solution

Beginning with Eq. (2.48),

EP D AP uOA C !EA A:E

A

(1)



and taking the derivative with respect to time, gives

ER D AR uOA C AP uOPA C !PEA AE C !EA A:EP

A





PO D !EA uOA and AEP is given by Eq. (1), so this expression becomes   R AE D AR uOA C AP !EA uO A C !PEA AE C ! EA AP uOA C !EA AE ; Noting that AP is a scalar and so it can be moved inside the cross product ! EA uOA and distributing the cross But, uA













product in the last term yields

ER D AR uOA C !EA AP uOA C !PEA AE C !EA AP uOA C !EA !E AE : Combining the two terms !A AP uA , we obtain the desired result, that is, O RE AE D AR uOA C 2!EA AP uOA C ! EP A AE C !EA !EA AE : A






















June 25, 2012


Dynamics 2e

179

Problem 2.123 The propeller shown has a diameter of 38 ft and is rotating with a constant angular speed of 400 rpm. At a given instant, a point P on the propeller is at rP .12:5 { 14:3 | / ft. Use Eq. (2.48) and the equation derived in Prob. 2.122 to compute the velocity and acceleration of P; respectively.

E D

OC

O

Solution

The position of point P is

E D rP x {O C rP y |O ;

rP where rP x

(1)

D 12:5 ft and rP y D 14:3 ft. Applying Eqs. (2.48 ) on p. 81 of the textbook, we have d jErP j vEP D rEPP D uO r C !E r rEP ; P

dt

O D E jE j

P

E

(2)



E

where urP rP = rP and where !rP is the angular velocity of rP and coincides with the angular velocity of the propeller so that !rP ! prop k; (3)

O

D

O

E D

jE j D

where !prop 400 rpm and k is a unit vector pointing in the positive ´ direction. We observe that d rP =dt 0 because P does not change its distance from the axis of rotation. Hence, substituting Eqs. (1) and (3) into Eq. (2), and carrying out the cross product, we have

E D!

vP

prop.rP y

O C rP x |O / )

E D.

{

O C 523:6 |O/ ft=s;

599:0 {

(4)

where we have used the fact that rP x 12:5 ft, rP y 14:3 ft, and ! prop applying the equation derived in Problem 2.122, we have that

D 400 rpm D 400 260 rad=s. Next,

D

vP

D

2 E D d dtjEr2P j C 2!Er d dtjErP j uO rE C !EP r rEP C !Er Because jErP j and ! E r are constant, Eq. (5) reduces to aEP D ! Er .!Er rEP /:

aP

P





P

P

P





E

.!rP

E

 rP /:

(5)

P

P



P

(6)



Substituting Eqs. (1) and (3) into Eq. (6), and carrying out the cross products, we have

aP

2 !prop.rP x

2

(7) O C rP y |O / ) aEP D .21;930 {O C 25;090 |O/ ft=s ; 2 where, again, we have used the fact that rP x D 12:5 ft, rP y D 14:3 ft, and ! D 400 rpm D 400 60 rad=s.

E D

{



prop




June 25, 2012


Solutions Manual

180

Problem 2.124

E D E D  OC O E D  OC O E D  OC O ED O E E E E

Consider the four points whose positions are given by the vectors rA 2 { 0 k m, rB 2 { 1 k m, rC 2 { 2 k m, and rD 2 { 3 k m. Knowing that the magnitude of these vectors is constant and that the angular velocity of these vectors at a given instant is ! 5 k rad=s, apply Eq. (2.48) to find the velocities vA , vB , vC , and vD . Explain why all the velocity vect ors are the same even though the position vectors are not.

 OC O

Solution

Using Eq. (2.48), the velocities of points A , B , C , and D are (1) E D ddtjErAj uO r C !Er rEA; vEB D d dtjErB j uO r C !E r rEB ; d jErC j vEC D uO r C ! E r rEC ; vED D d jEdtrD j uO r C !E r rED ; (2) dt where uO r D rEA =jErA j, u O r D rEB =jErB j, uO r D rEC =jErC j, uO r D rED =jErD j, and where, following the problem statement !E r D !E r D !E r D !E r D !E D 5 kO rad=s: (3) Since the magnitudes of rEA , rEB , rEC , and rED are constant, and in view of Eq. (3), Eqs. (1) and (2) reduce to vEA D ! E rEA; vEB D !E rEB ; (4) C C D D (5) vE D ! E rE ; vE D !E rE :         Recalling that rEA D 2 {O C 0 kO m, rEB D 2 {O C 1 kO m, rEC D 2 {O C 2 kO m, rED D 2 {O C 3 kO m, and !E D 5 kO rad=s, we can evaluate Eqs. (4) and (5) to obtain

vA

A

A

C

A

B



C



C

A

B

B

D

D



D

B

C

D









E D 10:00 |O m=s; vEB D 10:00 |O m=s; vEC D 10:00 |O m=s;

vA



E D 10:00 |O m=s:

vD

and

O

The velocities are the same because the given position vectors all have the same { component and because the points in question have the same distance from the axis of rotation.


June 25, 2012



Solutions Manual

182

Problem 2.126 When a wheel rolls without slipp ing on a stationary surface , the point O on the wheel that is in contact with the rolling surface has zero velocity. With this in mind, consider a nondeformabl e wheel rolling without slip on a flat stationary surface. The center of the wheel P is traveling to the right with a constant speed v 0 23 m=s. Letting R 0:35 m, determine the angular velocity of the wheel, using the stationary component system shown.

D

D

Solution

Since P moves parallel to the ground, we have that the velocity of P can be expressed as follows:

E D v0 {O:

vP

E

(1)

Letting rP denote the position of P relative to O , since the velocity of O is equal to zero at the instant rP . Hence, applying Eq. (2.48) on p. 81 of the textbook, we can write considered, we also know that vP

E D EP

E D d dtjErP j uO r C !Er rEP ;

vP

P

P

(2)



where !rP is the angular veloci ty of the vector rP . As both points O and P are on the wheel, we have that

E

jErP j D R Dconstant

E

and

E D !E

!rP

wheel

D!

O

wheel k:

(3)

In addition, at the instant shown, we also have

E D R |O:

rP

(4)

Substituting Eqs. (3) and (4) into Eq. (2), carrying out the cross product and simplifying, we have

E D

vP

O

!wheelR {:

(5)

Comparing Eqs. (1) and (5), we then conclude that

!wheel Recalling that v 0

D



v0 R

) !E

wheel

D



v0 k: R

O

(6)

D 23 m=s and R D 0:35 m, we can evaluate the last of Eqs. (6) to obtain !E D 65:71 kO rad=s: wheel




June 25, 2012


Dynamics 2e

183

Problem 2.127 The radar station at O is tracking the meteor P as it moves through the atmosphere. At the instant shown, the station measures the following data 21;000 ft,  40 ı , r for the motion of the meteor: r 22;440 ft=s, and  2:935 rad=s. Use Eq. (2.48) to determine the magnitu de and direction (relative to the xy coordinate system shown) of the velocity vector at this instant.

D

PD

D

PD

Solution

O

Referring to the figure at the right, the unit vector ur always points toward P . The unit vector u is perpendicular to ur and points in the direction of increasing  . Then, letting r denote the distance between P and the fixed r ur . Applying point O , we have that the position of P is described by rP

O

O

E D O

Eq. (2.48) on p. 81 of the textbook, we have

E D rP uO r C r !Er uO r ;

vP

E

(1)



O

where !r is the angular velocity of the unit vector ur . Since the angle  describes the orientation of rP , we have that

E

E D P kO ;

!r where k

(2)

O D uO r uO . Substituting Eq. (2) into Eq. (1) gives vEP D rP uO r C r P uO : 



(3)



Finding the magnitude of the vector, we then have that

q

jEvP j D rP 2 C r 2P 2 ) jEvP j D 65;590 ft=s; where we recalled that r D 21;000 ft, rP D 22;440 ft=s, and P D 2:935 rad=s. We now observe that uO r D cos  {O C sin  |O and uO D sin  {O C cos  |O; 

(4)







(5)

so that Eq. (3) can be rewritten as

„ ƒ‚ P … O C „P ƒ‚C P …

E D .rP cos 

vP



r  sin  / {

.r sin 

vx

O D .22;430 {O

r  cos  / |

vPy



O

61;640 | / ft=s:

(6)

Since vP is directed downw ard and to the right, the orientation of vP is  tan1 . vP y =vPx /:

E

Orientation of vP from x axis

E

where, again, r

D 21;000 ft, rP D

D

 tan1

ˇ rP

sin  r cos 

P

)

22;440 ft=s, and

E

D  C rrPP cos sin  

ˇ

j

70:01ı

E

Orientation of v from x axis

PD



j

D 70:01ı (cw),

(7)

2:935 rad=s.


June 25, 2012


Solutions Manual

184

Problem 2.128 The radar station at O is tracking the meteor P as it moves through the atmosphere. At the instant shown, the station measures the following data 21;000 ft,  40 ı , r for the motion of the meteor: r 22;440 ft=s, 2  187;500 ft=s2 , and  2:935 rad=s, r 5:409 rad=s . Use the equation derived in Prob. 2.122 to determine the magnitude and direction (relative to the xy coordinate system shown) of the acceleration vector at this instant.

PD

D

RD

D RD

PD

Solution

O

Referring to the figure at the right, the unit vector ur always points toward P . The unit vector u is perpendicular to ur and points in the direction of increasing  . The distance betwe en P and O is r so the position of P is rP r ur . Applying the equation derived in Problem 2.122, we can write

O

O

E D O aEP D rERP D rR uO r C 2! Er rP uO r C !EP r rEP C !E r !Er rEP  ; (1) where ! Er is the angular velocity of ErP . Since the an gle  describes the orientation of rEP , we have that !E r D P kO ; (2) O O where k D u O r uO . Because the direction of k is fixed, !EP r D R kO . Hence, substituting rEP D r uO r and Eq. (2) 











into Eq. (1) and simplifying, we have

E D .rR

aP



P O C .r R C 2rP P / uO :

r  2 / ur

(3)



Finding the magnitude of the vector, we then have that

q

jEaP j D rR r P 22 C r R C 2rP P 2 ) jEaP j D 19;300 ft=s ; where r D 21;000 ft, rP D 22;440 ft=s, P D 2:935 rad=s, rR D 187;500 ft=s , and R D Noting that u O r D cos  {O C sin  |O and uO D sin  {O C cos  |O , Eq. (3) becomes   r R C 2rPP sin   {O C rR r P2 sin  C r R C 2rPP cos   |O aEP D rR r P 2 cos  2







„

2





aP x

5:409 rad=s2 .





ƒ‚

(4)



…

„

ƒ‚

apy

…

2

) aEP D . 6599 {O C 18;130 |O/ ft=s ; (5) Since aEP is directed upward and to the left, the orientation of aEP is given by 180 ı tan1 .jaP y =aP x j/, i.e., rR r P2 sin  C r R C 2rPP cos  Orientation of aEP from x axis D 180 ı tan1 rR r P2 cos  r R C 2rPP sin  ) Orientation of aEP from x axis D 110:0ı .ccw/. where  D 40ı , r D 21;000 ft, rP D 22;440 ft=s, P D 2:935 rad=s, rR D 187;500 ft=s , and R D 







"ˇˇ



2

5:409 rad=s







ˇ#

2

.


June 25, 2012




Dynamics 2e

187

Problem 2.131

PD

4 ft=s. The end B of a robot arm is being extended with the constant rate r 0:4 rad=s and is constant, use Eq. (2.48) and the equation Knowing that  derived in Prob. 2.122 to determine the velocity and acceleration of B when r 2 ft. Express your answer using the component system shown.

PD

D

Solution

ED O

E

r ur . Differentiating r with The position of point B relative to the fixed point O , can be expressed as r respect to time according to Eq. (2.48) on p. 81 of the textbook, gives the velocity of B as follows: (1) E D rP uO r C !E r r uO r ; O where ! E r is the angular velocity of the vector rE. Letting k denote the unit vector perpendicular to the plane of motion such that kO D u O r uO , and observing that the angle  describes the orientation of the vector rE, we have !Er D P kO : (2)

vB







Substituting Eq. (2) into Eq. (1) and carrying out the cross-prod uct, we have

E D rP uO r C P kO

vB Given that r

2 ft, r

D

4 ft=s, and 

PD



O

r ur

) vEB D rP uO r C P r uO :

(3)



0:4 rad=s, we can evaluate the last of Eqs. (3) to obtain

PD vEB D .4:000 uO r C 0:8000 uO / ft=s: 

E D r uO r using

To obtain the acceleration of B , we compute the second time derivati ve of the position vector r the equation derived in Problem 2.122. This gives

E D rR uO r C 2!Er rP uO r C !PEr

aB





Using Eq. (2), we have

O C !E r !Er

r ur





O



r ur :

(4)

EP D R kO C P kPO D R kO ; P where kO D 0E because the motion is planar. Since rP and P are constant, rR D 0; and R D 0: !r

(5)

(6)

Using Eqs. (2), (5), and (6) we can simplify Eq. (4) to read

PO P kO P kO r ur ar r P 2 ur 2P r u :   E D PO C O ) ED O C PO Again, since r D 2 ft, rP D 4 ft=s, and P D 0:4 rad=s, we can evaluate the last of Eqs. (7) for to obtain aEB D . 0:3200 uO r C 3:200 uO / ft=s : aB

2  k  r ur













(7)

2


June 25, 2012


Solutions Manual

188

Problem 2.132 The end B of a robot arm is moving vertically down with a constant speed v0 2 m=s. Letting d 1:5 m, apply Eq. (2.48) to determine the rate at which r and  are changing when  37 ı .

D

D

D

Solution

Referring to the figure on the right, we begin by describing the position of point

O O

B relative to O using the . ur ; u / component system: rEB D r uO r : (1) P The velocity of B is vEB D rEB . Then, using Eq. (2.48 ) on p. 81 of the textbook, we have vEB D rP uO r C ! (2) E r r uO r ; where ! E r is the angular velocity of the vector rEB . Since the vector rEB rotates with the robotic arm, we have ! (3) Er D P kO : 

Substituting Eq. (3) into Eq. (2) we have

vB

E D rP uO r C P kO



r ur

) vEB D rP uO r C P r uO :

O

(4)



OO

Since point B is moving downward along a vertical line with speed v 0 , using the . {; | / component system, the velocity of B can also be described as follows:

E D

vB

O

v 0 | :

(5)

We now observe that

O D sin  uO r C cos  uO :

| Therefore, Eq. (5) can be rewritten as

E D

vB

(6)



v0 .sin 

O C cos  uO /:

ur

(7)



Equating the second of Eqs. (4) and Eq. (5) component by component, we have

P D v0 sin  and P r D v0 cos  : Recognizing that r cos  D d , i.e., r D d = cos  , we can solve Eqs. (8) for rP and P to obtain r



r

v0 sin 

(8)



and





v0 cos2 

:

(9)

PD PD d Finally, recalling that v 0 D 2 m=s,  D 37 ı , and d D 1:5 m, Eqs. (9) can be evaluated to obtain rP D 1:204 m=s and P D 0:8504 rad=s: 




June 25, 2012


Dynamics 2e

189

Problem 2.133 The end B of a robot arm is moving vertically down with a constant speed v0 6 ft=s. Letting d 4 ft, use Eq. (2.48) and the equation derived in 0 ı. Prob. 2.122 to determine r ,  , r , and  when 

D

D PPR

R

D

Solution

Referring to the figure on the right, the velocity is expressed both in terms of the

OO

O O

.{; | / component system and using the component system . ur ; u / along with Eq. (2.48) on p. 81 of the textbook. This gives

E D v0 |O and vE D rP uO r C P kO r uO r D rP uO r C P r uO : (1) Expressing . uO r ; uO / in terms of . {O; |O /, we have uO r D cos  {O C sin  |O and uO D sin  {O C cos  |O : (2) Substituting Eqs. (2) into the last of Eqs. (1) and collecting the {O and |O terms, we v













have

v

r cos 



r  sin  {

r sin 

r  cos  | :

(3)

E D P P  O C P C P  O Equating the {O and |O components of velocity given by the first of Eqs. (1) and Eq. (3), and keeping in mind that r D d = cos  , we have rP cos  r P sin  D 0 ) rP cos  P d tan  D 0 (4) P rP sin  C r  cos  D v0 ) rP sin  C P d D v0 : (5) Substituting  D 0 into Eqs. (4) and (5), we have 





P D0

r

and



PD



v 0

d

D

1:500 rad=s;

(6)

D D O O aE D rR uO r C 2P kO rP uO r C R kO r uO r C P kO .P kO r uO r / 2 ) aE D rR uO r C 2P rP uO C R r uO P r uO r : Alternatively, differentiating Eq. (3) with respect to time and rearranging terms, the acceleration expressed in the . {O; |O / component system is: aE D rR .cos  {O C sin  |O/ C 2P rP . sin  {O C cos  |O /

6 ft=s and d 4 ft. Using the equation derived in where we have used the following numerical data: v 0 Problem 2.122, the acceleration expressed in the . ur ; u / component system is: 









 




June 25, 2012


Solutions Manual

190

C R r. sin  {O C cos  |O/ P 2r.cos  {O C sin  |O/: Collecting {O and |O terms, then substituting r D d ,  D 0 , and the expressions in Eqs. (6), we have  v02   R  aE D0 D rR {O C  d |O : 

E

ˇ

E D 0E, so that

Since v is constant, a



v02

ı





(7)

d

and R 0; RD d D D where, again we have used the fact that v 0 D 6 ft=s and d D 4 ft.

r

9:000 ft=s2


June 25, 2012


Dynamics 2e

191

Problem 2.134 A micro spiral pump consists of a spiral channel attached to a stationary plate. This plate has two ports, one for fluid inlet and the other for outlet, the outlet being farther from the center of the plate than the inlet. The system is capped by a rotating disk. The fluid trapped between the rotating disk and stationary plate is put in motion by the rotation of the top disk, which pulls the fluid through the spiral channel. With this in mind, consider a channel with geometry given by the equation r  r0 , where  12 m is 146 m is the radius at the inlet, r is the distance from the spin axis, and  , called the polar slope, r 0 measured in radians, is the angular position of a point in the spiral channel. If the top disk rotates with a constant angular speed ! 30;000 rpm, and assuming that the fluid particles in contact with the rotating

D C

D D

D

disk essentially stuck the to it, determine and system acceleration of(which one such fluid with particle it is m. Express at r are170 answer usingthe thevelocity component shown rotates the when top disk).

D

Photo credit: “Design and Analysis of a Surface Micromachined Spiral-Channel Viscous Pump,” by M. I. Kilani, P. C. Galambos, Y. S. Haik, C. H. Chen, Journal of Fluids Engineering, Vol. 125, pp. 339–344, 2003.

Solution

Referring to the problem’s figure, we focus our attention on a single fluid particle moving along the channel. The unit vector ur always points from the srcin of the coordinate system to the particle so that the position r ur , where r is the distance of the particle from the spin axis. Using Eq. (2.48) on of the particle is r p. 81 of the textbook, we can express the velocity of a particle as

O

ED O

E D rP uO r C !E rE;

v

ED PO

PD

where !  k , with  can be rewritten as

 30;000 260

(1)



E

rad=s, is the angular velocity of r . Recalling that r

D  C r0, Eq. (1)

(2) E D P uO r P kO . C r0/ uO r ) vE D P uO r C P . C r0/ uO ; where we have used the fact that kO uO r D uO . Solving r D  C r 0 for  we find that  D .r r0 /=. Recalling that r 0 D 146 m and  D 12 m, for r D 170 m,  D 2:000 rad. Hence, given that P D 1000 rad=s, we can evaluate vE in the last of Eqs. (2) to obtain

v









 



v

E D .0:03770 uO r C 0:5341 uO / m=s: 

Differentiating the last of Eqs. (2) with respect to time, we have

E D R uO r C P uPO r C R . C r0/ uO C P 2 uO C P . C r0/ uOP :

a






(3)



June 25, 2012


Solutions Manual

192

O

E D P kO , applying Eq. (2.46) on p. 81 of the (4) OP D P kO uO D P uO r :

O

Observing that both ur and u rotate with angular velocity ! textbook, we have ur  k  u r  u and u



PO D P O O D P O Substituting Eqs. (4) into Eq. (3), with R D 0 because P is constant, we have aE D P 2 . C r0 / uO r C 2P 2 uO : Recalling that, for r D 170 m,  D 2:000 rad, and recalling that r0 D P D 30;000 260 rad=s, we can evaluate aE to obtain 









(5)



146 m, 



a

E D

1678 ur

D

12 m, and

O C 236:9 uO  m=s2: 


June 25, 2012


Dynamics 2e

193

Problem 2.135 A disk rotates about its center, which is the fixed point O . The disk has a straight channel whose centerline passes by O and within which a collar A is allowed to slid e. If, when A passes by O , the speed of A relative to 14 m=s and is increasing in the direction shown with a the channel is v 4 rad=s and rate of 5 m=s2 , determine the acceleration of A given that ! is constant. Express the answer using the component system shown, which rotates with the disk. Hint: Apply the equation derived in Prob. 2.122 to the vector describing the position of A relative to O and then let r 0.

D

D

D

Solution

E

OO

E

Let rA be the position of A relative to the fixed point O . Using the . {; | / component system, r can be written as rA r | : (1)

E D

O

Applying the equation derived in Problem 2.122, the acceleration of A is (2) E rP |O C !PEr rEA C !E r !Er rEA; where ! E r is the angular veloci ty of the vector rE. When A is at O , rE D 0E so that Eq. (2) can be simplified to aEA r D0 D rR |O 2!E r rP |O: (3)

E D rR |O

aA





2! r



ˇ

Now, we observe that





!r





! k:

(4)

E D O

Substituting Eq. (4) in Eq. (3), we have

ˇ





ˇ

(5) E r D0 D rR |O 2! kO rP |O ) aEA r D0 D 2! rP {O rR |O : Recalling that rP D v D 14 m=s, rR D 5 m=s , and ! D 4 rad=s, we can evaluate the last of Eqs. (5) to

aA













2



obtain

E

aA

ˇ

r D0

D.

2

O C 5:000 |O/ m=s :

112:0 {


June 25, 2012



Dynamics 2e

195

Problem 2.137 Assuming that the child shown is moving on the merry-go-round along a radial line, use the equation derived in Prob. 2.122 to determine the relation that ! , ! , r , and r must satisfy so that the child will not experience any sideways acceleration.

P

P

Solution

ED O

r ur , where r is the distance from the Using the component system shown, the position of the child is r spin axis and ur is the unit vector always pointing fro m the srcin of the system towar d the child. Using the equation derived in Problem 2.122, we can express the acceleration of the child takes in the following form:

O

(1) E D rR uO r C 2!E rP uO r C !PE rE !E .!E rE/; E , which is the angular velocity of the where we have recognized that the angular velocity of the vector rE is !

a











merry-go-round. Using the component system shown in the figure, we have that

E D ! kO

!

and

EP D !P kO ;

!

(2)

since the direction of the unit vector kO remains fixed. Recalling that rE D r uO r , and substituting Eqs. (2) into Eq. (1), we have aE D . rR ! 2 r/ uO r C .2! rP C r !/ (3) P uO q : Equation (3) shows that the component of the accelerati on in the direction of u O q is aq D .2! rP C !r/: (4) P 

Hence, in order for the child not to experience sideways acceleration, we must have

P C !rP D 0:

2! r


June 25, 2012


Solutions Manual

196

Problem 2.138 The mechanism shown is called a swinging block slider crank. First used in various steam locomotive engines in the 1800s, this mechanism is often found in door-closing systems. If the disk is rotating with 60 rpm, H 4 ft, R 1:5 ft, and r is the distance between B and O , a constant angular velocity  compute r and  when  90 ı . Hint: Apply Eq. (2.48) to the vector describ ing the position of B relative to O .

P

P

D

PD

D

D

Solution

E

We can express the velocity of B in two ways. First, as the time derivative of the position vector rB=A and second as the time derivative of the position vector rB=O . Referring to the figure in the problem statement, we can express these two position vectors as follows:

E

E D R uOB=A

rB=A

E

rB=O

and

D r uO S ;

(1)

where, as given the problem statement, r is the distance between B and O , and where we observe that the angular velocities of the unit vector in the above equations are

!uO B=A

E

D P kO

and

!uO S

E D P kO:

(2)



P D 0 since R is a constant, using Eq. (2.48) on p. 81 of the textbook, we have vEB D rEPB=A D P kO R uO B=A and vEB D rEPB=O D rP uO S P kO r uO S : Next, we observe that, for  D 90 ı , we have 1 uO B=A D {O; r D H 2 C R 2 ; and uO S D p .R {O C H |O/: R2 C H 2 Hence, observing that R



p





D 90ı, we have rP vEB D90 D p .R {O C H |O/ C H P {O R2 C H 2

(3)

(4)

Substituting Eqs. (4) into Eqs. (3), for 

E

vB

ˇ

 D90

ı

D R P |O

and

E

ˇ



ı



PO

R | :

(5)

Equating the two above expressions for vB component by component, we have

O W pR2rP RC H 2 C H P D 0; rP H |O W p R P D R P : R2 C H 2

{

(6) (7)




June 25, 2012


Dynamics 2e

197

P P

P

Equations (6) and (7) form a system of two equations in the two unknowns r and  (at  solution is RH  R2  r  D90 and   D90  : 2 2

D 90ı) whose

P P P D pR 2 C H 2 D R CH (8) Recalling that we have P D 60 rpm D 60.2 =60/ rad=s, H D 4 ft, R D 1:5 ft, we can evaluate the quantities

ˇ

ˇ

ı

ı

in Eqs. (8) to obtain

P

r

ˇ

 D90

ı

D 8:825 ft=s

and

P

ˇ

  D90

ı

D

0:7746 rad=s:


June 25, 2012


Solutions Manual

198

Problem 2.139 A sprinkler essentially consists of a pipe AB mounted on a hollow shaft. The water comes in the pipe at O and goes out the nozzles at A and B , causing the pipe to rotate. Assume that the particle s of water move through the pipe at a constant rate relative to the pipe of 5 ft=s and that the pipe AB is rotating at a constant angular velocity of 250 rpm. In all cases, express the answers using the right-handed and orthogonal component system shown. Determine the acceleration of the water particles when they are at d=2 from O (still within the horizontal portion of the pipe). Let d 7 in:

D

Solution

Referring to the figure at the right, we consider a water particle P that is in the horizontal part of the tube. Using the component system shown, the position of P is

E D r uO B :

rP

(1)

Then, applying Eq. (2.48) on p. 81 of the textbook, the velocity of P is

E D rP uO B C !E

vP



O

r uB ;

(2)

E

where the angular velocity of the arm is also the angular velocity of the vector rP as well as that of the unit vector uB . Differentiating Eq. (2) with respect to time, we have

O

E D rR uO B C rP uPO B C !EP

aP



O C !E rP uO B C !E

r uB





PO

r uB :

(3)

P

In Eqs. (2) and (3), r denotes the rate at which the water particles move relati ve to the arm. Therefore 5 ft=s and r 0 . Also, ! ! k , where ! 250 rpm is constant. In addition, using Eq. (2.46) on p. 81 of the textbook, we have that uB ! k  uB . Therefore, Eq. (3) can be simplified to

ED O D PO D O O aEP D 2rP ! uO C r! 2 uO B : (4) Recalling that rP D 5 ft=s and ! D 250 rpm D 250.2 =60/ rad=s, for r D d =2, where d D 7 in: D .7=12/ ft, PD

r

RD





we can evaluate the above expression to obtain

E D.

aP

O

261:8 uC 

199:9 uB / ft=s2 :

O


June 25, 2012


Dynamics 2e

199

Problem 2.140 A sprinkler essentially consists of a pipe AB mounted on a hollow shaft. The water comes in the pipe at O and goes out the nozzles at A and B , causing the pipe to rotate. Assume that the particle s of water move through the pipe at a constant rate relative to the pipe of 5 ft=s and that the pipe AB is rotating at a constant angular velocity of 250 rpm. In all cases, express the answers using the right-handed and orthogonal component system shown. 7 in:, Determine the acceleration of the water particles right before they are expelled at B . Let d ˇ 15 ı , and L 2 in: Hint: In this case, the vector describing the positio n of a water particle at B goes from O to B and is best written as r rB uB r´ k .

D

D

D

ED O C O

Solution

Referring to the figure at the right, we consider a water particle P that is in the slanted part of the tube. Using the compon ent system shown, the position of P is

E D Œd

rP



O C ` sin ˇ kO:

.L  `/ cos ˇ ç uB

(1)

Differentiating Eq. (1) with respect to time, we have (2) OP C `P sin ˇ kO; O where we have used the fact that d , L , and ˇ are constant as well as the fact that k does not change direction. We observe that `P is the rate at which the water particles move through the pipe, so that `P D 5 ft=s and `R D 0 because `P is constant. Using Eq. (2.46) on p. 81 of the textbook, we have that uPO B D ! kO uO B , where we have recognized that the angular velocity of the arm, i.e., ! E D ! kO , is also the angular velocity of the unit vector u O B . Hence, aEP can be rewritten as vEP D `P cos ˇ uO B !Œd .L `/ cos ˇ ç uO C C `P sin ˇ kO : (3)

E D `P cos ˇ uO B C Œd

vP



.L  `/ cos ˇ ç uB









Differentiating Eq. (3) with respect to time, we have (4) E D `P cos ˇ uPO B ! `P cos ˇç uO C !Œd .L `/ cos ˇç uOP C ; P O where we have used the fact that `, ! , d , L , ˇ , and k are constant. Using Eq. (2.46) on p. 81 of the textbook PO B D ! kO uO B D ! uO C and uPO C D ! kO uO C D ! uO B , we can rewrite aEP as again to write u aP 2! `P cos ˇ uC ! 2 Œd .L `/ cos ˇ ç uB : (5) O =60/ rad=s, ˇ D 15ı, and O d D 7 in: D .7=12/ ft, for Recalling that `P D 5 ft=s, !E DD250 rpm D 250.2 ` D L D 2 in: D .2=12/ ft, we can evaluate the above expression to obtain aEP D . 252:9 uO C 399:8 uO B / ft=s :

aP



























2


June 25, 2012


Solutions Manual

200

Problem 2.141 A particle P is moving along the curve C , whose equation is given by

y

2



x 2 .x  1/.2x  3/



D 4 x 2 C y 2





2

2x ;

at a constant speed vc . For any posi tion on the curv e C for which the radius of curvature is defined (i.e., not equal to infinity), what must be the angle  between the velocity vector v and the acceleration vector a?

E

E

Solution

The speed is constant. This tells us that there is no component of acceleration in the direction of velocity. Therefore, the angle  between v and a must be 90 ı .

E

E


June 25, 2012


Dynamics 2e

201

Problem 2.142 A particle P is moving along a path with the velocity sho wn. Is the sketch of the normal-tangential component system at P correct?

Solution

O

E

No, the unit vector u t must point in the direction of v .


June 25, 2012


Solutions Manual

202

Problem 2.143 A particle P is moving along a path with the velocity sho wn. Is the sketch of the normal-tangential component system at P correct?

Solution

O

No, the unit vector un must point toward the concave side of the curve.


June 25, 2012


Dynamics 2e

203

Problem 2.144 A particle P is moving along a straight line with the velocity and acceleration show n. What is wrong with the unit vectors shown in the figure?

Solution

O

The unit vector un is not defined for a straight line.


June 25, 2012


Solutions Manual

204

Problem 2.145 A particle P is moving along some path with the velocity and acceleration shown. Can the path of P be the straight line shown?

Solution

O

No, because the path is straig ht. It would need to be curved with a tangent at P coincident with u t and concavity on the side of un .

O


June 25, 2012


Dynamics 2e

205

Problem 2.146 The water jet of a fountain is let out at a speed v 0 of curvature of the jet at its highest point.

D 80 ft=s and at an angle ˇ D 60ı. Determine the radius

Solution

The tangent to the trajectory of the water jet at the highest point is horizo ntal. Therefore, the velocity at the highest point is completely horizontal and the normal direction coincides with the direction of gravity. We model the motion of the jet as projectile motion. This implies that the water particles have constant acceleration equal to the acceleration of gravity. In addition, the horizontal component of the velocity of the water particles is constant and therefore equal to the value it has when the jet is first emitted by the nozzle, namely, v 0 cos ˇ . Since at the highest point the acceleration, which is due to gravity, is along the normal direction, we have 2

an

D v D g

at

and

D 0;

(1)

where v is the speed of the water particles at the highest point on the trajectory. We have already argued that at the point in question the velocity vector is parallel to the horizontal direction. Therefore we must have

v

D v0 cos ˇ:

Substituting Eq. (2) into the first of Eqs. (1) and solving for

 Recalling that v 0

D

 , we have

v02 cos 2 ˇ g

(2)

:

(3)

D 80 ft=s, ˇ D 60ı and g D 32:2 ft=s , we can evaluate Eq. (3) to obtain  D 49:69 ft: 2


June 25, 2012


Solutions Manual

206

Problem 2.147 A telecommunications satellite is made to orbit the Earth in such a way as to appear to hover in the same point in the sky as seen by a person standing on the surface of the 1:385  108 ft and Earth. Assuming that the satellite’s orbit is circular with radius rg 1:008  104 ft=s, determine the knowing that the speed of the satellite is constant and equal to vg magnitude of the acceleration of the satellite.

D

D

Solution

Using normal-tangential components, the acceleration of the satellite is 2

E D vP uO t C v uO n;

a

(1)

O

O

where v is the speed,  is the radius of curvatu re of the path, and where u t and un are the unit vectors tangent and normal to the trajectory, respectively. We have

v

D v g D constant; vP D 0;



and

D rg :

(2)

Substituting Eqs. (2) into Eq. (1), we have 2

2

E D vrgg uO n ) jEaj D vrgg :

a Recalling that v g

D 1:008

 104 ft=s and r g

D 1:385

(3)

 108 ft, we can evaluate the last of Eqs. (3) to obtain 2

jEaj D 0:7336 ft=s :


June 25, 2012


Dynamics 2e

207

Problem 2.148

D

30 m. At the instant shown, the speed of the car A car travels along a city roundabout with radius  35 km=h and the magnitude of the acceleration of the car is 4:5 m=s2 . If the car is increasing its is v speed, determine the time rate of change of the speed of the car at the instant shown.

D

Solution

Using normal-tangential components, the acceleration of the car can be expressed as 2


a

O

(1)

O

where the unit vectors u t and un are tangent and perpendicu lar to the path, respectiv ely. Therefore, the magnitude of the acceleration is

jEaj D vP 2 C v2 2: (2) Solving the above equation for vP and recalling that vP > 0 because the car is increasing its speed, we have

s s

P D jEaj2

v

  2



v  

2

:

(3)

2

jEj D 4:5 m=s , v D 35 km=h D 35.1000=3600/ m=s, and  D 30 m, we can evaluate Eq. (3)

Recalling that a to obtain

2

P D 3:213 m=s :

v


June 25, 2012


Solutions Manual

208

Problem 2.149

D

A car travels along a city roundabout with radius  100 ft. At the instant shown, the speed of the car is v 25 mph and the speed is decreasing at the rate 8 ft=s2 . Determine the magnitude of the acceleration of the car at the instant shown.

D

Solution

Using normal-tangential components, the acceleration of the car can be expressed as 2


a

O

(1)

O

where the unit vectors u t and un are tangent and perpendicu lar to the path, respectiv ely. Therefore, the magnitude of the acceleration is 2

2

(2) jEaj D vP 2 C  v  : Recalling that vP D 8 ft=s , v D 25 mph D 25.5280=3600/ ft=s, and  D 100 ft, we can evaluate Eq. (2) to 

2

s

obtain

2

jEaj D 15:64 ft=s :


June 25, 2012


Dynamics 2e

209

Problem 2.150 Making the same assumptions stated in Example 2.15, consider the map of the Formula 1 circuit at Hockenheim in Germany and estimate the radius of curvature of the curves Südkurve and Nordkurve (at the locations indicate d in gold).

Photo credit: Courtesy of FIA

Solution

As was done in Example 2.15 on p. 95 of the textbook, we assume that by lateral G -force the Federation Internationale de l’Automobil e (FiA) that compiled the map in the problem statement really meant to provide a measurement of the acceleration norma l to the path of the racing cars expressed in “units of g ,” where g is the acceleration due to gravity. With this in mind, at the Südkurve the car is traveling at a speed of 150 km=h with an acceleration of 3:5g . Therefore, denoting by  Südkurve the radius of curvatu re of the Südkurve, we must have

a 

n Südkurve

D v

2

Südkurve Südkurve

)

Südkurve

D

150

2 1000 3600 m =s



3:5.9:81 m=s2 /

)

Südkurve

D 50:56 m.

Similarly, at the Nordkurve the car is traveling at a speed of 200 km=h with an acceleration of 3:4g . Therefore, denoting by  Nordkurve the radius of curvature of the Nordkurve, we have

a 

n Nordkurve

D

2 vNordkurve Nordkurve

)

Nordkurve

1000 3600 m

s

m

s2

2

200 = D 3:4.9:81 = /

)

Nordkurve


D 92:54 m:

June 25, 2012



Dynamics 2e

211

Problem 2.152 An aerobatics plane initiates the basic loop maneuver such that, at the bottom of the loop, the plane is going 140 mph, while subjecting the plane to approximately 4g of acceleration. Estimate the corresponding radius of the loop.

Solution

The acceleration of the airplane, expressed in normal tangential components, is 2


a

(1)

O

O

where v is the speed of the airplane, u t is the unit vector tangent to the path, un is the unit vector normal to the path, and  is the radius of curvat ure of the path. The change in speed as the airplane init iates the loop maneuver is negligible, so that, right at the beginning of the maneuver, we can simplify Eq. (1) to 2

E D v uO n:

a Since a

(2)

4g , from Eq. (2) we have

jEj D Recalling that v

2

4g

D v )

2



v : D 4g

(3) 2

D 140 mph D 140.5280=3600/ ft=s and g D 32:2 ft=s , we can evaluate  to obtain  D 327:3 ft:


June 25, 2012


Solutions Manual

212

Problem 2.153

D

The portion of a race track between points A (corresponding to x 0 ) and B is part of a parabolic curve described by the equation y  x 2 , where  is a constant. Let g denote the acceleration due to gravity. 180 mph experiences at A an acceleration Determine  such that a car driving at constant speed v 0 with magnitude equal to 1:5g .

D

D

Solution

A the tangent and normal Referring to the figure at the right, at point directions to the trajectory coincide with the x and the y axes, respectively. The expression of the acceleration in normal-tangential components is 2


a

(1)

where v is the speed and  is the radius of curvature. Since the speed is constant and equal to that v 0 and Eq. (1) reduces to

PD

v 0 , we have

2

E D v0 uO n:

a

(2)

We can determine  using Eq. (2.59) on p. 93 of the textbook, namely,



  2 3=2 D 1 Cd.dy=dx/ ) 2 y=dx 2

ˇ

ˇ



  2 3=2 D 1 C .22x/ ;

where, recalling that y  x 2 , we have used the fact that dy=dx the last of Eqs. (3) into Eq. (2) gives

D

ED

2 v02

a

1

C .2 x/23=2

jEj D 1:5g for x D 0, from Eq. (4) we have 2 v02 D 1:5g D 32 g )

Recalling that a

Recalling that g to obtain

32:2 ft=s2 and v0

D

180 mph

D



(3)

D 2 x and d 2y=dx 2 D 2 . Substituting

O

un :



(4)

D 4v3g2 :

(5)

0

180.5280=3600/ ft=s, we can evaluate the last of Eqs. (5)

D

D 0:3465

 10

3

ft1 :


June 25, 2012


Dynamics 2e

213

Problem 2.154

D

The portion of a race track between points A (corresponding to x 0 ) and B is part of a parabolic curve described by the equation y  x 2 , where  is a constant. Let g denote the acceleration due to gravity. 0:4103 ft1 , determine d such that a car driving at constant speed v0 180 mph experiences If  at B an acceleration with magnitude equal to g .

D

D

D

Solution

The expression of the acceleration in normal-tangential components is 2


a

(1)

where v is the speed and  is the radius of curvatu re. Since the speed is 0 and Eq. (1) reduces to constant and equal to v 0 , v

PD

2

E D v0 uO n:

a

(2)




  2 3=2 D 1 Cd.dy=dx/ ) 2 y=dx 2

ˇ

ˇ



  2 3=2 D 1 C .22x/ ;

(3)

where, recalling that y  x 2 , we have used the fact that dy=dx 2  x and d 2 y=dx 2 2  . Substituting the last of Eqs. (3) into Eq. (2), setting the magnitude of the result equal to the specified value of g , and d , we have letting x

D

D

D

D

2 v02

1 C .2 d / 

 Dg ) 2 3=2

0:4  103 ft1 , v0 Recalling that  evaluate the last of Eqs. (4) to obtain

D

d

D

1 2

s

2 v02 g



2=3 

1:

(4) 2

D 180 mph D 180.5280=3600/ ft=s, and g D 32:2 ft=s , we can d

D 831:0 ft:


June 25, 2012


Solutions Manual

214

Problem 2.155

D D

The portion of a race track between points A (corresponding to x 0 ) and B is part of a parabolic curve described by the equation y  x 2 , where  is a constant. Let g denote the acceleration due to gravity. 180 mph. Let a min and a max denote Suppose a car travels from A to B with a constant speed v 0 the minimum and maximum values of the magnitude of the acceleration, respectively. Determine a min if d 1200 ft and a max 1:5g .

D

D

jEj

jEj

jEj D

jE j

Solution

The expression of the acceleration in normal-tangential components is 2


a

(1)

where v is the speed and  is the radius of curvatu re. Since the speed is constant and equal to v 0 , v 0 and Eq. (1) reduces to

PD

2

E D v0 uO n:

a

(2)




  2 3=2 D 1 Cd.dy=dx/ ) 2 y=dx 2

ˇ

ˇ



  2 3=2 D 1 C .22x/ ;

where, recalling that y  x 2 , we have used the fact that dy=dx Eq. (3) into Eq. (2), we have

D

ED

a

2 v02 2 3=2

 1 C .2 x/ 

O

D 2 x and d 2y=dx 2 D 2 . Substituting

) jEaj D 

un

(3)

2 v02

1

C .2 x/23=2

:

From the last of Eqs. (4) we see that the magnitude of the acceleration is maximum at for x and becomes smaller as x increases. This implies that, for 0  x  d ,

jEaj D min

jEj

Setting a

max

2 v02 1



C .2 d /2 3=2

and

(4)

D 0, i.e., at A

jEaj D 2 v02:

(5)

max



from the last of Eqs. (5) equal to the specified maximu m value of 1:5g , we have

2 v02

D 1:5g D 23 g )



D 4v3g2 :

(6)

0


June 25, 2012


Dynamics 2e

215

Substituting the last of Eqs. (6) into the first of Eqs. (5), we then have that

jEaj D min

3g

"

2 1 Recalling that g Eq. (7) to obtain

C

2 3=2

 # 3gd 2v02

;

(7)

2

D 32:2 ft=s , d D 1200 ft, and v 0 D 180 mph D 180.5280=3600/ ft=s, we can evaluate 2

jEaj D 21:95 ft=s : min


June 25, 2012





Dynamics 2e

219

Problem 2.159

D

750 mph while performing a constant A jet is flying at a constant speed v 0 speed circular turn. If the magnitude of the acceleration needs to remain constant and equal to 9g , where g is the acceleration due to gravity, determine the radius of curvature of the turn.

3

3

Solution

Expressed in normal tangential components, the acceleration of the airplane is 2


a

(1)

O

O

where v is the speed of the airplane, u t is the unit vector tangent to the path in the direction of motion, un is the unit vector normal to the path and pointing toward the concave side of the path, and  is the radius of curvature of the path. Since the speed is constant, v 0 and Eq. (1) simplifies to

PD

2

E D v uO n;

a

(2)

jEj D 9g , from the above equation we have that

Recalling that a

9g Recalling that v

D

v2 

)



D

v2 9g :

(3) 2

D v 0 D 750 mph D 750.5280=3600/ ft=s and g D 32:2 ft=s , we have  D 4175 ft:


June 25, 2012


220

Solutions Manual

Problem 2.160 Particles A and B are moving in the plane with the same constant speed v , and their paths are tangent at P . Do these particles hav e zero acceleration at P ? If not, do these particles have the same acceleration at P ?

Solution

Without of the e for the two curves is not possible to answer the to question. If each P , then the curve hadknowledge zero curvature at curvatur acceleration of theitparticles at P would be equal zero because the particles are moving with constant speed . If the two curves at P had the same nonzero curvature, then the acceleration of the two particles at P would be the same. If the curves had differ ent nonzero curv ature at P , then the accelerations of the two particles would be different.


June 25, 2012


Dynamics 2e

221

Problem 2.161 Uranium is used in light water reactors to produce a controlled nuclear reaction for the generation of power. When first mined, uranium comes out as the oxide U 3 O8 , 0.7% of which is the isotope U-235 and 99.3% the isotope U-238. For it to be used in a nuclear reactor, the concentration of U-235 must be in the 3–5% range. The process of increasin g the percentage of U-235 is called enrichment, and it is done in a number of ways. One method uses centrifuges, which spin at very high rates to create artificial gravity. In these centrifuges, the heavy U-238 atoms concentrate on the outside of the cylinder (where the acceleration is largest), and the lighter U-235 atoms concentrate near the spin axis. Before centrifuging, the uranium is processed into gaseous uranium hexafluoride or UF 6 , which is then injected into the centrifuge. Assuming that the radius of the centrifuge is 20 cm and that it spins at 70;000 rpm, determine (a) The velocity of the outer surfa ce of the centrifu ge. (b) The acceleration in g experienced by an atom of uranium that is on the inside of the outer wall of the centrifuge.

Photo credit: Courtesy of the Department of Energy

Solution

Part (a). Since the speed v and the angular speed ! are related as v on the outer surface of the centrifuge is

v where we have used the fact that !

D ! )

v

D !, we have that the speed of points

D 1466 m=s,

(1)

D 70;000 rpm D 70;000.2=60/ rad=s and  D 20 cm D 0:2000 m.

Part (b). Under the assumpt ion that the centrifug e is spinning with a constant angula r speed, the only component of acceleration of a point on the wall of the centrifuge will be the normal component. Hence, the acceleration experienced by an atom at the inside outer wall of the centrifuge is 2

an

2

2

D v D .!g/ g D !g  g )

E D 1:096

a

 10

6

O

g un ,

O

where the unit vector un always points from a point on the periphery of the centrifuge toward the center of the 70;000 rpm centrifuge, and where we have used the expression of v in Eq. (1) along with the fact that ! 70;000.2=60/ rad=s,  20 cm 0:2000 m, and g 9:81 m=s2 .

D

D

D


D

D

June 25, 2012


Solutions Manual

222

Problem 2.162 Treating the center of the Earth as a fixed point, determine the magnitude of the acceleration of points on the surface of the earth as a function of the angle  6371 km as the radius of the Earth. shown. Use R

D

Solution

Let ! denote the angular speed of the Earth. Since the Earth undergoe s one full revolution per day, the angular speed ! is given by

!

rev  rad  D 11 day D .24 h2/.3600 D 43;200 rad =s: s=h/

(1)

Let  denote the distance between the point indicated on the figure and the axis of rotation of the Earth, i.e., a point on the surface of the Earth characterized by the angle  between the equator and the axis of rotation of the Earth. Then, we have we  R cos  : (2)

D

Next we observe that under the assumption that the angular speed of the Earth is constant, the only component of the accelerati on of the point in question is the normal component. Hence, we must have

a

Recalling that R obtain

a

v2

jE j D D  D n

 !2

D

R! 2 cos  :

(3)

D 6371 km D 6;371;000 m, using the value of ! in Eq. (1), Eq. (3) can be evaluated to

ˇ Eˇ D a

.33:69  103 cos  / m=s2 :


June 25, 2012


Dynamics 2e

223

Problem 2.163

D

An airplane is flying straight and level at a speed v 0 150 mph and with a constant time rate of increase 20 ft=s2 , when it starts to climb along a circular path with a radius of curvature  2000 ft. of speed v The airplane maintains v constant for about 30 s. Determine the acceleration of the airplane right at the start of the climb and express the result in the Cartesian component system shown.

PD

D

P

Solution

Using normal-tangential components, the acceleration can be expressed as 2


a

O

(1)

O

where the unit vectors u t and un are parallel and normal to the path, respectively. At the bottom of the loop, we have ut { and un | : (2)

O DO

O DO

Substituting Eqs. (2) into Eq. (1), and recalling that at the start of the loop

v

D v0, we have

2

E D vP {O C v0 |O:

a

(3)

2

P D 20 ft=s , v0 D 150 mph D 150.5280=3600/ ft=s, and  D 2000 ft, we can evaluate

Recalling that v Eq. (3) to obtain

2

E D .20:00 {O C 24:20 |O/ ft=s :

a


June 25, 2012



Dynamics 2e

225

Substituting the last of Eqs. (7) into Eq. (5) we have



sin

20 ft=s2 , v0 Recalling that v Eq. (8) for t 25 s to obtain

D

PD

2

 1  .v C vPt /  1 v t C vPt  {O v t C vP t    1  |O: v t C vP t C vP 1 v t C vPt  C .v C vP t /

E D vP cos

a

1 2

0



2

1 2

0

0



sin



1 2

0

2

0

2



cos





0

2

1 2

2

(8)

D 150 mph D 150.5280=3600/ ft=s, and  D 2000 ft, we can evaluate a

ED

.121:2 {

OC

230:0 | / ft=s2 :

O


June 25, 2012


Solutions Manual

226

Problem 2.165

D

An airplane is flying straight and level at a speed v 0 150 mph and with a constant time rate of increase 20 ft=s2 , when it starts to climb along a circular path with a radius of curvature  2000 ft. of speed v The airplane maintains v constant for about 30 s. Determine the acceleration of the airplane after it has traveled 150 ft along the path and express the result in the Cartesian component system shown.

PD

D

P

Solution

Using normal-tangential components, the acceleration can be expressed as 2

(1) E D vP uO t C v uO n; where the unit vectors uO t and uO n are parallel and normal to the path, respectively. Letting s D 0 correspond to v D v 0 , recalling that vP is constant, and using the

a

chain rule, we can write

P D v dv ) ds

v

v dv

Z

D vP ds )

v

v dv v0

D

Z

s

P

)

v ds 0

v2

D v02 C 2vPs:

(2)


E D vP uO t C v0 C 2vPs uO n;

a

(3)

Referring to the figure at the right of Eq. (1), we now observe that we can provide the solution to the problem by evaluating Eq. (3) for s 150 ft as long as we are able to express u t and un for s 150 ft in terms of { and | . To do so, we observe that

D

O

O

O D cos  {O C sin  |O

ut

and

O D

un

 sin 

O

D

O

O C cos  |O :

{

(4)

Substituting Eqs. (4) into Eq. (3) we can then express the acceleration of the airplane as follows:



E D vP cos 

a



v02

C 2vPs sin   {O C vP sin  C v02 C 2vPs cos   |O: 

(5)



Equation (5) implies that we will be able to provide the answer to the problem once we express  as a function of the path coordinate s . To this end, we observe that

s

s

D  )



D :

(6)


 s

E D vP cos

a





v02

C 2vPs sin s  {O C vP sin s  C v02 C 2vPs cos s  |O: 








(7)



June 25, 2012


Dynamics 2e

227

20 ft=s2 , v0 Recalling that v Eq. (7) for s 150 ft to obtain

D

PD

D 150 mph D 150.5280=3600/ ft=s, and  D 2000 ft, we can evaluate 2

E D .17:91 {O C 28:62 |O/ ft=s :

a


June 25, 2012


Solutions Manual

228

Problem 2.166 Suppose that a highway exit ramp is designed to be a circular 130 ft. A car begins to exit the highway at segment of radius  A while traveling at a speed of 65 mph and goes by point B with a speed of 25 mph. Compute the acceleration vector of the car as a function of the arc length s , assuming that the tangential component of the acceleration is constant between points A and B .

D

Solution

Using normal tangential components, we have that the acceleration is given by

E D a t uO t C an uO n;

a

O

(1)

O

where u t and un are the unit vectors tangent and normal to the path, respectively, and where, letting v denote the speed of the car, 2

at

D vP

and

an

D v ;

(2)

where  is the radius of curv ature of the path. The first of Eqs. (2) impl ies that the chan ge in speed is completely due to the tangential acceleration. Since this component of acceleration is assumed to be constant, the speeds at A and B are related via the following constant acceleration equation: 2 vB

2  vA

D 2at .sB

 sA /

)

at

D 2.svB2B

2  vA  sA /

:

(3)

The expression for the speed in terms of arc length s can be found again using constant acceleration equations, i.e.,

v 2 .s/

D vA2 C 2a t .s

 sA /

v 2 .s/

)

2  v2 A

D vA2 C vsBB

 sA

.s  sA /:

(4)

Combining Eqs. (1)–(4), we can then write 2  v2 A

E D 2.svBB

a

D

D



O C 1 sA / ut



vA2

2  v2 A

C vsBB

 sA

D

0, vA 65 mph 65.5280=3600/ ft=s, v B Recalling that sA  2 .65  / ft, we can express Eq. (5) as follows:

130 ft, and sB

D

D

ED

a



ft s2

ft s2



O

.s  sA / un :

(5)

D 25 mph D 25.5280=3600/ ft=s,  D s2

18:96 =  uO C 69:91 =  0:2917 s uO : t



n


June 25, 2012


Dynamics 2e

229

Problem 2.167 Suppose that a highway exit ramp is designed to be a circular 130 ft. A car begins to exit the highway at segment of radius  A while traveling at a speed of 65 mph and goes by point B with a speed of 25 mph. Compute the acceleration vector of the car as a function of the arc length s , assuming that between A and B the speed was controll ed so as to maintain constant the rate dv=ds .

D

Solution

Using normal-tangential components, the acceleration has the form 2


a

(1)

where v is the speed of the car and  is the radius of curvature of the path. To solve the problem, we need to express both v and v as a function of s . We are told that the quantity dv=ds is constant. We begin by determining this constant, which we denote by K . Since K dv=ds we can separate the variables s and v as follows:

P

dv

Z

D K ds )

D

vB

dv vA

D

Z

sB

)

K ds sA

vB

 vA

D K.sB

 sA /

)

K

D vsBB

 vA  sA

: (2)

Now that K is known, we repeat the integration process using generic upper limits of integration, that is,

dv

D K ds )

Z

v

D

dv vA

Z

s

K ds sA

)

v  vA

D K.s

 sA /:

(3)

Using the expression for K in the last of Eqs. (2), we have

v

D vA C vsBB

 vA  sA

.s  sA /:

(4)

P

To determine v as a function of s , we use the chain rule to write

ds : P D dv ) vP D dv ) vP D v dv dt ds dt ds We know observe that v is given in Eq. (4) and dv=ds D K is given in the last of Eqs. (2). v

(5)

Using these

consideration, we can go back to Eq. (1) and rewrite it as follows:





vA

C vsBB

2

 vA 

.s  sA /

v

B  vA

2

O C 1 vA C vsBB vsAA .s sA/ uO n: (6) sA sB sA Recalling that vA D 65 mph D 65.5280=3600/ ft=s, vA D 25 mph D 25.5280=3600/ ft=s,  D 130 ft, sA D 0 , sB D  =2 D 65  ft, we can write Eq. (6) as follows:  27:39 ft=s  C 0:08254 s s uO t aE D C 69:91 ft=s  0:4214 s s C 0:0006349 ft s s2 uO n: ED

a



ut



 





2

2



2


1



2

June 25, 2012


Solutions Manual

230

Problem 2.168

D

A water jet is ejected from the nozzle of a fountain with a speed v 0 12 m=s. 33ı , determine the rate of change of the speed of the water Letting ˇ particles as soon as these are ejected as well as the corresponding radius of curvature of the water path.

D

Solution

Using the normal-tangent ial component system shown in the figure at the right, we can express the acceleration of a water particle as it is emitted from the nozzle as follows: 2


a

(1)

D

O

O

where v v 0 is the speed of the particle as it leaves the nozzle, u t is the unit vector tangent to the path, un is the unit vector normal to the path, and  is the radius of curvature of the path. We assume that, as soon as a water particle leaves the nozzle, its motion is that of a projectile subjected only to the acceleration due to gravity. Therefore, the acceleration in question must be equal to

E D g.

a

O C cos ˇ uO n/:

 sin ˇ u t

(2)

Comparing Eqs. (1) and (2) we conclude that 2

PD

v

g sin ˇ

and

D



where we have used the fact that v v 0 . Recalling that v 0 evaluate the results in Eqs. (3) to obtain

PD

v

2

5:343 m=s

and

v0 ; D g cos ˇ

(3)

D 12 m=s, g D 9:81 m=s

2



and ˇ

D 33ı, we can

D 17:50 m:


June 25, 2012


Dynamics 2e

231

Problem 2.169 A water jet is ejected from the nozzle of a fountain with a speed v 0 . Letting ˇ the radius of curvature at the highest point on the water arch is 10 ft.

D 21ı, determine v0 so that

Solution

The water particles are in projectile motion after they are emitted from the nozzle. Therefore, their acceleration is vertically downward and equal to the acceleration due to gravity. At the highest point on the water arch the vertical component of the velocity of the water particles is equal to zero. As is the case in projectile motion, the horizontal component is constant and therefore equal to the value at the beginning of the motion, namely, v 0 cos ˇ . Because this component of velocity is positive and it is the only nonzero component at the instant considered, v 0 cos ˇ is also the value of the speed:

v

D v0 cos ˇ:

Recalling that speed and radius of curvature are related through the relation a n v 2 = and observing that at the highest point on the water arch the normal direction coincides with the direction of gravity, we have that an g , which implies

D

D

2

2

g

D v )

where we have used ˇ

g

D v0 cos

2ˇ

)

v0

D

r

g cos2 ˇ

)

v0

D 19:22 ft=s,

D 21ı, g D 32:2 ft=s , and  D 10 ft. 2


June 25, 2012


Solutions Manual

232

Problem 2.170

D

65 mph almost loses contact with the ground when it reaches the top of A car traveling with a speed v 0 the hill. Determine the radius of curvature of the hill at its top.

Solution

Using normal-tangential components, the acceleration of the car is 2

E D vP t C v uO n;

a

(1)

where v is the speed of the car,  is the radius of curvature of the path, and u t and un are the unit vectors tangent and normal to the path, respectively. The speed of the car is constant and equal to v 0 . Therefore, Eq. (1) simplifies to

O

O

2

E D v0 uO n:

a

(2)

If the car were to lose contact with the ground, the car would be in projectile motion and its acceleration would be equal to that due to gravity. At the top of the hill the tangent to the path is horizontal and the normal direction coincides with that of gravity. Therefore, in view of Eq. (2), at the top of the hill we would have

a

E D g uO n )

Recalling that v 0

g

D

v2 0 

)



D

v2 0 g:

(3)

2

D 65 mph D 65.5280=3600/ ft=s and g D 32:2 ft=s , we can evaluate  to obtain  D 282:2 ft:


June 25, 2012


Dynamics 2e

233

Problem 2.171

O A car is traveling at a constant speed over a hill. If, using a Cartesian coordinate system with srcin  at the top of the hill, the hill’s profile is described by the function y .0:003 m 1 /x 2 , where x and y are in meters, determine the minimum speed at which the car would lose contact with the ground at the top of the hill. Express the answer in km =h.

D

Solution


E D vP t C v uO n;

a

(1)

where v is the speed of the car,  is the radius of curvature of the path, and u t and un are the unit vectors tangent and normal to the path, respectively. For a car moving at constant speed, Eq. (1) simplifies to

O

O

2

E D v uO n:

a

(2)

We will denote by v min the minimum speed at which the car loses contact with the ground. If v were to exceed v min , the car would be in projectile motion and its acceleration would be equal to that due to gravity. At the top of the hill the tangent to the path is horizontal and the normal direction coincides with that of v min , at the top of the hill, i.e., for x 0 , we would have gravity. Therefore, in view of Eq. (2), for v

D

E D g uO n )

a

Recalling that the profile of the hill is y

g

D

D

v2

min

D  jx D 0 )

vmin

 .0:003 m 1 /x 2 , and

p D

g

jx D 0 :

(3)

recalling that

2 3=2

 we have

D

1 C .dy=dx / 

ˇ

d 2 y=dx 2

ˇ

  2 3=2 m x 2 D 1 C 0:006000 ) 0:006000 m 1

;

(4)

1 : D 0:006000 m Substituting the last of Eqs. (5) into the last of Eqs. (3) and recalling that g D 9:81 m=s , we have .x/

1

.0/

1

(5)

2

vmin

145:6 km=h;

D

where, as requested in the problem statemen t we have expressed the final answer in km =h.


June 25, 2012



Dynamics 2e

235

Problem 2.173

D

130 mph when it performs a turn A race boat is traveling at a constant speed v 0 with constant radius  to change its course by 90 ı as shown. The turn is performed while losing speed uniformly in time so that the boat’s speed at the end of the turn is vf 116 mph. If the magnitude of the acceleration is not allowed to exceed 2g , where g is the acceleration due to gravity, determine the tightest radius of curvatur e possible and the time needed to complete the turn.

D

Solution

Using normal tangentia l components, the accelerat ion of the boat is given by

E D a t uO t C an uO n;

a

O

(1)

O

where u t and un are unit vectors tangent and normal to the path. Denoting by  min the minimum value of the radius of curvature, a n is given by

D v

an

2

;

(2)

min

where v is the speed of the boat. The tangential acceleration a t is constant. Therefore, applying constant acceleration equations, we have

vf2



v02

D 2 at .sf

vf2

)

 s0 /

v02



D 2at





 min ; 2

(3)

D

. =2/min is the distance covered by the boat along its path while performing the turn. where sf  s0 Solving the above equation for a t , we obtain at

D

vf2

v02



min

:

(4)

jEj D

The magnitude of the acceleration must not exceed the value 2g . Recalling that a 2 write a n

C a2t D 4g 2, which, using Eqs. (2) and (4), gives v04

2  v2 2 0 f  2 2

v



r

q

a2t

C an2, we can

D 2g1 v04 C 12 vf2 v022: (5) 2 Recalling that v 0 D 130 mph D 130.5280=3600/ ft=s, vf D 116 mph D 116.5280=3600/ ft=s, and g D min

C

min

D 4g 2 )

min



32:2 ft=s2 , we can evaluate the last of Eqs. (5) to obtain min

D 565:7 ft:

We denote the time needed to perform the turn by tf  t0 . Because a t is constant, we must have vf v0 a t .tf  t0 /, which, using Eq. (4) and the last of Eqs. (5), after simplifi cation, gives

C

tf

 t0

D 2g.vfC v0/

where we have used fact that g 116.5280=3600/ ft=s.

r

v04

C 12 vf2



v02

2



)

tf

D

D 4:926 s,

2

D 32:2 ft=s , v0 D 130 mph D 130.5280=3600/ ft=s, and vf D 116 mph D


June 25, 2012


Solutions Manual

236

Problem 2.174 A truck enters an exit ramp with an initial speed v 0 . The ramp is a circul ar arc with radius  . Derive an expression for the magnitude of the acceleration of the truck as a function of the path coordinate s (and the parameters v 0 and  ) if the truck stops at B and travels from A to B with a constant rate of change of the speed with respect to s .

Solution

Using normal-tangential components, the acceleration of the truck is 2


a

(1)

O

O

where v is the speed of the truck,  is the radius of curvature of the path, and where u t and un are the unit vectors tangent and normal to the path, respectively. The problem statement states that the quantity dv=ds is constant. Letting  denote the constant in question, we can then write

D dv=ds )



 ds

D d v:

(2)

D

D

0 and sB Keeping in mind that the truck moves along a circular arc of radius  , let sA  =2 be the v0 and vB 0 be the values of values of the path coordinate s at A and B , respectively. Also, let vA speed for s sA and s sB , respectively. Then, we can integrate the last of Eqs. (2) as follows:

D

ZD

D

 =2

 ds

0

D

Z

0

1 2 

)

dv v0

D

v 0

)



D



D

2v0 : 

(3)

Now let v be the value of speed corresponding to the generic value s of the path coordinate. Then, substituting the last of Eqs. (3) into the last of Eqs. (2), we can integrate again as follows:

Z

s 

0

2v0 ds 

D

Z

v

dv v0

)



2v0 s 

Dv



v0

)

v

D v0



2v0 s: 

(4)

Now we recall that, using the chain rule, we can write

ds P D dv D dv ) vP D dv v ) vP D dt ds dt ds

v



2v0 2v0 v0  s ;  





(5)

D

where we have used the expression for v in the last of Eqs. (4), as well as the expression for  dv=ds in v 2 .v 2 =/2 . Therefore, using the last of the last of Eqs. (3). Referring to Eq. (1), we have that a Eqs. (4) and the last of Eqs.(5), after simplificati on, we have

p

jEj D P C

jEaj D 1



v0 

2v0 s 

s 4v  2 0 2 

C

v0 

2v0 s 



2

:


June 25, 2012



Solutions Manual

238 Carrying out the above integration, we have

8g ln 1 at

C avt 0tf D 2 :

D .vf

v0 /=tf , which, upon substituting into Eq. (7), gives



Because a t is constant, we must have a t

8gtf ln 1 vf  v0



Recalling that g

D

C vf v0

9:81 m=s2 , vf



D





 v 0

D 2 )

800 km=h

D

tf

(7)

 .vf v0 / : D 16g ln.vf =v 0 / 

800.1000=3600/ m=s, and v0

(8)

D

1100 km=h

D

1100.1000=3600/ m=s, from the last of Eqs. (8) we have tf

D 5:238 s:


June 25, 2012


Dynamics 2e

239

Problem 2.176

D D

70 mph. Using the Cartesian coordinate system A car is traveling over a hill with a constant speed v 0  shown, the hill’s profile is given by the function y .0:0005 ft 1 /x 2 , where x and y are measured in feet. At x 300 ft, the driver applies the brakes, causing a constant time rate of change of speed v 3 ft=s2 until the car arrives at O . Determine the distance traveled while applying the brakes along with the time to cover this distance. Hint: To compute the distance traveled by the car along the car’s path, observe that d s dx 2 dy 2 1 .dy=dx /2 dx , and that

D

PD

p C Dp C ZpC D pC

D

C 2 x 2 dx

1

x 2

1

C 2x 2

C 2C1 ln

2 2

C x C p1 C C x :

Solution



a

(1)

where v is the speed of the car,  is the radius of curvature of the path, and where u t and un are the unit vectors tangent and normal to the path, respectiv ely. Let t 0

0 the time at

which v v 0 and tf the time at which the car arrives at O . Using the path coordinate s (see figure at the sf  s0 the distance traveled between t 0 and tf . Since a t v is constant, we can right), we denote by d apply constant acceleration equations to write

OD O

DP D

D d

D v0tf C 12 at tf2 )

D

tf

q



1 v0 at



˙

q

v02

C 2a t d

:

(2)

3 ft=s2 < 0, we see that v 0 > v02 2a t d . This implies that there are two positive Recalling that a t real roots for tf . The physically acceptable root is the smaller of the two, which correspo nds to when the car arrives at O for the first time after applying the breaks. To interpret the second root we need to keep in mind that the car is still traveling to the right when it arrives at O for the first time. Hence, if a t < 0 is maintained after the car arrives at O , the car would keep slowing down while travel ing to the right until stopping and then to is would travel back and reach O a second time. For the purpose of the present calculati on, we are only interested in the first time that the car arrives at O , which is

D

tf

D



1 v0  at



C

v02

C 2a t d

:

(3)

q

All the quantities in Eq. (3) are known except for d , which is one of the unknowns of the problem. Hence, we now determine d and then we will use Eq. (3) to determine tf . One way to obtain d is to realize that

d

D

Z

sf

ds :

(4)

s0


June 25, 2012


Solutions Manual

240 From the problem figure, we have that

q

D

ds

dx 2

D

C dy 2:

(5)

D

y.x/ , we have dy .dy=dx/ dx , so that Eq. (5) Since the trajectory of the car is given in the form y implies that d s 1 .dy=dx/ 2 dx , and Eq. (4) can therefore be rewritte n as

D

D

where x0

dy=dx

D

Z qC xf

d

D

1

ZD p C 1

D

D

C 2 x 2 dx

x0

(6)

D

)

d

D )

x p 2

d

D



( s

1 v0 at

D

C

v02  2a t

D

1

C C 2x2 C 2C1 ln C x C

D

x0  2



qC 1

C 2 x02 

x q 0

2

1

C C 2x02 C 2C1 ln

1

xf

C 2x 2

1 ln C x0 2C





x0

q C C 1



C 2 x02 ; (7)

D

C x C q1 C C x ) :

 300 ft, v 0 70 mph 70.5280=3600/ ft=s, C Recalling that x 0 we can evaluate the last of Eqs. (7) and Eq. (8) to obtain

d

pC

D 0. Substituting the last of Eqs. (7) into Eq. (3), we have

where we have accounted for the fact that xf

tf

.dy=dx/ 2 dx ;

x0

0. Now we rec all that y .0:0005 ft1 /x 2 so that we can write and xf where C 0:001000 ft1 . Hence, we can write Eq. (6) as follows:

300 ft

C x ,

xf

d

pC

D 304:4 ft

and

tf

0

2 2 0

(8)

D 0:001000 ft , and a t D 1

 3 ft=s2 ,

D 3:106 s:


June 25, 2012


Dynamics 2e

241

Problem 2.177 Recalling that a circle of radius R and center at the srcin O of a Cartesian coordinate system with axes x and y can be expressed by the formula x 2 y 2 R 2 , use Eq. (2.59) to verify that the radius of curvature of this circle is equal to R.

C D

Solution

D

Equation (2.59) on p. 93 of the textbook tells us that, given a curve of the form y .x /, where x and y are Cartesian coordinates, then the radius of curvature of the curve as a function of x is 2 3=2



ç D Œ1 Cd.dy=dx/ 2 y=dx 2

ˇ

:

ˇ

(1)

To solve the problem we need to determine the quantities dy=dx and d 2 y=dx 2 for a circle. To do so, we start from the given equatio n for a circle with center at the srcin and radius R , i.e., x 2 y 2 R 2 . Differentiating this expression with respect to x , we have

2x

dy C 2y dx D0 )

C D

dy dx

x : y

D



D

x

(2)

Taking the derivative of Eq. (2) with respect to x , we have 2

d y dx 2

D

1

y

dy C yx2 dx )

2

d y dx 2

2

Cy

y3

2

;

(3)

where we have used the last of Eqs. (2) to obtain the last of Eqs. (3). Substituting the last of Eqs. (2) and the last of Eqs. (3) into Eq. (1), we have

ˇ ˇˇ

 2 23=2  2 23=2 D 1 Cx2x C=yy 2 y 3 )  D x x2CCy y 2 Recalling that x 2 C y 2 D R 2 , the last of Eqs. (4) gives 

ˇ

ˇˇˇ

ˇ

y3 y3

ˇˇ

)



D

x

2

y2

3=2

ˇ CC  ˇ x2

y2

:

(4)

3



D RR2 )



D R.


(5)

June 25, 2012


Solutions Manual

242

Problem 2.178 A particle P is moving along a path with the velocity sho wn. Discuss in detail whether or not there are incorrect elements in the sketch of the polar component system at P .

Solution

O

O

O

The unit vectors ur and u are both incorrect. The unit vector ur must be oriented along the radial line r and point away from the srcin. The unit vector u must be oriented perpendicular to ur and pointing in the direction of increasing  .

O


O

June 25, 2012


Dynamics 2e

243


Solution

O

Since u is perpendicula r to the radial line connecting P to O and it is directed in the direction of increasing  , the unit vector u is oriented correctly. The unit vector ur is parallel to the radial line connect ing P to O , but it is pointing toward O and this is incorrect.

O

O


June 25, 2012


Solutions Manual

244


Solution

O

O

The unit vector ur is oriented correctly, but u must be oriented opposite to the direction shown.


June 25, 2012


Dynamics 2e

245

Problem 2.181 A particle P is moving along a circle with cente r C and radius R in the direction shown. Letting O be the srcin of a polar coordinate system with the coordinates r and  shown, discuss in detail whether or not there are incorrect elements in the sketch of the polar component system at P .

Solution

O

O

Both ur and u are oriented as they should be.


June 25, 2012


Solutions Manual

246

Problem 2.182 A radar station is tracking a plane flying at a constant altitude with 550 mph. If at a given instant r 7 mi and a constant speed v 0 32 ı , determine the corresponding values of r ,  , r , and  . 

D

D

D PPR R

Solution

O

Referring to the figure at the right, the unit vector { described the direction of the airplane. Hence, we can give the following form to the velocity v of the airplane:

E

E D v 0 {O:

v

(1)

O

O

Using the polar component system with unit vectors ur and u , the velocity of the airplane can also be written as (2) E D rP uO r C r P uO : In order to compare the expressions for vE in Eqs. (1) and (2), we Or observe that the unit vector {O can be written in terms of uO r and uO as {O D cos  u

v







O

sin  u , so that Eq. (1)

can be rewritten as

v

v 0 cos  ur



v0 sin  u :

(3)

ED

O O P rP D v 0 cos  and r  D v0 sin  uO ) rP D v 0 cos  and P D .v0=r / sin  : (4) Recalling that v 0 D 550 mph D 550.5280=3600/ ft=s,  D 32 ı , and r D 7 mi D 36;960 ft, we can evaluate Equating Eq. (2) and Eq. (3) component by component, we have 





the last two of Eqs. (4) to obtain

P D 684:1 ft=s

r

and

PD



0:01157 rad=s:

EDE

E

0 because the airplane is flying at constant Denoting the acceleration of the airplane by a, we have a speed on a straight path. Therefore, recalling the expression of the acceleration in polar coordinates, we have

P O C .r R C 2rP P / uO ) rR r P 2 D 0 and r R C 2rP P D 0: (5) R P Solving the last two of Eqs. (5) for rR and  and using the expressions for rP and  in the last two of Eqs. (4) we have rR D .v 02 =r / sin2  and R D .v 0 =r /2 sin.2 /; (6) where we have used the following trigonometricıidentity: 2 sin  cos  D sin.2 /. Recalling again that v0 D 550 mph D 550.5280=3600/ ft=s,  D 32 , and r D 7 mi D 36;960 ft, we can evaluate Eqs. (6) to R

.r



r  2 / ur





obtain

2

R D 4:944 ft=s

r

and

R D 0:0004281 rad=s :



2


June 25, 2012




Dynamics 2e

249

Problem 2.185 At a given instant, the merry-go-round is rotating with an angular 18 rpm, and it is slowing down at a rate of 0:4 rad=s2 . velocity ! When the child is 2 :5 ft away from the spin axis, determine the time rate of change of the child’s distance from the spin axis so that the child experiences no transverse acceleration while moving along a radial line.

D

Solution

Using polar coordinates, the transverse component of acceleration is

a

D r R C 2rP P :

(1)

P

Setting a  equal to zero and solving for r , we have

R P D r2P :

r

P D D 18 rpm D 18.2=60/ rad=s, and R D ˛ D rP D 0:2653 ft=s:

! Recalling that  2:5 ft to obtain

r

D

(2)



0:4 rad=s2 ,


P

we can evaluate r for

June 25, 2012


Solutions Manual

250

Problem 2.186 At a given instant, the merry-go-round is rotating with an angular 18 rpm. When the child is 0:45 m away from the velocity ! spin axis, determine the second derivative with respect to time of the child’s distance from the spin axis so that the child experiences no radial acceleration.

D

Solution

Using polar coordinates, the transverse component of acceleration is

a

D rR



P

r  2:

(1)

R

Setting a r equal to zero and solving for r , we have

R D r P 2:

r

(2)

P D ! D 18 rpm D 18.2=60/ rad=s, we can evaluate rR for r D 0:45 m to obtain

Recalling that 

2

R D 1:599 m=s :

r


June 25, 2012



Solutions Manual

252

Problem 2.188 The polar coordinates of a particle are the following functi ons of time:

r

D r0 sin.t 3= 3/

and



D  0 cos.t = /;

D

1 s, and where t is time in seconds. where r 0 and  0 are constants,  Determine r 0 and  0 such that the velocity of the particle is completely in the radial direction for t 15 s and the corresponding speed is equal to 6 m=s.

D

Solution

In polar coordinates, the velocity vector is given by

E D rP uO r C r P uO :

v

(1)



Using the expressions for r and  given in the problem statement, we can rewrite Eq. (1) as follows: 2

3

3

E D 3r03t cos t 3 uO r r00 sin t 3 sin t uO : Recalling that  D 1 s, for t D 15 s, Eq. (2) reduces to vE D .403:8 s /r0 uO r .0:5211 s /r0 0 uO : v

 

  



1

1



(2)



(3)



From Eq. (3) we conclude that in order for the velocity to be completely in the radial direction, we must have

0

D 0:

(4)

Substituting the result in Eq. (4) into Eq. (2) we have

E D 3r03t

v

2

3

t 

cos

3

Enforcing the condition that v Eqs. (5) to obtain

O )

ur

v

D

ˇˇ

 ˇˇ

3r0 t 2 t3 cos 3 3 

ˇ

3

)

r0

D 3t 2 cos .tv3= 3/ :

ˇ

ˇ

(5)

D 6 m=s for t D 15 s, and recalling that  D 1 s, we can evaluate the last of r0

D 0:01486 m:


June 25, 2012




Dynamics 2e

255

Problem 2.191 During a given time interval, a radar station tracking an airplane records the readings

P D Œ449:8 cos  .t / C 11:78 sin  .t/ç mph; P D Œ11:78 cos  .t / 449:8 sin  .t/ç mph;

r.t/

r.t/  .t/



where t denotes time. Determine the speed of the plane. Furthermore, determine whether the plane being tracked is ascending or descending and the altitude) corresponding climbing plane’s expressed in ft=s.rate (i.e., the rate of change of the Solution

D P

DP

r and v r, In polar coordinates, the radial and transverse components of the velocity are vr respectively. We observe that the problem statement provides v r and v  as a function of time. With this in mind, we recall that the speed is given by v 2

v r2 2

v2

D

q

vr2

C v 2. Hence, using the information provide d, we have 

D C D rP C .r P /2  2 cos2  C .11:78/2 sin2  C 2.449:8/.11:78/ sin  cos  mph D .449:8/  C .11:78/2 cos2  C .449:8/2 sin2  2.449:8/.11:78/ sin  cos   mph: (1) Using the trigonometric identity sin 2 ˇ C cos2 ˇ D 1 , we can simplify the above expr ession to   v 2 D .449:8/2 C .11:78/2 mph ) v D 450:0 mph. (2) Letting y denote the elevation of the airplane, we have that y D r sin  . Taking the derivative of y with v



respect to time we have

P D rP sin  C r P cos  D Œ.449:8/ sin  cos  C .11:78/ sin2  C .11:78/ cos2  D 11:78 mph: Recalling that 1 mph D .5280=3600/ ft=s, we have that y



.449:8/ sin  cos  ç mph

The airplane is ascending at a rate of 17:28 ft=s.


June 25, 2012


Solutions Manual

256

Problem 2.192 The polar coordinates of a particle are the following functions of time: 2

D r0 1 C t and  D 0 t 2 ; where r 0 D 3 ft,  0 D 1:2 rad,  D 20 s, and t is time in seconds. Determine the velocity and the acceleration of the particle for t D 35 s and express the result using the polar component system formed by the unit vectors u O r and uO at t D 35 s. r







Solution

In polar coordinates, the velocity is given by


v

(1)



Using the expressions for r and  given in the problem statement, we can rewrite Eq. (1) as follows:

E D r0 uO r C 2r0t 0.t3 C  / uO :

v

(2)



In polar coordinates, the acceleration is given by

a

r E D R



P

 O C r R C 2rP P uO :

r  2 ur





(3)



Again using the expressions for r and  given in the problem statement, we can rewrite Eq. (1) as follows:

4r0 t 2 02 .t 5

C  / uO r C 2r030 .3t C  / uO : (4) Recalling that r 0 D 3 ft,  0 D 1:2 rad, and  D 20 s, for t D 35 s we can evaluate the expressions in Eqs. (2) ED

a





and (4) to obtain

E D .0:1500 uO r C 1:732 uO / ft=s

v



and

E D.

a

2

O C 0:1125 uO / ft=s :

0:3638 ur




June 25, 2012


Dynamics 2e

257

Problem 2.193 The polar coordinates of a particle are the following functions of time: 2

D r0 1 C t and  D 0 t 2 ; where r 0 D 3 ft,  0 D 1:2 rad,  D 20 s, and t is time in seconds. Determine the velocity and the acceleration of the particle for t D 35 s and express the result using the Cartesian component system formed by the unit vectors {O and |O .



r



Solution

In polar coordinates, the velocity is given by


v

(1)



Using the expressions for r and  given in the problem statement, we can rewrite Eq. (1) as follows:

E D r0 uO r C 2r0t 0.t3 C  / uO :

v

(2)



In polar coordinates, the acceleration is given by

E D rR

a



P  O C r R C 2rP P  uO :

r  2 ur

(3)



Again using the expressions for r and  given in the problem statement, we can rewrite Eq. (1) as follows: 2 2

a

(4) C  / uO r C 2r030 .3t C  / uO : We now observe that the unit vectors u O r and uO can be expressed in terms of the unit vectors {O and |O as follows: uO r D cos  {O C sin  |O and uO D sin  {O C cos  |O : (5)

ED

 4r0 t 0 5

.t









Substituting Eqs. (5) in Eq. (2), collecting terms, and using the given expression for  as a function of time, the velocity take s on the form

E D r03

v



2 0 2

t 

 2 cos

C / sin

 2t 0 .t

2

 t  0 2

OC r03

{



2t 0 .t

2 0 2

t 

C / cos

C 2 sin

2 0 2

 t 

O

| : (6)

Substituting Eqs. (5) in Eq. (4), collecting terms, and using the given expression for  as a function of time, we that the acceleration takes on the form

ED

a



2r0 0 2t 2 0 .t 5



2 0 2

t 

C  / cos

C  2.3t C  / sin

2 0 2

 t  {O t  2

2

C 2r050  2.3t C  / cos 20 2t 20.t C  / sin t 20 |O: (7) Recalling that r 0 D 3 ft,  0 D 1:2 rad, and  D 20 s, for t D 35 s we can evaluate the expressions in Eqs. (6)



 



and (7) to obtain

E D .0:7518 {O

v



O

1:568 | / ft=s and

2

E D .0:3705 {O C 0:08812 |O/ ft=s :

a


June 25, 2012


Solutions Manual

258

Problem 2.194 A particle is moving such that the time rate of change of its polar coordinates are

P D constant D 3 ft=s

r

and

P D constant D 0:25 rad=s:



Knowing that at time t D 0 , a particle has polar coordinates D 0:2 ft and  0 D 15ı, determine the position, velocity, and acceleration of the particle for t D 10 s. Express your answers in the polar component system formed by the unit vectors ur and u at t D 10 s. O O

r0



Solution

P

P

Since r and  are constant, they can be integrated with respect to time to obtain

D 0 C P t; (1) where r 0 and  0 are the values of r and  , respectively, for t D 0 . The position vect or is given by rE D r uO r ) rE D .r 0 C rP t/ uO r ; (2) where we have used the expression for r in the first of Eqs. (1). Recalling that r 0 D 0:2 ft and rP D 3 ft=s, for t D 10 s the position of the particle can be evaluated to obtain rE D 30:20 uO r ft : r

D r0 C rPt

and



The velocity of the particle is given by

vE D rP uO r C r P uO : P Recalling that rP and  are given and substituting the first of Eqs. (1) into Eq. (3), we have vE D rP uO r C .r0 C rP t /P uO : Recalling that r 0 D 0:2 ft, rP D 3 ft=s, P D 0:25 rad=s, for t D 10 s, Eq. (4) gives vE D .3 uO r C 7:550 uO / ft=s:

(3)



(4)





The acceleration of the particle is given by

E D rR

a



Substituting Eqs. (1) into Eq. (5) gives

a

P  O C r R C 2rP P  uO :

r  2 ur

.r0

rt/ 2 ur

(5)



2 r  u :

(6)

D 0:2 ft, rP D 3 ft=sE, PDD 0:25CradP =sP, forO tCD P10P Os the acceleration of the particle can be

Recalling that r 0 evaluated to obtain

E D.

a

2

O C 1:500 uO / ft=s :

1:888 ur




June 25, 2012


Dynamics 2e

259

Problem 2.195 A particle is moving such that the time rate of change of its polar coordinates are

P D constant D 3 ft=s

r

and

P D constant D 0:25 rad=s:



Knowing that at time t D 0 , a particle has polar coordinates D 0:2 ft and  0 D 15ı, determine the position, velocity, and acceleration of the particle for t D 10 s. Express your answers in the Cartesian component system formed by the unit vectors { and |O. O

r0

Solution

P

P

Since r and  are constant, they can be integrated with respect to time to obtain

D 0 C P t; where r 0 and  0 are the values of r and  , respectively, for t D 0 . The position vect or is given by rE D r uO r ) rE D .r 0 C rP t/ uO r ; r

D r0 C rPt

and



(1)

(2)

where we have the expression for r given in the first of Eqs. (1). The velocity of the particle is given by


v

P

P

(3)



Recalling that r and  are given and substituting the first of Eqs. (1) into Eq. (3), we have that the velocity becomes v r ur .r0 r t / u : (4)

EDPO C CP PO

The acceleration of the particle is given by

E D rR

a



P  O C r R C 2rP P  uO :

r  2 ur

Substituting Eqs. (1) into Eq. (5) gives

(5)



C rPt/P 2 uO r C 2rP P uO : (6) We now observe that we can express the unit vectors uO r and u O in terms of the unit vectors {O and |O as follows: uO r D cos  {O C sin  |O and uO D sin  {O C cos  |O : (7) ED

a

.r0









Using Eqs. (1), substituting Eqs. (7) into the last of Eqs. (2), Eq. (4), and Eq. (6), we have

E D .r 0 C rPt/cos0 C P t  {O C sin0 C P t  |O; (8) vE D rP cos 0 C P t .r0 C rP t /P sin 0 C P t {O C .r0 C rPt/Pcos0 C P t  C rP sin 0 C Pt  |O; (9)  2 P P P P aE D .r0 C rP t/ cos 0 C  t 2rP  sin 0 C  t {O 2r  cos   t  .r rt / 2 sin  t  | : (10)  0C P P 0C P C P PP 0 C P O ı Recalling that r 0 D 0:2 ft, rP D 3 ft=s,  D 0:25 rad=s, and  0 D 15 , for t D 10 s Eqs. (8)–(10) give rE D . 28:05 {O C 11:20 |O / ft; vE D . 5:585 {O 5:900 |O / ft=s; and aE D .1:197 {O 2:093 |O / ft=s : r


















2

June 25, 2012


Solutions Manual

260

Problem 2.196 A micro spiral pump consists of a spiral channel attached to a stationary plate. This plate has two ports, one for fluid inlet and another for outlet, the outlet being farther from the center of the plate than the inlet. The system is capped by a rotating disk. The fluid trapped between the rotating disk and the stationary plate is put in motion by the rotation of the top disk, which pulls the fluid through the spiral channel.  r0 , where r 0 146 m Consider a spiral channel with the geometry given by the equation r is the starting radius, r is the distance from the spin axis, and  , measured in radians, is the angular position 190 m, that the top disk of a point in the spiral channel. Assume that the radius at the outlet is r out rotates with a constant angular speed ! , and that the fluid particles in contact with the rotating disk are

D C D

D

essentially to of it.the Determine constant value (in rpm) such after 1:25 rev of the  and v ! 0:5 top disk, thestuck speed particlesthe in contact with thisthe disk is of m=s at the that outlet.

D


Solution

We denote by  out the value of the coordinate  corresponding to r the outlet. Since r  r0 , we have

D r , where r is the radial position of D C r r0 r D  C r0 )  D (1) )  D 5:602 m,  where we have used the fact that r D 190 m, r 0 D 146 m, and  D 1:25 rev D 1:25.2 / rad. Next, we recall that the velocity in polar coordinates is vE D rP u O r C r rP uO , so that the speed is  2 v D rP 2 C r P : (2) We can obtain rP by differentiating r with respect to time. This gives rP D  P : (3) Substituting Eq. (3) into Eq. (2) and solving for P , we have v P D (4) ) ! D .r r /2 C Œ.rv r / C  r ç2 ; 2 C r 2 out

out

out 

out

out

out

out

out



q

out

out  0

out  0

out 0

where we have used theRecalling fact that that where have used forrev ! r , r 190 mr, 0r, and 146  in mwe  expression out 1:25 the second of Eqs. (1). , and that for the out 0 1:25.2 / rad we must have v 0:5 m=s, the last of Eqs. (4) gives

p

D

D P DD C

p

!

D

D

D

D

D 2630 rad=s D 25;120 rpm:


June 25, 2012


Dynamics 2e

261

Problem 2.197 A micro spiral pump consists of a spiral channel attached to a stationary plate. This plate has two ports, one for fluid inlet and another for outlet, the outlet being farther from the center of the plate than the inlet. The system is capped by a rotating disk. The fluid trapped between the rotating disk and the stationary plate is put in motion by the rotation of the top disk, which pulls the fluid through the spiral channel.  r0 , where  12 m Consider a spiral channel with the geometry give n by the equation r 146 m is the starting radius, r is the distance from the spin axis, and  , is called the polar slope, r 0 measured in radians, is the angular position of a point in the spiral channel. If the top disk rotates with a constant angular speed ! 30;000 rpm, and assuming that the fluid particles in contact with the rotating

D C

D

D

D

disk are essentially stuckparticle to it, use the itpolar shown and determine the velocity and r 170 system acceleration of one fluid when is atcoordinate m.

D


Solution

In polar components, the velocity is given by

vE D rP uO r C r P uO : (1) P Since r D  C r0 and  D ! , Eq. (1) becomes vE D  ! uO r C r ! uO : (2) 2  6 Recalling that  D 12 m D 12 10 m and ! D 30;000 rpm D 30000 60 rad=s, for r D 170 m D 6 







170  10

m, Eq. (2) gives

E D .0:03770 uO r C 0:5341 uO / m=s:

v



For the acceleration we have

P  O C r R C 2rP P  uO : Recalling that P D ! D const., and recalling that r D  C r0 , we have rP D  !; rR D 0; and R D 0: E D rR

a



r  2 ur

(3)



(4)

Hence, Eq. (3) can be simplified to 

D

12 m Recalling that  170  106 m, Eq. (5) gives

D 12



2

2

a 106

2! uO : EmDandr!! uDO r C30;000 rpm D 30000 2

E D.

a



 60

rad=s, for r

D 170 m (5)D

2

O C 236:9 uO / m=s :

1678 ur




June 25, 2012



Dynamics 2e

263

Problem 2.199

D C

A particle moves along a spiral described by the equation r r 0  , where r 0 and  are constants, and where  is in radians. Assume that ˛t , where ˛ 0:15 rad=s2 and t is time expressed in seconds.  0:25 m and  0 for t 0, determine  such that, for If r t 10 s, the accel eration is comple tely in the radial direc tion. In addition, determine the value of the polar coordinates of the point for t 10 s.

PD D D

D

D

D

D

Solution

Since r

D 0:25 m and  D 0 for t D 0, using the equation of the spiral, we must have that r0 D 0:25 m:

(1)

Next, recalling that the acceleration has the expression

E D rR

a



P  O C r R C 2rP P  uO ;

r  2 ur

(2)



we proceed to determine expressions for the first and second time derivatives of the radial and transverse coordinates. The first time derivative of the transverse coordinate is given as ˛t . Hence, for this coordinate we have  ˛t  ˛: (3)

PD

) RD

In addition, since we will need it to properly compute the radial coordinate, recalling that we can integrate the first of Eqs. (3) with respect to time to obtain 1

D 0 for t D 0,

2

2 ˛t : (4)  To determine the time derivatives of the radial coordinate, we first substitute Eq. (4) into the equation of spiral and then differentiate with respect to time. This gives

D

r

D r0 C 12  ˛t 2 ) rP D  ˛t ) rR D  ˛:

(5)

Substituting Eqs. (3) and (5) into Eq. (2) gives

ED

a For t i.e.,

˛ 

 r0 ˛

2 2

t

3 4 1  2 ˛ t

 uO C ˛r r

0



C 25  ˛2t 2 uO :

(6)



D tf D 10 s (the subscript f stands for “final”) the transverse component of the acceleration vanishes,

2r 0 : C 25  ˛2tf2 D 0 )  D 5˛t 2 f Using Eq. (1), and recalling that ˛ D 0:15 rad=s and tf D 10 s, the last of Eqs. (7) gives  D 0:006667 m: ˛r 0

(7)



2

(8)



Substituting the last of Eqs. (7) into the first of Eqs. (5), and recalling that evaluate r and  for t 10 s to obtain

 is given by Eq. (4), we can

D

r

D 0:2000 m

and



D 7:500 rad:


June 25, 2012


Solutions Manual

264

Problem 2.200 A point is moving counterclockwise at constant speed v 0 along a spiral r 0  , where r 0 and  are constants described by the equation r with dimensions of length. Determine the expressions of the velocity and the acceleration of the particle as a function of  expressed in the polar component system shown.

D C

Solution

In polar coordinates, the expressions for the velocity is


v

(1)



observing that r is as a function of  is given in the problem statement, we need to determine expres sion for r and  as functions of  and the known quantities v 0 , r 0 , and  . To do so, we begin with differentiating the given expression for r with respect to time: r : (2)

P

P

PD P

Substituting Eq. (2) into Eq. (1) and keeping in mind that r equal to v 0 as follows:

D r0 C  , we enforce the condition that jEvj is v0

P C .r0 C  /2P 2 D v02 ) P D

 2 2

2

C .r0 C  /2 ;

(3)

where we have chosen the root with  > 0 to be consistent with the fact that the point is moving counterclockwise. Using the last of Eqs. (3) along with Eqs. (1) and (2), we can rewrite the velocity as a function of  and known constants:

P

ED

v

p

 v0

p

2

C .r0 C  /2 uO r C

C  / C .r0 C  /2 uO :

v0 .r0

p

2



To determine the acceleration, we recall that in polar coordinates the acceleration is given by

P  O C r R C 2rP P  uO : (4) We observe that we already have an expression for P and, through Eq. (2), a corresponding expression for rP . We now need to determine expressions for rR and R . For R we can write 2 P d P R D D dd P ) R D 2 v0 .r0 C  /2 2 ; (5) dt  C .r0 C  /   where we have used the last of Eqs. (3) to express P and to determine an expression for d P =d  . Next, differentiating Eq. (2) with respect to time we have rR D  R so that, using the last of Eqs. (5) we have  2 v02 .r0 C  / rR D  (6) 2 :  2 C .r0 C  /2 E D rR

a



r  2 ur




June 25, 2012


Dynamics 2e

265

D C

r 0  , using Eq. (2), the last of Eqs.(3), the last of Eqs. (5), and Eq. (6), after Finally, recalling that r simplification, we can rewrite Eq. (4) to obtain

ED

a



˚

C  /3  uO r C  v02 .r0 C  /2 C 2 2 C .r 0 C  /2 uO : 2 2 2 2 2 Œ 2 C .r0 C  /2 0 C 2r 0  C .1 C  / v02 .r0

r




June 25, 2012


266

Solutions Manual

Problem 2.201 A person driving along a rectilinear stretch of road is fined for speeding, having been clocked at 75 mph when the radar gun was pointing as shown. The driver claims that, because the radar gun is off to the side of the road instead of directly in front of his car, the radar gun overes timates his speed. Is he right or wrong and why?

Solution

The driver is wrong. The speed recorded by the radar gun is the component (or projection ) of the actual speed along the line connecting the radar gun and the moving car. As such, i.e., being a component, it can only be smaller than the true speed.


June 25, 2012



Solutions Manual

268

D

110 mph Recalling that vA 5280 ft, Eqs. (8) give

1 mi

D

P D

A

5280 3600

D 110 

ft=s, vB

0:001157 rad=s

5280 / ft=s, L D 50 ft, and d D D 175 mph D 175. 3600

P D

H

and

4:287 rad=s:

R

We now turn to the determination of  at A and H . The acceleration of the car is

E D ac .sin  uO r C cos  uO /: (9) 2 P R P In polar coordinates we have aE D rR r  uO r C r  C 2rP  uO , which, when compared to Eq. (9), implies     ac sin  D rR r P 2 and ac cos  D r R C 2rP P : (10) R Solving the second of Eqs. (10) for  , we have  1 R D ac cos  2v P sin  ; (11) a



















r

P

where we have used the expression for r in the first of Eqs. (5). Hence, observing that sin A using Eqs. (7), (8), the last of Eqs. (3), and the last of Eqs. (2), after simplification, we have

2 vB vA2 and RH D : (12) 2dL C  3600  ft=s, vB D 175 mph D 175. 5280 Recalling that vA D 110 mph D 110 5280 3600 / ft=s, L D 50 ft, and d D 1 mi D 5280 ft, Eqs. (12) give

R D

A



2 2 vB  vA dL .d=2/ 2 L2 2d 2





RA D

1:684  10

4

2

D d=.2rA/,

C .d=2/v2AC L2

rad=s2

and









2

0:07547 rad=s

:

RH D


June 25, 2012


Dynamics 2e

269

Problem 2.203 The radar station at O is tracking a meteor P as it moves through the atmosphere. At the instant shown, the station measures the following data 21;000 ft,  40 ı , r for the motion of the meteor: r 22;440 ft=s, 2  187;500 ft=s2 , and  2:935 rad=s, r 5:409 rad=s .

PD

D

RD

D RD

PD

(a) Determine the magnitude and direction (relati ve to the xy coordinate system shown) of the velocity vector at this instant. (b) Determine the magnitude and direction (relati ve to the xy coordinate system shown) of the acceleration vector at this instant.

Solution

Part (a).

Referring to the figur e at the right, the ve locity of the me teor is


v

(1)



Then the speed is

q

jEvj D rP 2 C r P 2 )

jEvj D 65;590 ft=s, (2) where we have used the following numerical data: rP D 22;440 ft=s, r D 21;000 ft, and P D 2:935 rad=s. To find the orientation of vE relative to the xy axes, we note that 





ur

O D cos  {O C sin  |O

and

so that Eq. (1) can be rewritten

E

sin  {

O C cos  |O;

O D

„ ƒ‚ P …  O C „P ƒ‚C P …  O D ˇˇ ˇˇ ˇˇ P E D D

E D rP cos 

v

u



r  sin  {

r sin 

vx

r  cos  |

(3)

O

O

.22;430 {  61;640 {/ ft=s:

(4)

vy

Since v points downward and to the right, its orientation from the x axis is Orientation of v from x axis where, again, we have used the fact we have

C r P cos   D 70:01ı; (5) rP cos  r P sin  that rP D 22;440 ft=s, r D 21;000 ft, and P D 2:935 rad=s. Therefore,  tan

vy vx

1

 tan

r sin 





E

Orientation of v from x axis Part (b).

1



ˇˇ



D 70:01ı (cw):

The acceleration of the meteor is

a

E D rR

E

Then the magnitude of a is

q

jEaj D rR





r  2 ur

P  O C r R C 2rP P  uO :

P 2 C r R C 2rP P /2 )

r2

(6)



2

jEaj D 19;300 ft=s ,


(7) June 25, 2012


Solutions Manual

270

PD

187;500 ft=s2 , r where we have used the following numerical data: r 22;440 ft=s,  2:935 rad=s, r 21;000 ft, and  5:409 rad=s2 . To determine the orientation of a relative to the xy system, we use Eqs. (3) to rewrite Eq. (6) as

RD

RD

D

 aE D rR

„



 rP

 2 cos 

ƒ‚

PD

E



r R C 2rP P/  {O C rR 

ax

 sin 

…



P

ƒ‚ C  R C …P P

r  2 sin 

„

r

O

2r  / cos  |

ay

D.

O C 18;130 {O/ ft=s:

6599 {

(8)

E

Since a is directed upward and to the left, its orientation from the x axis is

E

Orientation of a from x axis

D 180ı



D 180

tan1

ˇ a ˇ "ˇaˇ rR r P  ˇ rR r P  y

x

ı





 2 sin 



where, again, we have used the fact that 21;000 ft, and  5:409 rad=s2 . Hence, we have

RD

ˇˇ# ˇ

C r R C 2rP P / cos  D 110:0ı; (9) tan  2 cos  r R C 2rP P / sin  rR D 187;500 ft=s , rP D 22;440 ft=s, P D 2:935 rad=s, r D 1

E

Orientation of a from x axis

2







D 110ı (ccw):


June 25, 2012


Dynamics 2e

271

Problem 2.204

EP

The time derivative of the acceleration, i.e., a, is usually referred to as the jerk. Starting from Eq. (2.70), compute the jerk in polar coordina tes. Solution

Equation (2.70) on p. 105 of the textbook states

E D rR

a



where



P O C r R C 2rP P uO D ar uO r C a uO ;

r  2 ur



 

(1)



D r R C 2rP P : Differentiating with respect to time the second expression for aE in Eq. (1), we have aEP D aP r uO r C ar uOP r C aP uO C a uPO : ar

D rR



P





r2

a

and



Recalling that

OP D P kO uO r D P uO

ur



Eq. (3) can be rewritten as

and







(2)

(3)



OP D P kO uO D P uO r ;

u





(4)



EP D aP r P a  uO r C aP C P ar  uO :

a







(5)



Differentiating with respect to time Eqs. (2) gives

P D «r rPP 2

ar





PR

2r  

and

P D rP R C r « C 2rR P C 2rP R :

a

(6)

Substituting Eqs. (2) and (6) into Eq. (5) and simplifying, we have

 EP D «r

a



PR

3r  





PP O C

3r  2 u r

h « r 



i

P  C 3rR P C 3rP R uO :

3




June 25, 2012


Solutions Manual

272

Problem 2.205 The reciprocating rectilinear motion mechanism shown consists of a disk pinned at its center at A that rotates with a constant angular velocity !AB , a slotted arm CD that is pinned at C , and a bar that can oscillate within the guides at E and F . As the disk rotates, the peg at B moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the 30 ı , !AB 50 rpm, change in distance between C and B . Letting  R 0:3 ft, and h 0:6 ft, determine  and  , i.e., the angular velocity

D

P

D

R

D

D

and angular acceleration of the slotted arm CD , respectively.

Solution

Referring to the diagram at the right, we define two component systems: one consisting of the unit vectors ur and u , and the other consisting of the unit vectors u` and u . Both these component systems are polar, the first with coordinates r and  , and the R other with coordinates ` and  . The coordinate r const., whereas the coordinate ` varies with time. Because h 2R , geometry tells us that

O

when 

D 30ı,

O

O

D uO D uO r ; uO ` D uO ;

D D 

 



O

D ;

D h cos  : (1) A and C : ErB=A D r uO r and rEB=C D ` uO ` .

and

`

The position of B can be described relative to the fixed points Hence, the velocity of B can be given the following two expressions:

(2) E D rP uO r C r P uO D R!AB uO and vE D `P uO ` C `P uO ; where we accounted for the fact that r D R D const. and P D !AB . Setting the above two expressions of velocity equal to each other and using Eqs. (1), when  D 30 ı , we have R!AB uO D `P uO C .h cos  /P uO r ) `P D R!AB and P D 0 . (3)

v





















E

Similarly to v , the acceleration of B has the following two expressions:

    (4) 2 P  O C r R C 2rP P  uO D R!AB uO r and aE D `R `P 2 uO ` C `R C 2`PP uO ; where we recalled that r D R D const. and P D !AB D const. Setting the above two expres sions of acceleration equal to each other, using Eqs. (1) and the last of Eqs. (3), when  D 30 ı , we have 2 R!AB 2 R!AB uO r D `R uO C .h cos  /R uO r ) `R D 0 and R D : (5) h  aE D rR



r  2 ur















D

0:3 ft, !AB Recalling that R be evaluated to obtain





D 50 rpm

2 60

D 50  RD



rad=s, h

D 0:6 ft, and  D 2

15:83 rad=s

cos 

30 ı , the last of Eqs. (5) can

:


June 25, 2012


Dynamics 2e

273

Problem 2.206 As a part of an assembly process, the end effect or at A on the robotic arm needs to move the gear at B along the vertical line shown with some known velocity v0 and acceleration a 0 . Arm OA can vary its length by telescoping via internal actuators, and a motor at O allows it to pivot in the vertical plane. When  50 ı , it is required that v 0 8 ft=s (down) and that it be slowing down at a0 2 ft=s2 . Using h 4 ft, determine, at this instant, the values for r (the extensional acceleration) and  (the angular acceleration).

D D

R

D

D

R

Solution

Referring to the figure at the right, the length of the arm as a function of  is

r

D h= cos  :

(1)

The velocity of B can be expressed in both the Cartesian and polar component systems shown. Since B moves downward, this gives

E D

vB

O D rP uO r C r P uO :

v 0 |

(2)



We note that

O D sin  uO r C cos  uO ; (3) O C O D rP uO r C r P uO , PD PD v0 cos2  rP D v0 sin  and P D ; (4) h |



so that Eq. (2) can be written as v0 .sin  ur cos  u / which implies r v0 sin  and r  v0 cos  , i.e., 





where we have used Eq. (1). Since B is slowing down (in its downward motion), the acceleration of B , using both component systems, is

E D a0 |O D rR

aB



P  O C r R C 2rP P  uO ) a0 .sin  uO r C cos  uO / D rR

r  2 ur







where we have used Eq. (3). Equating components, we obtain

R

r



P D a0 sin 

r2

and

P  O C r R C 2rP P  uO ;

r  2 ur



R C 2rP P D a0 cos  :

r

(5)

(6)

Using the results from Eqs. (1) and (4), Eqs. (6) give

R D a0 sin  C cosh 

r





v0 cos2  h



2

2

) rR D a0 sin  C v0 cosh

3

)

2

R D 5:781 ft=s ,

r

and 2

R D a0 cos h

2





2.v0 sin  /



 v0 cos

2

2

3

2v0 cos 2 sin   ) R D a0 cos h h h h R )  D 1:421 rad=s , where we have used the following numerical data  D 50 ı , v 0 D 8 ft=s, a 0 D 2 ft=s , and h D 4 ft. 



cos 

 





2

2


June 25, 2012


Solutions Manual

274

Problem 2.207 As a part of an assembly process, the end effect or at A on the robotic arm needs to move the gear at B along the vertical line shown with some known velocity v0 and acceleration a 0 . Arm OA can vary its length by telescoping via internal actuators, and a motor at O allows it to pivot in the vertical plane. Letting v 0 and a 0 be positive if the gear moves and accelerates upward, determine expressions for r , r , r ,  , and  that are valid for any value of  .

PRP

R

Solution

Referring to the figure at the right, the length of the arm as a function of  is

r

D h= cos  :

(1)

The velocity of B can be expressed in both the Cartesian and polar component systems shown. Since v 0 > 0 when B moves upward, this gives

E D v0 |O D rP uO r C r P uO :

vB

(2)



We note that (3) O D sin  uO r C cos  uO ; O C O D rP uO r C r P uO , PD 0 PD 0 v0 cos2  rP D v 0 sin  and P D ; (4) h

|



cos  u / so that Eq. (2) can be written as v 0 .sin  ur v sin  and r  v cos  , i.e., which implies r



where we have used Eq. (1). Since a0 > 0 when B accelerates upward, the acceleration of B , in both component systems, is

E D a0 |O D rR

aB



P  O C r R C 2rP P  uO ) a0 .sin  uO r C cos  uO / D rR

r  2 ur







where used Eq. (3). Equating components, we obtain

R

r



P D a0 sin 

r2

and

P  O C r R C 2rP P  uO ;

r  2 ur



R C 2rP P D a0 cos  :

r

(5)

(6)

Using the results from Eqs. (1) and (4), Eqs. (6) give

R D a0 sin  C cosh 

r

v

0 cos

h

2



2

2

)

R D a0 sin  C v0 cosh

r

3

,

and

R D a0 cos2 



h



2v0 sin 

v

0 cos

h

2

  cos 

h

)

R D a0 cos2 



h



2v02 cos 3  sin  . h2


June 25, 2012



Solutions Manual

276

Since the speed of C is constant, the acceleration of C must always be directed toward the center of the cutting path. Therefore, when C is at D , we have

ED

a



v02 {: 

O

(8)

Substituting the first of Eqs. (2) into Eq. (8), we have

ED

a



v02 cos  ur 

2

 O C v0 sin uO : 

(9)



2

Now we recall that, in polar coordinates, we have a with Eq. (9), implies that, when C is at D

E D rR r P  uO r C r R C 2rP P  uO , which, by comparison v02 cos  v02 sin  rP P rR D r P 2 and R D 2 : (10) C  r r Substituting the expressions for r in Eq. (1), sin  and cos  in Eqs. (6), and for rP and P in Eqs. (7) into 







Eqs. (10), after simplification, when C is at D we have

RD

r



v02 .  d2



C d / d 2 C h 2 C d   C h2 C 2 C 2d 3=2

and

C h 2 2  : 2 d 2 C h 2 C  2 C 2d 

R D v02 





h d2






(11)

June 25, 2012


Dynamics 2e

277

Problem 2.209 In the cutting of sheet metal, the robotic arm OA needs to move the cutting tool at C counterclockwise at a constant speed v0 along a circular path of radius  . The center of the circle is located in the position shown relative to the base of the robotic arm at O . For all position s along the circular cut (i.e., for any value of  ), determine r , r ,  , r , and  as functions of the given quantities (i.e., d , h ,  , v 0 ). These quantities can be found “by hand,” but it is tedious, so you might consider using symbolic algebra software,

PPR

R

such as Mathematica or Maple.

Solution

Referring to the figure at the right, for a generic value of  we have

D

r

q

.h

C  sin  /2 C .d C  cos  /2: O

(1)

O |O D sin  uO r C cos  uO :

We note that the unit vectors { and | can be expressed as follows:

O D cos  uO r

{



O

sin  u

and



(2)

The velocity of C must be tangent to the cutt ing path. Since C moves counterclockwise around the cutting path and since the speed of C is v 0 , we must have

E D v 0.

v

 sin 

O C cos  |O /:

{

(3)


E D v 0.cos  sin  cos  sin  / uO r C v0.cos  cos  C sin  sin  / uO : (4) P Since in polar coordinates we have vE D rP u O r C r  uO , then from Eq. (4) we deduce that v0 rP D v 0 .sin  cos  cos  sin  / and P D .cos  cos  C sin  sin  /: (5) r Using geometry, we have sin  D .h C  sin  /=r and cos  D .d C  cos  /=r , which, in view of Eq. (1), v









give

d C  cos  C  sin  and cos  D : .d  cos  /2 .h  sin  /2 .d  cos  /2 C C C C C C Substituting Eqs. (1) and (6) into Eqs. (5), and simplifying, we have sin 

D

h

.h

p PDp C

r

 sin  /2

v0 .h cos   d sin  /

.d

 cos  /2

C .h C  sin  /2

(6)

p

and

PD



d2

v0 . C d cos  C h sin  / C h2 C 2 C 2d  cos  C 2h sin  :


(7)

June 25, 2012


Solutions Manual

278

Since the speed of C is constant, the acceleration of C must always be directed toward the center of the cutting path. Therefore, we have

ED

a



v02 .cos  { 

O C sin  |O /;

(8)

which, along with Eqs. (2), imply

ED

a



v02 .cos  cos  

2

C sin  sin  / uO r C v0 .cos  sin 



O

cos  sin  / u :

(9)


v02 h cos  d sin   C d cos  C h sin  C  sin  /2 C .d C  cos  /2 uO r C  .h C  sin  /2 C .d C  cos  /2 uO : (10) Now recall that in polar coordinates, we have aE D . rR r P 2 / uO r C .r R C 2rP P / uO , and therefore, by comparison

ED

a



v02 



p

p

.h







with the above expression, we can conclude that

R D r P 2

r and



v02 

R D 2 rP P C v02 r r



C d cos  C h sin  C C .d C  cos  /2 

p



.h

 sin  /2

h cos   d sin 

p

.h

C  sin  /2 C .d C  cos  /2 :

(11)

(12)

Substituting Eqs. (1) and (7) into Eqs. (11) and (12), after simplificati on, we have

C d cos  C h sin  /d 2 C h2 C d cos  C h sin   ; d 2 C h2 C 2C 2d  cos  C 2h sin 3=2 .h cos  d sin  / d 2 C h2 2 v2 R D 0  :  d 2 C h2 C  2 C 2d  cos  C 2h sin  2 RD

r



v02 . 






June 25, 2012


Dynamics 2e

279

Problem 2.210 The cam is mounted on a shaft that rotates about O with constant angular velocity ! cam. The profile of the cam is descr ibed by the funct ion `. / R 0 .1 0:25 cos3  /, where the angle  is measured relative to 3000 rpm the segment OA, which rotates with the cam. Letting ! cam and R 0 3 cm, determine the velocity and acceleration of the follower 33ı . Express the acceleration of the follower in terms of g , when  the acceleration due to gravity.

D

C

D

D D

Solution

90ı   . Therefore, letting r The point on the follower in contact with the cam corresponds to  O , we had that be the distance between the point on the follower in contact with the cam and point r `.  90 ı   / (  is measured in degrees), which gives

D

D

D

D R01 C 0:25 cos3.90ı  / D R01 C 0:25 sin3  ; (1) ı where we have used the fact that cos 90  D sin  . Next, we observe that the velocity and acceleration of the follower are vE D rP |O and aE D rR |O : (2) r





From Eq. (1), we have

P D 0:75R0P sin2  cos 

r and

(3)

R D 0:75R0R sin2  cos  C 0:75R0P 22 sin  cos2  sin3  : Recognizing that P D ! D constant, Eqs. (3) and (4) simplify to   rP D 0:75R0 ! sin2  cos  and rR D 0:75R0 ! 2 2 sin  cos 2  sin3  : r

(4)



cam

cam

(5)



cam

Substituting Eqs. (5) into Eqs. (2), we have

E D 0:75R0! sin2  cos  |O and aE D 0:75R0! 2 2 sin  cos2  sin3   |O: Recalling that R 0 D 3 cm D 0:03000 m, ! D 3000 rpm D 3000 260 rad=s, for  D 33ı, we have v

cam



cam



cam

v

ED

1:758 | m =s and

O

(6)

a

ED

136:9g | :

O

where we have expressed the acceleration in terms of the acceleration due to gravity g


2

D 9:81 m=s . June 25, 2012




Solutions Manual

282 Carrying out the integration, we have

K Recalling that !

D !:

(8)

D !f D 0:5 rad=s for  D .=4/ rad, from Eq. (8) we have K.=4/ D !f ) K D 4!f =:

(9)

Substituting the last of Eqs. (9) into Eq. (8) and recalling Eq. (6), we have

PD



4!f ; . rad/

(10)

where it is understood that  is expressed in radians. Then, differentiating Eq. (10) with respect to time we have

RD



4!f  . rad/

P ) R D

P

2

 4!  f

;

. rad/

(11)

where we have used Eq. (10) to express  , and where we note again that  is expressed in radians. Keeping in mind that the last of Eqs. (4) implies   =2, we can now rewrite Eqs. (5) using the results in Eqs. (10) and the last of Eqs. (11). This gives

PD



D

4!f  . rad/

and

RD



 4!  f

2

. rad/

;

(12)

where it is understood that  must be expressed in radians. Substituting Eqs. (12) in Eqs. (3), we have

v

fR D .4! j j  rad/

and 4

Recalling that !f

ˇ Eˇ D a

R

2

 4!  q

ı

f

. rad/

4

C  2:

(13)



D 0:5 rad=s, R D 4 in: D 12 ft,  D 32 D 32 180 rad, we can evaluate Eqs. (13) to obtain v D 0:1185 ft=s and aE D 0:08642 ft=s :

ˇˇ

2


June 25, 2012


Dynamics 2e

283

Problem 2.213 The mechanism shown is called a swinging block slider crank. First used in various steam locomotive engines in the 1800s, this mechanism is often found in door-closing systems. If the disk is rotating with 60 rpm, H 4 ft, a constant angular velocity  R 1:5 ft, and r denotes the distance between B and O , compute r ,  , r , and  when  90 ı .

D

PP R

PD R

D

D

Solution

Using the diagram at the right, the velocity of B at  and . ur ; u / component systems is, respectively,

O O

E D R P uO t

v

and

D 90ı in the .uO n; uO t /

E D rP uO r C r P uO

v

;

(1)

where the first of Eqs. (1) is true because B is in circular motion about A . We note that u t cos  ur  sin  u and un  sin  ur  cos  u ; where, for 

O D

D 90 ı,

O

O

p

D R=

C H2

O

O

p

(2)

D H = R2 C H 2 : Substituting the first of Eqs. (2) into the first of Eqs. (1), we have vE D R P .cos  uO r sin 

R2

O D

cos 

and

(3)

 sin 

compared to the second of Eqs. (1) component by component, yields

R cos 

r

P

RH 

r

r

O

u /, which, when

8:825 ft=s

(4)

R2

D P ) P D p C H2 ) P D R2 P R P sin  D r P ) P D ) P D 0:7746 rad=s, (5) 2 R C H2 p where we have used Eqs. (3), the fact that r D R2 C H 2 , and the following data: R D 1:5 ft, H D 4 ft, and P D 60 rpm D 60 260 rad=s. The acceleration of B at  D 90 ı in the two component systems is     aE D R P 2 uO n and aE D rR r P 2 uO r C r R C 2rP P uO ; (6) P

and













where, in the first of the above equations, we have used the fact that B is in uniform circular motion about A . Substituting the second of Eqs. (2) into the first of Eqs. (6) gives a R  2 .sin  ur cos  u /, which, when compared to the second of Eqs. (6) component by component, yields

ED

P 2 sin  D rR

R 

and



P

O C

p P ) rR D R4P 22 R2 C2H2 2 p R22P 2 2 ) R CH R CH

r2



2

18:23 ft=s

R3 H P 2 P R C C 2 R2 C H 22 )  D 9:779 rad=s , p where we have used Eqs. (3)–(5), the fact that r D R 2 C H 2 , and the following data: R D 1:5 ft, H D 4 ft, 2 and P D 60 rpm D 60 60 rad=s.

P 2 cos  D r R C 2rP P ) R D

R



RH  2 R2 H 2

RD

r

O



2




June 25, 2012


Solutions Manual

284

Problem 2.214 A satellite is moving along the elliptical orbit show n. Using the polar coordinate system in the figure, the satellite’s orbit is described by the equation

r. /

p2 D 2b2 a2 CabC2 .aa 2





b 2 cos  ; b 2 / cos.2 /



which implies the following identity

rr 00  2.r 0 /2  r 2 r3

D



a; b2

where the prime indicates differentiation with respect to  . Using r 2 this identity and knowing that the satellite moves so that K with K constant (i.e., according to Kepler’s laws), show that the radial component of acceleration is proportional to 1=r 2 , which is in agreement with Newton’s universal law of gravitation.

D P

Solution

We need to show that a r

D .constant/



  

K We are given r

1 r2

. First we will rewrite Kepler’s law as

D r 2P ) P D rK2 :

(1)

D r . / so we use the chain rule to write its derivative with respect to time as Kr 0 rP D r 0 P ) rP D 2 ;

(2)

r

P

where the prime denotes differentiation with respect to  , and where we have substitu ted Eq. (1) for  . Next we take the second derivative of r with respect to time to obtain

RD

r

2K

r3

2

r Pr C K r P 0



0

00

r2

P

) rR D Kr 2



r

00

r2



2.r 0 /2 ; r3



where we have factored K  out of the first of Eq. (3) and substituted Eq. (1) for ar r  r  2 and we have expressions for r , r , and  so we can write

DR

P

2

ar

D Kr 2

r

00

r2



2.r 0 /2 r3



r

R

P

K2 r4

)

2

ar

D Kr 2

00 

(3)

P

 . Now recall that

0 2  r2

 rr 2r  r3



:

(4)

We see that the expression in brackets is the identity giv en in the problem statement , so we will replace it with a=b 2 . Recalling that K , a , and b are constants, we prove the radial component of acceleration is proportional to 1=r 2 . 2

ar

D Kb2a

 1 

r2

:


June 25, 2012


Dynamics 2e

285

Problem 2.215

D

10;000 ft At a given instant, an airplane flying at an altitude h 0 begins its descent in preparation for landing when it is r.0/ 20 mi from the radar station at the destination’s airport. At that instant, 300 mph, the climb rate is constant the aiplane’s speed is v0 and equal to 5 ft=s, and the horizontal component of velocity is decreasing steadily at a rate of 15 ft=s2 . Determine the r ,  , r , and  that would be observed by the radar station.

D

D

PPR

R

Solution

When expressed in the Cartesian coordinate system shown, the velocity and acceleration of the airplane are

E D xP {O C yP |O

v where

PD

q

v02  y 2 ;

5 ft=s;

x



P

E D xR {O C yR |O ;

a

(1)

2

R D 15 ft=s ; yR D 0; (2) where it is understood that these values are at t D 0 . Using the polar y

PD

and

x

component system shown, the velocity and acceleration are

E D rP uO r C r P uO and aE D rR r P 2 uO r C r R C 2rP P  uO ; where the unit vectors u O r and uO are related to the unit vectors {O and |O as follows: {O D cos  uO r sin  uO and |O D sin  u O r C cos  uO : v





(3)









Using Eqs. (4), we can rewrite Eqs. (1) as follows: v . x cos  y sin  / ur

ED P

and

CP

(4)



O C . xP sin  C yP cos  / uO 

(5)



E D .xR cos  C yR sin  / uO r C . xR sin  C yR cos  / uO :

a



(6)



Equating the components of the first of Eqs. (3) and (5), we have

P D r 1. xP sin  C yP cos  /: (7) 1 At the instant considered in the problem, we have that  D sin .h=r/ . With this in mind, using the data in Eqs. (2) along with the fact that r D 20 mi D 20.5280/ ft, h D 10;000 ft, v 0 D 300 mph D 300 5280 3600 ft =s, P D xP cos  C yP sin 

r

and



and





we can evaluate Eqs. (7) to obtain:

PD

r

438:5 ft=s

P D 347:4

 10

6

rad=s:

Similarly, equating the components of the second of Eqs. (3) and (6), we have 1

2

r

R D xR cos  C yR sin  C r P





and  r x sin  y cos  2r  ; which can be evaluated with the help of Eqs. (7) and the data used in the evaluation of Eq. (7) to obtain

R D 14:95 ft=s

r

2

RD

and

RD





R

10:59  10

CR

6

P P

(8)

rad=s2 :


June 25, 2012



Dynamics 2e

287

Executing the above code we obtain the following plot:

We now determine the trajectory of the projectile starting from Eqs. (3) and (4) derived in Example 2.21 on p. 110 of the textbook, and subject to the initial conditions in Eqs. (9) and (10) of the example in question. For convenience, we repeat the equations we need here below:

rR r P 2 D R r  C 2rP P D 

g sin  ;

(7)

g cos  ;

(8)

with the initial conditio ns

D h; rP.0/ D v 0 sin  ; r.0/

D 2 rad; v0 P .0/ D cos  : h  .0/



(9) (10)

The above system of differential equations and initial conditions can be integrated with any appropriate mathematical software. The solution of these equations will be in terms of r and  as a function of time. To obtain the plot of the trajectory, we must resort to a parametric plot, i.e., a plot of the coordinates of the 2:491 s. To produce a plot that can be compared to the one shown above, we projectile for 0  t  tf must plot values of x and y corresponding to the values of r and  given by the numerical solution. We do so by observing that x r cos  and y r sin  : (11)

D

D

D

With the above in mind, we have used Mathematica with the following code to obtain a solution with time t going from 0 to 2:491 s:

The code above yields the following trajector y, which can be seen to be identical to the one obtained earlier (as expected). This solution s manual, in any print or electronic form, remain s the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplicat ion or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

June 25, 2012


288

Solutions Manual


June 25, 2012


Dynamics 2e

289

Problem 2.217 Reference frame A is translating relative to reference frame B . Both frames track the motion of a particle C . If at one instant the velocity of particle C is the same in the two frames, what can you infer about the motion of frames A and B at that instant? Solution

For two frames to measure the same velocity these frames must be at rest relative to one another at least at the instant at which the measurement is made. Therefore, we can conclude that frame A has zero velocity relative to frame B at the instant considered.


June 25, 2012


Solutions Manual

290

Problem 2.218

E

E

Reference frame A is translating relative to reference frame B with velocity vA=B and acceleration aA=B . A particle C appears to be stationary relative to frame A. What can you say about the ve locity and acceleration of particle C relative to frame B ? Solution

If the particle is stationary relative to frame A , then, relative to B , it moves just like frame A . Therefore, the velocity and acceleration of particle C relative to frame B are vA=B and aA=B .

E

E


June 25, 2012


Dynamics 2e

291

Problem 2.219

E

Reference frame A is translating relative to reference frame B with constant velocity vA=B . A particle C appears to be in uniform rectilinear motion relative to frame A . What can you say about the motion of particle C relative to frame B ? Solution

Because the relative velocity of frame A is constant, then A is in uniform rectilinear motion relative to B . Since the velocity of particle C is constant relative to A , and since the velocity of C relative to B is the sum of the velocity of A relative to B and of the velocity of C relative to A , then the velocity of C relative to be will also be constant. In turn, this implies that particle C will appear to be in uniform rectilinear motion relative to frame B .


June 25, 2012


Solutions Manual

292

Problem 2.220 A skier is going down a slope with moguls. Let the skis be short enough for us to assume that the skier’s feet are tracking the moguls’ profile. Then, if the skier is skilled enough to maintain her hips on a straight-line trajectory and vertically aligned over her feet, determine the velocity and acceleration of her hips relative to her feet when her speed is equal to 15 km=h. For the profile hI  0:15x 0:125 sin. x=2/ m, of the moguls, use the formula y.x/ where hI is the elevation at which the skier starts the descent.

D

C

Solution

The expressions of the vertical position of the skier’s hips and feet are denoted by yH and y F , respectively.

D .h I yF D hI

yH

 

x tan 8:53ı / m; 0:15x

C

(1)

 x 

 0:125 sin 2

m:

(2)

Now, calculating the relative position of her hips with respect to her feet,

yH=F

D yH



D

yF



.0:15  tan 8:53ı /x  0:125 sin

 x   2

m;

(3)

and taking the derivat ive of Eq. (3) with respect to time yields

vH=F

D xP



0:15  tan 8:53ı 

0:125 x cos 2 2

 

m=s:

(4)

However, x

3600 m =s. With this substituti on Eq. (4) becomes P D v0 cos 8:53ı and v0 D 15 km=h D 15 1000

vH=F

D

h

5:630  105  0:8091 cos

 x i  2

m=s:

(5)

The relative acceleration of the skier’s hips with respect to her feet can now be determined by differentiating Eq. (4) and substituting x v 0 cos 8:53ı . This yields

PD

aH=F

D 1:271xP sin

 x  2

m=s2

)

aH=F

D 5:237 sin

 x  2


m=s2 :

June 25, 2012


Dynamics 2e

293

Problem 2.221

E

E

Two particles A and B are moving in a plane with arbitrary velocity vectors vA and vB , respectively. Letting the rate of separation ( ROS) be defined as the component of the relative velocity vector along the line connecting particles A and B , determine a general expression for ROS. Express your result in terms rB  rA , where rA and rB are the position vectors of A and B , respectively, relative to some of rB=A chosen fixed point in the plane of motion.

E DE E

E

E

Solution

O

E

We begin by writing uB=A in terms of rB=A:

O

: D jErrEB=A B=Aj

E

D vEB vEA:

uB=A

(1)

The velocity of B relative to A is:

vB=A

(2)



The component of Eq. (2) in the direction of Eq. (1) is the with uB=A. This gives

E

ROS , which is therefore obtained by dotting vB=A

O

ROS

D vEB=A uO B=A ) 

ROS

B=A D vEB vEA jErrEB=A j. 




June 25, 2012


Solutions Manual

294

Problem 2.222 Airplanes A and B are flying along straight lines at the same altitude and with speeds vA vB 550 km=h, respectively. Determine the speed of A as perceived by B if  50 ı .

D

D

D 660 km=h and

3

3

3

3

Solution

Referring to the figure at the right, we use the Cartesian coordinate system shown to express the velocities of the airplanes A and B :

E D vA |O

vA

E D vB .sin  {O C cos  |O /:

vB

and

(1)

Using relative kinematics, the velocity of A as perceived by B is

vA=B

E D vEA vEB D 

vB

sin  {

O C .vA

 vB

cos  / | :

O E

3

3

(2)

3

3

The speed of A as perceived by B is the magnitude of the vector vA=B . Using the result in Eq. (2), we have

ˇ ˇ q

D vEA=B D vB2 sin2  C .vA vB cos  /2: (3) Recalling that vA D 660 km=h D 660 1000 3600 m =s, vB D 550 km=h D ı 550 1000 3600 m =s and  D 50 , we can evaluate the expressio n in Eq. (3) to obtain vA=B



vA=B

D 144:7 m=s:


June 25, 2012


Dynamics 2e

295

Problem 2.223 Three vehicles A, B , and C are in the positions shown and are moving with the indicated directions. We define the rate of separation ( ROS) of two particles P 1 and P2 as the component of the relative velocity of, say, P2 with respect to P 1 in the direction of the relative position vector of P 2 with respect to P 1 , which is along the line that connect s the two particles. At the given inst ant, determine the rates of separation ROSAB and ROSCB , that is, the rate of separation between A and B and between C and B . Let A B mph,use andthe Furthermore, treat v 60 mph, 55 and v Cdimensions 35 mph.shown the vehicles as vparticles in the figure.

D

D

D

Solution

We need to derive a convenient expression for the rate of separation. We begin by writing uB=A in terms of rB=A

O

E

O

uB=A

: D jErrEB=A B=A j

(1)

The velocity of B relative to A is:

E

vB=A

D vEB vEA:

(2)



The component of Eq. (2) in the direction of Eq. (1) is the

ROS, i.e.,

B=A D vEB vEA jErrEB=A j : (3) Now that we have a formula for the ROS, consider the . {O; |O / component

ROS

D vEB=A uO B=A ) 

ROS





system shown in the figure at the right.

ˇE ˇ D p rB=A

O

O

23 {  94 | ft

D .0:2377 {O 0:9713 |O/ ; (4) 2 ft C . 94/  42 {O C 65 |O ft rEC=B D .42/ D .0:5427 {O C 0:8399 |O/ : (5) 2 C .65/2 ft rC=B 5280 ft =s, v D 55 mph D 55 5280 ft =s, and v D 35 mph D 35 5280 ft =s, Recalling that vA D 60 mph D 60 3600 B C 3600 3600 and observing that vEA D vA |O , vEB D vB |O , and vEC D v C .cos 54ı {O C sin 54ı |O/, we have vEA D . 88:00 |O / ft=s; vEB D .80:67 |O / ft=s; vEC D .30:17 {O C 41:53 |O / ft=s: (6) rB=A

.23/2





ˇ ˇ p 




June 25, 2012


Solutions Manual

296 The rate of separation between vehicles A and B is found as ROSAB

D vEB vEA rrEB=A ) 



ˇ ˇ

ROSAB

B=A

D

163:8 ft=s:

D

16:50 ft=s:

Similarly, the rate of separation between vehicles C and B is found as ROSCB

D vEB vEC  rrEC =B ) 



ˇ ˇ C =B

ROSCB


(7)

June 25, 2012



Solutions Manual

298

O D sin 54ı {O

cos 54ı | , recalling that aC=A

O E D aEC aEA, and using Eqs. (5) and (3), we   2 aEC =A D vC = .sin 54ı {O cos 54ı |O/: (6) Recalling that v C D 42 km=h D 42 1000 m = s and  80 m, we can evaluate Eq. (6) to obtain D 3600 Observing that un have







E

aC=A

D .1:376 {O



1:000 | / m=s2 :

O


June 25, 2012



Solutions Manual

300

Problem 2.226

D

25 ft=s, During practice, a player P punts a ball B with a speed v 0 60ı , and at a height h from the gro und. Then the at an angle  player sprints along a straight line and catches the ball at the same height from the ground at which the ball was initially kicked. The length d denotes the horizontal distance between the player’s position at the start of the sprint and the ball’s position when the ball leaves the player’s foot. Also, let  t denote the time interval between the instant at which the ball leaves the player’s foot and the instant at which the player starts

D

sprinting. 3 ft and  t 0:2 s, and determine the average Assume that d speed of the player so that he catches the ball.

D

D

Solution

Referring to the figure at the right, we model the motion of the ball as projectile motion and since we are interested in computing the player’s average speed, we will model the player’s motion as rectilinear motion with constant speed. We adopt a Cartesian coordinate system with srcin at point O , which we choose as the point at which the ball leaves the player’s foot. Let tf (f stands for final) denote the time the playe r catches the ball. Then at time tf , the relative horizontal position of the player with respect to the ball must be equal zero:

xP =B .tf /

D0 )

xP .tf /  xB .tf /

D0 )

xP .tf /

D xB .tf /: (1)

Let t 0 be the time at which the ball leave s the player’s foot. Since the horizon tal component of the acceleration of the ball is equal to zero, the ball moves in the x direction with a velocity component that is constant and equal to v 0 cos  . Therefore, we have that

D

xB .tf /

D v0tf cos  :

(2)

Since the player starts sprinting  t after the ball is kicked, the motion of the player occurs during the time  t and ends at t tf . Therefore, denoting by  the time during which the interval that starts at time t player sprints, we have that  tf  t: (3)

D

D D

We also observe that the total distance traveled by the player, which we will denote by L , is given by

L

D xP .tf / C d )

L

D xB .tf / C d ;

(4)

where, in writing the last of Eqs. (4), we have used the last of Eqs. (1). By definition, the average speed of the player is now given by d v0 tf cos  L (5) .vP /avg .vP /avg :  tf  t

D

)

D C

This result tells us that the value of .vP /avg can be determined if we determine the value of tf . To this end we observe that since the player catches the ball at the same height from the ground at which the ball leaves the player’s foot, and given our choice of coordinate system, we can calculatetf as the time at which the ball This solution s manual, in any print or electronic form, remain s the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplicat ion or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

June 25, 2012


Dynamics 2e

301

D

0 . Recalling that the motion of the ball is projectile motion, we have that the comes back down to y coordinate of the ball as a function of time is given by

D v0t sin  12 gt 2; where g is the acceleration due to gravity. Setting yB D 0 , we obtain 2v0 sin  tf D 0 and tf D : g yB

(6)



The first root can be neglected since it coincides with the initial time.

y

(7)

Hence, substituting the second of

Eqs. (7) into the last of Eq. (5) and simplifying, we have

.vP /avg

2 D 2vdg0 Csinv0 sing2t ;

(8)



D

D

where we have used the trigonometric identity 2 sin  cos  sin 2 . Recalling that d 3 ft, g v0 25 ft=s,  60 ı , and  t 0:2 s, we can evaluate the above expression to obtain

D

D

D

.vP /avg

2

D 32:2 ft=s ,

D 17:30 ft=s:


June 25, 2012


Solutions Manual

302

Problem 2.227 A remote controlled boat, capable of a maximum speed of 10 ft=s in still water, is made to cross a stream with a width w 35 ft that is flowing with a speed v W 7 ft=s. If the boat starts from point O and keeps its orientation parallel to the cross-stream direction, find the location of point A at which the boat reaches the other bank while moving at its maximum speed. Furthermore, determine how much time the crossing requires.

D

D

Solution

Using a Cartesian coordinate system with its srcin at its velocity vB can be written as

O , as the boat is crossing the stream,

E

E D vEW C vEB=W D .7 {O C 10 |O/ ft=s; where vEW D 7 {O ft=s is the velocity of the water and vEB=W D 10 |O ft =s is the velocity of the boat relative to vB

the water. Using the the y component of velocity, the time of crossing is

t

D vwBy )

t

D 3:500 s,

(1)

D

35 ft. Since the x component of velocity is constant, using the crossing time in Eq. (1), we can where w calculate the downstream position of A as x

D vBx t )

x

D 24:50 ft.


June 25, 2012



Solutions Manual

304

D 0, are subject to the following initial conditions: x.0/ D 0 and y.0/ D 0:

which, given that the boat starts at O at time t

(8)

These equations can be integrated using appropriat e mathematical software . We have used Mathematica. Before presenting the code used to solve the problem, a remark about Eqs. (7) is necessary. Let’s suppose that the boat does make it to point A . In this case, the coordinat es of the boat would be .x 0; y w/. When this happens, the argument of the square roots in Eqs. (7) becomes equal to zero. In turn, because the square roots appear at the denominator of fractions, when the boat makes it to A , the equations suffer a division by zero that will cause the numerical software to fail. Now, when using a numerical method to solve Eqs. (7) we must specify the time interval we want the solution to cover. As just discussed, as soon as the boat makes it to A the numerical integ ration fails. Therefore, we need to use trial and error to find the maximum amount of time for which the equations can be integrated before numerical failure. However, this maximum amount of time will correspond to the time taken by the boat to reach A , which is one of the quantities we need to determine in this problem. With all the above in mind, we have integrated the equations in question for the 10 ft=s, using Mathematica with the boat traveling at the maximum speed relative to water, i.e., wB=W following code:

D

D

D

Notice that, using trial and error, we were able to integrate our equations only up to expressing this result to three significant figures, we will say that For vB=W

t

D 6:8627 s. Hence,

D 10 s, the boat reaches A in 6:863 s.

The solution obtained using Mathematica with the above code, can be plotted to depict the path followed by the boat. This path was plotted with the following code:


June 25, 2012


Dynamics 2e

305

Now we consider the case for which the speed of the boat is equal to the speed of the water current. We 7 ft=s. This time we find that the mathematical repeat the same calculations just described but with vB=W software does not fail for any amount of time, indicating that

D

When vB=W

D 7 s, the boat does not reach A:

We present our solution for t going from zero to 20 s and we plot both the path of the boat and the distance dAB x 2 .w  y/ 2 of the boat from A as a function of time. As can be seen from the plot of the path of the boat, the boat does reach the other side of the stream but when it does it points toward A while moving relative to the water with speed equal to the (absolute) speed of the water. As a result, an observer stationary

D

p

C

with the banks of the stream sees the boat become station ary. The fact that the boat no longer moves relati ve to the banks can be seen from the plot of the distance dAB , which, after about 9 s becomes constant (and remains different from zero).


June 25, 2012




Solutions Manual

308

Problem 2.231

D

An airplane is initially flying north with a speed v0 430 mph relative to 12 mph, forming the ground, while the wind has a constant speed v W 23 ı with the north-south direction. The airplane performs a an angle  course change of ˇ 75 ı eastward while maintaining a constant reading of the airspeed indic ator. Letting vP =A be the velocity of the airplane relative to the air and assuming that the airspeed indicator measures the magnitude of the component of vP =A in the direction of motion of the airplane, determine the speed of the airplane relative to the ground after

D

D

D

3

3

E

E

the course correction.

Solution

We will express the velocity of the airplane using a normal tangential component system. In this manner, the tangent direction will always be the direction of motion of the airplane. In addition, we will use subscripts 1 and 2 to denote quantities before and after the change in course, respectively. Before turning, the velocity of the airplane and wind relative to the ground are

E D v0 uO t

vP1

1

E D vW .

vW1

and

O C sin  uO n /:

 cos  u t1

(1)

1

The airspeed of the plane before the turn is

v1

D .vEP vEW / uO t ) 1



1



v1

1

D v 0 C vW cos  :

(2)

After turning, the velocity of the wind is

E D vW fcosŒ rad .ˇ C  /ç uO t C sinŒ rad .ˇ C  /ç uO n g D vW Œ cos.ˇ C  / uO t C sin.ˇ C  / uO n ç; where we have used the trigonometric identities cos. rad  / D cos  and sin . rad vW2





2



2

2

(3)

2







airspeed of the plane after the turn is

/

D sin  . The

D .vEP vEW / uO t ) v2 D vP C vW cos.ˇ C  /: Enforcing the condition that v 2 D v 1 , we have v0 C vW cos  D vP C vW cos.ˇ C  / ) vP D v 0 C vW Œcos  cos.ˇ C  /ç: Recalling that v 0 D 430 mph, v W D 12 mph,  D 23 ı , and ˇ D 75 ı , we can evaluate vP to obtain vP D 442:7 mph: v2

2



2



2

2

2

2



(4)

(5)

2

2


June 25, 2012




Dynamics 2e

311

Problem 2.234 An interesting application of the relative motion equations is the experimental determination of the speed at which rain falls. Say you perform an experiment in your car in which you park your car in the rain and measure the angle the falling rain makes on 20 ı . Next you drive your side window. Let this angle be  rest forward at 25 mph and measure the new angle  motion 70 ı that the rain makes with the vertical. Determine the speed of the falling rain.

D

D

Solution

O

O

In the figure shown, { and | are horizontal and vertical, respectively. Since rest is the orientation of the velocity of raindrops relative to the ground, the velocity in question is

E D

vR

O C cos  |O /;

vR .sin rest {

(1)

rest

E

where v R is the speed of the falling rain. The angle  motion describes the orientation of vR=C , the velocity of the raindrops relative to the moving car:

E

vR=C

D

O C cos 

vR=C .sin motion {

O

motion | /;

(2)

where vR=C is the speed of the raindrops as perceived by an observer moving with the car. Relative kinematics requires that vR vC vR=C ; (3)

E DE CE

where vC is the velocity of the car. Substituting Eqs. (1) and (2) into Eq. (3),

E

O

vR sin rest { 

O D vC {O

vR cos rest |



O

O

vR=C sin motion {  vR=C cos motion | :

(4)

Equating components, Eq. (4) yields the following system of two equations in two unknowns v R and v R=C : vR sin rest

D vC



vR=C sin motion

and



vR cos rest

D

vR=C

cos motion;

(5)

whose solution is

vR

D cos 

vC

rest tan motion 

Recalling that we have  rest Eqs. (6) to obtain

D 20ı, 

sin rest

motion

and

vR=C

D sin 

motion 

vC cos motion tan rest

:

(6)

5280 D 70ı, vC D 25 mph D 25 3600 ft =s, we can evaluate the first of

vrain

D 16:37 ft=s:


June 25, 2012


Solutions Manual

312

Problem 2.235 A woman is sliding down an incline with a constant acceleration of a0 2:3 m=s2 relative to the incline. At the same time the incline is accelerating to the right at 1:2 m=s2 relative to the ground. Letting  34ı and L 4 m and assuming that both the woman and the incline start from rest, determine the horizontal distance traveled by the woman with respect to the ground when she reaches the bottom of the slide.

D D

D

Solution

The xy and pq frames shown at the right are attached to the ground and to the incline, respecti vely. The srcins of these frames are chosen so that the x and p coordinates of the woman are both equal to zero at the initial time. We denote by aW and aW=I the accelerations of the woman relative to the ground and to the incline, respectively. Denoting by aI the acceleration of the incline relative to the ground, from the problem’s statement we have

E

E

E

E

aW=I

D a0 uOp

and

E D as {O:

aI

(1)

Then, using relative kinematics, we have

E D aEI C aEW=I : (2) Substituting Eqs. (1) into Eq. (2) and observing that u Op D cos  {O sin  |O , we obtain (3) aEW D .a s a0 cos  / {O a0 sin  |O; which implies that the x component of aEW is constant. Recalling that x W D 0 for t D 0 and that the woman aW









starts from rest, using constant acceleration equations, we have

D 12 .as

xW



a0 cos  /t 2 :

(4)

Denoting by tf and d the time taken by the woman to slide to the bottom of the incline and the corresponding horizontal distance traveled (with respect to the ground), respectively, we have

d

D

ˇˇ

1 .a 2 s

ˇˇ

a0 cos  /tf2 :

(5)

We observe that tf is also the time needed to travel the distance L in the p direction with the constant 0 and p 0 for t 0 , applying constant acceleration equations acceleration a 0 . Hence, recalling that p in the p direction we have

D

p

D 21 a0t 2 )

L

PD

D

D 21 a0tf2 )

tf


d Recalling that a s

2

D

ˇ

.as

2



D

p

2L=a0 :

(6)

a0 cos  /.L=a0 / : 34 ı , L

ˇ

(7)

D 1:2 m=s , a0 D 2:3 m=s ,  D D 4 m, we can evaluate Eq. (7) to obtain d D 1:229 m:


June 25, 2012


Dynamics 2e

313

Problem 2.236 The pendulum bob A swings about O , which is a fixed point, while bob B swings about A . Express the components of the acceleration of B relative to the Cartesian component system shown with srcin at the fixed point O in terms of L 1 , L 2 ,  ,  , and the necessary time derivatives of  and  .

Solution

Referring to the figure at the right, we use two polar component systems, one with unit vectors ur and u attached to A , the other with unit vectors uq and u attached to B . When expressed in terms of the unit vectors { and | , the unit vectors of the chosen polar component systems are:

O

O

O

O D sin  {O O D sin  {O

O

O

O

O uO D cos  {O C sin  |O; O uO D cos  {O C sin  |O :

ur



cos  | ;



(1)

uq



cos  | ;



(2)

O O

Using the . ur ; u / component system, and denoting by r and  the corresponding polar coordinates of A , the acceleration of A is

E D rR

aA



P  O C r R C 2rP P  uO :

r  2 ur

(3)



Recalling that r

D L1 D constant, Eq. (3) can be written P aEA D L1  2 uO r C L1 R uO ) aEA D L1R cos  P 2 sin   {O C L1R sin  C P 2 cos   |O; (4) where we used Eqs. (1) to express aEA in terms of {O and |O . Using the . uO q ; uO / component system, and denoting 







by q and  the polar coordinates of B relative to A , the acceleration of B relative to A is

E

aB=A

D qR



P  O C qR C 2qP P  uO

q  2 uq

:

(5)

D L2 D constant, Eq. (5) can be written aEB=A D L2 P 2 uO q C L2 R uO ) aEB=A D L2R cos  P 2 sin   {O C L2R sin  C P 2 cos   |O; (6) where we used Eqs. (2) to express a EB=A in terms of {O and |O. Relative kinematics demands that aEB D aEA C aEB=A. Therefore, using the last of Eqs. (4) and (6), we have aEB D L1 R cos  L1 P 2 sin  C L2 R cos  L2 P 2 sin  {O C L1R sin  C L1P 2 cos  C L2R sin  C L2P 2 cos   |O: Recalling that qB 










June 25, 2012


Solutions Manual

314

Problem 2.237 Revisit Example 2.24 in which the movie’s hero is traveling on train car A 18 m=s while the target B is moving at a constant with constant speed vA 40 m=s (so that aB 0). Recall that 4 s before an otherwise speed vB inevitable collision between A and B , a projectile P traveling at a speed of 300 m=s relative to A is shot toward B . Take advantage of the solution in Example 2.24 and determine the time it takes the projectile P to reach B and the projectile’s distance traveled.

D

D

D

Solution

Referring to Example 2.24 on p. 127 of the textbook and to the figure on the right, (Fig. 3 in Example 2.24), we denote by ˇ the angle A C B , so that ˇ 48:2ı . As was done in Example 2.24, we denote by d the distance .4 s/vA 72:00 m. To between A and C at the time of firing, so that d find the time taken by the projectile to hit the target, we observe that the distance traveled by the projectile in the x direction, is given by

O

D

D

dx

D

D d sin ˇ:

(1)

In Example 2.24, we had determined that the (absolute) velocity of the projectile was

E D .vA sin ˇ C vP =A cos  / {O C .vA cos ˇ vP =A sin  / |O: (2) Since vEP is constant, the time taken by the projectile to hit the target, which we denote by vP



 t , is simply equal

to d x in Eq. (1) divided by vP x , i.e.,

D vA sin ˇdCsinvPˇ=A cos  ; (3) which, recallin g that we have d D 72:00 m, ˇ D 48:2ı , vA D 18 m=s, vP =A D 300 m=s, and  D 64:40 ı t

(final result in Example 2.24), can be evaluated to obtain

D 0:3752 s: To find the distance traveled, we observe again that vEP is constant, so that the distance trav eled is .v sin ˇ C vP =A cos  /2 C .vA cos ˇ vP =A sin  /2 .d sin ˇ/ ; jEvP t j D A (4) vA sin ˇ C vP =A cos  where we have used the expressions for vEP in Eq. (2) and  t in Eq. (3). Since d D 72:00 m, ˇ D 48:2ı , vP =A D 300 m=s, and  D 64:40 ı , we can evaluate the right-han d side of Eq. (4) to obtain Distance traveled by P D 110:9 m: t

q




June 25, 2012


Dynamics 2e

315

Problem 2.238 Consider the following variation of the problem in Example 2.24 in which a movie hero needs to destroy a mobile robot B , except this time they are not going to collide at C . Assume that the hero is traveling on the train car A with 18 m=s, while the robot B travels at a constant speed constant speed vA vB 50 m=s. In addit ion, assume that at time t 0 s the train car A and the robot B are 72 and 160 m away from C , respectively. To prevent B from 0 s the hero fires a projectile P at B . If reaching its intended target, at t P can travel at a constant speed of 300 m=s relative to the gun, determine the

D

D

D

D

orientation to the the solution gun to hitˇ C has the type sinˇ thatAmust cos ˇbe given C cos   1, where  tan 1 A.

j

˙

j

D D

equation 1 .C DB . Hint: C sinAn cos  /,ofif 

Solution

We base the solution of this problem on the solution of Example 2.24 on p. 127 of the textbook. All the quantities used in this solution are defined in Example 2.24. We report here Eq. (14) from the Example, which remains valid under the conditions stated in this problem and which determines the value of the angle  that we want to determine:

.`  d cos ˇ/ cos  Dividing both sides of this equation by sin 

d 



D vsinP =Aˇ .vB d

d sin ˇ sin 

 vA `/ :

(1)

sin ˇ , we have

`  d cos ˇ cos  d sin ˇ

D vAv`P =AvdB d : 

(2)

The above equation is a transcendental equatio n in  whose solution can be obtained using the following technique. We consider the term multiplying the cos  on the left-hand side of the equation and we define an angle  such that `  d cos ˇ `  d cos ˇ : tan   tan 1 (3) d sin ˇ d sin ˇ

D

Then, recalling that tan  sin 



sin  cos 

cos 

)



D



D sin  = cos  , we can rewrite Eq. (2) as

D vAv`P =AvdB d ) 

sin  cos 



sin  cos 

)

B d / cos  D .vA` vPv=A d 

sin.

D



/

where we have used the trigonometric identity sin  cos   sin  cos  sin . now be solved for  to obtain .vA `  vB d / cos    sin1 :

D C



vP =Ad

B d / cos  ; D .vA` vPv=A d 

(4)

  /. The last of Eq. (4) can



(5)

Recalling that vA 160:0 m, d obtain

`

D

18 m=s, vB 50 m=s, vP =A 300 m=s, and where, using the results in Example 2.24, DD72:00 m, we Dcan evaluate  inDin the last of Eqs. (3) and then evaluate  in Eq. (5) to  D 63:57ı :


June 25, 2012


Solutions Manual

316

Problem 2.239 Consider the following variation of the problem in Example 2.24 in which a movie hero needs to destroy a mobile robot B . As was done in that problem, assume that the movie hero is traveling on the train car A with constant speed vA 18 m=s and that, 4 s before an otherwise inevitable collision at C , the hero fires a projectile P traveling at 300 m=s relative to A . Unlike Example 2.24, 10 m=s2 assume here that the robot B travels with a constant acceleration aB 20 m=s, where t 0 is the time of firing. Determine the and that v B .0/ orientation  of the gun fired by the hero so that B can be destroyed before the

D

D

D

D

collision at C .

Solution

The general strategy for the solution of moving target problems has been discussed in the Road Map of 0 be the time at which the projectile is fired, Example 2.24 on p. 127. According to this strategy, letting t there is a time tI > 0 such that rP =B .tI / rP .tI /  rB .tI / 0; (1)

D

E

DE

DE

E

that is, there is a time tI at which the projectil e and the target meet. To solve this problem we need to find the positions of the projectile and of the target as functions of time and then set them equal to each other as required by the above equation. Because A is moves at a constant speed along a straight line, then constant. Once P is fired, its velocity is also a constant given by

vP

E D vEA C vP =A uO P =A.0/;

E

vA is

(2)

where vP =A in known. Hence,

E

rP .t/

D rEP .0/ C vEA.0/ C vP =A uO P =A.0/ t:

(3)

D rEB .0/ C vEB .0/t C 21 aEB t 2:

(4)

Because B has a constant acceleration, we can use constant acceleration equations to write

E

rB .t /

Substituting Eqs. (3) and (4) into Eq. (1), we have

E

rP .0/

C vEA.0/ C vP =A uO P =A.0/ tI rEB .0/ vEB .0/tI 



E D 0E;

2 1  2 aB tI

(5)

Referring to the figure at the right (similar to Fig. 3 in Example 2.24 and showing the geometry at the time of firing) we have that rA .0/ rP .0/, so that Eq. (5) can be rewritten as

E

DE rEA=B .0/ C Œ vEA=B .0/ C vP =A uO P =A.0/çtI

 1 2 aB tI2

E D 0E:

(6)

The problem is solved when we are able to express all of the terms in Eq. (6) via known quantities and the only two unknowns of the problem, which are tI and the firing angle  . We therefore proceed to determine convenient expressions for each of the vectors in Eq. (6). This solution s manual, in any print or electronic form, remain s the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplicat ion or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

June 25, 2012


Dynamics 2e

317

D 0 in this problem is the same as that in the figure shown, we have rEA=B .0/ D d sin ˇ {O C .` d cos ˇ/ |O : Again referring to the geometry at t D 0 , and recalling that ˇ is known ( ˇ D 48:2ı ), we have vEA .0/ D vA uO C=A D vA .sin ˇ {O C cos ˇ |O / and vEB .0/ D vB .0/ |O ; Since the position at t



(7)



(8)

so that

E

vA=B .0/ Since t

D vA sin ˇ {O C .vA cos ˇ

O

 vB .0// | :

(9)

0 is the time of firing, we must have

D

O

uP =A.0/

D cos  {O



O

sin  | :

(10)

Finally, the problem statement and the geometry of the problem tell us that

E D aB |O:

aB

(11)

Substituting Eqs. (7) and Eqs. (9)–(11) into Eq. (6) and expressing the result on a component by component basis, we have d

sin ˇ

and

`  d cos ˇ

C ŒvA cos ˇ

C .vA sin ˇ C vP =A cos  /tI D 0

2 1  vB .0/  vP =A sin çtI  2 aB tI

D 0;

(12) (13)

18 m=s, vB .0/ 20 m=s, vP =A 300 m=s, aB 10 m=s2 , ˇ 48:2ı , and where, using the where vA results in Example 2.24, d 72:00 m and ` 160:0 m. Equations(12) and (13) form a system of two equations in the two unknowns  and tI that can be solved numerically with appropriate mathematical software. We have used Mathematica with the following code:

D

D D

D D

D

D

Notice that we have provided the root finding algorithm an initial guess for the solution consisting of the values  64:40 ı and tI 2 s, the first of which is the solution to the case discussed in Example 2.24 and the second is between the initial time and the time of collision. The above code yields the following solution:

D

D



D 65:88ı

and

tI

D 0:3947 s:


June 25, 2012



Dynamics 2e

319

Problem 2.241 The object in the figure is called a gun tackle, and it used to be very common on sailboats to help in the operation of front-loaded guns. If the end at A is pulled down at a speed of 1:5 m=s, determine the velocity of B . Neglect the fact that some portions of the rope are not vertically aligned.

Solution

By neglecting vertical misalignments, and using the y axis shown at the right, we can expresss the length of the cord as follows:

L

D yA C 2yB :

(1)

Since L is constant, taking its time derivative, we have

D yPA C 2yPB : Solving for yP B , and recalling that vEB D yPB |O, we have 0

vB

(2)

1  2 yA | :

(3)

E D P O Recalling that yPA D 1:5 m=s, we can evaluate the above expression to obtain

E D

vB

O m=s:

0:7500 |


June 25, 2012


Solutions Manual

320

Problem 2.242 The gun tackle shown is operated with the help of a horse. If the horse moves to the right at a constant speed of 7 ft=s, determine the velocity and acceleration of B when the horizontal distance from B to A is 15 ft. Except for the part of the rope between C and A, neglect the fact that some portions are not vertically aligned. Also neglect the change in the amount of rope wrapped around pulle y C as the horse moves to the right.

Solution

Using the coordinate system shown in the figure at the right, and denoting the length of the rope by L , we have

L

D 2yB C

q

xA2

C h2 :

(1)

Since L is constant, differentiating Eq. (1) with respect to time gives

0

P

D 2yPB C

Solving Eq. (2) for yB , we have

P : C h2

xA xA

q

xA2

(2)

P : (3) 2 C h2 Since vEB D yPB |O , recalling that h D 8 ft, xPA D 7 ft=s, and xA D 15 ft, we can evaluate Eq. (3) to obtain P D

yB

E D

vB

xA xA



q

xA2

O ft=s:

3:088 |

(4)

R D 0, we

To obtain the acceleration of B we differentiate Eq. (3) with respect to time and, observing that xA obtain

R D

yB

h2 xA2



2

P

x2

3=2

h2

:

(5)

E D yRB |O, again recalling that we have h D 8 ft, CxPA D 7 ft=s, and xA D 15 ft, we can evaluate Eq. (5)



Since aB to obtain

E D

aB

A



2

O ft=s :

0:3192 |


June 25, 2012


Dynamics 2e

321

Problem 2.243 The figure shows an inverted gun tackle with snatch block, which used to be common on sailboats. If the end at A is pulled at a speed of 1:5 m=s, determine the velocity of B . Neglect the fact that some portions of the rope are not vertically aligned.

Solution

Neglecting vertical misalignments, and using the Cartesian coordinate system shown at the right, we can express the length of the cord as follows: L yA 3yB : (1)

D C

Since L is constant, taking its derivative with respect to time, we have

0

D yPA C 3yPB :

(2)

E D yPB |O , solving Eq. (2) for yPB , we have

Recalling that vB

vB

1  3 yA | :

(3)

E D P O Recalling that yPA D 1:5 m=s, we can evaluate vEB to obtain E D

vB

O m=s:

0:5000 |


June 25, 2012


Solutions Manual

322

Problem 2.244 whip-upon-whip In maritime speak, the system in the figure is often called a purchase and is used for controlling certain types of sails on small cutters (by attaching point B to the sail to be unfurled). If the end of the rope at A is pulled with a speed of 4 m=s, determine the velocity of B . Neglect the fact tha t the segment of the rope between C and D is not vertically aligned, and assume that the slope of segment AC is constant.

Solution

Neglecting the vertical misalignment of the segment CD , assuming that the slope of the segment AC is constant, and using the coordinate system shown at the right, we can write the lengths of the two rope in the system as follows:

L1

D 2yB



yC

and

L2

D y C C sA:

(1)

Since L 1 and L 2 are constant, taking their time derivative we have (2) D yPC C sPA: Recalling that sPA D vA is known and eliminating yP C from Eqs. (2), we can solve for yB to obtain P yPB D 21 vA ) vEB D 21 vA |O : (3) Recalling that vA D 4 m=s, we have

0

D 2yPB yP C 

and

0





E D

vB

O m =s :

2:000 |


June 25, 2012


Dynamics 2e

323

Problem 2.245 The pulley system shown is used to store a bicycle in a garage. If the bicycle is hoisted by a winch that winds the rope at a rate v0 5 in:=s, determine the vertical speed of the bicycle.

D

Solution

Referring to the figure at the right, we have that pulleys B and C are at the same height. Also, A is a point on the branch of the rope that is being pulled in by the winch. The length of the rope between the left end of the system and point A can be written as follows:

L

D 4yB C sA:

(1)

Since L is constant, taking its time derivative with respect to time, and recalling that sA  v0 , we have

P 0 D 4 yPB C sPA ) yPB D

1  4 v0 :

(2)

5

Recalling that we have v 0 have

D 5 in: D 12 ft=s, and observing that the speed of the bicycle is equal to jyPB j, we v D 0:1042 ft=s: bicycle


June 25, 2012




Solutions Manual

326

Problem 2.248

3:7 m=s2 . Block A is released from rest and starts sliding down the incline with an acceleration a 0 Determine the acceleration of block B relative to the incline. Also, determine the time needed for B to move a distance d 0:2 m relative to A.

D

D

Solution

Expressing the length of the rope L in terms of the coordinates of A and B , we have L 3xA xB : (1)

D

C

Since L is constant, differentiating Eq. (1) twice with respect to time gives 0 3 xA xB : (2)

D R CR

Solving Eq. (2) with respect to xB , gives

R D 3RxRA D 3a0; (3) where we have used the fact that xRA D a0 . Recalling that a 0 D 3:7 m=s and that aEB D xRB {O, evaluating Eq. (3), we have xB





2

2

E D 11:10 {O m=s : Using the result in Eq. (3), we observe that xR B=A D xRB xRA D aB



@

D

 4a 0 constant. Therefore, xB=A, namely, the relative position of B with respect to A , can be expressed as a function of time as follows:

xB=A.t/

P

D xB=A.0/ C xP B=A.0/t



2a 0 t 2 ;

(4)

P

where xB=A.0/ and xB=A.0/ denote the initial values of xB=A and xB=A, respectively. Recalling that xB=A.0/ 0 since the blocks are released from rest, Eq. (4) simplifies to

P

D

xB=A.t /

D xB=A.0/



2a 0 t 2 :

Denoting by t d the time needed to travel the distance d , observing that d letting t t d in Eq. (5), we have

D Recalling that d

d

D 2a0 td2 ) 2

td

D

p

d=.2a 0 /:

(5)

D

xB=A.td /  xB=A.0/ , and

ˇ

D 0:2 m and a0 D 3:7 m=s , the last of Eqs. (6) can be evaluated to obtain td D 0:1644 s:


ˇ

(6)

June 25, 2012


Dynamics 2e

327

Problem 2.249 In the pulley system shown, the segment AD and the motion of A are not impeded by the load G . Assume all ropes are vertical ly aligned. 3 ft=s and Determine the velocity and acceleration of the load G if v0 a0 1 ft=s2 .

D

D

Solution

Referring to the figure at the right, and using the y coordinate shown, the length of the ropes labeled 1 , 2 , and 3 are

L1

D yG C 3yB ;

L2

D y G C yD



2yB ;

L3

D y G C yA



2yD :

(1)

Since L 1 , L 2 , and L 3 are constant, taking the time derivative of Eqs. (1) gives

0

yG

3yB ;

0

DP C P

yG

yD



DP CP

2yB ;

0

P

yG

yA  2yD :

D P CP

P

(2)

P P

Equations (2) forms a system of three equations in the three unknowns yB , yD , and yG whose solution is

P

P D 131 v0; yPD D 135 v0; yPG D 133 v0; (3) where we have used the fact that yPA D v 0 . Differentiating Eqs. (3) with respect to time, we have 1 5 3 yRB D 13 a0 ; yRD D 13 a0 ; yRG D 13 a0 ; (4) EG D yRG |O , and recalling that where we have used the fact that vP 0 D a 0 . Recalling that vEG D yPG |O and a v0 D 3 ft=s and a0 D 1 ft=s , we can evaluate the last of Eqs. (3) and (4) to obtain yB





2

E D

vG

O ft=s

0:6923 |

and

E D

aG

2

O ft=s :

0:2308 |


June 25, 2012


Solutions Manual

328

Problem 2.250 In the pulley system shown, the segment AD and the motion of A are not impeded by the load G . Assume all ropes are vertical ly aligned. The load G is initially at rest when the end A of the rope is pulled with the constant acceleration a0 . Determine a 0 so that G is lifted 2 ft in 4:3 s.

Solution

Referring to the figure at the right, and using the y coordinate shown, the lengths of the ropes labeled 1 , 2 , and 3 are

L1

D yG C 3yB ;

L2

D y G C yD



2yB ;

L3

D y G C yA



2yD :

(1)

Since L 1 , L 2 , and L 3 are constant, taking the time derivative of Eqs. (1) gives

0

D yPG C 3yPB ;

0

D yPG C yPD



P

2yB ;

0

D yPG C yPA



P

2yD :

Equations (2) form a system of three equations in the three unknowns and yG whose solution is

P

(2)

yB , yD ,

P P

(3) P D 131 v0; yPD D 135 v0; yPG D 133 v0; where we have used the fact that yPA D v 0 . Differentiating Eqs. (3) with respect to time, we have 1 5 3 yRB D 13 a0 ; yRD D 13 a0 ; yRG D 13 a0 ; (4) where we have used the fact that vP 0 D a 0 . The last of Eqs. (4) implies that the acceleration of G is constant.

yB





Therefore, using constant acceleration equations, we can write the following express ion of y G as a function of time: 3 yG .t / y G .0/ yG .0/t  26 a0 t 2 : (5)

D

D

CP

P

4 :3 s (f stands for final). Hence, recalling that G starts from rest and that therefore yG .0/ Let tf from Eq. (5) we have yG .tf /

D y G .0/

3 2  26 a0 tf

)

a0

D 0,

D 26ŒyG .0/3t 2 yG .tf /ç : 

(6)

f

Recalling that y G .0/  yG .tf /

D 2 ft and that tf D 4:3 s, we can evaluate the last of Eqs. (6) to obtain a0 D 0:9374 ft=s : 2


June 25, 2012


Dynamics 2e

329

Problem 2.251 A crate A is being pulled up an inclined ramp by a winch retracting the cord at a constant rate v 0 Letting h 1:5 ft, determine the speed of the crate when d 4 ft.

D

D

D 2 ft=s.

Solution

Referring to the problem’s figure, we begin by observing that the motion of the crate is rectilinear and, denoting the speed of the crate by vA , we have

ˇˇ p C C

vA

P

To determine d , we can write

L

Dw

D dP : h2

(1)

d 2;

(2)

where L denotes the length of the cord and w denotes the (constant) horizontal distance between the winch and the pulley. Recalling that L decreases at the rate v 0 , we have that

P D ph2d d d 2 : Solving Eq. (3) for dP and substituting the result in Eq. (1)C gives v 0

vA Therefore, recalling that h

D vd0

p

h2

(3)

C d 2:

(4)

D 1:5 ft and v0 D 2 ft=s, we can evaluate Eq. (4) for d D 4 ft to obtain vA D 2:136 ft=s:


June 25, 2012


Solutions Manual

330

Problem 2.252 A crate A is being pulled up an inclined ramp by a winch. The rate of winding of the cord is control led so as to hoist the crate up the incline with a constant speed v 0 . Letting ` denote the length of cord retracted by the winch per unit time, determine an expressi on for ` in terms of v 0 , h , and d .

P

P

Solution

Letting L denote the length of the cord and w denote the (constant) horizontal distance between the winch and the pulley, we can write

p

D w C h2 C d 2: (1) P D `P, differentiating Eq. (1) with respect to time, we have Recalling that L decreases at the rate `P, i.e., L d dP `P D p : (2) 2 h C d2 where dP is the time rate of change of the distance d . When the crate move s up the incline, dP < 0 and is equal in magnitude to the speed of the crate, i.e., dP D v0 . Substituting this relation into Eq. (2), we obtain the L







desired relation:

0 P D p dv : h2 C d 2

`


June 25, 2012



Solutions Manual

332

Problem 2.254

E

E

Let !BC denote the angular velocity of the relative position vector rC =B . As such, !BC is also the angular velocity of the connecting rod B C . Using the concept of

E

time derivative of a vector given in Section 2.4 on p. 80, determine the component of the relative velocity of C with respect to B along the direction of the connecting rod B C .

Solution

E D rEPC=B , where rEC=B D L uO C=B . Since L is P constant, we must have rEC =B D L uOP C =B . Using the formula for the time derivative PO C=B D !EBC uO C=B , so that of a unit vector, we have u vEC =B D L !EBC uO C=B : We now observe that ! EBC uO C=B is perpendicular to uO C=B . Therefore the component of vEC=B along the direction u O C=B is We being by observing that vC=B







E

Component of vC=B along B C

D 0:


June 25, 2012


Dynamics 2e

333

Problem 2.255 The piston head at C is constrained to move along the y axis. Let the crank AB be 2000 rpm, R 3:5 in:, rotating counterclockwise at a constant angular speed  5:3 in: Determine expressions for the velocity and acceleration of C as a and L function of  and the given parameters.

PD

D

D

Solution

Using the diagram at the right and the law of cosines, we have

L2

D R 2 C yC2

p

) yC D R cos  ˙ L2 R2 C R2 cos2  ) yC D R cos  ˙ L2 R2 sin2  ; (1) where we have used the identity sin2  D 1 cos2  . To determine the appropriate root, observe that for  D 0 we expect y C D R C L. Therefore, we have yC D R cos  C L2 R2 sin2  : (2) 

2RyC cos 



p



p





Differentiating Eq. (2) with respect to time and simplifying yields

yC

P D

E D P O

P

R2  cos 

P

R

 sin 

Cp D



L2 

yC | , and recalling that R 3:5 in: Recalling that vC  2000 rpm 2000 260 rad=s, we can evaluate Eq. (3) to obtain

D

E D

vC

 sin 



61:09

p ! "

2



(  cos 

1

C

E DR O D

R cos 

p

L2  R 2 sin2 

C sin 

2

L

E D

1:279  10



R 2 sin2 

D

D

PD

ft, and 

O

| ft =s:

3=2





3:5 12

5:3 12

ft, L

R sin 

p

L2  R 2 sin2  5:3 12

D 5:3 in: D

#)

:

(4) ft, and

0:2917 cos 

4

aC

D 5:3 in: D

R3 cos2  sin 

yC | , and recalling again that R 3:5 in: Recalling that aC  2000 rpm 2000 260  rad=s, we can evaluate Eq. (4) to obtain

PD

(3)

L

0:1951  0:08507 sin 

P D constant, differentiating Eq. (3), we have

yRC D R P 2

D

:

17:82 cos 

C

Recalling that 

2 R 2 sin  3:5 12 ft,

1

( "  Cp cos 

C sin 

! p

0:1951  0:08507 sin2  cos2  sin 

0:02481 0:1951 0:08507 

sin2 



3=2



0:2917 sin 

0:1951  0:08507 sin2 


#)

| ft =s2 :

O

June 25, 2012



Dynamics 2e

335

Problem 2.257 In the cutting of sheet metal, the robotic arm OA needs to move the cutting tool at C counterclockwise at a constant speed v0 along a circular path of radius  . The center of the circle is located in the position shown relative to the base of the robotic arm at O . For all position s along the circular cut (i.e., for any value of  ), determine r and  as functions of the given quantities (i.e., d , h,  , v 0 ). These quantities can be found by hand, but it is very tedious, so you might consider using symbolic algebra software,

R

R

such as Mathematica or Maple.

Solution

Referring to the figure at the right, the coordinates C are

D d C  cos  D r cos  and yC D h C  sin  D r sin  : (1) 2 C y 2 , we have Observing that r 2 D x C C r D .d C  cos  /2 C .h C  sin  /2 : (2) xC

q

Differentiating Eqs. (1) with respect to time, we have

r cos  r  sin 

  sin 

and

 cos 

r sin  r  cos  :

P

DP P P D P C P (3) Equations (3) can be viewed as a system of two equations in the two unknowns P and rP whose solution is P D . P =r/. cos  cos  sin  sin  /; (4) rP D  P .sin  cos  cos  sin  /: 



Next, observe that we have sin 

D h C r sin  ;

cos 

D d C r cos  ;

and

P D v 0 =:



(5)

Hence, substituting Eqs. (2) and (5) into Eqs. (4) and simplifying , we have

v0 . C d cos  C h sin  / v0 .h cos  d sin  / (6) C  cos  /2 C .h C  sin  /2 and rP D .d C  cos  /2 C .h C  sin  /2 : To find expressions for rR and R we must take the time derivative of Eqs. (6) and then replace P with its

PD





p

.d

expressions in the last of Eqs. (5). Doing so, after simplification, yields the following expres sions:

C d cos  C h sin  /.d 2 C h2 C  d cos3=2C h sin  / ;  .d C  cos  /2 C .h C  sin  /2 2 2 v0 .d C h2 2 /.d sin  h cos  / R D  2 :  .d C  cos  /2 C .h C  sin  /2 RD

r



v02 .








June 25, 2012


Solutions Manual

336

Problem 2.258

D

At the instant shown, block A is moving at a constant speed v 0 3 m=s to the 2:3 m. Using h 2:7 m, determine how much time is needed to left and w lower B 0:75 m from this position.

D

D

Solution

Referring to the figure at the right, we will use the Cartesian coordinate system shown with srcin at the fixed pulley O . The length of the rope can be descri bed as

L

D xA C yB C

q

xA2

C h2 :

(1)

We will denote quantities evaluated at the initial and final positions of the system by the subscripts 1 and 2 , respectively. Because the length of the rope is constant, we set the expressions of the length of the rope corresponding to the initial and final positions of the system equal to each other. Initially, we have xA1 w . In addition, we have yB2  yB1 d 0:75 m. Hence, referring to Eq. (1), we have

D

w

D D

C yB1 C

p

w2

q C C p p  C C C

C h2 D xA2 C yB2 C xA2

)  ) xA22 C d



d

w



2 xA2

h2

w

w2

w2

h2

2

h2

2xA2 d

This last equation can be solved for xA2 to obtain



h2  d xA2

D



2 d





x2



 D

q Cp

h2

A2



w

w2



C h2 D xA22 C h2:

p w 2 C h 2 2 pw 2 C h 2  :

(2)

w

w

(3)

The expression above gives the position of A corresponding to the final position achieved by B as given by the problem statement . Because A and B are connected by an inextensible rope that is assumed not to go slack, the time taken by B to achieve its final position will be equal to the time taken by A to achieve its final position. With this in mind, letting t denote the time to be determined, and recallin g that the velocity of A is constant, we then have

h2  d

w xA2

D xA1

)

t

w

pw 2 C h 2

2

D v0 2v0d w pw2 C h2 ) t D 0:1556 s, where we have used the fact that xA1 D w , and where we have used the following numerical data: v 0 D 3 m=s, h D 2:7 m, d D 0:75 m, and w D 2:3 m. 

v0 t










June 25, 2012


Dynamics 2e

337

Problem 2.259

D

D D

At the instant shown, h 10 ft, w 8 ft, and block B is moving with a speed v0 5 ft=s and an acceleration a 0 1 ft=s2 , both downward. Determine the velocity and acceleration of block A .

D

Solution

We will use the Cartesian coordinate system shown at the right with srcin at the fixed pulley O . The length of the rope can be described as

L

D xA C yB C

q

xA2

C h2 :

(1)

Since the length of the rope is constant, differentiating the above equation with respect to time, we have

q q

yPB xA2 C h2 P 0 D xPA C yPB C xPA D : ) (2) xA2 C h2 xA C xA2 C h2 Since v EA D xPA {O, recalling that yPB D v0 D 5 ft=s and h D 10 ft, we can evaluate the last of Eqs. (2) for x D w D 8 ft to obtain xA xA



q

E D

vA

O ft=s:

3:078 {

Taking the derivative with respect to time of the last of Eqs. (2), after simplification, we have

R D

xA

E DR O

R

qC q C C

yB 

xA

xA2

h2

xA2

h

C 2

P

qC q  C C

x

A

xA2

xA2

h2

h2

2

:

(3)

2

P D D 5 ft=s, yRB D a 0 D 1 ft=s , and h D 10 ft, we can evaluate the

xA { , recalling that yB v0 Since aA w 8 ft to obtain last of Eq. (3) for x

D D



2 x  yB A

E D

aA

2 0:8931 { ft =s :

O


June 25, 2012


Solutions Manual

338

Problem 2.260 As a part of a robotics competition, a robotic arm with a rigid open hand at C is to be designed so that the hand catches an 0 egg without breaking it. The egg is released from rest at t from point A . The arm, initial ly at rest in the position show n, starts moving when the egg is released . The hand must catch the egg without any impact with the egg. This can be done by specifying that the hand and the egg must be at the same position at the same time with identica l velocities. A student proposes

D

R

to this itusing a constant which fair bits  forthe of do work) is found that thevalue arm of catches egg (after at t a0:4391 for  13:27 rad=s2 . Using these values of t and  , determine the acceleration of both the hand and the egg at the time of catch. Then, explain whether or not using a constant value of  , as has been proposed, is an acceptable strategy.

RD

R

D

R

Solution

Referring to the diagram at the right, we will use a Cartesian coordinate system with its srcin at O . As far as the acceleration of the egg is concerned, up to the time of catch the acceleration of the egg is

R D

ye

2

9:81 m=s

;

(1)

where the subscript e stands for ‘egg’. As far as the determination of the acceleration of point C is concerned, we begin by observing 0 because the arm that since C is constant, and since C .0/ starts from rest, we can use the constant accelerati on relations to provide the expression of the angular coordinate  C as a function of time: C .t/  C .0/ 21 C t 2 :

R

P

D

D

C R

(2)

Since the trajectory of C is a straight vertical line,  C .t/ is related to y C .t/ as follows:

yC .t/

D d tanŒC .t /ç:

(3)

Differentiating Eq. (3) twice with respect to time, we have

D d RC sec2 C .t/ C 2d PC2 .t / sec2 C .t / tan C .t/ ) yRC .t / D d sec 2 C .t/RC C 2PC2 .t / tan C .t/: (4) To evaluate the expression above, in addition to RC (which is given), we need the values of  C and PC at the R

yC .t/

time of catch. To determine these values, we begin by denoting by tf the time of catch, where the subscript f stands for ‘final.’ Since the egg is falling under the action of gravity, using constant acceleration equations, for t tf , the y coordinate of the egg is given by

D

ye .tf /

D y e .0/

1 2  2 gtf ;


(5) June 25, 2012


Dynamics 2e

339

D

D

where y e .0/ 0:6 m is the initial vertical position of the egg. At the time of catch we must have y e .tf / yC .tf /. Therefore, setting t tf in Eq. (3), replacing yC .tf / with the right-hand side of Eq. (5), and solving for  C .tf /, we have

D

C .tf /

D tan1

"

2ye .0/  gtf2 2d

#

:

(6)

P

Next, we determine an expression for C .tf /. To do so, we different iate Eq. (2) and evaluate it for t

P

C .tf /

D tf :

D RC tf :

(7)

Substituting Eq. (7) into the last of Eqs. (4), at the time of catch, we have

D d sec 2 C .tf /RC C 2RC2 tf2 tan C .tf /: (8) Recalling that tf D 0:4391 s, d D 0:5 m, g D 9:81 m=s , and y e .0/ D 0:6 m, we can evaluate  C .tf / in Eq. (6) (this gives  C .tf / D 34:66 ı ) and then use it, along with RC D 13:27 rad=s , in Eq. (8) to obtain R

yC .tf /

2





R

yC

ˇ

t D0:4391 s

D

2

44:51 m=s

2

:

jR j jR j

Since, at the time of catch, the acceleration yC > ye , the arm and egg will only be in contact for an instant and will then separate again. Consequently, the proposed strategy is not acceptable for catching the egg.


June 25, 2012


Solutions Manual

340

Problem 2.261 Referring to the problem of a robot arm catc hing an egg (Prob. 2.260), the strategy is that the arm and the egg must have the same velocity and the same position at the same time for the arm to gently catc h the egg. In addition, what should be true about the accelerations of the arm and the egg for the catch to be successful after they rendezvous with the same velocity at the same position and time? Describe what happ ens if the accelerations of the arm and egg do not match.

Solution

R

R

After they rendez vous, the relati ve acceleration of the arm with respect to the egg must be zero. If ye < yC then the arm and egg will separate right after the catch . If ye > yC the egg will experience a jerk.

R

R


June 25, 2012






Dynamics 2e

345

Problem 2.266

D

30 m=s and a A glider is descending with a constant speed v 0 constant descent rate of 1 m=s along a helical path with a constant 400 m. Determine the time the glider takes to complete radius R a full 360 ı turn about the axis of the helix (the ´ axis).

D

Solution

Referring to the figure at the right, we adopt a cylindrical coordinate system with srcin on the ground and the ´ axis coinciding with the axis of the helical path. Using the polar component system corresponding to the chosen coordinate system, we can write the velocity of the airplane as follows:

E D RP uO R C RP uO C ´P uO ´:

v

(1)



PD ED PO CPO

0 and Since the radius of the helix is constant, we have that R Eq. (1) simplifies to v R  u ´ u´ :

(2)

Since the speed of the airplane is constant and equal v 0 , using Eq. (2) we can write

v02

P

D R 2P 2 C ´P 2;

P

which we can solve for  since ´ is known and equal to

PD˙



q

(3)

1 m=s:

v02  ´2 R

P

:

(4)

Of the two roots in Eq. (4) we will select the positive one to match the direction of motion of the glider indicated in the problem’s figure. We now observe that since v 0 , ´, and R are constant,  is also a constant. Therefore, letting t denote the time needed to complete a full 360ı arc around the helix, t can be computed by simply dividing the measure of the angle 360 ı in radians, i.e., 2  , by the angular veloci ty  . This gives

P

P

P

t

Recalling that R

400 m, v 0

D

D

30 m=s, and ´

D

PD t

2 R

q

:

(5)

v02  ´2

P

1 m=s, we

can evaluate Eq. (5) to obtain

D 83:82 s:


June 25, 2012


Solutions Manual

346

Problem 2.267

D

An airplane is flying horizontally at a constant speed v0 320 mph while 1500 rpm. If the its propellers rotate at a constant angular speed ! 14 ft, determine the magnitude of the propellers have a diameter d acceleration of a point on the periphery of the propeller blades.

D

D

Solution

Referring to the figure at the right, we define a fixed cylindrical coordinate system with the ´ axis coinciding with the propeller’s shaft, R direction perpendicular to the shaft and going from the ´ axis to the point whose acceleration we want to measure, and such that the triad . uR ; u ; u´ / is right-handed . Next, we recall that in cylindrical coordinates the acceleration is given by the following formula:

O O O

E D RR

a



P  O C RR C 2RP P  uO C ´R uO ´:

R  2 uR

(1)



Using the problem’s given information, we have

R

D 21 d D 7:000 ft; RP D 0; RR D 0; P D ! D 1500 rpm D 157:1 rad=s; ´R D 0:

(2)

Substituting the above information in the formula for the acceleration we have

ED

a

P 2 uO R D .

R 

1:728  10

5

ft=s2 / uR :

O

(3)

The magnitude of the above vector is 2

a

ˇ Eˇ D

17;280 ft=s :


June 25, 2012





Solutions Manual

350

Problem 2.271 A golfer chips the ball on a flat, level part of a golf course as shown. Letting ˛ 23 ı , ˇ 41 ı , and the initial speed be v 0 6 m=s, determine the x and y coordinates of the place where the ball will land.

D

D

D

Solution

D 0, ay D 0, and a´ D g. Hence x D x 0 C xP 0 t; y D y 0 C yP0 t; and ´ D ´ 0 C ´P 0 t 12 gt 2 : At time t D 0 the ball is at the srcin of our coordinate syst em so x0 D y 0 D ´ 0 D 0 . The initial compone nts For projectile motion ax





of the velocity are

P D v 0 cos ˇ cos ˛; yP0 D v0 cos ˇ sin ˛;

x0

and

P D v 0 sin ˇ:

´0

Using the equations written so far, we have that the motion of the ball is described by the following equations:

x

D .v0 cos ˇ cos ˛/t;

y

D .v 0 cos ˇ sin ˛/t;

´

and

D .v0 sin ˇ/t

1 2  2 gt :

To determine the location of landing, we observe that the ´ coordinate of the landing spot must be ´ Next we find the time corresponding to ´ 0 , i.e.,

D

´

D .v0 sin ˇ/t

1 2  2 gt

D0 )

t

D 0.

D 2v0 gsin ˇ :

The x and y components of the position corresponding to this time are

 2v ˇ   2v g ˇ 

xland

D .v 0 cos ˇ cos ˛/

yland

D .v 0 cos ˇ sin ˛/

0 sin

0 sin

g

2

)

xland

cos ˛ D v0 sin 2ˇ D 3:345 m; g

)

yland

sin ˛ D v0 sin 2ˇ D 1:420 m; g

2

6 m=s, ˛ 23 ı , ˇ 41 ı , and g where we have used the following numerical data: v 0 summary, the location of the ball’s landing spot is identified by the following coordinates

D

xland

D 3:345 m;

yland

D 1:420 m;

D

and

D

´land

2

D 9:81 m=s . In

D 0:


June 25, 2012


Dynamics 2e

351

Problem 2.272 Relative to the cylindrical coordinate system shown, with srcin at

D C D

O , the radial and ´ coordinates of point G are R d .L=2/ cos ˇ 0:5 m and L and ´ .L=2/ sin ˇ , respectively, where d 0:6 m. The shaft CD rotates as shown with a constant angular velocity !s 10 rad=s, and the angle ˇ varies with time as follows: ˇ ˇ0 sin.2!t/ , where ˇ0 0:3 rad, ! 2 rad=s, and t is time in 3s seconds. Determine the velocity and the acceleration of G for t (express the result in the cylindrical component system . uR ; u ; u´ /,

D

D

D

D

D O O O



with u

D D

O D uO ´ uO R ).

Solution

The problem gives the radial and vertical coordinates of point G , which are

R

D d C 12 L cos ˇ

´

and

D

1  2 L sin ˇ:

(1)

D

ˇ 0 sin.2!t/ . Substituting the expression for ˇ In addition, the angle ˇ is given as a function of time: ˇ into Eqs. (1) we have the R and ´ coordinates of G directly as a function of time: R

D d C 12 L cosŒˇ0 sin.2!t/ ç

´

and

D

1  2 L sinŒˇ0 sin .2!t/ ç:

(2)

Now we recall that, in cylindrical coordinates, the velocity is given by

v

R uR

R  u

´ u´ :

(3)

ED P O C PO CP O The quantities RP and ´P can be computed by differentiating Eqs. (2) with respect to time. This process, while possibly tedious, is straightforward and gives

PD ´P D

R

Lˇ0 !

sinŒˇ0 sin.2!t/ ç cos.2!t/;

(4)

Lˇ0 !

cosŒˇ0 sin.2!t/ ç cos.2!t/:

(5)

P

As far as  is concerned, we observe that the whole system rotate s in the  direction with a constant angular velocity ! s :  !s constant: (6)

PD D Recalling that d D 0:5 m, L D 0:6 m, ! s D 10 rad=s, ˇ 0 D 0:3 rad, and ! D 2 rad=s, for t D 3 s, we can evaluate the first of Eqs. (2) as well as Eqs. (4)–(6) to obtain

R

D 0:7961 m; RP D 0:04869 m=s; ´P D

0:2999 m=s;

and

Using the results in Eqs. (7) we can evaluate Eq. (3) to obtain the velocity at

E D .0:04869 uO R C 7:961 uO

v

 

P D 10:00 rad=s: t D 3 s: 

(7)

O

0:2999 u´ / m=s:

In cylindrical coordinates, the acceleration is given by

E D RR

a



P  O C RR C 2RP P  uO C ´R uO ´:

R  2 ur

(8)




June 25, 2012


Solutions Manual

352

P P

R R

R

We have already computed the terms R , R ,  . Hence, we only need to determine the terms R ,  , and ´ to compute the acceleration of point G . Differentiating Eqs. (4)–(6) with respect to time we have

˚ ˚

P D 2Lˇ0! 2 sin.2!t/ sinŒˇ0 sin.2!t/ ç ˇ0 cosŒˇ0 sin.2!t/ ç cos2.2!t/ ; (9) R D 0 (10)  ´P D 2Lˇ0 ! 2 sin.2!t/ cosŒˇ0 sin.2!t/ ç C ˇ0 sinŒˇ0 sin.2!t/ ç cos2 .2!t/ ; (11) Again recalling that d D 0:5 m, L D 0:6 m, ! s D 10 rad=s, ˇ 0 D 0:3 rad, and ! D 2 rad=s, for t D 3 s, we can evaluate Eqs. (9)–(11) for t D 3 s wo obtain R



2

R

RD

0:1798 m=s

2

;

R D 0;



and

´

RD

0:8120 m=s

Hence, using Eqs. (7) and (12), we can evalua te the expression in Eq. (8) for t

E D.

a

O C 0:9738 uO

79:79 uR

 

:

(12)

D 3 s to obtain

0:8120 u´ / m=s2 :

O


June 25, 2012



Solutions Manual

354

v v

D r P D vE uO ) P D p2p2`v02 C h2 cos  .cos  sin  / D r P sin  D vE uO ) P D sin  p2pv02`2 C h2 .sin  C cos  /; 

(9)











(10)

450 mph 450 5280 where we have used the expression for r in Eq. (3). Recalling that v 0 3600 ft =s, ` 5 mi 5.5280/ ft, and h 10;000 ft, and using Eqs. (4), we can evaluate Eqs. (8)–(1 0) to obtain

D

D

D

P D 0; P D 0;

r

and

PD



D

D

0:01768 rad=s:

The second of Eqs. (1) states that all of the components of the acceleration are equal to zero. Therefore, we have

D rR r P2 r P 2 sin2  D 0 ) rR D r P2 C r P 2 sin2  ; (11) r P a D r R C 2rP P r P 2 sin  cos  D 0 ) R D P 2 sin  cos  2 r P (12) r P a D r R sin  C 2rP P sin  C 2r P P cos  D 0 ) R D 2 P 2P P .tan  /1 : (13) r Recalling again that v 0 D 450 mph D 450 5280 3600 ft =s, ` D 5 mi D 5.5280/ ft, and h D 10;000 ft, and using ar

















Eqs. (4), as well as Eqs. (8)–(10), we can evaluate Eqs. (11)–( 13) to obtain

R D 11:27 ft=s ; R D 78:10

r

2

 10

6

rad=s2 ;

and

R D 0:




June 25, 2012


Dynamics 2e

355

Problem 2.274 A carnival ride called the octopus consists of eight arms that rotate about the ´ axis at the constant angular 6 rpm. The arms have a length L 22 ft and form an angle  with the ´ axis. Assuming that velocity  70:5ı ,  1 25:5ı , and ! 1 rad=s, determine  varies with time as  .t /  0 1 sin !t with  0 the magnitude of the accelera tion of the outer end of an arm when  achieves its maximum value.

PD

D

D C

D

D

D

Photo credit: © Gary L. Gray

Solution

D 0 C 1 sin !t is maximum when sin !t D 1: D 0 C 1 for !t D .=2/ rad:

Since  0 and  1 are positive, the function  .t /



D

max

(1)

Next, we recall that the components of acceleration in spherical coordinates are

D rR r P2 r P 2 sin2  ; a D r R C 2rP P r P 2 sin  cos  ; a D r R sin  C 2rP P sin  C 2r P P cos  : To use these equations we need the values of r , rP , rR , P , R , P , and R for  D  r D L D costant ) rP D 0 and rR D 0: ar





(2)



 

max .

First we observe that (3)

Using the given  .t/, we have

P D 1! cos !t



P

R

RD



and

1 !

2

sin !t:

(4)

Finally, for  and  we have

P D 6 rpm D constant ) R D 0: (5) Using the results in Eqs. (3)–(5), for  D  D 0 C 1 the components of acceleration become ar D LP 2 sin2 .0 C 1 /; a D L1 ! 2 LP 2 sin.0 C 1 / cos.0 C 1 /; a D 0; (6) where we have used the fact that, for  D  , sin !t D 1 and cos !t D 0 . Recalling that the magnitude of the acceleration is jEaj D ar2 C a 2 C a 2 , using Eqs. (6), for  D  , we have aE D  LP 2 sin2.0 C 1/2 C  L1! 2 LP 2 sin.0 C 1/ cos.0 C 1/2; which, recalling that  0 D 70:5ı D 70:5 180 rad ,  1 D 25:5ı D 25:5 180 rad , ! D 1 rad=s, L D 22 ft, and P D 6 rpm D 6 2 rad=s, can be evaluated to obtain 

max







q





max





max



max







ˇ ˇ q



 60

ˇE ˇ D amax

12:36 ft=s2 :


June 25, 2012


Solutions Manual

356

Problem 2.275

ˇ A particle is moving over the surface of a right cone with angle K , where K is a constant. The and under the constraint that R 2  ´ tan ˇ . Determine the expressions equation describing the cone is R for the velocity and the acceleration of the particle in terms of K , ˇ , ´ , and the time derivatives of ´ .

PD D

Solution

We use the the cylindrical coordinate system implied by the problem’s figure. Next we recall that the general expressions for the velocity and acceleration in cylindrical coordinates are as follows:

E D RP uO R C RP uO C ´P uO ´

P  O C RR C 2RP P  uO C ´R uO ´: (1) The problem is solved by determining all of the terms in the expressions for vE and aE and then substituting the terms in question into the above equations . We begin by determining the terms related to the coordinate R . We are told that R D ´ tan ˇ . Hence, v



and

E D RR

a



R  2 uR



recalling that ˇ is constant, we have

R

D ´ tan ˇ ) RP D ´P tan ˇ ) RR D ´R tan ˇ:

(2)

Next we consider the terms related to  . Specifically, we start with the constraint equation R 2  obtain

K

K D R 2P ) P D RK2 D ´2 tan ) R D 2ˇ



P

2K ´ : ´3 tan2 ˇ

P D K , and (3)

Substituting the first two of Eqs. (2) and the second of Eqs. (3) into the first of Eqs. (1), we have

K u E D ´P tan ˇ uO R C ´ tan O C ´P uO ´: ˇ

v



Substituting Eqs. (2) and the last two of Eqs. (3) into the second of Eqs. (1), we have



E D ´R tan ˇ

a



K2 ´3 tan3 ˇ



O C ´R uO ´:

uR


June 25, 2012


Dynamics 2e

357

Problem 2.276 Solve Prob. 2.275 for general surfaces of revolution; that is, R is no longer equal to ´ tan ˇ , but is now an arbitrary function of ´, that is, R f.´/ . The expressions you need to find will cont ain K , f.´/ , derivatives of f .´/ with respect to ´, and derivatives of ´ with respect to time.

D

Solution

We use the cylindrical coordinate system implied by the problem’s figure. Next we recall that the general expressions for the velocity and acceleration in cylindrical coordinates are as follows:

E D RP uO R C RP uO C ´P uO ´

P  O C RR C 2RP P  uO C ´R uO ´: (1) The problem is solved by determining all of the terms in the expressions for vE and aE and then substituting the terms in question into the above equations . We begin by determining the terms related to the coordinate R . Since we are told that R D f.´/ , and keeping in mind that ´ D ´.t/ , using the chain rule, we have   2 dR d´ RP D ) RP D ´P df and RR D d ´P df ) RR D ´R df C ´P 2 d f2 : (2) v



d´ dt

and

E D RR

a

d´



R  2 uR

dt



d´

d´

d´

Next we consider the terms concerning the coordinate  . Recalling that we have the constra int equation

K

D R 2P . Hence, we can write K d P d´ df K D R 2 P ) P D 2 ´P : ) R D D f 32K f .´/ d´ dt .´/ d´ 

(3)

P

P

D

f.´/ , substituting the expression for R from Eqs.(2) along with the expression for  Recalling that R from Eqs. (3) into the first of Eqs. (1) we have K E D df ´P uO r C uO C ´P uO ´ : d´ f.´/

v



P

D

R

f.´/ , substituting the expressions for R and R from Eqs.(2) along with the Again recalling that R expressions for  and  from Eqs. (3) into the second of Eqs. (1) we have

P

R

2

E D ´R C ´P 2 dd´f2

a





K2 uR f 3 .´/



O C ´R uO ´:


June 25, 2012


Solutions Manual

358

Problem 2.277

D

35 ft, In a racquetball court, at point P with coordinates xP 16 ft, and ´P 1 ft, a ball is imparted a speed v0 90 mph 63 ı and ˇ 8 ı (ˇ is the and a direction defined by the angles  angle formed by the initial velocity vector and the xy plane). The ball bounces off the left vertical wall to then hit the front wall of the court. Assume that the rebound off the left vertical wall occurs such that (1) the component of the ball’s velocity tangent to the wall before and after rebou nd is the same and (2) the component of yP

D

D

D

D

D

velocity normal the walltoright is equal magnitude and opposite in to direction the after same impact component of in velocity right before impact. Accounting for the effect of gravity, determine the coordinates of the point on the front wall that will be hit by the ball after rebounding off the left wall. Solution

From P to the left wall the racquetball undergoes projectile motion and therefore the components of the 0 , ay 0 , and a´ g. In turn, the coordinate s ball’s acceleration in the given coordinate system are ax of the ball as a function of time are

D

D

D

D xP C xP 0t; y D yP C yP0t; and ´ D ´P C ´P 0t 12 gt 2; where xP D 35 ft, yP D 16 ft, and ´P D 1 ft. Next, the initial compone nts of velocity are xP 0 D v0 cos ˇ sin  ; yP0 D v0 cos ˇ cos  ; and ´P 0 D v 0 sin ˇ: x

(1)





(2)



The first part of the motion of the ball is described by

x

D xP



.v0 cos ˇ sin  /t;

y

D yP



.v0 cos ˇ cos  /t;

´

and

Letting the subscript lw stand for ‘left wall’, we must have y.tlw /

D ´P C .v0 sin ˇ/t

1 2  2 gt :

(3)

D 0 so that

D 0 ) t D v0 cosyˇP cos  D 0:2696 s: (4) The corresponding x and ´ coordinates at t D t are x D xP .v0 cos ˇ sin  /t (5) ) x D 3:598 ft; 1 ´ D ´P C .v0 sin ˇ/t gt ´ 4:783 ; ) D ft (6) 2 where, in Eqs. (4)–(6), we have used the fact that x P D 35 ft, yP D 16 ft, ´P D 1 ft, v0 D 90 mph D ı ı 90 5280 3600 ft =s, ˇ D 8 , and  D 63 . yP



.v0 cos ˇ cos  /tlw

lw

lw



lw

lw

lw 

lw

lw

lw

lw

After impact the x and y components of velocity are

P D

xlw

v0 cos ˇ sin 

and

P D v 0 cos ˇ cos  :

ylw

The ´ component of velocity after impact is calculated with the constant acceleratio n equation v

P D v 0 sin ˇ

´lw



(7)

D v0 C at .

gt lw:


(8) June 25, 2012


Dynamics 2e

359

After the impact with the left wall, the ball is again in projectile motion. Hence, the equations describ ing the motion of the ball after the impact are

D .v 0 cos ˇ cos  /t; and ´ D ´ C .v0 sin ˇ gt /t 12 gt 2; (9) where we have “reset” the time variable so that t D 0 now corresponds to when the ball bounces off the the left wall. We are now ready to consider the impact of the ball with the front wall. Letting the subscript fw stand for ‘front wall’, we must have x .t / D 0 , so that x x Dx .v0 cos ˇ sin  /t D 0 ) t D D 0:03089 s: (10) v0 cos ˇ sin  x

Dx

lw 

.v0 cos ˇ sin  /t;

y



lw

lw



fw

fw

lw

lw 

fw

fw

The corresponding values of the y and ´ coordinates are

yfw ´fw

D .v 0 cos ˇ cos  /t D ´ C .v0 sin ˇ

fw



lw

2 gt lw/tfw  12 gt fw

) )

yfw ´fw

D 1:833 ft; D 5:067 ft:

(11) (12)

In summary, the coordinates of the point on the front wall that is impacted by the ball, we have

xfw

D 0;

yfw

D 1:833 ft;

and

´fw

D 5:067 ft:


June 25, 2012


Solutions Manual

360

Problem 2.278 An airplane is being tracked by a radar station at A . At the instant t 0, the following data is recorded: r 15 km,  80ı , 15ı , r 350 km=h,  0:003 rad=s.  0:002 rad=s,  If the airplane is flying to keep each of the spherical velocity components constant for a few minutes, determin e the spherical components of the airplane’s acceleration when t 30 s.

D D

D

PD

PD

D

PD D

Solution

In spherical coordinates, the components of acceleration are

D rR r P2 r P 2 sin2  ; a D r R C 2rP P r P 2 sin  cos  ; a D r R sin  C 2rP P sin  C 2r P P cos  : To solve this problem, we must provide the values of r , rP , rR ,  , P , R , P , and R for t D 30 s. ar





(1)



 

We will now use the assumption that the velocity components are constant to determine the value of the quantities just listed. The expression of the velocity in spherical coordinates is

E D rP uO r C r P uO C r P sin  uO D v r uO r C v uO C v uO :

v











(2)



Then, under the assumption that v r is constant, for the radial coordinate r we have

vr .0/

D rP.0/ D constant ) rR.t/ D 0

r.t/

and

D r.0/ C rP.0/t:

(3)

1000

Recalling that r.0/ at t 30 s we have

D

D 15 km D 15;000 m and that rP.0/ D 350 km=h D 350 3600 m=s, then, using Eqs. (3),

D 17920; rP.30 s/ D 97:22 m=s; and rR.30 s/ D 0: (4) Next we proceed to determine the values of  , P , and R at t D 30 s. To do so, referring to Eq. (2), we start from the consideration that v D r P . Hence, using the expression for r in the last of Eqs. (3) we have r.0/ P .0/ r.0/ rP.0/ P .0/ v D r P D constant D r.0/ P .0/; ) P D ) R D Œr.0/ r.0/ C rP.0/t C rP.0/t ç2 : (5) The last two of Eqs. (5) will allow us to compute P and R at t D 30 s. However, we also need the value of  at t D 30 s. To compute such a value we now proceed to integrat e the second of Eqs. (5) with respect to time. r.30 s/







This gives

P D ddt D r.0/r.t/P .0/ )



 .t/   .0/

D

Z

t 0

P CP

r.0/  .0/ dt r.0/ r.0/t

)

 .t /

P.0/ t D  .0/ C r.0/rP.0/P .0/ ln 1 C rr.0/



:

(6)

Using the last of Eqs. (6) and the last two of Eqs. (5) we then have

 .30 s/

D 76:86ı; P .30 s/ D

0:001674 rad=s;

and

R

 .30 s/

D 9:086

 10


6

rad=s2 ;

(7)

June 25, 2012


Dynamics 2e

361

D 15 km D 15;000 m, rP.0/ D 350 km=h D Next we proceed to determine the values of P , and R at t D 30 s. We start from the fact that v D r P sin 

where we have used the following numerical data: r.0/ 350 1000 80 ı , and  .0/ 0:002 rad=s. 3600 m =s,  .0/

P D

D



is assumed to be constant. Hence, we have

P

r.t/  .t / sin  .t /

P D r.0/ P .0/ sin  .0/ ) P .t / D r.0/r.t/ .0/sinsin .t/.0/ P ) R D r.0/  .0/ sin  .0/fPr.t/2 sin 2.t / C r.t/Œ cos  .t/çP .t /g : 

r .t/ sin  .t/

Hence, recalling that r.0/ 15 km 15;000 m, r.0/ 350 km=h 0:003 rad=s, and using Eqs. (4) and (7), for t 30 s, we have

D

P

 .30 s/

D

D

D 0:002540 rad=s

P D

and

R

 .30 s/

D

ar

D

2

0:1598 m=s

;

a

D

2

0:1884 m=s

;

D 350 1000 3600 m =s,  .0/ D 80 ı , P .0/ D 2

0:00001279 rad=s

In conclusion, substituting Eqs. (4), (7), and (9) into Eqs. (1), for t

(8)

:

(9)

D 30 s we have

and

a

2

D 0:2232 m=s :


June 25, 2012


Solutions Manual

362

Problem 2.279 An airplane is being tracked by a radar station at A . At the instant t 0 , the following data is recorded: r 15 km,  80 ı ,  ı 15 , r 350 km=h,  0:002 rad=s,  0:003 rad=s. If the airplane is flying to keep each of the spherical velocity components constant, plot the trajectory of the airplane for 0 < t < 150 s.

D

D PD

PD

PD

D

D

Solution

To plot the trajectory of the airplane we need to find expressions for the coordinates of the airplane as a function of time. Clearly, this must be done while enforcing the condition that the velocity components remain constant during the time interval of interest. For this purpose, we consider the general expression for the velocity in spherical coordinates , which is

E D rP uO r C r P uO C r P sin  uO D v r uO r C v uO C v uO :

v











(1)



Then, under the assumption that v r is constant, for the radial coordinate r we have

D rP.0/ D constant ) r.t/ D r.0/ C rP.0/t; (2) where r.0/ D 15 km and that rP.0/ D 350 km=h. Next we proceed to determine an expression for  .t/. To do so, referring to Eq. (1), we start from the consideration that v D r P . Hence, using the expressi on for r in the last of Eqs. (2) we have r.0/ P .0/ v D r P D constant D r.0/ P .0/; ) P D : (3) vr .0/





r.0/

r.0/t

CP

Then, to determine  .t/ we now proceed to integrate the last of Eqs. (3) with respect to time. This gives

P D ddt D r.0/r.t/P .0/ )



 .t/   .0/

D

Z

t 0

P CP

r.0/  .0/ dt r.0/ r.0/t

 rP.0/  t ; (4) D  .0/ C r.0/rP.0/P .0/ ln 1 C r.0/ where r.0/ D 15 km D 15;000 m, rP.0/ D 350 km=h,  .0/ D 80 ı , and P .0/ D 0:002 rad=s. Next we try and provide an expression for  .t /. We start from the fact that v D r P sin  is assumed to )

 .t /





be constant. Hence, we have

P

r.t/  .t / sin  .t /

P D r.0/ P .0/ sin  .0/ ) P .t / D r.0/r.t/ .0/sinsin .t/.0/ :

(5)

Although the expressions for r.t/ and  .t/ are currently known, even if one were to substitute these expressions into the above equation, we would obtain an expression for  that cannot be integrated with

P

respect to time in closed-form. Hence, we must proceed to integrate with respect to time numerically. This can be done with appropriate mathematical softwa re. We have used Mathematica as described below. Once

r.t/ ,  .t/, and  .t / are known, in order to plot the trajectory of the airplane, we need to transform the spherical coordinates into corresponding Cartesian coordinates. We do so using the following relations:

x

D r sin  cos  ;


(6) June 25, 2012


Dynamics 2e

363

y ´

D r sin  sin  ; D r cos  :

(7) (8)

The code provided below first defines the known parameters of the problem. Then the functions giving r.t/ and  .t/ are stated, followed by the instructions necessary to determine  .t/. Once the solution for  .t/ is obtained, the spherical coordinates of the airplane are transformed into corresponding Cartesian coordinates and the resulting trajectory is plotted for 0 < t < 150 s. Referring to the plot of the trajectory, the airplane is moving from the upper right to the lower left.

Executing the above code yields the following plot (except for the labels “start” and “finish”):

start

finish


June 25, 2012


Solutions Manual

364

Problem 2.280 The velocity and acceleration of point P expressed relative to frame A at some time t are

E

vP =A

D .12:5 {OA C 7:34 |OA/ m=s

and

E

aP =A

D .7:23 {OA



3:24 |A / m=s2 :

O

Knowing that frame B does not move relative to frame A , determine the expressions for the velocity and acceleration of P with respect to frame B . Verify that the speed of P and the magnitude of P ’s acceleration are the same in the two frames.

Solution

To express the velocity and acceleration vectors relative to the reference frame B , the base vectors of frame A need to be expressed in terms of the base vectors of frame B . Because of the orientation of frame B relative to A , we have

O D cos.15ı/ {OB C sin.15ı/ |OB

{A

and

O D

|A

 sin.15

ı

O C cos.15ı/ |OB ;

/ {B

(1)

Substituting the expressions in Eqs. (1) into the given expres sion for the velocity of P , we have

E

vP =B

D

˚

Œ12:5 cos.15ı /  7:34 sin.15ı /ç {B

O C Œ12:5 sin.15ı/ C 7:34 cos.15ı/ç |OB  m=s:

E

The magnitude of vP =B is

q D

(2)

C Œ12:5 sin.15ı/ C 7:34 cos.15ı/ç m=s 12:5 2 Œcos2 .15ı / C sin2 .15ı /ç C 7:342 Œcos2 .15ı / C sin2 .15ı /ç m=s D 12:52 C 7:342 m=s D vP =A; (3) where we have used the trigonometric identity sin2 ˛ C cos2 ˛ D 1 . Equation (3) states that the speed of P vB=P

Œ12:5 cos.15ı /  7:34 sin.15ı /ç2

q D

p

as seen by the two frames is the same. Evaluating Eqs. (2) and (3), we have

E

vP =B

D .10:17 {OB C 10:33 |OB / m=s

and

vP =B

D vP =A D 14:50 m=s:

Substituting Eqs. (1) into the given expression for the acceleration of P , we have

E

aP =B

D

˚

Œ7:23 cos.15ı /

C 3:24 sin.15ı/ç {OB C Œ7:23 sin.15ı/



3:24 cos.15ı /ç |B m=s2 :

E

The magnitude of aP =B is


O



(4)

June 25, 2012


Dynamics 2e

365

ˇE ˇ D q q D

Œ7:23 cos.15ı /

2

C 3:24 sin.15ı/ç2 C Œ7:23 sin.15ı/ 3:24 cos.15ı/ç m=s 7:232 Œcos2 .15ı / C sin2 .15ı /ç C 3:242 Œcos2 .15ı / C sin2 .15ı /ç m=s D 7:232 C 3:242 m=s D aEP =A ; (5) where we have used the trigonometric identity sin 2 ˛ C cos2 ˛ D 1 . Equation (5) states that the magnitude of aP =B



2

p

ˇ ˇ

2

the acceleration of P as seen by the two frames is the same. Evaluatin g Eqs. (4) and (5), we have

E

aP =B

D .7:822 {OB



1:258 |B / m=s2

O

and

E

aP =B

2

D aEP =A D 7:923 m=s :

ˇ ˇ ˇ ˇ


June 25, 2012


Solutions Manual

366

Problem 2.281 The motion of a point P with respect to a Cartesian coordinat e system is described 2 t { Œ4 ln.t 1/ 2t 2 ç | ft, where t is time expressed in seconds. by r 4 s and t 2 6 s. Then find the time Determine the average velocity between t 1 t for which the x component of P ’s velocity is exactly equal to the x component of P ’s average velocity between times t1 and t 2 . Is it poss ible to find a time at which P ’s velocity and P ’s average velocity are exactly equal? Explain why. Hint: Velocity is a vector.

ED f p OC

C C

Og

D

N

D

Solution

The average velocity over the time interval Œt 1 ; t2 ç is

E D rE.t2t2/ rtE1.t1/ : 

vavg

E

(1)



D 4 s and t2 D 6 s, we have vE D .0:4495 {O C 20:67 |O / ft=s;

Using the given function for r.t/ , for t 1

(2)

avg

E

Next, we compute the velocity of P by differentiating the position vector r.t/ with respect to time. This gives

1 

E D pt {O C

v

4t

C 1 C4 t |O



ft=s:

(3)

N

To determine t we then need to solve the equation

 

N D p1 N t

vx .t /

N

ft=s

D 0:4495 ft=s;

(4)

which, can be solved for t to obtain

N D 4:949 s:

t

(5)

Substituting the result in Eq. (5) to compute the y component of the velocity vector in Eq. (3), we have

N D 20:47 ft=s ¤ .v

vy .t /

avg/y :

(6)

This result implies that In general, it is not possible to find a time instant in an interval Œt 1 ; t2 ç for which the velocity and the average velocity are equal.

N

While it is always possible for a scalar function to find a value of time t at which the function is equal to its

average a time interval containing comes to a vector function, finding a common time that t , when itnot works forover every scalar component is, in general, possible.

N


June 25, 2012


Dynamics 2e

367

Problem 2.282 The figure shows the displacement vector of a point P between two time instants t 1 and t 2 . Is it possible for the vector vavg shown to be the average velocity of P over the time interval Œt1 ; t2 ç?

E

Solution

No. vavg must have the same direction as  r .t 1 ; t2 /.

E

E


June 25, 2012


Solutions Manual

368

Problem 2.283 A dynamic fracture model proposed to explain the behavior of cracks propagating at high velocity views the crack path as a wavy path . In this model, a crack tip appearing to travel along a straight path actually travels at roughly the speed of sound along a wavy path. Let the wavy path of the crack tip be described by the function y h sin.2 x= /, where h is the amplitude of the crack tip fluctuations in the direction perpendicular to the crack plane and  is the corresponding period. Assume that the crack tip travels along the

D

s (e.g., the speed of sound). wavy path a constant speed Find theatexpressio n for the vx component of the crack tip velocity as a function of v s ,  , h , and x .

Solution

E D xP {O C yP |O . Hence, letting

Using the coordinate system shown , the velocity of the crack tip has the form v

vs be the speed of the crack, we must have vs Now recalling that y

q

D xP 2 C yP 2:

D h sin.2 x=/, we have

(1)

dy dx P D dx ) yP D 2h xP cos dt

2 x 

y

 

:

(2)


vs

s

D xP 2 C

 2h xP  2 x   

cos

 

2

s

D jPxj 1 C

 2h  2 x   cos 

 

2

;

(3)

which can be solved to find that the x component of the velocity is

PD

x

vs 

q

2

C

 ;

4h 2  2 cos2 2x

P

where we have chosen x to be positive since the crack is assumed to propagate in the positive x direction.


June 25, 2012


Dynamics 2e

369

Problem 2.284 A dynamic fracture model proposed to explain the behavior of cracks propagating at high velocity views the crack path as a wavy path . In this model, a crack tip appearing to travel along a straight path actually travels at roughly the speed of sound along a wavy path. Let the wavy path of the crack tip be described by the function y h sin.2 x= /, where h is the amplitude of the crack tip fluctuations in the direction perpendicular to the crack plane and  is the corresponding period. Assume that the crack tip travels along the

D

wavy path at a constant speed v s (e.g., the speed of sound). Denote the apparent crack tip velocity by v a , and define it as the average value of the x component of the crack velocity, that is,

va

D 1

Z



vx dx: 0

D

In dynamic fracture experiments on polymeric materials, va 2vs =3, vs is found to be close to 800 m=s, and  is of the order of 100 m. What value of h would you expect to find in the experiments if the wavy crack theory were confirmed to be accurate? Solution

To solve this problem we need to first determine the x component of the velocity of the crack tip. Once, we determine an expression for v x , then we will solve the equation stated in the problem numeri cally for h . We begin by observing that, using the coordinate system shown, the veloci ty of the crack tip has the form

v

E D xP {O C yP |O . Hence, letting vs be the speed of the crack, we must have vs D xP 2 C yP 2 : Now recalling that y D h sin.2 x= /, we have  2 x  dy dx 2h xP yP D ) yP D cos : dx dt  

q

(1)

(2)


vs

s

x2

D P C

 2h xP  2 x   

cos

 

s

2

D jPxj 1 C

 2h  2 x   cos 

 

2

;

(3)

which can be solved to find that the x component of the velocity is

vx

vs 

D xP D

P

2

C

;

4h 2  2 cos2 2x

q

(4)

 

where we have chosen x to be positive since the crack is assumed to propagate in the positive x direction. Now, that v x is known, using the formula for v a given in the problem statement, we have

va

D 1

Zq 

0

vs  2

C

  dx D

4h 2  2 cos2 2x

Zq 

0

vs 2

C

  dx :

(5)

4h 2  2 cos2 2x


June 25, 2012


Solutions Manual

370

2 800 m=s, v a 100 m. Hence, since the variable x in Eq. (5) has Recall that we have v s 3 vs , and  the role of dummy variable of integration, Eq. (5) is an equation in the single unknown h . The equation in question is an integral equation and can be solved numerically using appropriate mathematical software. We have used Mathematica with the code given below . As is often the case with the numerical solut ion of equations, we had to supply Mathematica with an initial guess for the solution. Since the quantity h is expected to be of the same order of a few  m, our guess for h was set to 1 m.

D

D

D

The execution of the code above yields the following result:

h

D 29:43

 10

6

m:


June 25, 2012


Dynamics 2e

371

Problem 2.285 The motion of a peg sliding within a rectilinear guide is controlled by an actuator in such a way that the peg’s acceleration takes on the form x a0 .2 cos 2! t  ˇ sin !t/, where t is time, a 0 3:5 m=s2 , ! 0:5 rad=s, and ˇ 1:5. Determine the total distance tra veled by the a 0 ˇ=! 0:3 m=s. peg during the time interval 0 s  t  5 s if x.0/ When compared with Prob. 2.57, why does the addition of 0:3 m=s in the initial velocity turn this into a problem that requires a computer to solve?

RD

D

D

P D

D

C

Solution

We begin by obtaining an expression for the velocity of the peg as a function of time. Since the acceleration is given as a function of time, we can integrate it as follows:

v

D xP .0/ C

Z

t 0

a0 .2 cos 2! t



ˇ sin !t/dt

)

v

D a0 !ˇ C 0:3 m=s C a0

2

2!

sin 2! t

C !ˇ cos !t





a0

ˇ ; (1) !

which we can simplify to obtain

v

D a!0 Œsin 2! t C ˇ cos !t ç C 0:3 m=s:

(2)

To find the distance traveled, we need to establish if and when the peg switches direction during the time interval considered. This can be easily done by using any appropriate mathematical software that can plot the function v over the time interval considered. Recalling that a 0 have used Mathematica with the following code:

D 3:5 m=s , ! D 0:5 rad=s, and ˇ D 1:5, we 2

The above code yields the following plot:

D

3 s. Hence, we need to solve the As can be seen from the above plot, the velocity changes sign near t 0 to determine where exactly the sign switch occurs. Because the expressi on we have for v equation v

D


June 25, 2012


Solutions Manual

372

D

0 cannot be solved analytically. Hence, we will solve it includes the constant term 0:3 m, the equation v numerically. This can be done with any appropriate mathematical software. As is common with root finding algorithms, we need to provide a guess for the solution. Based on the graph presented above, we set our guess to t 3 s and then use the following Mathematica code:

D

which yields the following solution:

t

D 3:166 s:

(3)

Using this result, and letting d denote the distance traveled, we have

d

D

Z

3:166 s 0

a

0

!

cos !t.2 sin !t



Cˇ/C0:3 m=s

dt 

Z

5s 3:166 s

a

0

!

cos !t.2 sin !t



Cˇ/ C0:3 m=s

dt: (4)

The above integrals can be computed either numerica lly or analytically. Since this is a computer problem, we have chosen to carry out the integration numerically. We have used Mathematica with the following code:

The execution of the above code, yields the following result:

d

D 52:81 m:


June 25, 2012


Dynamics 2e

373

Problem 2.286 The acceleration of an object in rectilinear free fall while immersed in a linear viscous g  Cd v= m, where g is the acceleration of gravity, C d is a constant fluid is a 0 drag coefficient, v is the object’s velocity, and m is the object’s mass. Letting v 0 for t 0 , where s is position and t is time, determine the position as a and s function of time.

D

D

D

D

Solution

D dv=dt , we can separate the time and

The acceleration is given as a function of velocity. Recalling that a velocity variables as follows:

dt

Z

D dva )

Z

t

dt 0

D

v 0

Cd m

t

D1



)

t

Cd v mg

)

v

 mg

D Cd

s

v dt

D

0

ZD

where we have enforced the condition that s simplifying, we have

s

D mg C2 d



1

Cd v ; mg



(1)



Cd t

D

Cd

0 for t

Cm



e

1

0

Z D

Cd m

Cd m

t



v . This

:

(2)

D ds=dt , so that we can separate the

t

mg

ds

)



1  e

Now that we have velocity as a function of time, we recall that v variables s and t by writing

ds

D Cmd ln

D 0 for t D 0 . We now solve the last of Eqs. (1) for

where we have enforced the condition that v gives

e

dv g  Cd v= m



e

Cd m

t

 dt;

(3)

0. Carrying out the above integration and t



1



:


June 25, 2012


Solutions Manual

374

Problem 2.287 Heavy rains cause a stretch of road to have a coefficient of friction that changes as a function of location. Under these conditions, the acceleration of a car skidding while trying to stop can be approximated by s .k  cs/g , where  k is the friction coefficient under dry conditions, g is the acceleration of gravity, and c , with units 0:5, of m 1 , describes the rate of fricti on decrement. Let  k c 0:015 m1 , and v 0 45 km=h, where v 0 is the initial velocity of the car. Determine the time it will take the car to stop and the

RD

D

D

D

percent increase stopping time with respect to dry conditions, c 0 . in i.e., when Hint:

D

Zp

1 1

C

x2

dx



D log x C

pC 1



x2 :

Solution

We are given the acceleration as a function of position. We will first relate the acceleration to the velocity using the chain rule, and then relate the velocity to position:

Z

Z

v s D ddssP ds ) a D ddssP sP ) a ds D sP d sP ) v sP d sP D 0 .k cs/ g ds; (1) dt where v D sP, v D v 0 for s D 0 , and where we have used the given expression of the acceleration in that last

a

0





of Eqs. (1). Carrying out the integration, we have s

g 1 2 2 v2  1 2 v0

D 2c .cs



 k /2

ˇ

v

)

0

D

g c

rr

c g v02

C .cs



k /2  2k :

(2)

This result will be needed later. We now determine the expression for the stopping position of the car, which 0 for s sf in Eq. (2), and solving for sf , we have we will denote by sf (f stands for final). Setting v

sf

D

D ck ˙ c1

s

D

2k



cv 02 : g

(3)

Only the smaller of the two roots in Eq. (3) makes physical sense (once the car stops, it will not stop at an increased position). Hence, we select the following root:

sf

D

k c



1 c

s

2k



cv 02 : g

(4)

Note that we could have identified the correct the solution to choose by evaluating the two roots numerically. 0:5, c 0:015 m1 , g 9:81 m=s2 , and v0 45 km=h 45 1000 Recalling that k 3600 , the roots corresponding to the minus and plus signs are sf 26:31 m and sf 40:35 m, respectively. We now go back to Eq. (2) and proceed to determine the relatio n between position and time. To do so, we ds=dt . Observing that we have the velocity as a function of position, we can separate the recall that v variables s and t as follows: d t ds=v . Hence, using Eq. (2), we have

D

D

D

D D

Z D Dr Z q tf

dt

0

c g

D

sf

0

D

ds c 2 g v0

C .cs



k /2  2k

D

;


(5)

June 25, 2012


Dynamics 2e

375

D

D 0 and we have denoted by tf the time at which the

0 for t where we have enforced the condition that s car comes to a stop. To carry out the integration, we let A

D gc v02



2k ;

(6)

and we perform the following change of variables of integration:

D cs

x For s

0 we have x

D

k

and for s

D



)

k

sf we have x

D 1 tf D p cg

Z

c sf

cs  D k

f

D c1 d x:

ds

k



(7)

k . Hence, Eq. (5) becomes

pAdxC x 2 :

(8)

Using the hint, we obtain

tf

1 ln x cg



Dp

p ˇ C C ˇ A

x2

csf k

)

k

tf

csf 1 ln cg



Dp

k

 k

which, recallin g the definition of A in Eq. (6), after simplification yields

tf

D p1cg ln v csfc=g k 

0

p



Realling that c 0:015 m1 , g 9:81 m=s2 ,  k 0:5 , and v 0 sf in Eq. (4) and then eva luate tf in Eq. (10) to obtain

D

D

D

tf

D PDP C 0 D v0

C

qC q C C A

.csf

A



2k

:

k /2

;

(9)

(10)

k

D 45 km=s D 45 1000 3600 m =s, we can evaluate

D 5:839 s:

(11)

RD

0 , the acceleration of the car would be s k g. Hence, we can use the With dry conditions, i.e., for c constant acceleration equation s s0 ac t to find the stopping time. This gives 

k gtf

)

tf

D 2:548 s:

(12)

Comparing this result to that in Eq. (11), we find the percent increase to be Stopping time percent increase

D tf

wet

 tfdry

tfdry



100

)

Stopping time percent increase


D 129:1%.

June 25, 2012


Solutions Manual

376

Problem 2.288 The acceleration of a particle of mass m suspended by a linear spring with spring constant k and unstretched length L 0 (when the spring length is equal to L 0 , the spring exerts g  .k=m/.x  L0 /. Assuming that at t 0 no force on the particle) is given by x 0 m, derive the expression of the particle’s the particle is at rest and its position is x position x as a function of time. Hint: A good table of integrals will come in handy.

RD

D

D

Solution

We have acceleration as a function of position and we integrate it as follows:

Z

D ddxxP dx ) dt

a

xP

P PD

xdx 0

Z

x



g

0

k x  L0 m



r

k 2 kL0 x x x 2gx 2m m Now that we have x as a function of x , we relate it to time as follows: 1 2 2x

P D gx

C



) PD

P

P D dx ) dt

x

Now we let A

D mk

 t

2kL0 m

D

Z

dt 0

D



C 2g D 2

rZ m k

x

pAx



Making use of the trigonometric identity 1 tan2

r ! t 2

k m

C gkm

L0

t

)

D2

t

D Ax x ) 

D A sin2

r ! t 2

k m



Finally, recalling that A is the quantity 2 L0

x

0

x

dx 2kL0 m

:

C 2g

r

m tan1 k





k  m x2

p x pA x x : 

C tan2  D sec2  , we have that

"  r !#

x sec2

Now making use of the trigonometric identity sin 2 . =2/

x

Zr 

D

0 x: C 2kL m

and we rewrite the expression for t as follows:



x2

k 2 x m

x

dx x

P )

0

 dx

x 0

Z

t



 dx :

)

x

t 2

k m

D 12 .1



D A tan2

ˇˇ r ! 0

t 2

cos  /, we can express x as

" r !#

D A2

k : m

1  cos

k t m

:

C gkm , the expression for x can be written as

D L0 C gkm 1



r k t :

cos

m


June 25, 2012


Dynamics 2e

377

Problem 2.289 In a movie scene involving a car chase, a car goes over the top of a ramp 18 ı and ˇ 25 ı , determine the at A and lands at B below. Letting ˛ speed of the car at A if the car is to be airborne for a full 3 s. Furthermore, determine the distance d covered by the car during the stunt, as well as the impact speed and angle at B . Neglect aerodynamic effects. Express your answer using the U . S . Customary system of units.

D

D

Solution

This is a projectile problem and we begin by relating the jump speed v 0 to the velocity of the car and then to the car’s trajectory. Referring to the figure at the right, the initial velocity and acceleration in the .x ; y/ coordinate system are

P D v0 cos.˛ C ˇ/; yP0 D v0 sin.˛ C ˇ/; xR D g sin ˇ; yR D g cos ˇ:

x0

(1) (2)



Using constant accele ration equations , we have that the coordinates of the (airborne) car as a function of time are

D v0 cos.˛ C ˇ/t C 21 g sin ˇ t 2; y D v 0 sin.˛ C ˇ/t 21 g cos ˇ t 2 : We want to satisfy the condition that y D 0 at t D tf D 3 s, where the subscript f stands for flight. x

(3) (4)



gtf cos ˇ

2

1

0

(5) D v0 sin.˛ C ˇ/tf 2 g cos ˇ tf ) v0 D 2 sin.˛ C ˇ/ : Recalling that g D 32:2 ft=s , tf D 3 s, ˇ D 25ı , and ˛ D 18ı , we can evaluate the result in the last of 

2

Eqs. (5) to obtain

v0

D 64:19 ft=s:

We now observe that the distance d is equal to the valu e of x at time t for v 0 in the last of Eqs. (5), we have that the distance d is given by

d

D tf . Using Eq. (3) and the expressio n

gtf cos ˇ cos .˛ C ˇ/tf C 12 g sin ˇ tf2 ) D 2 sin .˛ C ˇ/

d

D 202:1 ft.

(6)

To find the impact speed and the impact angle, we need to determine the velocit y at impact. To do so, we v 0 ac t to obtain use the constant acceleration equation v

D C

E D Œv0 cos.˛ C ˇ/ C g sin ˇ tf ç {O C Œv0 sin.˛ C ˇ/

vi

v

) Ei D

O

g cos ˇ tf ç |



cos ˇ

gt f

2

tan.˛

sin ˇ {

C ˇ/ C

O



1 2

gt cos ˇ | ; (7) f

where we have used the expression for v 0 in the last of Eqs. (5). Therefore, the impact speed is

vi

D gtf

s

cos ˇ



O

2

C sin ˇ C 41 cos2 ˇ: 2 tan.˛ C ˇ/


(8) June 25, 2012


Solutions Manual

378

Recalling that g obtain

D 32:2 ft=s , tf D 3 s, ˇ D 25ı, and ˛ D 18ı, we can evaluate the result in Eq. (8) to vi D 98:08 ft=s: 2

We now compute the impact angle of the car with the ground at B measuring it with respect to the slope. This angle is given by Impact angle

D tan1

25 ı and ˛

Recalling that ˇ

D

v 

iy

vix

D tan1



1 2

cos ˇ



cos ˇ

C sin ˇ 2 tan.˛ C ˇ/

1

 

:

(9)

18 ı , we can evaluate the above result to obtain

D

Impact angle

D 26:51ı measured clockwise from the slope:


June 25, 2012


Dynamics 2e

379

Problem 2.290 Consider the problem of launching a projectile a distance R from O to D with a known launch speed v 0 . It is probably clear to you that you also need to know the launch angle  if you want the projectile to land exactly at R . But it turns out that the condition determining whether or not v 0 is large enough to get to R does not depend on  . Determine this condition on v 0 . Hint: Find v 0 as a function of R and  , and then remember that the sine function is bounded by 1. Solution

For projectile motion we have the initial velocity and acceleration in the .x ; y/ coordinate system, i.e.,

P D v 0 cos  ; yP0 D v0 sin  ; xR D 0; yR D g:

x0

(1) (2)



Using the constant acceleration equations, we obtain

x For t

D v 0 cos  t

and

y

D v0 sin  t

1 2  2 gt :

(3)

D tD we have y.tD / D 0. Using this fact in the second of Eqs. (3), yields 2v0 sin  tD D : g

Then substituting Eq. (4) into the first of Eqs. (3) and setting

R

D

v02 g

x

(4)

R , we obtain

D ) v02 D singR2 ;

2 sin  cos 

D

(5)

j

where we have used the trigonometric identity 2 sin  cos  sin 2 . Since we must have sin 2 have a corresponding condition on v 0 of the following form:

v0



p

j



1, we

gR:


June 25, 2012



Dynamics 2e

381

Problem 2.292 A roller coaster travels over the top A of the track section shown with a 60 mph. Compute the largest radius of curvature  at A such speed v that the passengers on the roller coaster will experience weightlessness at A.

D

Solution

To experience weightlessness, the passengers must be in free fall, i.e., their acceleration must be equal to the acceleration of gravity . Since the tangent to the trajectory at A is horizontal, at A the direction of acceleration must be toward C , i.e., it must be completely in the normal direction. With this in mind, using normal-tangential components and recalling that the normal acceleration is given by v 2 = , we must have

v2 

2

Dg )

where we have used the fact that v



D vg )



D 240:5 ft,

5280 ft =s and g D 32:2 ft=s . D 60 mph D 60 3600 2


June 25, 2012


Solutions Manual

382

Problem 2.293 Determine, as a function of the latitude  , the normal acceleration of the point P on the surface of the Earth due to the spin !E of Earth about its axis. In addition, determin e the normal accelerati on of the Earth due to its rotation about the Sun. Using these results, determine the latitude above which the acceleration due to the orbital motion of the Earth is more significant than the acceleration due 6371 km for the to the spin of the Earth about its axis. Use R E mean radius of the Earth, and assume the Earth’s orbit about the Sun

D

8

is circular with radius R O

D 1:497

 10

km.

Solution

Assuming that the center of the Earth is fixed and the Earth rotates as shown, the point P moves around a R E cos . The speed of vP P !E cos . circle centered at the spin axis of the Earth with radius P Therefore, the normal acceleration of P due to rotation is

D

2

aP n

2

2

2

D vPP D !ERREEcoscos  )

aP n

D

D ! E2 RE cos  D 0:03369 cos  m=s , 2

(1)

2 where we have used the data R E 6371 km 6371  103 m and ! E 1 rev=day rad =s. We .24/.3600/ now model the motion of the Earth around the Sun as a circular motion along a circle centered at the Sun with radius R O and angular speed ! O . The speed of P due to this motion is the same as that of the Earth, namely, v E R O !O . The corresponding normal accelera tion is

D

D

D

D

D

2

aE n

2

2

D vEO D !ORRO O )

where we have used the data !O

1497  108 m.

aEn

D 1 rev=year D

D !O2 RO D 0:005942 m=s , 2

2 rad =s .365/.24/.3600/

and RO

D 1:497

(2)  108 km

D

The latitude  at which the acceleration due to the motion of the Earth around the Sun is equal to that due to the Earth’s spin about its own axis is 2 !E RE cos 

Recalling that R E

D 6371 km D 6371 R O D 1:497

2 rad =s, and .365/.24/.3600/

obtain

D !O2 RO )



D cos1

2 O O 2R !E E

! R 

:

(3)

2 D 1 rev=day D .24/.3600/ rad =s, ! O D 1 rev=year D D 1497 108 m, we can evaluate the last of Eqs. (3) to  D 79:84ı :

 103 m,



!E

 108 km




June 25, 2012


Dynamics 2e

383

Problem 2.294

D

A car is traveling over a hill with a speed v 0 160 km=h. Using the Cartesian coordinate system shown,  .0:003 m 1 /x 2 , where x and y are measured in the hill’s profile is described by the function y 100 m, the driver realizes that her speed will cause her to lose contact with the ground meters. At x once she reaches the top of the hill at O . Verify that the driver’s intuition is correct, and determine the minimum constant time rate of change of the speed such that the car will not lose contact with the ground at O . Hint: To compute the distance traveled by the car along the car’s path, observe that ds dx 2 dy 2 1 .dy=dx/ 2 dx and that

D

D

p

C

D

pC

1

ZpC

D

x

C 2 x 2 dx

1

D 2 1 C C 2x2 C 2C lnC x C

p

1

pC

C 2x 2 :



Solution

The minimum speed to lose contact with the ground is such that 2 vmin 

2 vmin

Dg )

D g.0/:

(1)

To calculate the radius of curvature at the srcin of the coordinate system indicated in the problem’s figure we use the following equation: 2 3=2

.x/

D

Therefore Since v 0

2 3=2

1 C .dy=dx/ 

ˇ

d 2 y=dx 2

)

ˇ

vmin

.0/

D

1 C .0:006x/  0:006

D pg D 40:44 m=s:

ˇˇ

x D0

D 166:7 m:

(2)

(3)

D 160 km=h D 44:44 m=s, we conclude that

The car will lose contact with the ground.

PD

a c such that the car does not lose contact. Since Now we have to find the minimum constant value of v we need to relate a change in speed to a change in position, we can use the constant acceleration equation v 2  v02 2ac .s  s0 / with s 0 0 and vf v min , where s is the path coordinate along the profile of the hill and the subscript f stands for ‘final.’ This gives

D

D

D

2 2 vmin  v0

D 2ac s )

ac

Dv

2

min 

v02

2sf

:

(4)

To evaluate this equation we need to express the path coordinate s in terms of the Cartesian coordinates x and y . Taking advantage of the hint given in the problem, we can write

Z qC 0

sf

D

100 m

1

.0:006x/ 2 dx

D

x p 2

1

C C 2x2 C 2C1 ln C x C 1 C C 2x2



p




0 100 m

D 105:7 m; (5) June 25, 2012


Solutions Manual

384 where C and v min

D 0:006 m . Using this result in Eq. (4), along with the fact that v0 D 160 km=h D 44:44 m=s D 40:44 m=s (see Eq. (3)), we have 

1

ac

D

2

1:606 m=s

:


June 25, 2012



Solutions Manual

386 which can be solved for  to obtain



D  2rad ln

f

 

)

0



D  4rad ln

v  f

v0

where we have used Eq. (1) and the following property of logarithms: ln x n that  d =ds . Next recalling that an 8g and an v 2 =, we can write

D

v2

D 8g

D )

D

dv ds

2v

We now observe that v dv is also the quantity v ds

dv dt

D 8g dds )

v

dv ds

;

(6)

D n ln x. From Eq. (2) we see

D 4g :

(7)

dv=dt . Therefore, we can write

PD

D 4gdv : (8) Letting t D 0 correspond to the beginning of the maneuver, we have v D v 0 for t D 0 and v D vf for t D tf D 4g  )

dt

so that we can integrate the last of Eqs. (8) as follows:

Z

tf

dt 0

D

Z

vf v0

dv 4g 

)

tf

D vf4gv0 ) 

tf

v0 / rad f ; D .v16g ln.vf =v0 / 

(9)

9:81 m=s2 , v0 1100 km=h where we have used the last of Eqs. (6). Recalling that we have g 1000 1100 1000 m = s, and v 800 km = h 800 m = s, we can evaluate the last of Eqs. (9) to obtain f 3600 3600

D

D

D

tf

D

D

D 5:238 s:


June 25, 2012


Dynamics 2e

387

Problem 2.296 The mechan ism show n is called a swinging block slider cran k. First used in various steam locomotive engines in the 1800s, this mechanism is often foun d in door-closing systems. Let H 1:25 m, R 0:45 m, and r denote the distance between B and O . Assuming that the speed of B is constant and equal to 5 m=s, determine r ,  , r , and  when  180 ı .

D

D

P P R

D

R

Solution

We start by defining a polar coordinate system with srcin at O and radial direction along the segment OB so that the coordinate r measures the distance from O to B . We take as transverse coordinate the angle  . Observe that for 0 <  < 180 ı , r is growing. For  180ı , r achieves its maximum value r max R H , and for 180ı <  < 360 ı , r decreases. Therefore, 180 ı must be equal to the rate of change of r for  zero, i.e.,

D C

D

D

P

r

ˇ

 D180

ı

D 0:

(1)

In polar coordinates, the velocity of B is expressed as

E D rP uO r C r P uO

vB For 

D

:

(2)

180 ı , recalling that B is moving counterclockwise with the constant speed

v 0 , we have

D v0 uO )  D180 D R Cv0 H : Recalling that v 0 D 5 m=s, R D 0:45 m, and H D 1:25 m, we can evaluate the last of Eqs. (3) to obtain P D180 D 2:941 rad=s: Eˇ

vB

 D180

ı

Dr

max

 Pˇ

  D180

ı

ˇ

 uO







ı



Pˇ



ı



(3)



Next we recall that the general expression of the acceleration in polar coordinates is

aB

(4) E D .rR r P2/ uO r C .r R C 2rP P/ uO : Recalling that, for  D 180 ı , r D R C H , rP D 0 , and P takes on the expression in the last of Eqs. (3), Eq. (4) reduces to   v02 aEB D180 D rR D180 uO r C Œ.R C H /R D180 ç uO : (5) RCH

ˇ



ı

ˇ





ı





ˇ



ı




June 25, 2012


Solutions Manual

388

Since B is in uniform circula r motion along a circle with center at A , the acceleration of B is always directed toward A and, for  180 ı , we must have

D

E

aB

ˇ

 D180

ı

D



v02 ur : R

O

(6)

Setting Eqs. (5) and (6) equal to each other component by component, we have

R

r and

ˇ

 D180

ı



v02

D RCH



.R

v02 R

) rR

ˇ

 D180

ı

D v02 R C1 H





1 R



(7)

(8) C H /R D180 D 0: Recalling that v 0 D 5 m=s, R D 0:45 m, and H D 1:25 m, we can evaluate the last of Eqs. (7) and Eq. (8) to obtain rR D180 D 40:85 m=s and R D180 D 0:

ˇ



ı



ˇ



2

ı

ˇ



ı


June 25, 2012


Dynamics 2e

389

Problem 2.297 The cam is mounted on a shaft that rotates about O with constant angular velocity ! cam. The profile of the cam is descr ibed by the funct ion `. / R0 .1 0:25 cos3  /, where the angle  is measured relative 3 cm, to the segment OA, which rotates with the cam. Letting R0 determine the maximum value of angular velocity !max such that the maximum speed of the follower is limited to 2 m=s. In addition, compute the smallest angle  min for which the follower achieves it maximum speed.

D

C

D

Solution

Let y denote the position of the follow er when in contact with the cam. In addition, let f (where the subscript f stands for follo wer) denote the value of  that identifies the radial line on the cam that goes from 90 ı   . Keeping in point O to the follower. Hence, we have that the relation between f and  is f mind that y describes the velocity of the follower, we have

D

P

y

D `.f / D R0.1 C 0:25 cos3 f / ) yP D dd`f ddtf :

P D P and P D !

Since f



cam ,

(1)

we have

y

P D P dd`f ) yP D 

!cam

d` : d f

(2)

Since ! cam is constant, y is maximum when dd` is maximum. Hence, taking the derivative of `. f / with f respect to f , we have

P

d` d f

D

0:75R0 cos

2

f sin f :

(3)

To maximize dd` , we differentiate the above quantity with respect to f and set the result equal to 0. This f gives

d 2` d f2

D 1:5R0 cos f sin2 f



0:75R0 cos3 f

cos2 f

D0 )

)



2 sin2 f

ˇ Dp ˇP ˇD p

sin f yP max

1=3 and

D0

ˇ

cos 2 f yP max

D 2=3:

(4)

Thus, the maximum magnitude of Eq. (3) and the maximum magnitude of Eq. (2) are, respectively,

d` df

jP jDv

Setting ymax

max

ˇ ˇ

R0 max

D 2p 3

!camR0

and

and solving for ! cam, we have

!max

Dv

max 2

p3

R0

)

ymax

!max

2 3 :

D 230:9 rad=s,


(5)

(6) June 25, 2012


Solutions Manual

390

D

D

D

2 m=s and R 0 3 cm 0:03000 m. The minimum angle  for which where we have used the data v max 90 ı   and that, by the first of the follower achieves its maximum speed is obtained by recalling that f Eqs. (4), we have sin f

D

p

1=3

)




D sin1

p ˇ 1=3

yP max

min

D 90ı



D



)

min

D 90ı



sin1

p

1=3;

(7)

D 54:74 ı:


June 25, 2012



Solutions Manual

392

P

Recalling that  is given in Eq. (3), and substituting Eqs. (4) and (6) into Eq. (5), we obtain

RD





v02 ŒR 2  .RL R 2 / cos ç 2 cos 3  sin  R2 .R L  R cos  /4

C



C

(

v02 .RL R 2 / sin  cos2  R2 .R L  R cos  /2

C

)

R sin  ŒR2  .RL R 2 / cos ç 2 : (7) .R L  R cos  /3

C

C

C

1000 137 m, L 15 m, and v 0 210 km=h 210 3600 m =s, for  30 ı , again we Recalling again that R ı 64:04 ), and then evaluate  in Eq. (7) to can first evaluate  from the second of Eqs. (1) (this gives  obtain

D

D

D

RD



D

0:2380 rad=s

2

D

D

R

:


June 25, 2012


Dynamics 2e

393

Problem 2.299

D

A plane is initially flying north with a speed v 0 430 mph relative to the ground 12 mph in the north-south direction. while the wind has a constant speed v W 0:45 mi. Assume that the The plane performs a circular turn with radius of  airspeed indicator on the plane measures the absolute value of the component of the relative velocity of the plane with respect to the air in the direction of motion. Then determine the value of the tangential component of the airplane’s acceleration when the airplane is halfway through the turn, assuming that the airplane maintains constant the reading of the airspeed indicator.

D

D

3

3

Solution

Let the subscripts P and W denote quantities pertaining to the airplane and wind, respectively. Then, referring to the figure on the right and using a normal tangential component system, the velocity of the airplane and wind are

E D vP uO t

vP

and

E D vW .

vW

 cos 

O C sin  uO n/:

ut

(1)

Therefore the velocity of the airplane relati ve to the wind is

E

vP =W

D .vP C vW cos  / uO t



O

vW sin  un :

(2)

E

Now we recall that the quantity measured by the airspeed indicator is the component of vP =W that is in the direction of motion, i.e., vai vP =A u t vP vW cos  ;

D jE



O jD C

where the subscript ‘ ai’ stands for ‘airspeed indicator.’ Now, recalling that the measure of  in radians is given by 

D s =, we can rewrite v D

ai

as follows:

vai

D vP C vW cos s :

(3)

D C

Observing that for  0 we have v ai v 0 vW and recalling that v ai is maintained constant along the turn, we can solve Eq. (3) for vP as a function of s to obtain

vP .s/

D v 0 C vW



1  cos

s : 



(4)

The tangential component of acceleration is the time derivative of Eq. (4). Hence, we can write

D dvdsP ds D vP dvdsP ) at D v0 C vW 1 cos s vW sin s : (5) dt p For s= D .  =4/ rad, we have sin .s= / D cos .s=  / D 2=2. Therefore, midway through the turn, we have p2 v p2 W a t D v0 C vW 1 : (6) 2 2



at

"

D







 !# 

D

D

430 mph 430 5280 = v 12 mph Recalling that v0 3600 ft s, W 0:45.5280/ ft, we can evaluate the above expression to obtain at

D

12 5280 = 3600 ft s, and 

D

0:45 mi

D

2

D 3:330 ft=s :


June 25, 2012


Solutions Manual

394

Problem 2.300 A fountain has a spout that can rotate about O and whose angle ˇ is controlled so as to vary with time ˇ0 Œ1 sin 2 .!t/ ç, with ˇ0 15ı and ! 0:4 rad=s. The length of the spo ut is according to ˇ L 1:5 ft, the water flow through the spout is constant, and the water is ejected at a speed v 0 6 ft=s, measured relative to the spout. Determine the largest speed with which the water particles are released from the spout.

D

D

C

D

D

D

Solution

We define a polar coordinate system with srcin at O and transverse coordinate coinciding with ˇ . The expression of the velocity of the water particle s as they leave the spout is given by

E D rP uO r C LˇP uO ˇ )

v

where, based on the problem statement,

v

q

D rP 2 C .Lˇ/P 2:

(1)

P D 2ˇ0! sin !t cos !t D ˇ0! sin.2!t/; (2) where we have used the trigonometric identity 2 sin x cos x D sin .2x/ . Substituting Eqs. (2) into the last of P D v 0 D 6 ft=s

r

and

ˇ

Eqs. (1) we obtain the following expression for the speed:

v

D

q

v02

C ŒLˇ0! sin.2!t/ ç2:

(3)

From the above expression, recalling that v0 , ˇ0 , and ! are constant, we see that v is maximum when sin.2!t/ is maximum. Since the sine function has maximum value equal to 1, we have

q

C L2ˇ02! 2 ) v D 6:020 ft=s, (4) where we have used the following numerical data: v 0 D 6 ft=s, L D 1:5 ft, ˇ 0 D 15 ı , and ! D 0:4  rad =s. vmax

D

v02

max


June 25, 2012


Dynamics 2e

395

Problem 2.301 A fountain has a spout that can rotate about O and whose angle ˇ is controlled so as to vary with time ˇ0 Œ1 sin 2 .!t/ ç, with ˇ0 15ı and ! 0:4 rad=s. The length of the spo ut is according to ˇ L 1:5 ft, the water flow through the spout is constant, and the water is ejected at a speed v 0 6 ft=s, measured relative to the spout. Determine the magnitude of the acceleration immediately before release when ˇ 15 ı .

D

D

C

D

D

D

D

Solution

We define a polar coordinate system with srcin at O and transverse coordinate coinciding with ˇ . The expression of the acceleration in polar coordinates is

E D .rR

a



P O C .r ˇR C 2rP ˇ/P uO ˇ :

r ˇ 2 / ur

(1)

Now, recalling that v 0 is constant, when the water particles leave the spout we have

D L; rP D v0; rR D 0; ˇP D 2ˇ0! sin !t cos !t D ˇ 0! sin.2!t/; and ˇR D 2! 2ˇ0 cos.2!t/; (2) where we have used the trigonometric identity 2 sin x cos x D sin.2x/ . Now, we observe that ˇ 0 D 15ı , and therefore for ˇ D 15ı , we must have ˇ D ˇ0 and sin.!t/ D 0, which implies sin.2!t/ D 0 and cos.2!t/ 1 . Hence, for ˇ 15 ı , we have D D r D L; rP D v 0 ; rR D 0; ˇP D 0; and ˇR D 2! 2 ˇ0 ; (3) so that, for ˇ D 15 ı , the acceleration takes the form aE D 2L! 2 ˇ0 uO ˇ : (4) ı Consequently, for ˇ D 15 , we have r

jEaj D 2L! 2ˇ0 )

2

jEaj D 1:240 ft=s , where we have used the following numerical data: L D 1:5 ft, ˇ 0 D 15 ı , and ! D 0:4  rad=s.


(5)

June 25, 2012


Solutions Manual

396

Problem 2.302 A fountain has a spout that can rotate about O and whose angle ˇ is controlled so as to vary with time ˇ0 Œ1 sin 2 .!t/ ç, with ˇ0 15ı and ! 0:4 rad=s. The length of the spo ut is according to ˇ L 1:5 ft, the water flow through the spout is constant, and the water is ejected at a speed v 0 6 ft=s, measured relative to the spout. Determine the highest position reached by the resulting water arc.

D

D

C

D

D

D

Solution

Referring to the figure at the right, in the solution of this problem we will use two coordinate systems. The first is a polar coordinate system with srcin at O and transverse coordinate ˇ . Let ˇ be the value of ˇ corresponding to the trajectory of the water jet achieving the maximum height. Then the second is a fixed Cartesian coordinate system with srcin coinciding with the position of the mouth of the spout corresponding to ˇ . Letting v denote the speed of the water particles at the spout’s mouth, then the velocity of the water particles at the spout’s mouth is

N

N

E D v.cos ˇ {O C sin ˇ |O /:

v

(1)

Once the water particles leave the spout, they are in projectile motion and the trajectory with the maximum height is determine d both by the vertical component of the speed, i.e,

vy

D v sin ˇ;

(2)

and the vertical position of mouth of the water spout. To derive an expression for the speed v of the water particles as they leave the spout, we use the chosen polar coordinate system, for which the velocity at the mouth of the spout is expressed as

E D rP uO r C LˇP uO ˇ )

v

where, based on the problem statement,

v

q

D rP 2 C .Lˇ/P 2:

(3)

P D 2ˇ0! sin !t cos !t D ˇ0! sin.2!t/; (4) where we have used the trigonometric identity 2 sin x cos x D sin .2x/ . Substituting Eqs. (2) into the last of P D v 0 D 6 ft=s

r

and

ˇ

Eqs. (3) we obtain the following expression for the speed:

v

D

q

v02

C ŒLˇ0! sin.2!t/ ç2:

(5)

Consequently, the vertical component of velocity at the spout is

vy

D

q

v02

C ŒLˇ0! sin.2!t/ ç2 sin ˇ0.1 C sin2 !t/:


(6) June 25, 2012


Dynamics 2e

397

D C

h 1 h2 where h 1 is the height of the mouth of the spout with respect to the base of the spout, Now let h and where h 2 is the maximum elevation of the water jet measured from the horizontal line going through the mouth of the spout. Hence, for h 1 we have

D L sin ˇ D L sin ˇ0.1 C sin2 !t /: By contrast, h2 is found using the constant acceleration equation v 2 D v02 C 2a.s direction. Recalling that vy D 0 for y D h 2 , we have h1

0

D



v02

q

C ŒLˇ0! sin.2!t/ ç2 sin

D 2g1

n

v02

C sin2 !t /



which can be solved for h 2 to obtain

h2

ˇ0 .1

C ŒLˇ0! sin.2!t/ ç2

o

sin2 ˇ0 .1



 

(7)  s0 /

in the vertical

2 

2gh 2 ;

C sin2 !t/:

(8)

Consequently, the expression for the height of the water jet as a function of time is

h

D L sin ˇ0.1 C sin2 !t / C 2g1

n

v02

C ŒLˇ0! sin.2!t/ ç2

o

sin2 ˇ0 .1



C sin2 !t /:

(9)

This function needs to be maximized. This can be done by differentiating h with respect to time and setting the result equal to zero. Doing so, after simplificati on, yields the following equation:

ˇ0 ! sin.2!t/ 2Lg cos ˇ0 .1 2g

n

C sin2 !t/ C 2Lˇ0 cos.2!t/ sin2 ˇ0.1 C sin2 !t/ C Œv02 C L2ˇ02 C ! 2 sin2.2!t/ ç sin 2ˇ0.1 C sin2 !t/ D 0 (10)   Recalling that we have L D 1:5 ft, ˇ 0 D 15 ı , ! D 0:4  rad=s, and g D 32:2 ft=s , and although this may



o

2

require plotting the terms within braces as a function of time, it turns out that the term within braces can never be equal to zero. Hence, the solution of the above equation reduc es to the solution of the equation

D 0 ) t D 0 ˙ n! and t D 2! ˙ n! ; with n D 0 ;1;2;:: : (11) Since the function ˇ D ˇ 0 .1 C sin2 !t/ is at a maximum for t D 2! ˙ n! the function h will also achieve its maxima for t D 2! ˙ n! . In addition, since the function h is a periodic function with period ! , the values of h for t D 2! ˙ n! are all identical to one another and we can therefore evaluate h by simply letting t D 2! , i.e., for n D 0 . Hence, recalling that L D 1:5 ft, ˇ 0 D 15ı , ! D 0:4  rad=s, and g D 32:2 ft=s , for t D 2! D 1:250 s, we can evaluate h in Eq. (9) to obtain sin.2!t/















max



2



hmax

D 0:8898 ft:


June 25, 2012




Solutions Manual

400

Problem 2.305 Block B is released from rest at the position shown, and it has a 5:7 m=s2 . Determine the constant acceleration downward a 0 velocity and acceleration of block A at the instant that B touches the floor.

D

Solution

The length of the rope is

q

C yA2 C yB : (1) For t D 0, we have that yA .0/ D l w . Letting tf represent the L

D

d2



time at the final position, the length of the rope at the initial and final positions is

L

d2

D

C yA2 .0/CyB .0/

and

L

D

d2

C yA2 .tf /CyB .tf /:

(2) Equating these two expressions for L and rearranging terms, we have

q

q

d2

C yA2 .tf /

q q D

d2

C yA2 .0/



D 0: D h, we obtain d 2 C yA2 .0/ C h2 :

ŒyB .tf /  yB .0/ç

(3)

Ssquaring both sides of the above equation, with yB .tf /  yB .0/

d2

C yA2 .tf / D d 2 C yA2 .0/



2h

Solving the above equation for yA .tf / , we have

r

(4)

q

C yA2 .0/ D 0:1864 m; (5) where we have used the following numerical data: yA .0/ D l w D 3:75 m, h D 2 m, and d D 2:5 m. Next, differentiating Eq. (1) with respect to time and solving for yPA , we have yA yPA yPB 0 yB yA d 2 y2: (6) 2 2 D d C yA C P ) P D yA CA Now, we use the constant acceleration equation sP 2 D sP02 C 2ac .s s0 / to find yPB after B has traveled a distance h . This gives 2 yPB D 2a0h ) yPB D 2a0h D 4:775 m=s; (7) yA .tf /

D

yA2 .0/

q

C h2



2h d 2





q

q



p


June 25, 2012


Dynamics 2e

401

a0 5:7 m=s2 and h 2 m. Then, using Eq. (5) and the last of where we have used the fact that yB Eqs. (7), and recalling that d 2:5 m, we can evaluate Eq. (6) to obtain for t tf to obtain

D

R D D P

D

yA .tf /

D

64:22

D

m=s:

(8)

Now we differentiate Eq. (6) with respect to time to obtain

R D yyARB 

yA

q

d2

C yA2 C yPByy2PA A

q

d2

C yA2



q P CP

yB yA

d2

yA2

:

Recalling that d 2:5 m and yB a 0 5:7 m=s2 , and using the (full precision) value of yA .tf / in Eq. (5), the (full precision) value of yA .tf / in Eq. (8), and the (full precision) value of yB .tf / in the last of Eqs. (7), we can evaluate the above expression for t tf to obtain

D

P

R D D

P

D

R

yA .tf /

D

2

22;080 m=s

:


June 25, 2012


Solutions Manual

402

Problem 2.306

D

At a given instant, an airplane is flying horizontally with speed v0 290 mph and acceleration a 0 12 ft=s2 . At the same time, the airplane’ s 1500 rpm while accelerating at propellers rotate at an angular speed ! a rate ˛ 0:3 rad=s2 . Knowing that the propeller diameter is d 14 ft, determine the magnitude of the acceleration of a point on the periphery of the propellers at the given instant.

D

D

D

D

Solution

Using a cylindrical coordinat e system with srcin at the propeller’s axis of rotation and of motion, the general expression for acceleration is

´ axis in the direction

E D RR RP 2 uO R C RR C 2RP P  uO C ´R uO ´: For the propeller tip we have R D d =2, RP D 0 , RR D 0 , ´R D a 0 , P D ! , and R D ˛ . Thus, the accelera tion is a



ED

a





d! 2 uR 2

O C d˛2 uO C a0 uO ´; 

and the magnitude of the acceleration is

ˇEˇ D s a



where we have used the data a 0

d! 2 2

2

  d˛  C

2

2

C a02 )

ˇ Eˇ D a

172;700 ft=s2 ,

D 12 ft=s , ! D 1500 rpm D 1500 260 rad=s, ˛ D 0:3 rad=s , and d D 14 ft. 2




2

June 25, 2012


Dynamics 2e

403

Problem 2.307 A golfer chips the ball as show n. Treating ˛ , ˇ , and the initial speed v 0 as given, find an expression for the radius of curvature of the ball’s trajectory as a function of time and the given parameters. Hint: Use the Cartesian coordinate system shown to determine the acceleration and the velocity of the ball. Then reexpress these quantities, using normal-tangential components.

Solution

This is 3 -D projectile motion. We will follow the hint and develop equations both in Cartersian components as well as in normal-tangential components. In Cartesian coordinates the components of acceleration are

R D 0; yR D 0;

x

RD

´

and

g:

OOO

O O O

Using constant acceleration equations, the velocity vector expressed in the . {; | ; k/ and . u t ; un ; ub / component systems are

E D v 0 cos ˇ cos ˛ {O C v0 cos ˇ sin ˛ |O C .v0 sin ˇ vE D v uO t ; v



O

gt / k;

where, v is the speed, which can be given the form

v

D

q

vx2

C vy2 C v´2 )

v

D

q

v02

C g 2t 2



2v0 gt sin ˇ;

(1)

and where the tangent unit vector u t can be related to the base vectors of the Cartesian component system as follows: v0 cos ˇ cos ˛ v0 cos ˇ sin ˛ v0 sin ˇ  gt ut { | k: (2)

O

O D

OC

v

OC

v

O

v

To express the acceleration in normal-tangential components, we will need an expression for the time derivative of the speed. Hence, we proceed to differentiate v with respect to time, to obtain

PD

v

g2t

q

v02

C



v0 g sin ˇ

g2t 2 

2v0 gt sin ˇ

OOO

D gv gt





v0 sin ˇ :

(3)

O O O

The acceleration vector expressed in the . {; | ; k/ and . u t ; un ; ub / component systems are

v2

E D vP uO t C  uO n

a

and

ED

a

O

g k :

Equating the above expressions for the acceleration, we obtain an expression for the normal unit vector



un

O D

 2 v

Using Eq. (3), we can rewrite Eq. (4) as follows:

O D

un



g 1 .g t v2 v



vP uO C g kO : t



O C kO

v0 sin ˇ/ u t

(4)



:


June 25, 2012


Solutions Manual

404

O O D 1, we obtain the following equation :  1 2 1 C 2 .gt v0 sin ˇ/ 2 C .g t v0 sin ˇ/ kO uO t :

Now, recalling that we must have un un 

1

D

2 g2

v4





v



v

(5)



O O

Observe that the last term on the right-hand side of Eq. (5) requires the computation of the product k u t . v0 sin ˇ gt This computation can be done by using Eq. (2), which shows that k u t . Consequently, Eq. (5) v becomes 

O O D 

2 2

1

D vg4



1

C v12 .gt



v0 sin ˇ/2 

2 .gt v2



v0 sin ˇ/2



2 2

D vg6

v 2  .gt





Keeping in mind that the above expression is to be solved for , the term v 2  .gt 1  sin2  as follows: simplified using Eq. (1) and the trigonometric identity cos 2 

D

v 2  .gt



v0 sin ˇ/ 2

D v2



g2t 2

C 2gtv 0 sin ˇ



v02 sin 2 ˇ

D v02



v0 sin ˇ/2 :



(6)



v0 sin ˇ/2 can be

v02 sin 2 ˇ

D v 02.1



sin2 ˇ/

D v02 cos2 ˇ:

(7)

This result allows us to rewrite Eq. (6) as follows: 2 2

1

D vg6

v02 cos 2 ˇ:

Solving Eq. (6) for  , we have

(8)

3



D gv0vcos ˇ :

(9)

Finally, substituting the expression for v in Eq. (1), we have 2



2 2

0 gt sin ˇ/ D .v0 C g tgv0 2v cos ˇ

3=2

:


June 25, 2012


Dynamics 2e

405

Problem 2.308 A carnival ride called the octopus consists of eight arms that rotate about the ´ axis with a constant angular 6 rpm. The arms have a length L 8 m and form an angle  with the ´ axis. Assuming that velocity  70:5ı ,  1 25:5ı , and ! 1 rad=s, determine  varies with time as  .t /  0 1 sin !t with  0 the magnitude of the accelera tion of the outer end of an arm when  achieves its minimum value.

PD

D

D C

D

D

D

Photo credit: © Gary L. Gray

Solution

Using a spherical coordinate system the components of acceleration of a point C at the end of an arm with constant length L are

ar a a

D L12! 2 cos2 !t LP 2 sin2.0 C 1 sin !t/; D L! 21 sin !t LP 2 sin.0 C 1 sin !t / cos.0 C 1 sin !t/; D 2L1! P cos !t cos.0 C 1 sin !t/: 

(1)





(2)



(3)

D C 1 sin !t . Because the mininum of the sine

To determine the minimum value of  , recall that  0 function is equal to 1, then the minimum valu e of  is

min

D 0



1

D 45ı D 41  rad;

(4)

D 70:5ı and  D 25:5ı. We observe that  occurs when sin.!t/ D 1, !t D 270 ı D 32  rad: (5) Evaluating the acceleration components for  D  D 41  rad and ! t D 32  rad, we have   ar D 21 LP 2 ; a D L ! 2 1 21 P 2 ; and a D 0: (6) Consequently, the magnitude of the acceleration for  D  is jE aj D ar2 C a2 C a2, which gives

where we have used the data  0 i.e., when



E

a

ˇˇ

min

ˇ

min

min





D L 21 P 4 C ! 412

q



P

! 2  2 1

where we have used the following numerical data: L  1 25:5ı 25:5 180 rad.

D

D



min



min

min

)

a

E

ˇˇ

q





2

min

D 2:534 m=s ,

D 8 m, P D 6 rpm D 6 260 rad=s, ! D 1 rad=s, and 


June 25, 2012


Solution Manual for Engineering Mechanics Statics and Dynamics 2nd Edition by Plesha

Recommend Documents