DYNAMICS solution chapter11

CHAPTER 11

PROBLEM 11.1 The motion of a particle is defined by the relation x = 1.5t 4 − 30t 2 + 5t + 10, where x and t are expressed in meters and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle when t = 4 s.

SOLUTION Given:

x = 1.5t 4 − 30t 2 + 5t + 10 dx = 6t 3 − 60t + 5 dt dv = 18t 2 − 60 a= dt v=

Evaluate expressions at t = 4 s. x = 1.5(4) 4 − 30(4) 2 + 5(4) + 10 = −66 m

x = −66.0 m �

v = 6(4)3 − 60(4) + 5 = 149 m/s

v = 149.0 m/s �

a = 18(4) 2 − 60 = 228 m/s 2 �

a = 228.0 m/s 2 ��

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3

PROBLEM 11.2 The motion of a particle is defined by the relation x = 12t 3 − 18t 2 + 2t + 55, where x and t are expressed in meters and seconds, respectively. Determine the position and the velocity when the acceleration of the particle is equal to zero.

SOLUTION Given:

x = 12t 3 − 18t 2 + 2t + 5 dx = 36t 2 − 36t 36t + 2 dt dv a= = 72t − 36 dt v=

Find the time for a = 0. 72t − 36 = 0 � t = 0 0.5 s

Substitute into above expressions. x = 12(0.5 12(0.5) 12 (0.5))3 − 18(0.5) 2 + 2(0. 2(0.5) 2( 0.5) 5) + 5 = 3

v = 36(0.5 36(0.5) 36 (0.5)) 2 − 36(0.5)) + 2 = −7 m/s

x = 3.00 m �

v = −7.00 m/s ��

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4

PROBLEM 11.3 5 3 5 2 t − t − 30t 30t + 8 x, where x and t are expressed in 3 2 meters and seconds, respectively. Determine the time, the position, and the acceleration when v = 0.

The motion of a particle is defined by the relation x =

SOLUTION We have

5 5 x = t 3 − t 2 − 30t + 8 3 2

Then

v=

dx = 5t 2 − 55tt − 30 dt

and

a=

dv = 10t − 5 dt

When v = 0: or At t = 3 s:

5t 2 − 5t − 30 = 5(t 2 − t − 6) = 0 t = 3 s and and t = −2 s (Reject)

t = 3.00 s

5 3 5 2 x3 = (3 (3)) − ((3) 3) − 30(3)) + 8 3 2

or

a3 = 10(3) − 5

or a3 = 25.0 m/s2

x3 = −59.5 m


5

PROBLEM 11.4 The motion of a particle is defined by the relation x = 6t 2 − 8 + 40 cos π t , where x and t are expressed in meters and seconds, respectively. Determine the position, the velocity, and the acceleration when t = 6 s.

SOLUTION We have

x = 6t 2 − 8 + 40 cos π t

Then

v=

dx = 12t − 40π sin π t dt

and

a=

dv = 12 − 40π 2 cos π t dt

At t = 6 s:

x6 = 6(6) 2 − 8 + 40 cos 6π

or

x6 = 248 m

v6 = 12(6) − 40π sin 6π

or

v6 = 72.0 m/s

a6 = 12 − 40π 2 cos 6π

or

a6 = −383 m/s2

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6

PROBLEM 11.5 The motion of a particle is defined by the relation x = 6t 4 − 22tt 3 − 12t 2 + 3t + 33, where x and t are expressed in meters and seconds, respectively. Determine the time, e, the position, and the velocity when a = 0.

SOLUTION We have

x = 6t 4 − 2t 3 − 12t 12t 2 + 3t + 3

Then

v=

dx = 24t 3 − 6t 2 − 224 4t + 3 dt

and

a=

dv = 72t 2 − 12t 12t − 24 dt

When a = 0:

72t 2 − 12t − 24 = 12(6 12(6t (6t 2 − t − 2) = 0 (3t − 2)(2 2)(2t + 1) 1) = 0

or t=

or At t =

2 s: 3

2 1 s and and t = − s (Reject Re ) Reject 3 2 4

3

2

�2� �2� �2� �2� x2/3 = 6 � � − 2 � � − 12 � � + 3 � � + 3 3 3 3 � � � � � � �3� 3

t = 0.667 s �

or

x2/ 2/3 = 0.259 m �

2

�2� �2� �2� v2/3 = 24 � � − 6 � � − 2244 � � + 3 3 3 � � � � �3�

or v2/3 = −8.56 m/s �


7

PROBLEM 11.6 The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x is expressed in meters and t in seconds. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION x = 2t 3 − 15t 2 + 24t + 4 dx = 6t 2 − 30t + 24 v= dt dv = 12t − 30 a= dt

(a)

v = 0 when

6t 2 − 30t + 24 = 0 6(t − 1)(t − 4) = 0

(b)

a = 0 when

For t = 2.5 s:

12t − 30 = 0

t = 1.000 s or t = 4.00 s

�

t = 2.5 s

x2.5 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4 x2.5 = +1.500 m �

To find total distance traveled, we note that v = 0 when t = 1 s:

x1 = 2(1)3 − 15(1)2 + 24(1) + 4 x1 = +15 m

For t = 0,

x0 = +4 m

Distance traveled From t = 0 to t = 1 s: From t = 1 s to t = 2.5 s:

x1 − x0 = 15 − 4 = 11 m → x2.5 − x1 = 1.5 − 15 = 13.5 m ←

Total distance traveled = 11 m + 13.5 m = 24.5 m

�

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8

PROBLEM 11.7 The motion of a particle is defined by the relation x = t 3 − 6t 2 − 36t − 40, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the velocity, the acceleration, and the total distance traveled when x = 0.

SOLUTION We have

x = t 3 − 6t 2 − 36t − 40

Then

v=

dx = 3t 2 − 12t − 36 dt

and

a=

dv = 6t − 12 dt

(a)

3t 2 − 12t − 36 = 3(t 2 − 4t − 12) = 0

When v = 0:

(t + 2)(t − 6) = 0

or

t = −2 s (Reject) and t = 6 s

or (b)

t 3 − 6t 2 − 36t − 40 = 0

When x = 0: Factoring

(t − 10)(t + 2)(t + 2) = 0 or t = 10 s

Now observe that

0 6s

and at t = 0: t = 6 s:

t = 6.00 s

t t

6 s:

v

0

10 s:

v

0

x0 = −40 m x6 = (6)3 − 6(6)2 − 36(6) − 40 = −256 m

t = 10 s:

Then

v10 = 3(10) 2 − 12(10) − 36

or v10 = 144.0 m/s

a10 = 6(10) − 12

or

a0 = 48.0 m/s2

| x6 − x0 | = | −256 − (−40) | = 216 m x10 − x6 = 0 − (−256) = 256 m

Total distance traveled = (216 + 256) m = 472 m

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9

PROBLEM 11.8 The motion of a particle is defined by the relation x = t 3 − 9t 2 + 24t − 8, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION We have

x = t 3 − 9t 2 + 24t − 8

Then

v=

dx = 3t 2 − 18t + 24 dt

and

a=

dv = 6 t − 18 dt

(a)

3 t 2 − 18t + 24 = 3(t 2 − 6t + 8) = 0

When v = 0:

(t − 2)(t − 4) = 0

or

or t = 2.00 s and t = 4.00 s (b)

When a = 0:

6t − 18 = 0 or t = 3 s x3 = (3)3 − 9(3)2 + 24(3) − 8

At t = 3 s: First observe that 0

t

2 s: 2s

t

3 s:

v

0

v

0

or

x3 = 10.00 m

Now At t = 0:

x0 = −8 m

At t = 2 s:

x2 = (2)3 − 9(2) 2 + 24(2) − 8 = 12 m

Then

x2 − x0 = 12 − (−8) = 20 m | x3 − x2 | = |10 − 12| = 2 m

Total distance traveled = (20 + 2) m = 22.0 m


10

PROBLEM 11.9 The acceleration of a particle is defined by the relation a = −8 m/s 2 . Knowing that x = 20 m when t = 4 s and that x = 4 m when v = 16 m/s, determine (a) the time when the velocity is zero, (b) the velocity and the total distance traveled when t = 11 s.

SOLUTION dv = a = −8 m/s 2 dt

We have

� dv = � −8 dt + C

Then

C = constant

v = −8t + C (m/s)

or

dx = v = −8t + C dt

Also x

t

At t = 4 s, x = 20 m:

�

or

x − 20 = [−4t 2 + Ct ]t4

20

dx =

� (−8t + C ) dt 4

x = −4t 2 + C (t − 4) + 84 (m)

or When v = 16 m/s, x = 4 m:

16 = −8t + C � C = 16 + 8t 4 = −4t 2 + C (t − 4) + 84

Combining Simplifying

0 = −4t 2 + (16 + 8t )(t − 4) + 80 t 2 − 4t + 4 = 0 t=2s

or

C = 32 m/s v = −8t + 32 (m/s)

and

x = −4t 2 + 32 t − 44 (m)

(a)

When v = 0:

(b)

Velocity and distance at 11 s.

−8t + 32 = 0

or t = 4.00 s �

v11 = −(8)(11) + 32

At t = 0:

v11 = −56.0 m/s �

x0 = −44 m

t = 4 s:

x4 = 20 m

t = 11 s:

x11 = −4(11) 2 + 32(11) − 44 = −176 m

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No of this this Manual Manual may may be be displayed, displayed, PROPRIETARY No part part of reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced you are are aa student student using using this this Manual, Manual, distribution to to teachers teachers and and educators educators permitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If distribution If you you are are using using it it without without permission. permission. you

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11

PROBLEM 11.9 (Continued) Now observe that

Then

0 � t � 4 s:

v�0

4 s � t � 11 s:

v�0

x4 − x0 = 20 − ( −44) = 64 m | x11 − x4 | = | − 176 − 20| = 196 m

Total distance traveled = (64 + 196) m = 260 m

��


12

PROBLEM 11.10 The acceleration of a particle is directly proportional to the square of the time t. When t = 0, the particle is at x = 24 m. Knowing that at t = 6 s, x = 96 m and v = 18 m/s, express x and v in terms of t.

SOLUTION a = kt 2

We have

k = constant

dv = a = kt 2 dt

Now v

t

At t = 6 s, v = 18 m/s:

�

or

1 v − 18 = k (t 3 − 216) 3

18

dv =

� kt dt 2

6

1 v = 18 + k (t 3 − 216)(m/s) 3

or

dx 1 = v = 18 + k (t 3 − 216) dt 3

Also x

t�

1

�

or

1 �1 � x − 24 = 18t + k � t 4 − 216t � 3 �4 �

24

dx =

� ��18 + 3 k (t

� − 216) � dt �

At t = 0, x = 24 m:

0

3

Now At t = 6 s, x = 96 m: or Then or and or

1 �1 � 96 − 24 = 18(6) + k � (6) 4 − 216(6) � 3 �4 � k=

1 m/s 4 9

1 � 1 �� 1 � x − 24 = 18t + � �� t 4 − 216t � 3 � 9 �� 4 � x(t ) =

1 4 t + 10t + 24 108

�

1�1� v = 18 + � � (t 3 − 216) 3� 9� v(t ) =

1 3 t + 10 27

�


13

13

PROBLEM 11.11 The acceleration of a particle is directly proportional to the time t. At t = 0, the velocity of the particle is v = 16 m/s. Knowing that v = 15 m/s and that x = 20 m when t = 1 s, determine the velocity, the position, and the total distance traveled when t = 7 s.

SOLUTION a = kt

We have

dv = a = kt dt

Now At t = 0, v = 16 m/s: or

v

dv =

16

v − 16 =

or

kt dt

1 2 kt 2 1 2 k t (m/s) 2 1 2 k(1 s ) 2

k = −2 m/ s3

and

v = 16 − t 2

dx = v = 16 − t 2 dt

Also

or

0

15 m/s = 16 m/s +

or

At t = 1 s, x = 20 m:

t

v = 16 +

or At t = 1 s, v = 15 m/s:

k = constant

x

20

dx =

t

1

(16 − t 2 ) dt

1 x − 20 = 16t − t 3 3

t 1

1 13 x = − t 3 + 16t + (m) 3 3

Then At t = 7 s:

When v = 0:

v7 = 16 − (7) 2

or v7 = −33.0 m/s

1 13 x7 = − (7)3 + 16(7) + 3 3

or

x7 = 2.00 m

16 − t 2 = 0 or t = 4 s


14

PROBLEM 11.11 (Continued)

At t = 0:

x0 =

13 3

1 13 x4 = − (4)3 + 16(4) + = 47 m 3 3

t = 4 s:

Now observe that

Then

0

t

4 s:

v

0

4s

t

7 s:

v

0

13 = 42.67 m 3 | x7 − x4 | = |2 − 47| = 45 m x4 − x0 = 47 −

Total distance traveled = (42.67 + 45) m = 87.7 m

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced distribution distribution to toteachers teachersand andeducators educatorspermitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you you are are aastudent studentusing usingthis thisManual, Manual, you you are are using using itit without without permission. permission.

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15

PROBLEM 11.12 The acceleration of a particle is defined by the relation a = kt 2 . (a) Knowing that v = −32 m/s when t = 0 and that v = +32 m/s when t = 4 s, determine the constant k. (b) Write the equations of motion, knowing also that x = 0 when t = 4 s.

SOLUTION a = kt 2 dv = a = kt 2 dt

(1)

t = 0, v = −32 m/s and t = 4 s, v = +32 m/s 32

(a)

−32

dv =

4 0

kt 2 dt

1 32 − (−32) = k (4)3 3

(b)

k = 3.00 m/s4

Substituting k = 3 m/s4 into (1) dv = a = 3t 2 dt

t = 0, v = −32 m/s:

v −32

dv =

t 0

a = 3t 2

3t 2 dt

1 v − (−32) = 3(t )3 3

v = t 3 − 32

dx = v = t 3 − 32 dt t = 4 s, x = 0:

x 0

dx =

t 4

t

(t 3 − 32) dt ; x =

1 4 t − 32t 4 4

1 4 1 t − 32t − (4) 4 − 32(4) 4 4 1 x = t 4 − 32t − 64 + 128 4 x=

x=

1 4 t − 32t + 64 4


16

PROBLEM 11.13 The acceleration of a particle is defined by the relation a = A − 6 t 2 , where A is a constant. At t = 0, the particle starts at x = 8 m with v = 0. Knowing that at t = 1 s, v = 30 m/s, determine (a) the times at which the velocity is zero, (b) the total distance traveled by the particle when t = 5 s.

SOLUTION a = A − 6t 2

We have

dv = a = A − 6t 2 dt

Now At t = 0, v = 0:

�

v 0

t

� ( A − 6t 0

2

)dt

30 = A(1) − 2(1)3

At t = 1 s, v = 30 m/s:

A = 32 m/s 2

or

and v = 32t − 2t 3

dx = v = 32t − 2t 3 dt

Also At t = 0, x = 8 m:

�

x

8

dx =

t

� (32t − 2t )dt 3

0

1 x = 8 + 16t 2 − t 4 (m) 2

or When v = 0:

32t − 2t 3 = 2t (16 − t 2 ) = 0 t = 0 and t = 4.00 s

or (b)

dv =

v = At − 2t 3 (m/s)

or

(a)

A = constant

At t = 4 s:

1 x4 = 8 + 16(4)2 − (4)4 = 136 m 2

t = 5 s:

1 x5 = 8 + 16(5) 2 − (5)4 = 95.5 m 2

�


17

17

PROBLEM 11.13 (Continued) Now observe that

Then

0 � t � 4 s:

v�0

4 s � t � 5 s:

v�0

x4 − x0 = 136 − 8 = 128 m

| x5 − x4 | = |95.5 − 136| = 40.5 m

Total distance traveled = (128 + 40.5) m = 168.5 m

�


18

PROBLEM 11.14 It is known that from t = 2 s to t = 10 s the acceleration of a particle is inversely proportional to the cube of the time t. When t = 2 s, v = −15 m/s, and when t = 10 s, v = 0.36 m/s. Knowing that the particle is twice as far from the origin when t = 2 s as it is when t = 10 s, determine (a) the position of the particle when t = 2 s, and when t = 10 s, (b) the total distance traveled by the particle from t = 2 s to t = 10 s.

SOLUTION a=

We have

k t3

k = constant

dv k =a= 3 dt t

Now At t = 2 s, v = −15 m/s:

�

v

−15

dv =

�

t

v − (−15) = −

or

v=

or At t = 10 s, v = 0.36 m/s:

0.36 =

k

2 t3

k2 2

dt �1 1 � � 2− � (2) 2 � �t

k�1 1 − 2 �� 4 t 2

k�1 1 − 2 � 2 � 4 10

or

k = 128 m ⋅ s

and

v =1−

(a)

� � − 15 �

64 (m/s) t2

64 dx = v =1− 2 dt t

We have

64 � dt + C 2 � �

�

� dx = � ��1 − t

Then

x=t+

or Now x2 = 2 x10 :

� � − 15 (m/s) �

2+

C = constant

64 + C (m) t

64 64 � � + C = 2 �10 + +C� 2 10 � �

or

C = 1.2 m

and

x=t+

64 + 1.2 (m) t


19

19


At t = 2 s:

x2 = 2 +

t = 10 s:

64 + 1.2 2

x10 = 10 +

64 + 1.2 10

Note: A second solution exists for the case x2 � 0, x10 � 0. For this case, C = −22 x2 = 11

and (b)

When v = 0:

1−

x2 = 35.2 m �

or

x10 = 17.60 m �

4 m 15

11 13 m, x10 = −5 m 15 15

64 = 0 or t = 8 s t2

At t = 8 s:

x8 = 8 +

Now observe that 2 s � t � 8 s:

v�0

8 s � t � 10 s:

v�0

Then

or

64 + 1.2 = 17.2 m 8

| x8 − x2 | = |17.2 − 35.2| = 18 m x10 − x8 = 17.6 − 17.2 = 0.4 m

Total distance traveled = (18 + 0.4) m = 18.40 m

�

Note: The total distance traveled is the same for both cases.


20

PROBLEM 11.15 The acceleration of a particle is defined by the relation a = −k/x. It has been experimentally determined that v = 15 m/s when x = 0.6 m and that v = 9 m/s when x = 1.2 m. Determine (a) the velocity of the particle when x = 1.5 m, (b) the position of the particle at which its velocity is zero.

SOLUTION a=

vdv − k = x dx

Separate and integrate using x = 0.6 m, v = 15 m/s. v 15

vdv = − k

x 0.6

v

dx x

1 2 = − k ln x v 2 15

x

0.6

x 1 2 1 v − (15) 2 = − k ln 2 2 0.6

(1)

When v = 9 m/s, x = 1.2 m 1 2 1 1.2 (9) − (15) 2 = −k ln 2 2 0.6

Solve for k. k = 103.874 m2/s2

(a)

Substitute

2 2 k = 103.874 m /s and x = 1.5 m into (1).

1 2 1 1.5 v − (15) 2 = −103.874 ln 2 2 0.6 v = 5.89 m

(b)

For v = 0, 1 x 0 − (15) 2 = −103.874 ln 2 0.6 ln

x = 1.083 0.6 x = 1.772 m

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced you are are aastudent studentusing usingthis thisManual, Manual, distribution to toteachers teachersand andeducators educatorspermitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you distribution you are are using using itit without without permission. permission. you

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21

PROBLEM 11.16 A particle starting from rest at x = 1 m is accelerated so that its velocity doubles in magnitude between x = 2 m and x = 8 m. Knowing that the acceleration of the particle is defined by the relation a = k[ x − (A/x)], determine the values of the constants A and k if the particle has a velocity of 29 m/s when x = 16 m.

SOLUTION We have

v v

When x = 1 m, v = 0:

0

dv A =a=k x− x dx

vdv =

x 1

k x−

A dx x

1 2 1 v = k x 2 − A ln x 2 2

or

=k

or

1 2 1 x − A ln x − 2 2

1 2 1 1 v8 = k (8) 2 − A ln 8 − = k (31.5 − A ln 8) 2 2 2

x = 8 m:

or

1

1 2 1 1 3 v2 = k (2) 2 − A ln 2 − =k − A ln 2 2 2 2 2

At x = 2 m:

Now

x

1 2 v 2 8 1 2 v 2 2

v8 = 2: v2

= (2) 2 =

k (31.5 − A ln 8) k ( 32 − A ln 2 )

6 − 4 A ln 2 = 31.5 − A ln 8 25.5 = A(ln 8 − 4 ln 2) = A(ln 8 − ln 24 ) = A ln

1 2 A = −36.8 m2

or When x = 16 m, v = 29 m/s:

Noting that We have

1 1 25.5 1 (29) 2 = k (16) 2 − ln(16) − 1 2 2 2 ln ( 2 ) ln(16) = 4 ln 2 and ln 841 = k 236 −

1 = − ln(2) 2

ln 25.5 = 4 ln(2) − 1 − ln(2) k = 1.832 s −2

or


22

22

PROBLEM 11.17 A particle oscillates between the points x = 40 mm and x = 160 mm with an acceleration a = k(100 – x), where a and x are expressed in mm/s2 and mm, respectively, and k is a constant. The velocity of the particle is 18 mm/s when x = 100 mm and is zero at both x = 40 mm and x = 160 mm. Determine (a) the value of k, (b) the velocity when x = 120 mm.

SOLUTION (a)

v

We have When x = 40 mm, v = 0:

�

v

0

dv = a = k (100 − x) dx

vdv =

�

x

40

k (100 − x) dx x

or

1 2 1 � � v = k �100 x − x 2 � 2 2 � 40 �

or

1 2 1 � � v = k �100 x − x 2 − 3200 � 2 2 � �

When x = 100 mm, v = 18 mm/s:

1 1 � � (18) 2 = k �100(100) − (100) 2 − 3200 � 2 2 � � k = 0.0900 s −2 �

or (b)

When x = 120 mm:

1 2 1 � � v = 0.09 �100(120) − (120) 2 − 3200 � = 144 2 2 � � v = ±16.97 mm/s ��

or


23

23

PROBLEM 11.18 A particle starts from rest at the origin and is given an acceleration a = k/( x + 4)2 , where a and x are expressed in mm/s2 and m, respectively, and k is a constant. Knowing that the velocity of the particle is 4 m/s when x = 8 m, determine (a) the value of k, (b) the position of the particle when v = 4.5 m/s, (c) the maximum velocity of the particle.

SOLUTION (a)

We have When x = 0, v = 0: or When x = 8 m, v = 4 m/s:

v

�

v

0

dv k =a= dx ( x + 4)2

vdv =

�

x

0

k dx ( x + 4) 2

1 2 1� � 1 v = −k � − � 2 � x+4 4� 1 2 1� � 1 (4) = −k � − � 2 �8+4 4� k = 48 m3 /s 2 �

or (b)

When v = 4.5 m/s:

1 1� � 1 (4.5) 2 = −48 � − � 2 � x+4 4� x = 21.6 m �

or (c)

Note that when v = vmax , a = 0. Now a → 0 as x → ∞ so that 1 2 1 � �1 �1� vmax = 48lnx →−1∞ � − = 48 � � � 2 �4 x+4� �4�

vmax = 4.90 m/s ��

or


24

PROBLEM 11.19 A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact the equipment experiences an acceleration of a = − kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.

SOLUTION a=

vdv = −k x dx

Separate and integrate.

�

vf

v0

vdv = −

�

xf

0

k x dx

1 2 1 2 1 v f − v0 = − kx 2 2 2 2

xf 0

1 = − k x 2f 2

Use v0 = 4 m/s, x f = 0.02 m, and v f = 0. Solve for k. 1 1 0 − (4) 2 = − k (0.02) 2 2 2

k = 40,000 s −2

Maximum acceleration. amax = −kxmax : ( −40,000)(0.02) = −800 m/s 2 a = 800 m/s 2

��


25

25

PROBLEM 11.20 Based on experimental observations, the acceleration of a particle is defined by the relation a = −(0.1 + sin x/b), where a and x are expressed in m/s2 and meters, respectively. Knowing that b = 0.8 m and that v = 1 m/s when x = 0, determine (a) the velocity of the particle when x = −1 m, (b) the position where the velocity is maximum, (c) the maximum velocity.

SOLUTION v

We have When x = 0, v = 1 m/s:

�

v

1

dv x � � = a = − � 0.1 + sin 0.8 �� dx �

vdv =

�

x

0

x � � − � 0.1 + sin � dx 0.8 � � x

x � 1 2 � (v − 1) = − �0.1x − 0.8 cos � 2 0.8 � �0

or

x 1 2 v = −0.1x + 0.8 cos − 0.3 2 0.8

or (a)

When x = −1 m:

−1 1 2 v = −0.1(−1) + 0.8 cos − 0.3 2 0.8 v = ± 0.323 m/s �

or (b)

When v = vmax, a = 0: or

(c)

x � � − � 0.1 + sin =0 0.8 �� x = −0.080 134 m

x = −0.0801 m �

When x = −0.080 134 m: 1 2 −0.080 134 vmax = −0.1(−0.080 134) + 0.8 cos − 0.3 2 0.8 = 0.504 m 2 /s 2 vmax = 1.004 m/s �

or


26

PROBLEM 11.21 Starting from x = 0 with no initial velocity, a particle is given an acceleration a = 0.8 v 2 + 49, where a and v are expressed in m/s2 and m/s, respectively. Determine (a) the position of the particle when v = 24 m/s, (b) the speed of the particle when x = 40 m.

SOLUTION v

We have v

dv = a = 0.8 v 2 + 49 dx

vdv

�

or

� v 2 + 49 � = 0.8 x � �0

or

v 2 + 49 − 7 = 0.8 x

(a)

0

v 2 + 49

=

�

x

When x = 0, v = 0:

0

0.8 dx

v

When v = 24 m/s:

242 + 49 − 7 = 0.8x x = 22.5 m �

or (b)

When x = 40 m:

v 2 + 49 − 7 = 0.8(40) v = 38.4 m/s �

or


27

27

PROBLEM 11.22 The acceleration of a particle is defined by the relation a = − k v , where k is a constant. Knowing that x = 0 and v = 81 m/s at t = 0 and that v = 36 m/s when x = 18 m, determine (a) the velocity of the particle when x = 20 m, (b) the time required for the particle to come to rest.

SOLUTION (a)

We have

v

dv = a = −k v dx v dv = − k dx

so that v

�

or

2 3/ 2 v [v ]81 = −kx 3

or

2 3/ 2 [v − 729] = − kx 3

When x = 18 m, v = 36 m/s:

v dv =

81

�

x

When x = 0, v = 81 m/s:

0

− k dx

2 (363/ 2 − 729) = − k (18) 3 k = 19 m/s 2

or Finally When x = 20 m:

2 3/ 2 (v − 729) = −19(20) 3 v3/ 2 = 159

or (b)

We have

dv = a = −19 v dt

At t = 0, v = 81 m/s:

�

v

dv

81

v

=

t

� −19dt 0

or

v 2[ v ]81 = −19t

or

2( v − 9) = −19t

When v = 0:

v = 29.3 m/s �

2(−9) = −19t t = 0.947 s ��

or


28

PROBLEM 11.23 The acceleration of a particle is defined by the relation a = − 0.8 v, where a is expressed in m/s2 and v in m/s. Knowing that at t = 0 the velocity is 40 m/s, determine (a) the distance the particle will travel before coming to rest, (b) the time required for the particle to come to rest, (c) the time required for the particle to be reduced by 50 percent of its initial value.

SOLUTION a=

(a)

vdv = −0.8v dx

dv = −0.8dx

Separate and integrate with v = 40 m/s when x = 0. v 40

dv = −0.8

x 0

dx

v − 40 = −0.8 x

Distance traveled. For v = 0, (b) Separate.

x=

−40 −0.8

a=

dv = −0.8v dt

x = 50.0 m

x dv = − 0.8dt 40 v 0 ln v − ln 40 = −0.8t v

ln

v = −0.8t 40

t = 1.25 ln

40 v

For v = 0, we get t = ∞. t=∞

(c)

For v = 0.5(40 m/s) = 20 m/s, t = 1.25 ln

40 = 0.866 s 20

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29

29

PROBLEM 11.24 A bowling ball is dropped from a boat so that it strikes the surface of a lake with a speed of 8 m/s. Assuming the ball experiences a downward acceleration of a = 10 − 0.9v2 when in the water, determine the velocity of the ball when it strikes the bottom of the lake.

SOLUTION v0 = 8 m/s,

x − x0 = 10 m

a = 10 − 0.9v 2 = k (c 2 − v 2 )

10 = 11.111 m2/s2 0.9 c = 3.3333 m/s

k = 0.9 m−1 and c 2 =

Where

Since v0

c, write a=v

dv = − k (v 2 − c 2 ) dx

vdv = − k dx v − c2 2

v

Integrating,

1 ln(v 2 − c 2 ) = − k ( x − x0 ) 2 v0 ln

v2 − c2 = −2k ( x − x0 ) v02 − c 2

v 2 − c2 = e−2 k ( x − x0 ) v02 − c 2

(

)

v 2 = c 2 + v02 − c 2 e −2 k ( x − x0 ) = 11.111 + [(8) 2 − 11.111] e− (2)(0.9)(10) = 11.111 + 8.05 × 10−7 = 11.111 m2/s2 v = 3.33 m/s PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced you are areaastudent studentusing usingthis thisManual, Manual, distributionto toteachers teachersand andeducators educatorspermitted permittedby byMcGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you distribution you are areusing using itit without without permission. permission. you 30

30

PROBLEM 11.25 The acceleration of a particle is defined by the relation a = 0.4(1 − kv), where k is a constant. Knowing that at t = 0 the particle starts from rest at x = 4 m and that when t = 15 s, v = 4 m/s, determine (a) the constant k, (b) the position of the particle when v = 6 m/s, (c) the maximum velocity of the particle.

SOLUTION (a)

dv = a = 0.4(1 − kv) dt

We have

dv = 0 1 − kv v

�

At t = 0, v = 0:

t

� 0.4dt 0

1 − [ln(1 − kv)]v0 = 0.4t k

or or

ln(1 − kv) = −0.4kt

At t = 15 s, v = 4 m/s:

ln(1 − 4k ) = −0.4k (15)

(1)

= −6k k = 0.145703 s/m

Solving yields

k = 0.1457 s/m �

or (b)

v

We have

vdv = 0 1 − kv

�

When x = 4 m, v = 0:

v

�

x

4

0.4dx

v 1 1/k =− + 1 − kv k 1 − kv

Now Then

dv = a = 0.4(1 − kv) dx

� 1 1 �− + � dv = 0 � k k (1 − kv) �

�

v�

�

x

4

0.4 dx

v

or

� v 1 � x � − k − k 2 ln(1 − kv) � = 0.4[ x]4 � �0

or

�v 1 � − � + 2 ln(1 − kv) � = 0.4( x − 4) �k k �

When v = 6 m/s:

� � 6 1 ln(1 − 0.145 703 × 6) � = 0.4( x − 4) −� + 2 � 0.145 703 (0.145 703) �

or

0.4( x − 4) = 56.4778 x = 145.2 m �

or


31

31

PROBLEM 11.25 (Continued) (c)

The maximum velocity occurs when a = 0. a = 0: 0.4(1 − kvmax ) = 0 vmax =

or

1 0.145 703 vmax = 6.86 m/s �

or An alternative solution is to begin with Eq. (1). ln(1 − kv) = −0.4kt v=

Then Thus, vmax is attained as t

1 (1 − k −0.4 kt ) k

∞ vmax =

1 k

as above.�


32

PROBLEM 11.26 A particle is projected to the right from the position x = 0 with an initial velocity of 9 m/s. If the acceleration of the particle is defined by the relation a = −0.6v3/ 2 , where a and v are expressed in m/s2 and m/s, respectively, determine (a) the distance the particle will have traveled when its velocity is 4 m/s, (b) the time when v = 1 m/s, (c) the time required for the particle to travel 6 m.

SOLUTION (a)

v

We have When x = 0, v = 9 m/s:

v

�

9

dv = a = −0.6v3/2 dx

v − (3/2) dv =

�

x

0

−0.6dx

or

2[v1/2 ]9v = −0.6 x

or

x=

1 (3 − v1/ 2 ) 0.3

When v = 4 m/s:

x=

1 (3 − 41/ 2 ) 0.3 x = 3.33 m �

or (b)

dv = a = −0.6v3/2 dt

We have v

t

When t = 0, v = 9 m/s:

�

or

−2[v − (1/2) ]v9 = −0.6t

or When v = 1 m/s:

9

v − (3/2) dv =

1 v 1 1

� −0.6dt 0

−

1 = 0.3t 3

−

1 = 0.3t 3 t = 2.22 s �

or (c)

We have

(1)

1 v

−

1 = 0.3t 3 2

or Now

9 � 3 � v=� = � (1 + 0.9t ) 2 � 1 + 0.9t � dx 9 =v= dt (1 + 0.9t ) 2


33

33


�

At t = 0, x = 0:

x 0

dx =

9 dt 0 (1 + 0.9t ) 2

�

t

t

1 � � 1 x = 9 �− � � 0.9 1 + 0.9t � 0

or

1 � � = 10 �1 − � � 1 + 0.9t � 9t = 1 + 0.9t

When x = 6 m:

6=

9t 1 + 0.9t t = 1.667 s �

or An alternative solution is to begin with Eq. (1). x=

1 (3 − v1/ 2 ) 0.3

dx = v = (3 − 0.3x) 2 dt

Then Now At t = 0, x = 0:

�

x

0

dx = (3 − 0.3 x) 2

t

� dt 0

x

or

1 � 1 � x = t= � � 0.3 � 3 − 0.3x � 0 9 − 0.9 x

Which leads to the same equation as above.


34

PROBLEM 11.27 Based on observations, the speed of a jogger can be approximated by the relation v = 7.5(1 − 0.04x)0.3, where v and x are expressed in km/h and kilometers respectively. Knowing that x = 0 at t = 0, determine (a) the distance the jogger has run when t = 1 h, (b) the jogger’s acceleration in m/s2 at t = 0, (c) the time required for the jogger to run 6 km.

SOLUTION (a)

dx = v = 7.5(1 − 0.04 x)0.3 dt

We have

x

At t = 0, x = 0: or

0

dx = (1 − 0.04 x)0.3

t 0

7.5dt

1 1 [(1 − 0.04 x)0.7 ]0x = 7.5t − 0.7 0.04 1 − (1 − 0.04 x)0.7 = 0.21t

or or

x=

1 [1 − (1 − 0.21t )1/0.7 ] 0.04

At t = 1 h:

x=

1 {1 − [1 − 0.21(1)]1/0.7 } 0.04 x = 7.15 km

or (b)

We have

(1)

a=v

dv dx

d [7.5(1 − 0.04 x)0.3 ] dx 2 0.3 = 7.5 (1 − 0.04 x) [(0.3)(−0.04)(1 − 0.04 x) −0.7 ] = 7.5(1 − 0.04 x)0.3

= −0.675(1 − 0.04 x) −0.4

At t = 0, x = 0:

1h a0 = −0.675 km/h × 3600 s 2

× (1000 m) a0 = −52.1 × 10 −6 m/s2

or (c)

2

From Eq. (1)

t=

When x = 6 mi:

t=

1 [1 − (1 − 0.04 x)0.7 ] 0.21

1 {1 − [1 − 0.04(6)]0.7 } 0.21 = 0.83229 h t = 49.9 min

or

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35

35

PROBLEM 11.28 Experimental data indicate that in a region downstream of a given louvered supply vent, the velocity of the emitted air is defined by v = 0.18v0 /x, where v and x are expressed in m/s and meters, respectively, and v0 is the initial discharge velocity of the air. For v0 = 3.6 m/s, determine (a) the acceleration of the air at x = 2 m, (b) the time required for the air to flow from x = 1 to x = 3 m.

SOLUTION (a)

dv dx 0.18v0 d � 0.18v0 � = x dx �� x ��

We have

a=v

=−

When x = 2 m:

a=−

0.0324v02 x3

0.0324(3.6) 2 (2)3 a = −0.0525 m/s 2 �

or (b)

0.18v0 dx =v= x dt

We have From x = 1 m to x = 3 m:

�

3

1

xdx =

�

t3

t1

0.18v0 dt

3

or

�1 2 � � 2 x � = 0.18v0 (t3 − t1 ) � �1

or

(t3 − t1 ) =

1 2

(9 − 1)

0.18(3.6) t3 − t1 = 6.17 s ��

or


36

PROBLEM 11.29 The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as a=

−9.81 [1 + ( y/6.37 × 106 )]2

where a and y are expressed in m/s2 and meters, respectively. Using this expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) 540 m/s, (b) 900 m/s, (c) 12,000 m/s.

SOLUTION We have

v

dv =a=− dy y = 0,

When

9.81

(

y 6.37 × 106

1+

)

2

v = v0

y = ymax , v = 0 0

Then

v0

ymax

(1 +

0

−9.81 y 6.37 × 106

v02 = 124.98 × 106 1 −

or

ymax =

or

dy

v0 = 540 m/s:

ymax =

v0 = 900 m/s:

ymax

0

1 1+

ymax 6.37 × 106

v02 19.62 −

v02 6.37 × 106

(540)2 19.62 −

(540) 2 6.37 × 106

ymax = 14897 m

or (b)

)

2

1 1 − v02 = −9.81 −6.37 × 106 2 1 + 6.37y× 106

or

(a)

v dv =

ymax =

(900)2 19.62 −

(900) 2 6.37 × 106

ymax = 41554 m

or


37

37


(c)

v0 = 12,000 m/s:

ymax =

(12 000)2 19.62 −

(12 000) 2 6.37 × 106

ymax = −4019 m

or

The velocity 12,000 m/s is approximately the escape velocity vR from the earth. For vR ymax → ∞

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No of this this Manual Manual may may be be displayed, displayed, PROPRIETARY No part part of reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced distribution If you distribution to to teachers teachers and and educators educators permitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If you are are aa student student using using this this Manual, Manual, you you are are using using it it without without permission. permission. 38

38

PROBLEM 11.30 The acceleration due to gravity of a particle falling toward the earth is a = − gR 2 /r 2, where r is the distance from the center of the earth to the particle, R is the radius of the earth, and g is the acceleration due to gravity at the surface of the earth. If R = 6370 km, calculate the escape velocity, that is, the minimum velocity with which a particle must be projected vertically upward from the surface of the earth if it is not to return to the earth. (Hint: v = 0 for r = ∞.)

SOLUTION We have

v

dv gR 2 =a=− 2 dr r r = R, v = ve

When

r = ∞, v = 0

Then or or or

0

ve

vdv =

∞

R

−

gR 2 dr r2

1 1 − ve2 = gR 2 2 r

∞ R

ve = 2 gR = (2 × 0.00981 km/s2 × 6370 km)1/2

ve = 11,179 m/s

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced distribution to toteachers teachersand andeducators educatorspermitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you you are are aastudent studentusing usingthis thisManual, Manual, distribution you are are using using itit without without permission. permission. you

39

39

PROBLEM 11.31 The velocity of a particle is v = v0 [1 − sin (π t/T )]. Knowing that the particle starts from the origin with an initial velocity v0 , determine (a) its position and its acceleration at t = 3T , (b) its average velocity during the interval t = 0 to t = T .

SOLUTION (a)

� dx � π t �� = v = v0 �1 − sin � � � dt � T ��

We have At t = 0, x = 0:

�

x

0

dx =

t

�v 0

0

� � π t �� 1 − sin � � � dt � T �� t

or

� T � π t �� x = v0 �t + cos � � � � T �� 0 � π � T � πt � T � = v0 �t + cos � � − � �T � π� � π

At t = 3T :

� T � π × 3T � T � xBT = v0 �3T + cos � �− � π � T � π� � 2T � � = v0 � 3T − π �� x3T = 2.36 v0T �

or a=

Also At t = 3T :

π πt dv d � � � π t �� = �v0 �1 − sin � � � � = −v0 cos dt dt � � T T � T ��

a3T = −v0

π T

cos

π × 3T T a3T =

or (b)

(1)

π v0 T

�

Using Eq. (1) At t = 0:

T T� � x0 = v0 �0 + cos(0) − � = 0 π π� �


40


At t = T :

Now

� T � πT xT = v0 �T + cos � π � T � 2T � � = v0 � T − π �� = 0.363v0T vave =

� T� �−π � � �

xT − x0 0.363v0T − 0 = ∆t T −0 vave = 0.363v0 �

or


41

41

PROBLEM 11.32 The velocity of a slider is defined by the relation v = v′ sin (ωn t + φ ). Denoting the velocity and the position of the slider at t = 0 by v0 and x0 , respectively, and knowing that the maximum displacement of the slider is 2 x0 , show that (a) v′ = v02 + x02ωn2 2 x0ωn , (b) the maximum value of the velocity occurs when x = x0 [3 − (v0 /x0ωn ) 2 ]/2.

(

)

SOLUTION (a)

v0 = v′ sin (0 + φ ) = v′ sin φ

At t = 0, v = v0 : Then

2

cos φ = v′ − v02 v′ dx = v = v′ sin (ωn t + φ ) dt

Now x

t

At t = 0, x = x0 :

�

or

� 1 � cos (ωn t + φ ) � x − x0 = v′ � − � ωn �0

x0

dx =

� v′ sin (ω t + φ )dt n

0

t

or

x = x0 +

v′

ωn

[cos φ − cos (ωn t + φ )]

Now observe that xmax occurs when cos (ωn t + φ ) = −1. Then xmax = 2 x0 = x0 +

Substituting for cos φ or

x0 =

v′

ωn

[cos φ − ( −1)]

2 2 � v′ �� v′ − v0 � + 1 � ωn � v1 � �

x0ωn − v1 = v′2 − v02

Squaring both sides of this equation x02ωn2 − 2 x0ωn + v′2 = v′2 − v02

or

v′ =

v02 + x02ωn2 2 x0ωn

Q. E. D.


42

PROBLEM 11.32 (Continued) (b)

First observe that vmax occurs when ωn t + φ = π2 . The corresponding value of x is v′ � � π �� cos φ − cos � � � ωn � � 2 �� v′ cos φ = x0 +

xvmax = x0 +

ωn

Substituting first for cos φ and then for v′ xvmax = x0 +

2

v′ − v02 v′

v′

ωn

1/ 2

2 � � 1 �� v02 + x02ωn2 � 2 = x0 + �� − v0 � � ωn �� 2 x0ωn � � � 1 v04 + 2v02 x02ωn2 + x04ωn4 − 4 x02ωn2 v02 = x0 + 2 x0ωn2

(

= x0 + = x0 +

1

(

� x 2ω 2 − v 2 0 2 � 0 n 2 x0ωn �

)

1/ 2

1/ 2 2�

) ��

x02ωn2 − v02 2 x0ωn2

2 x0 � � v0 � � = �3 − � � � 2 � � x0ωn � � � �

Q. E. D.


43

43

PROBLEM 11.33 A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows that she traveled 0.2 km while accelerating. Determine (a) the acceleration of the car, (b) the time required to reach 99 km/h.

SOLUTION (a)

Acceleration of the car. v12 = v02 + 2a( x1 − x0 ) a=

Data:

v12 − v02 2( x1 − x0 )

v0 = 45 km/h = 12.5 m/s v1 = 99 km/h = 27.5 m/s x0 = 0 x1 = 0.2 km = 200 m a=

(b)

(27.5) 2 − (12.5) 2 (2)(200 − 0)

a = 1.500 m/s 2 �

Time to reach 99 km/h. v1 = v0 + a (t1 − t0 ) v1 − v0 a 27.5 − 12.5 = 1.500 = 10.00 s

t1 − t0 =

t1 − t0 = 10.00 s ��


44

PROBLEM 11.34 A truck travels 220 m in 10 s while being decelerated at a constant rate of 0.6 m/s2. Determine (a) its initial velocity, (b) its final velocity, (c) the distance traveled during the first 1.5 s.

SOLUTION (a)

x = x0 + v0t +

Initial velocity.

1 2 at 2

x − x0 1 − at t 2 220 1 = − (−0.6)(10) 10 2

v0 =

(b)

Final velocity.

v = v0 + at v = 25.0 + ( −0.6)(10)

(c)

v0 = 25.0 m/s

vf = 19.00 m/s

Distance traveled during first 1.5 s. x = x0 + v0t +

1 2 at 2

= 0 + (25.0)(1.5) +

1 (−0.6)(1.5) 2 2

x = 36.8 m


45

45

PROBLEM 11.35 Assuming a uniform acceleration of 3 m/s2 and knowing that the speed of a car as it passes A is 50 km/h, determine (a) the time required for the car to reach B, (b) the speed of the car as it passes B.

SOLUTION (a)

Time required to reach B. vA = 50 km/h = 13.89 m/s,

xA = 0,

xB = 50 m,

a = 3 m/s2

1 2 at 2 1 50 = 0 + 13.89t + (3) t2 2 xB = x A + v A t +

1.5t 2 + 13.89t − 50 = 0

t= =

Rejecting the negative root. (b)

(13.89)2 − (4)(1.5)(−50) (2)(1.5)

−13.89 ±

−13.89 ± 22.2 3 t = 2.78 s

t = 2.78 s

Speed at B. vB = vA + at = 13.89 + (3)(2.78) = 22.23 m/s vB = 80 km/h

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46

PROBLEM 11.36 A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 m at the end of the powered portion of the flight and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that g = 9.81 m/s2, determine (a) the speed v1 of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.

SOLUTION (a)

1 2 at 2

We have

y = y1 + v1t +

At tland ,

y=0

Then

0 = 89.6 m + v1(16 s) +

1 (−9.81 m/s2)(16 s)2 2 v1 = 72.9 m/s

or (b)

v 2 = v12 + 2a( y − y1 )

We have At

y = ymax , v = 0

Then

0 = (72.9 m/s)2 + 2(−9.81 m/s2)(ymax − 89.6) m ymax = 360 m

or


47

47

PROBLEM 11.37 A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs with constant velocity. If the sprinter’s time for the first 35 m in 5.4 s, determine (a) his acceleration, (b) his final velocity, (c) his time for the race.

SOLUTION 0 � × � 35 m, a = constant

Given:

35 m � × � 100 m, v = constant

At t = 0, v = 0 when

x = 35 m, t = 5.4 s

Find: (a)

a

(b)

v when x = 100 m

(c)

t when x = 100 m

(a)

We have At t = 5.4 s: or

x = 0 + 0t + 35 m =

1 2 at 2

0 � x � 35 m

for

1 a(5.4 s)2 2

a = 2.4005 m/s 2 a = 2.40 m/s 2 �

(b)

First note that v = vmax for 35 m � x � 100 m. Now

(c)

v 2 = 0 + 2a( x − 0)

for

0 � x � 35 m

When x = 35 m:

2 vmax = 2(2.4005 m/s 2 )(35 m)

or

vmax = 12.9628 m/s

We have When x = 100 m:

vmax = 12.96 m/s �

x = x1 + v0 (t − t1 )

35 m � x � 100 m

for

100 m = 35 m + (12.9628 m/s)(t2 − 5.4) s t2 = 10.41 s �

or


48

PROBLEM 11.38 A small package is released from rest at A and moves along the skate wheel conveyor ABCD. The package has a uniform acceleration of 4.8 m/s 2 as it moves down sections AB and CD, and its velocity is constant between B and C. If the velocity of the package at D is 7.2 m/s, determine (a) the distance d between C and D, (b) the time required for the package to reach D.

SOLUTION (a)

For

A→ B

and

C→D v 2 = v02 + 2a( x − x0 )

we have Then,

2 = 0 + 2(4.8 m/s 2 )(3 − 0) m vBC

at B

= 28.8 m 2 /s 2 2 vD2 = vBC + 2aCD ( xD − xC )

and at D

d = 2.40 m �

or For

A→ B

and

C → D, v = v0 + at

we have Then

d = xd − xC

(7.2 m/s)2 = (28.8 m 2 /s 2 ) + 2(4.8 m/s 2 )d

or (b)

(vBC = 5.3666 m/s)

A→ B 5.3666 m/s = 0 + (4.8 m/s 2 )t AB t AB = 1.11804 s

or and or

C→D

7.2 m/s = 5.3666 m/s + (4.8 m/s 2 )tCD tCD = 0.38196 s


49

49

PROBLEM 11.38 (Continued) Now, for we have

B → C, xC = xB + vBC t BC

or

3 m = (5.3666 m/s)t BC

or

t BC = 0.55901 s

Finally,

t D = t AB + t BC + tCD = (1.11804 + 0.55901 + 0.38196) s t D = 2.06 s �

or


50

PROBLEM 11.39 A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant speed. Believing that the driver of the automobile might be intoxicated, the officer starts his car, accelerates uniformly to 90 km/h in 8 s, and, maintaining a constant velocity of 90 km/h, overtakes the motorist 42 s after the automobile passed him. Knowing that 18 s elapsed before the officer began pursuing the motorist, determine (a) the distance the officer traveled before overtaking the motorist, (b) the motorist’s speed.

SOLUTION

(vP )18 = 0

(a)

(vP )26 = 90 km/h = 25 m/s

(vP ) 42 = 90 km/h = 25 m/s

Patrol car: For 18 s � t � 26 s: At t = 26 s:

vP = 0 + aP (t − 18) 25 m/s = aP (26 − 18) s

or

aP = 3.125 m/s 2

Also,

xP = 0 + 0(t − 18) −

At t = 26 s: For 26 s � t � 42 s: At t = 42 s:

(xP ) 26 =

1 aP (t − 18) 2 2

1 (3.125 m/s 2 )(26 − 18) 2 = 100 m 2

xP = ( xP )26 + (vP ) 26 (t − 26) (xP ) 42 = 100 m + (25 m/s)(42 − 26)s = 500 m ( xP )42 = 0.5 km �

(b)

For the motorist’s car:

xM = 0 + vM t

At t = 42 s, xM = xP :

500 m = vM (42 s)

or

vM = 11.9048 m/s vM = 42.9 km/h �

or


51

51

PROBLEM 11.40 As relay runner A enters the 20-m-long exchange zone with a speed of 12.9 m/s, he begins to slow down. He hands the baton to runner B 1.82 s later as they leave the exchange zone with the same velocity. Determine (a) the uniform acceleration of each of the runners, (b) when runner B should begin to run.

SOLUTION (a)

For runner A: At t = 1.82 s:

x A = 0 + (v A )0 t +

1 a At 2 2

20 m = (12.9 m/s)(1.82 s) +

1 a A (1.82 s) 2 2 a A = −2.10 m/s 2 �

or Also At t = 1.82 s:

v A = (v A ) 0 + a A t (v A )1.82 = (12.9 m/s) + ( −2.10 m/s 2 )(1.82 s) = 9.078 m/s

For runner B:

vB2 = 0 + 2aB [ xB − 0]

When

xB = 20 m, vB = v A : (9.078 m/s) 2 + 2aB (20 m)

or

aB = 2.0603 m/s 2 aB = 2.06 m/s 2 �

(b)

For runner B:

vB = 0 + aB (t − t B )

where t B is the time at which he begins to run. At t = 1.82 s: or

9.078 m/s = (2.0603 m/s 2 )(1.82 − t B ) s t B = −2.59 s

Runner B should start to run 2.59 s before A reaches the exchange zone.

�


52

PROBLEM 11.41 Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 0.5 m/s2 and that B has a constant deceleration of 0.3 m/s2, determine (a) when and where A will overtake B, (b) the speed of each automobile at that time.

SOLUTION aA = +0.5 m/s2

aB = −0.3 m/s2

| v A |0 = 36 km/h = 10 m/s

A overtakes B

| vB |0 = 54 km/h = 15 m/s

Motion of auto A: v A = (v A )0 + a At = 10 + 0.5t

(1)

1 1 xA = (xA)0 + (vA)0t + aA t2 = 0 + 10t + (0.5)t2 2 2

(2)

vB = (vB )0 + aB t = 15 − 0.3t

(3)

Motion of auto B:

x B = ( x B ) 0 + ( vB ) 0 t +

(a)

1 1 aB t 2 = 24 + 15t + (−0.3)t2 2 2

(4)

A overtakes B at t = t1 . x A = xB : 10t + 0.25t12 = 24 + 15t1 − 0.15t12 0.4t12 − 5t1 − 24 = 0 t1 = −3.65 s and t1 = 16.2 s

Eq. (2):

x A = 10(16.2) + 0.25(16.2)2

t1 = 16.2 s x A = 227.6 m

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53

53


(b)

Velocities when

t1 = 16.2 s

Eq. (1):

vA = 10 + 0.5(16.2) vA = 18.1 m/s

Eq. (3):

vA = 65.2 km/h

vB = 15 − 0.3(16.2) vB = 10.14 m/s

vB = 36.5 km/h →

PROPRIETARY MATERIAL. MATERIAL. © 2009 The The McGraw-Hill McGraw-Hill Companies, be displayed, displayed, PROPRIETARY © 2010 Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of or used used beyond beyond the the limited limited reproduced of the the publisher, publisher, or distribution to teachers by McGraw-Hill for their their individual individual course course preparation. preparation. IfIf you you are are aa student distribution to teachers and and educators educators permitted permitted by McGraw-Hill for student using using this this Manual, Manual, you are using using itit without without permission. permission. you are 54

54

PROBLEM 11.42 In a boat race, boat A is leading boat B by 36 m and both boats are traveling at a constant speed of 168 km/h. At t = 0, the boats accelerate at constant rates. Knowing that when B passes A, t = 8 s and v A = 216 km/h, determine (a) the acceleration of A, (b) the acceleration of B.

SOLUTION (a)

We have

v A = (v A ) 0 + a A t (v A )0 = 168 km/h = 46.67 m/s

At t = 8 s: Then

v A = 216 km/h = 60 m/s 60 m/s = 46.67 m/s + aA (8s) a A = 1.67 m/s2

or (b)

We have

x A = ( x A ) 0 + (v A ) 0 t +

and

xB = 0 + (vB )0 t +

At t = 8 s:

x A = xB

1 a At 2 2

1 aB t 2 2

( x A )0 = 36 m (vB )0 = 46.67 m/s

1 (1.67 m/s2)(8s)2 2 1 = (46.67 m/s)(8s) + aB (8s)2 2

36 m + (46.67 m/s)(8 s) +

aB = 2.8 m/s2

or


55

55

PROBLEM 11.43 Boxes are placed on a chute at uniform intervals of time t R and slide down the chute with uniform acceleration. Knowing that as any box B is released, the preceding box A has already slid 6 m and that 1 s later they are 10 m apart, determine (a) the value of t R , (b) the acceleration of the boxes.

SOLUTION

Let tS = 1 s be the time when the boxes are 10 m apart. a A = aB = a; ( x A )0 = ( xB )0 = 0; (v A )0 = (vB )0 = 0.

Let (a)

1 2 at 2

For

t > 0, x A =

For

t > t R , xB =

At

t = tR , xA = 6 m

At

t = t R + tS ,

1 a(t − t R ) 2 2 6=

1 2 at R 2

(1)

x A − xB = 10 m

1 1 a(t R + t S ) 2 − a(t R + tS − t R )2 2 2 1 2 1 1 = at R + at R tS + at S2 − atS2 = 6 + atRtS 2 2 2 10 − 6 4 at R = = = 4 m/s tS 1 10 =

(2)

Dividing Equation (1) by Eq. (2), 1 2

at R2

at R

(b)

=

1 6 tR = 2 4

t R = 3.00 s

Solving Eq. (2) for a, a=

4 = 1.33 m/s2 3

a = 1.33 m/s2

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56

PROBLEM 11.44 Two automobiles A and B are approaching each other in adjacent highway lanes. At t = 0, A and B are 1 km apart, their speeds are v A = 108 km/h and vB = 63 km/h, and they are at Points P and Q, respectively. Knowing that A passes Point Q 40 s after B was there and that B passes Point P 42 s after A was there, determine (a) the uniform accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time.

SOLUTION (a)

x A = 0 + (v A )0 t +

We have At t = 40 s:

1 a At 2 2

1000 m = (30 m/s)(40 s) +

(v A )0 = 108 km/h = 30 m/s

1 a A (40 s)2 2 a A = −0.250 m/s 2 �

or x B = 0 + ( vB ) 0 t +

Also, At t = 42 s:

1000 m = (17.5 m/s)(42 s) +

(vB )0 = 63 km/h = 17.5 m/s 1 aB (42 s) 2 2

aB = 0.30045 m/s 2

or (b)

1 aB t 2 2

aB = 0.300 m/s 2 �

When the cars pass each other x A + xB = 1000 m

Then

or Solving

1 2 (−0.250 m/s)t AB + (17.5 m/s)t AB 2 1 2 + (0.30045 m/s 2 )t AB = 1000 m 2

(30 m/s)t AB +

2 2 0.05045t AB + 95t AB − 2000 = 0

t = 20.822 s and t = −1904 s t � 0 � t AB = 20.8 s �


57

57


We have

vB = ( vB ) 0 + a B t

At t = t AB :

vB = 17.5 m/s + (0.30045 m/s 2 )(20.822 s) = 23.756 m/s vB = 85.5 km/h �

or


58

PROBLEM 11.45 Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a constant speed of 96 km/h. At t = 0, A starts and accelerates at a constant rate a A , while at t = 5 s, B begins to slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other x = 88 m and v A = vB , determine (a) the acceleration a A , (b) when the vehicles pass each other, (c) the distance d between the vehicles at t = 0.

SOLUTION

For t

vA = 0 + aAt

0:

xA = 0 + 0 + 0

t

5 s:

1 aA t 2 2

xB = 0 + (vB )0 t (vB )0 = 90 km/h = 25 m/s

At t = 5 s:

xB = (25 m/s)(5 s) = 125 m

For t

1 vB = (vB )0 + aB (t − s ) aB = − a A 6

5 s:

xB = ( xB ) S + (vB )0 (t − s ) +

Assume t

1 aB (t − s ) 2 2

5 s when the cars pass each other.

At that time (t AB ), v A = vB :

a At AB = (25 m/s) −

x A = 88 m:

88 m =

aA (tAB − 5) 6

1 2 a At AB 2


59

59

PROBLEM 11.45 (Continued) a A( 76 t AB − 56 )

Then

1 2

=

25 88

12.5t 2AB − 102.7t AB + 73.3 = 0

or

t AB = 0.79 s and

Solving (a)

a At A2 B

With t AB

5 s,

88 m =

tAB = 7.4 s

1 a A (7.4 s)2 2 a A = 3.2 m/s2

or (b)

t AB = 7.00 s

From above

Note: An acceptable solution cannot be found if it is assumed that t AB (c)

We have

5 s.

d = x + ( xB )t AB = 88 m + 125 m + (25 m/s)(7.4 s) s +

1 1 − × 3.2 m/s2 (7.4 s)2s2 2 6 d = 383 m

or

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60

PROBLEM 11.46 Two blocks A and B are placed on an incline as shown. At t = 0, A is projected up the incline with an initial velocity of 8 m/s and B is released from rest. The blocks pass each other 1 s later, and B reaches the bottom of the incline when t = 3.4 s. Knowing that the maximum distance from the bottom of the incline reached by block A is 6 m and that the accelerations of A and B (due to gravity and friction) are constant and are directed down the incline, determine (a) the accelerations of A and B, (b) the distance d, (c) the speed of A when the blocks pass each other.

SOLUTION (a)

We have

v A2 = (v A )02 + 2a A [ x A − 0]

When

x A = ( x A ) max , v A = 0

Then or

0 = (8 m/s)2 + 2aA (6 m) a A = −5.3 m/s2 a A = 5.3 m/s2

or Now

x A = 0 + (v A )0 t +

and

xB = 0 + 0t +

1 a At 2 2

1 aB t 2 2

At t = 1 s, the blocks pass each other. ( x A )1 + ( xB )1 = d

At t = 3.4 s, xB = d : Thus or

( x A )1 + ( xB )1 = ( xB )3.4

(8 m/s)(1 s) + +

or

1 ( −5.3 m/s2)(1 s)2 2 1 1 aB (1 s) 2 = aB (3.4 s) 2 2 2

aB = 1.01 m/s2

a B = 1.01 m/s2


61

61


(b)

At t = 3.4 s, xB = d :

d=

1 (1.01 m/s2)(3.4 s)2 2 d = 5.8 m

or (c)

We have

v A = (v A )0 + aA t

At t = 1 s:

v A = 8 m/s + (−5.3 m/s)(1 s) v A = 2.7 m/s

or

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No of this this Manual Manual may may be be displayed, displayed, PROPRIETARY No part part of reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced distribution If you distribution to to teachers teachers and and educators educators permitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If you are are aa student student using using this this Manual, Manual, you you are are using using it it without without permission. permission. 62

62

PROBLEM 11.47 Slider block A moves to the left with a constant velocity of 6 m/s. Determine (a) the velocity of block B, (b) the velocity of portion D of the cable, (c) the relative velocity of portion C of the cable with respect to portion D.

SOLUTION From the diagram, we have x A + 3 yB = constant

Then

v A + 3vB = 0

(1)

and

a A + 3aB = 0

(2)

(a)

Substituting into Eq. (1)

6 m/s + 3vB = 0

v B = 2 m/s ↑ �

or (b)

From the diagram

yB + yD = constant

Then

vB + vD = 0

v D = 2 m/s ↓ ��

(c) �

From the diagram

x A + yC = constant

Then

v A + vC = 0

Now

vC/D = vC − vD = (−6 m/s) − (2 m/s) = − 8 m/s

vC = −6 m/s

vC/D = 8 m/s ↑ ��


63

63

PROBLEM 11.48 Block B starts from rest and moves downward with a constant acceleration. Knowing that after slider block A has moved 400 mm its velocity is 4 m/s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 2 s.

SOLUTION From the diagram, we have x A + 3 yB = constant

Then

v A + 3vB = 0

(1)

and

a A + 3aB = 0

(2)

(a)

Eq. (2): a A + 3aB = 0 and a B is constant and positive � a A is constant and negative Also, Eq. (1) and (vB )0 = 0 � (v A )0 = 0 v A2 = 0 + 2a A [ x A − ( x A )0 ]

Then When |∆ x A | = 0.4 m:

(4 m/s) 2 = 2a A (0.4 m) a A = 20 m/s 2 → ��

or Then, substituting into Eq. (2): −20 m/s 2 + 3aB = 0 aB =

or

20 m/s 2 3

a B = 6.67 m/s 2 ↓ ��


64


We have

vB = 0 + a B t

At t = 2 s:

� 20 � m/s 2 � (2 s) vB = � � 3 � v B = 13.33 m/s ↓ ��

or yB = ( yB )0 + 0 +

Also At t = 2 s: or�

yB − ( yB )0 =

1 aB t 2 2

1 � 20 � m/s 2 � (2 s) 2 � 2 �� 3 � y B − (y B )0 = 13.33 m ↓ ��

�


65

65

PROBLEM 11.49 The elevator shown in the figure moves downward with a constant velocity of 4.5 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d) the relative velocity of the counterweight W with respect to the elevator.

SOLUTION Choose the positive direction downward. (a)

Velocity of cable C. yC + 2 yE = constant vC + 2vE = 0 vE = 4.5 m/s

But, or (b)

vC = −2vE = −9 m/s

vC = 9 m/s ↑

Velocity of counterweight W. yW + yE = constant vW + vE = 0 vW = −vE = −9 m/s

(c)

vW = 9 m/s ↑

Relative velocity of C with respect to E. vC/E = vC − vE = (−9 m/s) − (+4.5 m/s) = −13.5 m/s vC/E = 13.5 m/s ↑

(d)

Relative velocity of W with respect to E. vW/E = vW − vE = (−4.5 m/s) − (4.5 m/s) = −9 m/s vW/E = 9 m/s ↑


66

PROBLEM 11.50 The elevator shown starts from rest and moves upward with a constant acceleration. If the counterweight W moves through 10 m in 5 s, determine (a) the accelerations of the elevator and the cable C, (b) the velocity of the elevator after 5 s.

SOLUTION We choose Positive direction downward for motion of counterweight. yW =

At t = 5 s,

1 aW t 2 2

yW = 10 m 10 m =

1 aW (5 s) 2 2

aW = 0.4 m/s2

(a)

Accelerations of E and C. Since Thus: Also, Thus:

(b)

aW = 0.4 m/s2 ↓

yW + yE = constant vW + vE = 0, and aW + aE = 0 aE = − aW = −(0.4 m/s2),

a E = 0.4 m/s2 ↑

yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0 aC = −2aE = −2(−0.4 m/s2) = + 0.8 m/s2,

aC = 0.8 m/s2 ↓

Velocity of elevator after 5 s. vE = (vE )0 + aE t = 0 + (−0.4 m/s2)(5 s) = −2 m/s

( v E )5 = 2 m/s ↑


67

67

PROBLEM 11.51 Collar A starts from rest and moves upward with a constant acceleration. Knowing that after 8 s the relative velocity of collar B with respect to collar A is 600 mm/s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 6 s.

SOLUTION From the diagram 2 y A + yB + ( yB − y A ) = constant

Then

v A + 2 vB = 0

(1)

and

a A + 2aB = 0

(2)

(a)

Eq. (1) and (v A )0 = 0

(vB )0

Also, Eq. (2) and a A is constant and negative

a B is constant

and positive v A = 0 + a At

Then

vB = 0 + a B t

vB/A = vB − v A = (aB − a A )t

Now

1 aB = − a A 2

From Eq. (2)

3 vB/A = − a At 2

So that

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68


At t = 8 s:

3 600 mm/s = − a A (8 s) 2 a A = 50 mm/s2 ↑

or and then

1 aB = − (−50 mm/s2) 2 a B = 25 mm/s2 ↑

or (b)

At t = 6 s:

vB = (25 mm/s2)(6 s) v B = 150 mm/s ↓

or Now At t = 6 s:

yB = ( yB )0 + 0 + y B − ( y B )0 =

1 aB t 2 2

1 (25 mm/s2)(6 s)2 2 y B − (y B )0 = 450 mm ↓

or

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69

69

PROBLEM 11.52 In the position shown, collar B moves downward with a velocity of 300 mm/s. Determine (a) the velocity of collar A, (b) the velocity of portion C of the cable, (c) the relative velocity of portion C of the cable with respect to collar B.

SOLUTION From the diagram 2 y A + yB = ( yB − y A ) = constant

Then

v A + 2 vB = 0

(1)

and

a A + 2aB = 0

(2)

(a)


v A + 2(300 mm/s) = 0 v A = 600 mm/s ↑

or (b)

From the diagram

2 y A + yC = constant

Then

2v A + vC = 0 2(−600 mm/s) + vC = 0

Substituting

vC = 1.2 m/s ↓

or (c)

vC/B = vC − vB

We have

= (1200 mm/s) − (300 mm/s) vC/B = 900 mm/s ↓

or

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70

PROBLEM 11.53 Slider block B moves to the right with a constant velocity of 300 mm/s. Determine (a) the velocity of slider block A, (b) the velocity of portion C of the cable, (c) the velocity of portion D of the cable, (d) the relative velocity of portion C of the cable with respect to slider block A.

SOLUTION

From the diagram

xB + ( xB − x A ) − 2 x A = constant

Then

2vB − 3v A = 0

(1)

and

2aB − 3a A = 0

(2)

Also, we have

− xB − x A = constant vD + v A = 0

Then (a)


2(300 mm/s) − 3v A = 0 v A = 200 mm/s → ��

or (b)

From the diagram Then Substituting

(3)

xB + ( xB − xC ) = constant 2vB − vC = 0 2(300 mm/s) − vC = 0 vC = 600 mm/s → �

or


71

71


From the diagram Then Substituting

( xC − x A ) + ( xB − x A ) = constant vC − 2v A + vD = 0 600 mm/s − 2(200 mm/s) + vD = 0 v D = 200 mm/s ← �

or (d)

We have

vC/A = vC − v A = 600 mm/s − 200 mm/s vC/A = 400 mm/s → ��

or


72

PROBLEM 11.54 At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slider block A has moved 240 mm to the right its velocity is 60 mm/s, determine (a) the accelerations of A and B, (b) the acceleration of portion D of the cable, (c) the velocity and change in position of slider block B after 4 s.

SOLUTION

xB + ( xB − x A ) − 2 x A = constant

From the diagram Then

2vB − 3v A = 0

(1)

and

2aB − 3a A = 0

(2)

(a)

First observe that if block A moves to the right, v A → and Eq. (1) � v B → . Then, using Eq. (1) at t = 0 2(150 mm/s) − 3(v A )0 = 0 (v A )0 = 100 mm/s

or

Also, Eq. (2) and aB = constant � a A = constant v A2 = (v A )02 + 2a A [ x A − ( x A )0 ]

Then When x A − ( x A )0 = 240 mm:

(60 mm/s) 2 = (100 mm/s) 2 + 2a A (240 mm)

or

aA = −

40 mm/s 2 3 a A = 13.33 mm/s 2 ← ��

or


73

73

PROBLEM 11.54 (Continued) Then, substituting into Eq. (2) � 40 � 2aB − 3 � − mm/s 2 � = 0 � 3 � aB = −20 mm/s 2

or (b)

a B = 20.0 mm/s 2 ← ��

From the solution to Problem 11.53 vD + v A = 0

Then Substituting

aD + a A = 0 � 40 � mm/s 2 � = 0 aD + � − 3 � � a D = 13.33 mm/s 2 → �

or (c)

We have

v B = ( vB ) 0 + a B t

At t = 4 s:

vB = 150 mm/s + ( −20.0 mm/s 2 )(4 s) v B = 70.0 mm/s → �

or Also At t = 4 s:

y B = ( y B ) 0 + ( vB ) 0 t +

1 aB t 2 2

yB − ( yB )0 = (150 mm/s)(4 s) +

1 (−20.0 mm/s 2 )(4 s) 2 2 y B − (y B )0 = 440 mm → ��

or


74

PROBLEM 11.55 Block B moves downward with a constant velocity of 20 mm/s. At t = 0, block A is moving upward with a constant acceleration, and its velocity is 30 mm/s. Knowing that at t = 3 s slider block C has moved 57 mm to the right, determine (a) the velocity of slider block C at t = 0, (b) the accelerations of A and C, (c) the change in position of block A after 5 s.

SOLUTION

From the diagram 3 y A + 4 yB + xC = constant

Then

3v A + 4vB + vC = 0

(1)

and

3a A + 4aB + aC = 0

(2)

v B = 20 mm/s ↓ ;

Given:

( v A )0 = 30 mm/s ↑

(a)

Substituting into Eq. (1) at t = 0 3(−30 mm/s) + 4(20 mm/s) + (vC )0 = 0 vC = 10 mm/s

(b)

xC = ( xC )0 + (vC )0 t +

We have At t = 3 s:

or 1 aC t 2 2

57 mm = (10 mm/s)(3 s) +

1 aC (3 s) 2 2

aC = 6 mm/s 2

�

( vC )0 = 10 mm/s → �

or

aC = 6 mm/s 2 → ��

v B = constant → aB = 0

Now


75

75

PROBLEM 11.55 (Continued) Then, substituting into Eq. (2) 3a A + 4(0) + (6 mm/s 2 ) = 0 a A = −2 mm/s 2

(c)

We have At t = 5 s:

y A = ( y A )0 + (v A )0 t +

or 1 a At 2 2

y A − ( y A )0 = (−30 mm/s)(5 s) + = −175 mm

a A = 2 mm/s 2 ↑ �

1 (−2 mm/s 2 )(5 s) 2 2 y A − (y A )0 = 175 mm ↑ �

or


76

PROBLEM 11.56 Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s 2 . Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider C after 3 s.

SOLUTION

From the diagram 3 y A + 4 yB + xC = constant

Then

3v A + 4vB + vC = 0

(1)

and

3a A + 4aB + aC = 0

(2)

(vB ) = 0,

Given:

a A = constent (aC ) = 75 mm/s 2 → v B = 480 mm/s ↓

At t = 2 s, (a)

vC = 280 mm/s →

Eq. (2) and a A = constant and aC = constant � aB = constant vB = 0 + a B t

Then At t = 2 s:

480 mm/s = aB (2 s) aB = 240 mm/s 2

�

or

a B = 240 mm/s 2 ↓ ��

or

a A = 345 mm/s 2 ↑ �

Substituting into Eq. (2) 3a A + 4(240 mm/s 2 ) + (75 mm/s 2 ) = 0 a A = −345 mm/s


77

77


vC = (vC )0 + aC t

We have At t = 2 s:

280 mm/s = (vC )0 + (75 mm/s)(2 s) vC = −130 mm/s

�

or

( vC )0 = 130 mm/s → ��

or

(v A )0 = 43.3 mm/s ↑ �

Then, substituting into Eq. (1) at t = 0 3(v A )0 + 4(0) + (130 mm/s) = 0 v A = −43.3 mm/s

(c)

We have At t = 3 s:

xC = ( xC )0 + (vC )0 t +

1 aC t 2 2

xC − ( xC )0 = (130 mm/s)(3 s) + = −728 mm

1 (75 mm/s 2 )(3 s) 2 2

or

xC − (xC )0 = 728 mm → ��


78

PROBLEM 11.57 Collar A starts from rest at t = 0 and moves downward with a constant acceleration of 140 mm/s2. Collar B moves upward with a constant acceleration, and its initial velocity is 160 mm/s. Knowing that collar B moves through 400 mm between t = 0 and t = 2 s, determine (a) the accelerations of collar B and block C, (b) the time at which the velocity of block C is zero, (c) the distance through which block C will have moved at that time.

SOLUTION From the diagram − y A + ( yC − y A ) + 2 yC + ( yC − yB ) = constant

Then

−2v A − vB + 4vC = 0

(1)

and

−2a A − aB + 4aC = 0

(2)

(v A ) 0 = 0

Given:

(a A ) = 140 mm/s2 ↓ ( v B )0 = 160 mm/s ↑ a B = constant ↑

y − (yB )0 = 400 mm ↑

At t = 2 s (a)

We have At t = 2 s:

y B = ( y B ) 0 + ( vB ) 0 t +

1 aB t 2 2

−400 mm = (−160 mm/s)(2 s) + aB = −40 mm/s2 or

1 aB (2 s)2 2

a B = 40 mm/s2 ↑

Then, substituting into Eq. (2) −2(140 mm/s2) − (−40 mm/s2) + 4aC = 0 aC = 60 mm/s2 or

aC = 60 mm/s2 ↓

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79

79


Substituting into Eq. (1) at t = 0 −2(0) − (−160 mm/s) + 4(vC )0 = 0 or (vC )0 = −40 mm/s vC = (vC )0 + aC t

Now

(c)

When vC = 0:

0 = (−40 mm/s) + (60 mm/s2)t

or

t=

2 3

t = 0.667 s

yC = ( yC )0 + (vC )0 t +

We have At t =

2 s 3

s:

yC − (yC )0 = (−40 mm/s)

1 aC t 2 2

2 1 2 s + (60 mm/s2) s 3 3 2

= −13.3 mm

or

2

y C − (y C )0 = 13.3 mm ↑

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80

PROBLEM 11.58 Collars A and B start from rest, and collar A moves upward with an acceleration of 60t 2 mm/s. Knowing that collar B moves downward with a constant acceleration and that its velocity is 160 mm/s after moving 640 mm, determine (a) the acceleration of block C, (b) the distance through which block C will have moved after 3 s.

SOLUTION From the diagram − y A + ( yC − y A ) + 2 yC + ( yC − yB ) = constant

Then

−2v A − vB + 4vC = 0

(1)

and

−2a A − aB + 4aC = 0

(2)

(v A )0 = 0 ( vB ) 0 = 0

Given:

a A = 3t 2 mm/s2 ↑ a B = constant ↓ y B − (y B )0 = 640 mm ↓,

When

v B = 160 mm/s

(a)

We have

vB2 = 0 + 2aB [ yB − ( yB )0 ]

When yB − ( yB )0 = 640 mm: (160 mm/s)2 = 2aB (640 mm) aB = 20 mm/s2

or Then, substituting into Eq. (2)

−2(−20t 2 mm/s2) − (20 mm/s2) + 4aC = 0 aC = 5(1 − 6t 2) mm/s2

or (b)

Substituting into Eq. (1) at t = 0 −2(0) − (0) + 4(vC )0 = 0 or (vC )0 = 0 dvC = aC = 5(1 − 6t 2) dt

Now

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81

81

PROBLEM 11.58 (Continued) vC

At t = 0, vC = 0:

dvC =

0

t

0

(5 − 30t 2)dt

or

vC = 5t − 10t 3

Thus,

vC = 0 5t(1 − 2t 3) = 0

At

t = 0,

or

t = 0.707 s

Therefore, block C initially moves downward (vC

yC

( yC )0

dyC =

t

0

(5t − 10t 3)dt

5 2 4 (t − t ) 2

or

yC − ( yC )0 =

At t = 0.707 s

5 yC − ( yC )0 = [(0.707)2 − (0.707)4] = 0.625 mm 2

At t = 3 s:

5 yC − ( yC )0 = [(3) 2 − (3) 4 ] = −180 mm 2

Total distance traveled

0).

dyC = vC = 5t − 10t 3 dt

Now At t = 0, yC = ( yC )0 :

0) and then moves upward (vC

= (0.625) + −180 − 0.625 = 181.25 mm = 181 mm

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82

PROBLEM 11.59 The system shown starts from rest, and each component moves with a constant acceleration. If the relative acceleration of block C with respect to collar B is 60 mm/s 2 upward and the relative acceleration of block D with respect to block A is 110 mm/s 2 downward, determine (a) the velocity of block C after 3 s, (b) the change in position of block D after 5 s.

SOLUTION From the diagram 2 y A + 2 yB + yC = constant

Cable 1: Then

2v A + 2vB + vC = 0

(1)

and

2a A + 2aB + aC = 0

(2)

( yD − y A ) + ( yD − yB ) = constant

Cable 2: Then and

(3)

− a A − aB + 2aD = 0

(4)

At t = 0, v = 0; all accelerations constant; aC/B = 60 mm/s 2 ↑, aD/A = 110 mm/s 2 ↓

Given: (a)

− v A − vB + 2 vD = 0

We have

aC/B = aC − aB = −60 or aB = aC + 60

and

aD/A = aD − a A = 110 or a A = aD − 110

Substituting into Eqs. (2) and (4) Eq. (2):

2(aD − 110) + 2( aC + 60) + aC = 0 3aC + aD = 100

or Eq. (4):

−(aD − 110) − ( aC + 60) + 2aD = 0 − aC + aD = −50

or

(5)

(6)


83

83

PROBLEM 11.59 (Continued) Solving Eqs. (5) and (6) for aC and aD aC = 40 mm/s 2 aD = −10 mm/s 2

Now

vC = 0 + aC t

At t = 3 s:

vC = (40 mm/s 2 )(3 s) vC = 120 mm/s ↓ ��

or (b)

We have At t = 5 s:

yD = ( yD )0 + (0)t + yD − ( yD )0 =

1 aD t 2 2

1 ( −10 mm/s 2 )(5 s) 2 2 y D − (y D )0 = 125 mm ↑ ��

or


84

PROBLEM 11.60* The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if aB � 10 mm/s 2 , (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

SOLUTION From the diagram 2 y A + 2 yB + yC = constant

Cable 1: Then

2v A + 2vB + vC = 0

(1)

and

2a A + 2aB + aC = 0

(2)

Cable 2: ( yD − y A ) + ( yD − yB ) = constant Then

− v A − vB − 2 vD = 0

(3)

and

− a A − aB + 2aD = 0

(4)

t =0

Given: At

v=0 (y A )0 = ( yB )0 = ( yC )0

All accelerations constant at t = 2 s yC/A = 280 mm ↑ vB/A = 80 mm/s ↓

When

y A − ( y A )0 = 160 mm ↑ yB − ( yB )0 = 320 mm ↓ aB � 10 mm/s 2


85

85

PROBLEM 11.60* (Continued)

(a)

We have

y A = ( y A )0 + (0)t +

1 a At 2 2

and

yC = ( yC )0 + (0)t +

1 aC t 2 2

Then

yC/A = yC − y A =

1 (aC − a A )t 2 2

At t = 2 s, yC/A = −280 mm: −280 mm =

or

1 (aC − a A )(2 s) 2 2

aC = a A − 140

(5)

Substituting into Eq. (2) 2a A + 2aB + (a A − 140) = 0

or

1 a A = (140 − 2aB ) 3

Now

vB = 0 + a B t

(6)

v A = 0 + a At vB/A = vB − v A = (aB − a A )t

Also When

�

yB = ( yB )0 + (0)t +

1 aB t 2 2

v B/A = 80 mm/s ↓ : 80 = (aB − a A )t ∆y A = 160 mm ↓ : 160 =

1 a At 2 2

∆yB = 320 mm ↓ : 320 =

1 aB t 2 2

(7)

1 (aB − a A )t 2 2

Then

160 =

Using Eq. (7)

320 = (80)t or t = 4 s

Then

160 =

1 a A (4)2 2

or

a A = 20 mm/s 2 ↓ �

and

320 =

1 aB (4) 2 2

or

a B = 40 mm/s 2 ↓ ��

Note that Eq. (6) is not used; thus, the problem is over-determined.


86

PROBLEM 11.60* (Continued) Alternative solution: v A2 = (0) + 2a A [ y A − ( y A )0 ]

We have

vB2 = (0) + 2aB [ yB − ( yB )0 ]

Then

vB/A = vB − v A = 2aB [ yB − ( yB )0 ] − 2a A [ y A − ( y A )0 ]

When

v B/A = 80 mm/s ↓ :

80 mm/s = 2 � aB (320 mm) − a A (160 mm) � � �

or

20 = 2

(

200 B − 100 A

)

(8)

Solving Eqs. (6) and (8) yields a A and a B . (b)

Substituting into Eq. (5) aC = 20 − 140 = −120 mm/s 2

and into Eq. (4) −(20 mm/s 2 ) − (40 mm/s 2 ) + 2aB = 0

or

aD = 30 mm/s 2

Now

vC = 0 + aC t

When vC = −600 mm/s:

− 600 mm/s = (−120 mm/s 2 )t t =5s

or

yD = ( yD )0 + (0)t +

Also At t = 5 s:

yD − ( yD )0 =

1 aD t 2 2

1 (30 mm/s 2 )(5 s)2 2 y D − (y D )0 = 375 mm ↓ ��

or


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87

PROBLEM 11.61 A

B

x d

A subway car leaves station A; it gains speed at the rate of 2 m/s2 for 6 s and then at the rate of 3 m/s2 until it has reached the speed of 24 m/s. The car maintains the same speed until it approaches station B; brakes are then applied, giving the car a constant deceleration and bringing it to a stop in 6 s. The total running time from A to B is 40 s. Draw the a−t, v−t, and x−t curves, and determine the distance between stations A and B.

SOLUTION Acceleration-Time Curve. Since the acceleration is either constant or zero, the a−t curve is made of horizontal straight-line segments. The values of t2 and a4 are determined as follows:

a(m/s2) 4 3

0 , t , 6:

2 1 0 –1

34 40 6 t2

t(s) a4

–2 –3 –4

Change in v 5 area under a – t curve v6 2 0 5 (6 s)(2 m/s2) 5 12 m/s 6 , t , t2: Since the velocity increases from 12 to 24 m/s, Change in v 5 area under a – t curve 24 m/s 2 12 m/s 5 (t2 2 6)(3 m/s2) t2 5 10 s t2 , t , 34: Since the velocity is constant, the acceleration is zero. 34 , t , 40: Change in v 5 area under a – t curve a4 5 24 m/s2 0 2 24 m/s 5 (6 s)a4 The acceleration being negative, the corresponding area is below the t axis; this area represents a decrease in velocity. Velocity-Time Curve. Since the acceleration is either constant or zero, the v−t curve is made of straight-line segments connecting the points determined above.

v(m/s)

Change in x 5 area under v−t curve

24 12 0

6 10

34 40

t(s)

0 6 10 34

, , , ,

t t t t

, , , ,

6: 10: 34: 40:

x6 2 0 x10 2 x6 x34 2 x10 x40 2 x34

5 5 5 5

1 2 (6)(12) 5 36 1 2 (4)(12 1 48)

m 5 72 m (24)(24) 5 576 m 1 2 (6)(24) 5 72 m

Adding the changes in x, we obtain the distance from A to B: d 5 x40 2 0 5 756 m

x(m)

d 5 756 m

756 m

0

6 10

34 40

◀

Position-Time Curve. The points determined above should be joined by three arcs of parabola and one straight-line segment. In constructing the x−t curve, keep in mind that for any value of t the slope of the tangent to the x−t curve is equal to the value of v at that instant.

t(s)

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PROBLEM 11.62 For the particle and motion of Problem 11.61, plot the v −t and x −t curves for 0 t 20 s and determine (a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate.

SOLUTION (a)

t = 0, v0 = −3.6 m/s,

Initial conditions:

x0 = 0

Change in v equals area under a −t curve: v0 = −3.6 m/s 0

t

4 s:

v4 − v0 = (0.6 m/s2)(4 s) = +2.4 m/s

v4 = −1.2 m/s

4s

t

10 s:

v10 − v4 = (1.2 m/s2)(6 s) = +7.2 m/s

v10 = +6 m/s

10 s

t

12 s:

v12 − v10 = (−1 m/s2)(2 s) = −2 m/s

v12 = +4 m/s

12 s

t

20 s:

v20 − v12 = (−1 m/s2)(8 s) = −8 m/s

v20 = −4 m/s

Change in x equals area under v −t curve: x0 = 0 0

t

4 s:

4s

t

5 s:

5s

t

10 s:

12 s

t

10 s:

1 (−3.6 −1.2)(4) = −9.6 m 2 1 x5 − x4 = (−1.2)(1) = −0.6 m 2 1 x10 − x5 = (+6)(5) = +15 m 2 1 x12 − x10 = (+6 + 4)(2) = +10 m 2 x4 − x0 =

x4 = −9.6 m x5 = −10.2 m x10 = +4.8 m x12 = +14.8 m

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16 s

t

12 s:

20 s

t

16 s:

1 (+4)(4) = +8 m 2 1 x20 − x16 = (−4)(4) = −8 m 2 x16 − x12 =

x16 = +22.8 m x20 = +14.8 m

From v −t and x −t curves, we read (a)

At t = 10 s:

vmax = +6 m/s

(b)

At t = 16 s:

xmax = +22.8 m

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PROBLEM 11.63 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 mm at t = 0, (a) construct the a −t and x −t curves for 0 t 50 s, and determine (b) the total distance traveled by the particle when t = 50 s, (c) the two times at which x = 0.

SOLUTION (a)

at = slope of v −t curve at time t

From t = 0 to t = 10 s:

v = constant

−20 − 60 = −5 mm/s2 26 − 10

t = 10 s to t = 26 s:

a=

t = 26 s to t = 41 s:

v = constant

t = 41 s to t = 46 s:

a=

t = 46 s:

a=0

a=0

−5 − (−20) = 3 mm/s2 46 − 41

v = constant

a=0

x2 = x1 + (area under v −t curve from t1 to t2 )

At t = 10 s:

x10 = −540 + 10(60) + 60 mm

Next, find time at which v = 0. Using similar triangles tv = 0 − 10 60

At

t = 22 s: t = 26 s: t = 41 s: t = 46 s: t = 50 s:

=

26 − 10 80

or

tv = 0 = 22 s

1 (12)(60) = 420 mm 2 1 x26 = 420 − (4)(20) = 380 mm 2 x41 = 380 − 15(20) = 80 mm x22 = 60 +

20 + 5 = 17.5 mm 2 x50 = 17.5 − 4(5) = −2.5 mm x46 = 80 − 5

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From t = 0 to t = 22 s: Distance traveled = 420 − (−540) = 960 mm t = 22 s to t = 50 s: Distance traveled = |− 2.5 − 420| = 422.5 mm Total distance traveled = (960 + 422.5) mm = 1382.5 mm = 1.383 mm

(c)

Using similar triangles Between 0 and 10 s:

(t x = 0 )1 − 0 10 = 540 600 (t x = 0 )1 = 9.00 s

or Between 46 s and 50 s:

(t x = 0 )2 − 46 4 = 17.5 20 (t x =0 ) 2 = 49.5 s

or

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PROBLEM 11.64 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 mm at t = 0, (a) construct the a −t and x −t curves for 0 t 50 s, and determine (b) the maximum value of the position coordinate of the particle, (c) the values of t for which the particle is at x = 100 mm.

SOLUTION (a)

at = slope of v −t curve at time t

From t = 0 to t = 10 s:

v = constant

−20 − 60 = −5 mm/s2 26 − 10

t = 10 s to t = 26 s:

a=

t = 26 s to t = 41 s:

v = constant

t = 41 s to t = 46 s:

a=

t = 46 s:

a=0

a=0

−5 − ( −20) = 3 mm/s2 46 − 41

v = constant

a=0

x2 = x1 + (area under v −t curve from t1 to t2 )

At t = 10 s:

x10 = −540 + 10(60) = 60 mm

Next, find time at which v = 0. Using similar triangles tv = 0 − 10 60

At

t = 22 s: t = 26 s: t = 41 s: t = 46 s: t = 50 s:

=

26 − 10 80

or

tv = 0 = 22 s

1 (12)(60) = 420 mm 2 1 x26 = 420 − (4)(20) = 380 mm 2 x41 = 380 − 15(20) = 80 mm x22 = 60 +

20 + 5 = 17.5 mm 2 x50 = 17.5 − 4(5) = −2.5 mm x46 = 80 − 5

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Reading from the x −t curve

(c)

Between 10 s and 22 s

xmax = 420 mm

100 mm = 420 mm − (area under v−t curve from t, to 22 s) mm 1 100 = 420 − (22 − t1 )(v1 ) 2

or or

(22 − t1 )(v1 ) = 640

Using similar triangles v1 60 = 22 − t1 12

Then or

v1 = 5(22 − t1 )

or

(22 − t1 )[5(22 − t1 )] = 640 t1 = 10.69 s and t1 = 33.3 s

We have

10 s

t1

22 s

t1 = 10.69 s

Between 26 s and 41 s: Using similar triangles 41 − t2 15 = 20 300

or t2 = 40 s


94

PROBLEM 11.65 A parachutist is in free fall at a rate of 200 km/h when he opens his parachute at an altitude of 600 m. Following a rapid and constant deceleration, he then descends at a constant rate of 50 km/h from 586 m to 30 m, where he maneuvers the parachute into the wind to further slow his descent. Knowing that the parachutist lands with a negligible downward velocity, determine (a) the time required for the parachutist to land after opening his parachute, (b) the initial deceleration.

SOLUTION Assume second deceleration is constant. Also, note that 200 km/h = 55.555 m/s, 50 km/h = 13.888 m/s

(a)

Now ∆x = area under v −t curve for given time interval Then or

� 55.555 + 13.888 � (586 − 600) m = −t1 � � m/s 2 � �

t1 = 0.4032 s (30 − 586) m = −t2 (13.888 m/s)

or

t2 = 40.0346 s 1 (0 − 30) m = − (t3 )(13.888 m/s) 2

or

t3 = 4.3203 s ttotal = (0.4032 + 40.0346 + 4.3203) s ttotal = 44.8 s �

or (b)

We have

ainitial = =

∆ vinitial t1 [−13.888 − (−55.555)] m/s 0.4032 s

= 103.3 m/s 2 ainitial = 103.3 m/s 2 ↑ �

or

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PROBLEM 11.66 A machine component is spray-painted while it is mounted on a pallet that travels 4 m in 20 s. The pallet has an initial velocity of 80 mm/s and can be accelerated at a maximum rate of 60 mm/s2. Knowing that the paining process requires 15 s to complete and is performed as the pallet moves with a constant speed, determine the smallest possible value of the maximum speed of the pallet.

SOLUTION First note that (80 mm/s) (20 s) � 4000 mm, so that the speed of the pallet must be increased. Since vpaint = constant, it follows that vpaint = vmax and then t1 � .5 s. From the v −t curve, A1 + A2 = 4000 mm and it is seen that (vmax ) min occurs when � v − 80 � a1 � = max � is maximum. t1 � �

Thus,

(vmax − 80)mm/s = 60 mm/s t1(s) t1 =

or and

� 80 + vmax t1 � 2 �

(vmax − 80) 60

� � + (20 − t1 ) (vmax ) = 4000 �

Substituting for t1 (vmax − 80) � 80 + vmax � 60 2 �

vmax − 80 � � � vmax = 4000 � + � 20 − 60 ��

2 vmax − 2560vmax + 486400 = 0

Simplifying Solving

vmax = 207 mm/s

and

For

vmax = 207 mm/s,

t1 � 5 s

vmax = 2353 mm/s,

t1 � 5 s

vmax = 2353 mm/s

(vmax )min = 207 mm/s �

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PROBLEM 11.67 A temperature sensor is attached to slider AB, which moves back and forth through 60 cm. The maximum velocities of the slider are 12 cm/s to the right and 30 cm/s to the left. When the slider is moving to the right, it accelerates and decelerates at a constant rate of 6 cm/s2; when moving to the left, the slider accelerates and decelerates at a constant rate of 20 cm/s2. Determine the time required for the slider to complete a full cycle, and construct the v – t and x−t curves of its motion.

SOLUTION The v −t curve is first drawn as shown. Then ta =

vright aright

=

12 cm/s =2s 6 cm/s2

vleft 30 cm/s = aleft 20 cm/s = 1.5 s

td =

A1 = 60 cm

Now or

[(t1 − 2)s](12 cm/s) = 60 cm t1 = 7 s

or

A2 = 60 cm

and or

{[(t2 − 7) − 1.5] s}(30 cm/s) = 60 cm

or

t2 = 10.5 s tcycle = t2

Now

tcycle = 10.5 s

We have xii = xi + (area under v −t curve from ti to tii ) At

t = 5 s:

1 (2) (12) = 12 cm 2 x5 = 12 + (5 − 2)(12)

t = 7 s:

= 48 cm x7 = 60 cm

t = 2 s:

t = 8.5 s: t = 9 s: t = 10.5 s:

x2 =

1 x8.5 = 60 − (1.5)(30) 2 = 37.5 cm x9 = 37.5 − (0.5)(30) x10.5

= 22.5 cm =0

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PROBLEM 11.68 A commuter train traveling at 40 km/h is 3 km from a station. The train then decelerates so that its speed is 20 km/h when it is 0.5 km from the station. Knowing that the train arrives at the station 7.5 min after beginning to decelerate and assuming constant decelerations, determine (a) the time required for the train is travel the first 2.5 km, (b) the speed of the train as it arrives at the station, (c) the final constant deceleration of the train.

SOLUTION Given: At t = 0, v = 40 km/h x = 0; when x = 2.5 km, v = 20 km/h; at t = 7.5 min, x = 3 km; constant decelerations The v −t curve is first drawn is shown. (a)

A1 = 2.5 km

We have or

(t1 min)

40 + 20 1h km/h × = 2.5 km 2 60 min t1 = 5 min

or (b)

A2 = 0.5 km

We have or

(7.5 − 5) min ×

20 + v2 1h km/h × = 0.5 km 2 60 min v2 = 4 km/h

or (c)

We have

afinal = a12 =

(4 − 20) km/h 1000 m 1 min 1h × × × 60s 3600 s (7.5 − 5) min km afinal = −0.03 m/s2

or

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PROBLEM 11.69 Two road rally checkpoints A and B are located on the same highway and are 12 km apart. The speed limits for the first 8 km and the last 4 km of the section of highway are 100 km/h and 70 km/h, respectively. Drivers must stop at each checkpoint, and the specified time between Points A and B is 8 min 20 s. Knowing that a driver accelerates and decelerates at the same constant rate, determine the magnitude of her acceleration if she travels at the speed limit as much as possible.

SOLUTION Given:

(vmax ) AC = 100 km/h, (vmax )CB = 70 km/h; v A = vB = 0; t AB = B min, 20 s; | a | = constant; v = vmax as such as possible

The v −t curve is first drawn as shown, where the magnitudes of the slopes (accelerations) of the three inclined lines are equal. Note: At

8 min 20 s =

5 h 36

t1 , x = 8 km 5 h, x = 12 km 36

Denoting the magnitude of the accelerations by a, we have a=

100 ta

a=

30 tb

a=

70 tc

where a is in km/h2 and the times are in h. Now A1 = 8 km:

1 1 (t1 )(100) − (ta )(100) − (tb )(30) = 8 2 2


99

100


Substituting

1 � 100 � 1 � 30 � 100t1 − � (100) − � � (30) = 8 � 2� a � 2� a � t1 = 0.08 +

or Also A2 = 4 km: Substituting

1 � 5 � � 36 − t1 � (70) − 2 (tc )(70) = 4 � � 1 � 70 � � 5 � � 36 − t1 � (70) − 2 � a � (70) = 4 � � � � t1 =

or Then or

54.5 a

0.08 =

103 35 − 1260 a

54.5 103 35 − = a 1260 a

1000 m � 1 h � a = 51.259 km/h × ×� � km � 3600 s �

2

2

a = 3.96 m/s 2 �

or


100

PROBLEM 11.70 In a water-tank test involving the launching of a small model boat, the model’s initial horizontal velocity is 6 m/s, and its horizontal acceleration varies linearly from −12 m/s 2 at t = 0 to −2 m/s 2 at t = t1 and then remains equal to −2 m/s 2 until t = 1.4 s. Knowing that v = 1.8 m/s when t = t1 , determine (a) the value of t1 , (b) the velocity and the position of the model at t = 1.4 s.

SOLUTION Given:

v0 = 6 m/s; for 0 � t � t , a × t ;

for

t1 � t � 1.4 s a = −2 m/s 2 ;

at

t = 0 a = −12 m/s 2 ; at t = t1 a = −2 m/s 2 , v = 1.8 m/s 2

The a −t and v −t curves are first drawn as shown. (a)

We have

vt1 = v0 + A1 � 12 + 2 � 2 1.8 m/s = 6 m/s − (t1 s) � � m/s � 2 �

or

t1 = 0.6 s ��

or (b)

We have

v1.4 = vt1 + A2

or

v1.4 = 1.8 m/s − (1.4 − 0.6)5 × 2 m/s 2 v1.4 = 0.20 m/s �

or

Now x1.4 = A3 + A4 , where A3 is most easily determined using integrating. Thus, for 0 � t � t1 :

a=

−2 − ( −12) 50 t − 12 = t − 12 0.6 3

dv =a dt 50 = t − 12 3

Now


101

102


At t = 0, v = 6 m/s: or

�

v

6

dv =

t � 50

�

� �� 3 t − 12 �� dt 0

v=6+

25 2 t − 12t 3

We have

25 dx = v = 6 − 12t + t 2 3 dt

Then

A3 =

�

xt1

0

dx =

�

0.6

0

(6 − 12t +

25 2 t )dt 3

0.6

25 � � = �6t − 6t 2 + t 3 � = 2.04 m 9 �0 �

Also Then

� 1.8 + 0.2 � A4 = (1.4 − 0.6) � � = 0.8 m 2 � �

x1.4 = (2.4 + 0.8) m x1.4 = 2.84 m �

or


102

PROBLEM 11.71 A car and a truck are both traveling at the constant speed of 56 km/h; the car is 12 m behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 12 m in front of the truck, and then resume the speed of 56 km/h. The maximum acceleration of the car is 1.5 m/s2 and the maximum deceleration obtained by applying the brakes is 6 m/s2. What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 80 km/h? Draw the v − t curve.

SOLUTION Relative to truck, car must move a distance: Allowable increase in speed:

∆x = 4.8 + 12 + 15 + 12 = 43.8 m ∆vm = 80 − 56 = 24 km/h = 6.7 m/s

Acceleration Phase:

t1 = 6.7/1.5 = 4.5 s

1 A1 = (6.7)(4.5) = 15.1 m 2

Deceleration Phase:

t3 = 6.7/6 = 1.1 s

1 A3 = (6.7)(1.1) = 7.4 m 2

But: ∆x = A1 + A2 + A3 :

43.8 m = 15.1 + (6.7)t2 + 7.4 ttotal = t1 + t2 + t3 = 4.5 s + 3.2 s + 1.1 s = 8.8 s

t2 = 3.2 s t F = 8.8 s


103

104

PROBLEM 11.72 Solve Problem 11.71, assuming that the driver of the car does not pay any attention to the speed limit while passing and concentrates on reaching position B and resuming a speed of 56 km/h in the shortest possible time. What is the maximum speed reached? Draw the v − t curve.

SOLUTION Relative to truck, car must move a distance: ∆x = 4.8 + 12 + 15 + 12 = 43.8 m

∆vm = 1.5t1 = 6t2; ∆x = A1 + A2 :

t2 =

43.8 m =

1 (∆vm )(t1 + t2 ) 2

43.8 m =

1 1 (1.5t1 ) t1 + t1 2 4

1 t1 4

t12 = 46.72 t1 = 6.835 s t2 =

1 t1 = 1.709 4

ttotal = t1 + t2 = 6.835 + 1.709

t F = 8.54 s

∆vm = 1.5t1 = 1.5(6.835) = 10.35 m/s = 36.9 km/h Speed vtotal = 56 km/h, vm = 56 km/h + 36.9 km/h

vm = 92.9 km/h

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104

PROBLEM 11.73 An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s2 until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the elevator begins to move, a man standing 12 m above the initial position of the top of the elevator throws a ball upward with an initial velocity of 20 m/s. Determine when the ball will hit the elevator.

SOLUTION Given:

At t = 0 vE = 0; For 0 � vE � 7.8 m/s, aE = 1.2 m/s 2 ↑; For vE = 7.8 m/s, aE = 0;

At t = 2 s, vB = 20m/s ↑ The v −t curves of the ball and the elevator are first drawn as shown. Note that the initial slope of the curve for the elevator is 1.2 m/s 2 , while the slope of the curve for the ball is − g (−9.81 m/s 2 ).

The time t1 is the time when vE reaches 7.8 m/s. Thus,

vE = (0) + aE t

or

7.8 m/s = (1.2 m/s 2 )t1

or

t1 = 6.5 s

The time ttop is the time at which the ball reaches the top of its trajectory. Thus, or or

vB = (vB )0 − g (t − 2) 0 = 20 m/s − (9.81 m/s 2 ) (t top − 2) s ttop = 4.0387 s


105

106

PROBLEM 11.73 (Continued) Using the coordinate system shown, we have 0 � t � t1 :

�1 � yE = −12 m + � aE t 2 � m �2 �

At t = t top :

yB =

and

yE = −12 m +

At and at t = t1 ,

1 (4.0387 − 2) s × (20 m/s) 2 = 20.387 m 1 (1.2 m/s 2 )(4.0387 s) 2 2 = −2.213 m

t = [2 + 2(4.0387 − 2)] s = 6.0774 s, yB = 0 yE = −12 m +

1 (6.5 s) (7.8 m/s) = 13.35 m 2

The ball hits the elevator ( yB = yE ) when ttop � t � t1 . For t � t top :

�1 � yB = 20.387 m − � g (t − t top ) 2 � m 2 � �

Then, when

y B = yE 1 20.387 m − (9.81 m/s 2 ) (t − 4.0387) 2 2 1 = −12 m + (1.2 m/s 2 ) (t s) 2 2

or Solving

5.505t 2 − 39.6196t + 47.619 = 0 t = 1.525 s and t = 5.67 s t = 1.525 s �

Choosing the smaller value


106

PROBLEM 11.74 The acceleration record shown was obtained for a small airplane traveling along a straight course. Knowing that x = 0 and v = 60 m/s when t = 0, determine (a) the velocity and position of the plane at t = 20 s, (b) its average velocity during the interval 6 s � t � 14 s.

SOLUTION

Geometry of “bell-shaped” portion of v −t curve

The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of v −t diagram is: = ∆x = 6 m

(a)

v20 = 50 m/s �

When t = 20 s: x20 = (60 m/s) (20 s) − (shaded area) = 1200 m − 6 m

(b)

From t = 6 s to t = 14 s:

x20 = 1194 m �

∆t = 8 s ∆x = (60 m/s) (14 s − 6 s) − (shaded area) = (60 m/s)(8 s) − 6 m = 480 m − 6 m = 474 m vaverage =

∆x 474 m = 8s ∆t

vaverage = 59.25 m/s �


107

108

PROBLEM 11.75 Car A is traveling on a highway at a constant speed (vA)0 = 95 km/h and is 115 m from the entrance of an access ramp when car B enters the acceleration lane at that point at a speed (vB)0 = 25 km/h. Car B accelerates uniformly and enters the main traffic lane after traveling 60 m in 5 s. It then continues to accelerate at the same rate until it reaches a speed of 95 km/h, which it then maintains. Determine the final distance between the two cars.

SOLUTION Given:

(v A )0 = 95 km/h, (vB )0 = 25 km/h; at t = 0, ( x A )0 = −115 m, (xB)0 = 0;

at t = 5 s,

xB = 60 m; for 25 km/h

95 km/h,

vB

aB = constant; for vB = 95 km/h,

First note

aB = 0 95 km/h = 26.4 m/s 25 km/h = 6.9 m/s

The v −t curves of the two cars are then drawn as shown. Using the coordinate system shown, we have at t = 5 s, xB = 60 m:

(5 s)

(6.9 + vB ) S m/s = 60 m 2 (vB )S = 17.1 m/s

or Then, using similar triangles, we have

(26.4 − 6.9) m/s (17.1 − 6.9) m/s = (= aB) t1 5s

or

t1 = 9.6 s

Finally, at t = t1 xB/A = xB − x A = (9.6 s)

6.9 + 26.4 m/s 2

−[−115 m + (9.6 s)(26.4 m/s)] xB/A = 21.4 m

or

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108

PROBLEM 11.76 Car A is traveling at 64 km/h when it enters a 48 km/h speed zone. The driver of car A decelerates at a rate of 4.8 m/s2 until reaching a speed of 48 km/h, which she then maintains. When car B, which was initially 18 m behind car A and traveling at a constant speed of 72 km/h, enters the speed zone, its driver decelerates at a rate of 6 m/s2 until reaching a speed of 45 km/h. Knowing that the driver of car B maintains a speed of 45 km/h, determine (a) the closest that car B comes to car A, (b) the time at which car A is 21 m in front of car B.

SOLUTION (v A )0 = 64 km/h; For 48 km/h

Given:

vA

64 km/h;

aA = −4.8 m/s2; For vA = 48 km/h, aA = 0; ( x A/B )0 = 18 m, (vB)0 = 72 km/h; when xB = 0, aB = −6 m/s2; for vB = 45 km/h aB = 0 64 km/h = 17.8 m/s 72 km/h = 20 m/s

First note

48 km/h = 13.3 m/s 45 km/h = 12.5 m/s

At t = 0

The v −t curves of the two cars are as shown. At

t = 0:

car A enters the speed zone

t = (t B )1 :

car B enters the speed zone

t = tA:

car A reaches its final speed

t = tmin :

v A = vB

t = (t B )2 :

car B reaches its final speed

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109

110

PROBLEM 11.76 (Continued) (a)

aA =

We have

−4.8 m/s2 =

or

(v A )final − (v A )0 tA (13.3 − 17.8) m/s tA

t A = 0.938 s

or Also

18 m = (t B )1 (vB )0

or

18 m = (t B )1 (20 m/s) aB =

(vB )final − (vB )0 (t B ) 2 − (t B )1

−6 m/s2 =

(12.5 − 20) m/s [(t B )2 − 0.9] s

and or

(t B )1 = 0.9 s

or

(tB)2 = 2.15 s

Car B will continue to overtake car A while vB

v A . Therefore, ( x A/B ) min will occur when v A = vB ,

which occurs for (t B )1

tmin

(tB ) 2

For this time interval v A = 13.3 m/s vB = (vB )0 + aB [t − (tB )1 ]

Then

13.3 m/s = 20 m/s + (−6 m/s2)(tmin − 0.9) s

at t = tmin :

tmin = 2.017 s

or

Finally ( x A/B ) min = ( x A )tmin − ( xB )tmin = tA

(v A )0 + (v A )final + (tmin − t A )(v A )final 2

− ( xB )0 + (t B )1 (vB )0 + [tmin − (t B )1 ] = (0.938 s)

(vB )0 + (v A )final 2

17.8 + 13.3 m/s + (2.017 − 0.938) s × (13.3 m/s) 2

− −18 m + (0.9 s)(20 m/s) + (2.017 − 0.9) s ×

20 + 13.3 m/s 2

= (14.586 + 14.351) m − (−18 + 18 + 18.598) m = 10.339 m

or ( x A/B ) min = 10.3 m


110


(b)

Since ( x A/B ) [Note (t B ) 2

18 m for t

tmin , it follows that x A/B = 21 m for t

tmin ]. Then, for t

(t B )2

(t B )2

x A/B = ( x A/B ) min + [(t − tmin )(v A )final ] − [(t B )2 − (tmin )]

or

(v A )final + (vB )final + [t − (t B ) 2 ](vB )final 2

21 m = 10.339 m + [(t − 2.017) s × (13.3 m/s)] − (2.15 − 2.017) s ×

13.3 + 12.5 m/s + (t − 2.15) s × (12.5) m/s 2 t = 15.4 s

or

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111

112

PROBLEM 11.77 A car is traveling at a constant speed of 54 km/h when its driver sees a child run into the road. The driver applies her brakes until the child returns to the sidewalk and then accelerates to resume her original speed of 54 km/h; the acceleration record of the car is shown in the figure. Assuming x = 0 when t = 0, determine (a) the time t1 at which the velocity is again 54 km/h, (b) the position of the car at that time, (c) the average velocity of the car during the interval 1 s � t � t1.

SOLUTION Given:

t = 0, x = 0, v = 54 km/h; for t = t1 ,

At

v = 54 km/h

First note (a)

54 km/h = 15 m/s vb = va + (area under a −t curve from ta to tb )

We have Then

at

t = 2 s:

v = 15 − (1)(6) = 9 m/s

t = 4.5 s:

1 v = 9 − (2.5)(6) = 1.5 m/s 2

t = t1:

15 = 1.5 +

1 (t1 − 4.5)(2) 2 t1 = 18 s ��

or (b)

Using the above values of the velocities, the v −t curve is drawn as shown.


112

PROBLEM 11.77 (Continued) Now

x at t = 18 s x18 = 0 + Σ (area under the v −t curve from t = 0 to t = 18 s)

� 15 + 9 � = (1 s)(15 m/s) + (1 s) � � m/s � 2 � 1 � � + �(2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s) � 3 � � 2 � � + �(13.5 s)(1.5 m/s) + (13.5 s)(13.5 m/s) � 3 � � = [15 + 12 + (3.75 + 6.25) + (20.25 + 121.50)] m = 178.75 m

(c)

First note

or

x18 = 178.8 m �

x1 = 15 m x18 = 178.75 m

Now

vave =

∆x (178.75 − 15) m = 9.6324 m/s = ∆t (18 − 1) s vave = 34.7 km/h ��

or


113

114

PROBLEM 11.78 As shown in the figure, from t = 0 to t = 4 s, the acceleration of a given particle is represented by a parabola. Knowing that x = 0 and v = 8 m/s when t = 0, (a) construct the v −t and x–t curves for 0 � t � 4 s, (b) determine the position of the particle at t = 3 s. (Hint: Use table inside the front cover.)

SOLUTION Given: (a)

At

t = 0, x = 0, v = 8 m/s

We have

v2 = v1 + (area under a −t curve from t1 to t2)

and

x2 = x1 + (area under v −t curve from t1 to t2)

Then, using the formula for the area of a parabolic spandrel, we have 1 at t = 2 s: v = 8 − (2)(12) = 0 3 1 t = 4 s: v = 0 − (2)(12) = −8 m/s 3

The v −t curve is then drawn as shown.

Note: The area under each portion of the curve is a spandrel of Order no. 3. (2)(8) =4m 3 +1 (2)(8) t = 4 s: x = 4 − =0 3 +1

Now at t = 2 s: x = 0 +

The x −t curve is then drawn as shown.


114


We have At t = 3 s: a = −3(3 − 2)2 = −3 m/s 2 1 v = 0 − (1)(3) = −1 m/s 3 (1)(1) x =4− 3 +1 x3 = 3.75 m ��

or


115

116

PROBLEM 11.79 During a manufacturing process, a conveyor belt starts from rest and travels a total of 0.4 m before temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ±1.6 m/s2 per second, determine (a) the shortest time required for the belt to move 0.4 m, (b) the maximum and average values of the velocity of the belt during that time.

SOLUTION Given:

(a)

At

t = 0, x = 0, v = 0; xmax = 0.4 m;

when

x = xmax , v = 0;

da dt

= 1.6 m/s2 max

Observing that vmax must occur at t = 12 tmin , the a −t curve must have the shape shown. Note that the magnitude of the slope of each portion of the curve is 1.6 m/s2/s.

We have

at t = ∆t : t = 2∆t :

v =0+

1 1 (∆t )(amax ) = amax ∆t 2 2

vmax =

1 1 amax ∆t + ( ∆t )( amax ) = amax ∆t 2 2

Using symmetry, the v −t is then drawn as shown.

Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have (2∆t )(vmax ) = xmax vmax = amax ∆t

2amax ∆t 2 = xmax


116


Now

or Then

amax = 1.6 m/s2/s so that ∆t 2(1.6∆t m/s3) ∆t2 = 0.4 m ∆t = 0.5 s tmin = 4∆t tmin = 2.00 s

or (b)

We have

vmax = amax ∆t = (1.6 m/s2/s × ∆t)∆t = 1.6 m/s2/s × (0.5 s)2 vmax = 0.4 m/s

or Also

vave =

∆x 0.4 m = ∆t total 2.00 s vave = 0.2 m/s

or


117

118

PROBLEM 11.80 An airport shuttle train travels between two terminals that are 2.6 km apart. To maintain passenger comfort, the acceleration of the train is limited to ±1.2 m/s2, and the jerk, or rate of change of acceleration, is limited to ±0.2 m/s2 per second. If the shuttle has a maximum speed of 32 km/h, determine (a) the shortest time for the shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.

SOLUTION xmax = 2.6 km; |amax | = 1.2 m/s2

Given: da dt

First note (a)

= 0.2 m/s2/s; vmax = 32 km/h max

32 km/h = 8.9 m/s

To obtain tmin, the train must accelerate and decelerate at the maximum rate to maximize the time for which v = vmax. The time ∆t required for the train to have an acceleration of 1.2 m/s2 is found from da dt

or or

= max

amax ∆t

1.2 m/s2 0.2 m/s2/s ∆t = 6 s

∆t =

Now, da = constant dt

after 5 s, the speed of the train is

v5 =

1 (∆t )(amax ) 2

or

v5 =

1 (6 s)(1.2 m/s2) = 3.6 m/s 2

since

Then, since v5 < vmax, the train will continue to accelerate at 1.2 m/s2 until v = vmax. The a−t curve must then have the shape shown. Note that the magnitude of the slope of each inclined portion of the curve is 0.2 m/s2/s.


118

PROBLEM 11.80 (Continued) Now

at t = (12 + ∆t1 ) s, v = vmax : 1 (6 s)(1.2 m/s2) + (∆t1 )(1.2 m/s2) = 8.9 m/s 2

2

or

∆t1 = 1.42 s

Then

at t = 6 s: t = 7.42 s:

1 (6)(1.2) = 3.6 m/s 2 v = 3.6 + (1.42)(1.2) = 5.3 m/s

t = 13.42 s:

v = 5.3 +

v=0+

1 (6)(1.2) = 8.9 m/s 2

Using symmetry, the v −t curve is then drawn as shown.

Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have 2 (1.42 s)

3.6 + 5.3 m/s 2

+ (12 + ∆t2) s × (8.9 m/s) = 2600 m

or

∆t2 = 278.7 s

Then

tmin = 4(6 s) + 2(1.42 s) + 278.7 s = 305.54 s tmin = 5.09 min

or (b)

We have

vave =

∆x 2.6 km 3600 s = × ∆t 305.54 s 1h vave = 30.6 km/h

or

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119

120

PROBLEM 11.81 The acceleration record shown was obtained during the speed trials of a sports car. Knowing that the car starts from rest, determine by approximate means (a) the velocity of the car at t = 8 s, (b) the distance the car has traveled at t = 20 s.

SOLUTION Given: a −t curve; at 1.

t = 0, x = 0, v = 0

The a −t curve is first approximated with a series of rectangles, each of width ∆t = 2 s. The area (∆t )(aave ) of each rectangle is approximately equal to the change in velocity ∆v for the specified interval of time. Thus, ∆v � aave ∆t

where the values of aave and ∆v are given in columns 1 and 2, respectively, of the following table. 2.

Noting that v0 = 0 and that v2 = v1 + ∆v12

where ∆v12 is the change in velocity between times t1 and t2, the velocity at the end of each 2 s interval can be computed; see column 3 of the table and the v −t curve. 3.

The v −t curve is next approximated with a series of rectangles, each of width ∆t = 2 s. The area (∆t )(vave ) of each rectangle is approximately equal to the change in position ∆x for the specified interval of time. Thus,

∆x � vave ∆t

where vave and ∆x are given in columns 4 and 5, respectively, of the table. 4.

With x0 = 0 and noting that x2 = x1 + ∆x12

where ∆x12 is the change in position between times t1 and t2, the position at the end of each 2 s interval can be computed; see column 6 of the table and the x −t curve.


120


(a)

At t = 8 s, v = 32.58 m/s

(b)

At t = 20 s

v = 117.3 km/h �

or

x = 660 m ��


121

122

PROBLEM 11.82 Two seconds are required to bring the piston rod of an air cylinder to rest; the acceleration record of the piston rod during the 2 s is as shown. Determine by approximate means (a) the initial velocity of the piston rod, (b) the distance traveled by the piston rod as it is brought to rest.

SOLUTION a −t curve; at

Given: 1.

t = 25, v = 0

The a −t curve is first approximated with a series of rectangles, each of width ∆t = 0.25 s. The area (∆t)(aave) of each rectangle is approximately equal to the change in velocity ∆v for the specified interval of time. Thus, ∆v � aave ∆t

where the values of aave and ∆v are given in columns 1 and 2, respectively, of the following table. 2.

v(2) = v0 +

Now and approximating the area

�

2 0

�

2 0

a dt = 0

a dt under the a −t curve by Σaave ∆t ≈ Σ∆v, the initial velocity is then

equal to v0 = −Σ∆v

Finally, using v2 = v1 + ∆v12

where ∆v12 is the change in velocity between times t1 and t2, the velocity at the end of each 0.25 interval can be computed; see column 3 of the table and the v −t curve. 3.

The v −t curve is then approximated with a series of rectangles, each of width 0.25 s. The area (∆t )(vave ) of each rectangle is approximately equal to the change in position ∆x for the specified interval of time. Thus ∆x ≈ vave ∆t

where vave and ∆x are given in columns 4 and 5, respectively, of the table.


122

PROBLEM 11.82 (Continued) 4.

With x0 = 0 and noting that x2 = x1 + ∆x12

where ∆x12 is the change in position between times t1 and t2, the position at the end of each 0.25 s interval can be computed; see column 6 of the table and the x −t curve.

(a)

We had found

(b)

At t = 2 s

v0 = 1.914 m/s � x = 0.840 m ��


123

124

PROBLEM 11.83 A training airplane has a velocity of 37.8 m/s when it lands on an aircraft carrier. As the arresting gear of the carrier brings the airplane to rest, the velocity and the acceleration of the airplane are recorded; the results are shown (solid curve) in the figure. Determine by approximate means (a) the time required for the airplane to come to rest, (b) the distance traveled in that time.

SOLUTION Given: a −v curve: v0 = 37.8 m/s

The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on the figure). For uniformly accelerated motion v22 = v12 + 2a( x2 − x1 ) v2 = v1 + a(t2 − t1 )

v22 − v12 2a v2 − v1 ∆t = a

∆x =

or

For the five regions shown above, we have Region

v1 , m/s

v2 , m/s

a, m/s2

∆ x, m

∆t , s

1

37.8

36

− 3.8

17.5

0.474

2

36

30

− 9.9

20

0.606

3

30

24

−13.7

11.8

0.438

4

24

12

−16.2

13.3

0.741

5

12

0

−17.4

4.1

0.690

66.7

2.949

Σ

(a)

From the table, when v = 0

t = 2.95 s

(b)

From the table and assuming x0 = 0, when v = 0

x = 66.7 m


124

PROBLEM 11.84 Shown in the figure is a portion of the experimentally determined v − x curve for a shuttle cart. Determine by approximate means the acceleration of the cart (a) when x = 0.25 m, (b) when v = 2 m/s.

SOLUTION Given: v − x curve

First note that the slope of the above curve is

dv . dx

a=v

(a)

Now

dv dx

When

x = 0.25 m, v = 1.4 m/s

Then

a = 1.4 m/s

1 m/s 0.34 m a = 4.1 m/s2

or (b)

When v = 2 m/s, we have a = 2 m/s

1 m/s 0.7 m a = 2.86 m/s2

or

Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary that the same scale be used for the x and v axes (e.g., 1 cm = 0.5 m, 1 cm = 0.5 m/s). In the above solution, ∆v and ∆x were measured directly, so different scales could be used. PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY No part reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced you are are aa student student using using this this Manual, Manual, distribution to to teachers teachers and and educators educators permitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you distribution you are are using using itit without without permission. permission. you

125

126

PROBLEM 11.85 Using the method of Section 11.8, derive the formula x = x0 + v0t + 12 at 2 for the position coordinate of a particle in uniformly accelerated rectilinear motion.

SOLUTION The a −t curve for uniformly accelerated motion is as shown.

Using Eq. (11.13), we have x = x0 + v0t + (area under a −t curve) (t − t )

� 1 � = x0 + v0t + (t × a) � t − t � � 2 � 1 2 = x0 + v0 t + at Q.E.D. 2

�


126

PROBLEM 11.86 Using the method of Section 11.8, determine the position of the particle of Problem 11.61 when t = 14. PROBLEM 11.61 A particle moves in a straight line with the acceleration shown in the figure. Knowing that it starts from the origin with v0 = −18 m/s, (a) plot the v − t and x − t curves for 0 t 20 s, (b) determine its velocity, its position, and the total distance traveled after 12 s.

SOLUTION

x0 = 0 v0 = −18 m/s

When t = 14 s: x = x0 + v0t + ΣA(t1 − t ) = 0 − (18 m/s)(14 s) + [(3 m/s2)(4 s)](12 s) + [(6 m/s2)(6 s)](7 s) + [(−5 m/s)(4 s)](2 s) x14 = −252 m + 144 m + 252 m − 40 m

x14 = +104 m

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127

128

PROBLEM 11.87 While testing a new lifeboat, an accelerometer attached to the boat provides the record shown. If the boat has a velocity of 2 m/s at t = 0 and is at rest at time t1, determine, using the method of Section 11.8, (a) the time t1, (b) the distance through which the boat moves before coming to rest.

SOLUTION The area under the curve is divided into three regions as shown. (a)

First note

ta 0.75 = 18 22.5

or

ta = 0.60 s

Now

v = v0 +

t 0

a dt

where the integral is equal to the area under the a −t curve. Then, with v0 = 2 m/s, vt1 = 0

(b)

We have

0 = 2 m/s +

or

t1 = 2.32 s

1 1 (0.6 s )(18 m/s2) − (0.15 s)(4.5 m/s2) − (t1 − 0.75) s × 4.5 m/s2) 2 2 t1 = 2.32 s

From the discussion following Eq. (11.13) and assuming x = 0, we have x = 0 + v0 t1 + ΣA(t1 − t )

where A is the area of a region and t is the distance to its centroid. Then for t1 = 2.32 s x = (2 m/s)(2.32 s) + −

1 (0.6 s)(18 m/s2) (2.32 − 0.2) s 2

1 (0.15 s)(4.5 m/s2) (2.32 − 0.70) s 2

− [(1.57 s)(4.5 m/s2)] 2.32 − 0.75 +

1 × 1.57 s 2

= [4.64 + (11.448 − 0.547 − 5.546)] m x = 10 m

or

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128

PROBLEM 11.88 For the particle of Problem 11.63, draw the a −t curve and determine, using the method of Section 11.8, (a) the position of the particle when t = 52 s, (b) the maximum value of its position coordinate. PROBLEM 11.63 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 mm at t = 0, (a) construct the a −t and x −t curves for 0 t 50 s, and determine (b) the total distance traveled by the particle when t = 50 s, (c) the two times at which x = 0.

SOLUTION a=

We have where

dv dt

dv dt

is the slope of the v −t curve. Then t=0

from

to t = 10 s:

v = constant

t = 10 s to t = 26 s: a = t = 26 s to t = 41 s:

t > 46 s:

−20 − 60 = −5 mm/s2 26 − 10

v = constant

t = 41 s to t = 46 s: a =

a=0

a=0

−5 − (−20) = 3 mm/s2 46 − 41

v = constant

a=0

The a −t curve is then drawn as shown. (a)

From the discussion following Eq. (11.13), we have

x = x0 + v0t1 + ΣA(t1 − t )

where A is the area of a region and t is the distance to its centroid. Then, for t1 = 52 s x = −540 mm + (60 mm/s)(52 s) + {−[(16 s)(5 mm/s2)](52 − 18) s + [(5 s)(3 mm/s2)](52 − 43.5)s} = [−540 + (3120) + (−2720 + 127.5)] mm x = −12.50 mm

or

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129

130


Noting that xmax occurs when v = 0 ( dx = 0 ), it is seen from the v–t curve that xmax occurs for dt 10 s t 26 s. Although similar triangles could be used to determine the time at which x = xmax (see the solution to Problem 11.63), the following method will be used. For

10 s

t1

26 s, we have x = −540 + 60t1 − [(t1 − 10)(5)]

1 (t1 − 10) mm 2

5 = −540 + 60t1 − (t1 − 10) 2 2

When x = xmax : or Then

dx = 60 − 5(t1 − 10) = 0 dt (t1 ) xmax = 22 s 5 xmax = −540 + 60(22) − (22 − 10)2 2 xmax = 420 mm

or

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130

PROBLEM 11.89 The motion of a particle is defined by the equations x = 4t 3 − 5t 2 + 5t and y = 5t 2 − 15t , where x and y are expressed in millimeters and t is expressed in seconds. Determine the velocity and the acceleration when (a) t = 1 s, (b) t = 2 s.

SOLUTION x = 4t 3 − 5t 2 + 5t dx vx = = 12t 2 − 10t + 5 dt dvx ax = = 24t − 10 dt

(a)

y = 5t 2 − 15t dy vy = = 10t − 15 dt dv y ay = = 10 dt

When t = 15, vx = 7 mm/s, v y = −5 mm/s v = 7 2 + (−5) 2 = 8.60 φ = tan −15/ 7 = 35.5° v = 8.60 mm/s

ax = 14 mm/s 2

35.5° �

a y = 10 mm/s 2

a = 142 + 102 = 17.20 mm/s 2 � 10 � φ = tan −1 � � = 35.5° � 14 � a = 17.20 mm/s 2

� (b)

35.5° ��

When t = 25, vx = 33 mm/s, v y = 5 mm/s v = 33.4 mm/s

8.62° �

ax = 38 mm/s 2 a y = 10 mm/s 2 a = 39.3 mm/s 2

�

14.74° ��


131

132

PROBLEM 11.90 The motion of a particle is defined by the equations x = 2 cos π t and y = 1 − 4 cos 2π t , where x and y are expressed in meters and t is expressed in seconds. Show that the path of the particle is part of the parabola shown, and determine the velocity and the acceleration when (a) t = 0, (b) t = 1.5 s.

SOLUTION We have

x = 2 cos π t

Then

y = 1 − 4(2 cos 2 π t − 1)

� x� = 5 − 8� � �2�

y = 1 − 4 cos 2π t

2

y = 5 − 2x2

or

Q.E.D.

Now

vx =

dx = −2π sin π t dt

vy =

and

ax =

dvx = −2π 2 cos π t dt

ay =

(a)

At t = 0: vx = 0

dy = 8π sin 2π t dt dv y dt

= 16π 2 cos 2π t v=0 �

vy = 0 ax = −2π 2 m/s 2

a y = 16π 2 m/s 2 a = 159.1 m/s 2

or (b)

�

82.9° �

At t = 1.53: vx = −2π sin(1.5π ) = 2π m/s v y = 8π sin(2π × 1.5) = 0 v = 6.28 m/s → �

or ax = −2π 2 cos(1.5π ) = 0 a y = 16π 2 cos(2π × 1.5) = −16π 2

a = 157.9 m/s 2 ↓ ��

or


132

PROBLEM 11.91 The motion of a particle is defined by the equations x = t 2 − 8t + 7 and y = 0.5t 2 + 2t − 4, where x and y are expressed in meters and t in seconds. Determine (a) the magnitude of the smallest velocity reached by the particle, (b) the corresponding time, position, and direction of the velocity.

SOLUTION x = t 2 − 8t + 7 vx = dx/dt = 2t − 8

y = 0.5t 2 + 2t − 4 vy = t + 2

ax = dvx /dt = 2

a y = dv y /dt = 1

v 2 = vx2 + v 2y = (2t − 8) 2 + (t + 2) 2

When v is minimum, v2 is also minimum; thus, we write d (v 2 ) = 2(2t − 8)(2) + 2(t + 2) = 2(4t − 16 + t + 2) = 0 dt = 2(5t − 14) = 0

t = 2.80 s �

When t = 2.8 s : x = (2.8)2 − 8(2.8) + 7 y = 0.5(2.8) 2 + 2(2.8) − 4 vx = 2(2.8) − 8 vx = −2.40 m/s � � v y = 2.8 + 2 v y = +4.80 m/s �

x = −7.56 m � y = 5.52 m � v = 5.37 m/s

63.4° ��


133

134

PROBLEM 11.92 The motion of a particle is defined by the equations x = 4t − 2 sin t and y = 4 − 2 cos t, where x and y are expressed in meters and t is expressed in seconds. Sketch the path of the particle, and determine (a) the magnitudes of the smallest and largest velocities reached by the particle, (b) the corresponding times, positions, and directions of the velocities.

SOLUTION x = 4t − 2 sin t

We have t, s

x, m

y, m

0

2.0

4.28

4.0

π

12.57

6.0

3π 2

20.8

4.0

2π

25.1

2.0

0

π 2

(a)

x = 4t − 2 sin t

We have

y = 4 − 2 cos t

y = 4 − 2 cos t

dx = 4 − 2 cos t dt

vy =

dy = 2 sin t dt

Then

vx =

Now

v 2 = vx2 + v 2y = (4 − 2 cos t ) 2 + (2 sin t ) 2 = 20 − 16 cos t

By observation, for vmin , cos t = 1, so that 2 vmin =4

or

vmin = 2 m/s

2 vmax = 36

or

vmax = 6 m/s

cos t = 1

or

t = 2nπ s

for vmax , cos t = −1, so that

(b)

When v = vmin :

where n = 0, 1, 2, . . . Then

x = 4(2nπ ) − 2 sin (2nπ )

or

x = 8nπ m

y = 4 − 2(1)

or

y=2m

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134

PROBLEM 11.92 (Continued) Also,

vx = 4 − 2(1) = 2 m/s v y = 2 sin (2nπ ) = 0 θv min = 0 →

When v = vmax :

or t = (2n + 1)π s

cos t = −1

where n = 0, 1, 2, Then

x = 4(2n + 1)π − 2 sin (2n + 1)π y = 4 − 2(−1)

Also,

or

x = 4(2n + 1)π m or

y=6 m

vx = 4 − 2(−1) = 6 m/s v y = 2 sin (2n + 1)π = 0 θvmax = 0 →


135

136

PROBLEM 11.93 The motion of a particle is defined by the position vector r = A(cos t + t sin t )i + A(sin t − t cos t ) j, where t is expressed in seconds. Determine the values of t for which the position vector and the acceleration are (a) perpendicular, (b) parallel.

SOLUTION We have

r = A(cos t + t sin t )i + A(sin t − t cos t ) j

Then

v=

and

a=

(a)

dr = A( − sin t + sin t + t cos t )i dt + A(cos t − cos t + t sin t ) j = A(t cos t )i + A(t sin t ) j dv = A(cos t − t sin t )i + A(sin t + t cos t ) j dt

When r and a are perpendicular, r ⋅ a = 0 A[(cos t + t sin t )i + (sin t − t cos t ) j] ⋅ A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0

(b)

or

(cos t + t sin t )(cos t − t sin t ) + (sin t − t cos t )(sin t + t cos t ) = 0

or

(cos2 t − t 2 sin 2 t ) + (sin 2 t − t 2 cos 2 t ) = 0

or

1− t2 = 0

or

t =1 s �

or

t=0 �

When r and a are parallel, r × a = 0 A[(cos t + t sin t )i + (sin t − t cos t ) j] × A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0

or Expanding

[(cos t + t sin t )(sin t + t cos t ) − (sin t − t cos t )(cos t − t sin t )]k = 0 (sin t cos t + t + t 2 sin t cos t ) − (sin t cos t − t + t 2 sin t cos t ) = 0 2t = 0

or


136

PROBLEM 11.94 The damped motion of a vibrating particle is defined by the position vector r = x1[1 − 1/(t + 1)]i + ( y1e−π t/ 2 cos 2π t ) j, where t is expressed in seconds. For x1 = 30 mm and y1 = 20 mm, determine the position, the velocity, and the acceleration of the particle when (a) t = 0, (b) t = 1.5 s.

SOLUTION We have

1 � � −π t/2 r = 30 �1 − cos 2π t ) j � i + 20(e � t +1�

Then

v=

= 30

1 � π � i + 20 � − e −π t/2 cos 2π t − 2π e−π t/2 sin 2π t � j 2 (t + 1) 2 � �

= 30

� 1 �1 �� i − 20π �e −π t/ 2 � cos 2π t + 2 sin 2π t � � j 2 (t + 1) �2 ��

a=

and

dr dt

dv dt

� π � 2 �1 � i − 20π � − e−π t/2 � cos 2π t + 2 sin 2π t � + e−π t/2 (−π sin 2π t + 4 cos 2π t ) � j 3 (t + 1) �2 � � 2 � 60 i + 10π 2 e−π t/2 (4 sin 2π t − 7.5 cos 2π t ) j =− (t + 1)3

= −30

(a)

At t = 0:

� 1� r = 30 �1 − � i + 20(1) j � 1�

r = 20 mm ↑ �

or � �1 �1� �� v = 30 � � i − 20π �(1) � + 0 � � j �1� �� 2

or

60 a = − i + 10π 2 (1)(0 − 7.5) j (1)

v = 43.4 mm/s

a = 743 mm/s 2

or

46.3° �

85.4° �


137

138


(b)

At t = 1.5 s:

1 � � −0.75π r = 30 �1 − (cos 3π ) j � i + 20e � 2.5 � = (18 mm)i + ( −1.8956 mm) j

or

r = 18.10 mm

6.01° �

v = 5.65 mm/s

31.8° �

a = 70.3 mm/s 2

86.9° �

30 �1 � i − 20π e−0.75π � cos 3π + 0 � j 2 (2.5) 2 � � = (4.80 mm/s)i + (2.9778 mm/s) j

v=

or a=−

60 i + 10π 2 e−0.75π (0 − 7.5 cos 3π ) j (2.5)3

= (−3.84 mm/s 2 )i + (70.1582 mm/s 2 ) j

or


138

PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described

by the particle is a conic helix.)

SOLUTION We have

r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k

Then

v=

dr = R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k dt

and

a=

dv dt

(

)

= R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i

(

)

+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k

(

)

(

)

= R −2ωn sin ωn t − ωn2 t cos ωn t i + R 2ωn cos ωn t − ωn2 t sin ωn t k

Now

v 2 = vx2 + v y2 + vz2

= [ R(cos ωn t − ωn t sin ωn t )]2 + (c)2 + [ R(sin ωn t + ωn t cos ωn t )]2

(

)

= R 2 � cos 2 ωn t − 2ωn t sin ωn t cos ωn t + ωn2 t 2 sin 2 ωn t � + sin 2 ωn t + 2ωn t sin ωn t cos ωn t + ωn2 t 2 cos 2 ωn t � + c 2 �

(

(

)

)

= R 2 1 + ωn2 t 2 + c 2

(

Also,

)

v = R 2 1 + ωn2 t 2 + c 2 �

or a 2 = ax2 + a 2y + az2

(

) cos ω t − ω t sin ω t ) � �

2

= � R −2ωn sin ωn t − ωn2 t cos ωn t � + (0) 2 � �

( (

2

2 + � R 2ωn n n n � = R 2 � 4ωn2 sin 2 ωn t + 4ωn3t sin ωn t cos ωn t + ωn4t 2 cos 2 ωn t � + 4ωn2 cos 2 ωn t − 4ωn3t sin ωn t cos ωn t + ωn4 t 2 sin 2 ωn t � �

(

(

= R 2 4ωn2 + ωn4 t 2

)

)

)

a = Rωn 4 + ωn2 t 2 �

or�


139

140

PROBLEM 11.96* The three-dimensional motion of a particle is defined by the position vector r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k , where r and t are expressed in meters and seconds, respectively. Show that the curve described by the particle lies on the hyperboloid (y/A)2 − (x/A)2 − (z/B)2 = 1. For A = 3 and B = 1, determine (a) the magnitudes of the velocity and acceleration when t = 0, (b) the smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other.

SOLUTION We have

r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k

or

x = At cos t cos t =

Then

x At

sin t =

or y A

2

Then

y A

2

or

x A

2

t2 =

x A

2

−1 =

−

x A

2

−

z = Bt sin t

z Bt

t2 = x At

2

z B

2

+

z B

2

+

cos 2 t + sin 2 t = 1

Now

(a)

y = A t2 +1

z B

+

z Bt

y A

2

−1

2

=1

2

=1

Q.E.D.

With A = 3 and B = 1, we have v=

dr t = 3(cos t − t sin t )i + 3 j + (sin t + t cos t )k 2 dt t +1

( ) t

and

t 2 + 1 − t t 2 +1 dv = 3( − sin t − sin t − t cos t )i + 3 j a= dt (t 2 + 1) + (cos t + cos t − t sin t )k 1 = −3(2 sin t + t cos t )i + 3 2 j + (2 cos t − t sin t )k (t + 1)3/2


140

PROBLEM 11.96* (Continued) v = 3(1 − 0)i + (0) j + (0)k

At t = 0:

v = vx2 + v y2 + vz2 v = 3 m/s

or a = −3(0)i + 3(1) j + (2 − 0)k

and

a 2 = (0) 2 + (3) 2 + (2) 2 = 13

Then

a = 3.61 m/s2

or (b)

If r and v are perpendicular, r ⋅ v = 0

(

)

[(3t cos t )i + 3 t 2 + 1 j + (t sin t )k ] ⋅ [3(cos t − t sin t )i + 3

or

(3t cos t )[3(cos t − t sin t )] + (3 t 2 + 1) 3

Expanding

t t2 +1

t t2 + 1

j + (sin t + t cos t )k ] = 0

+ (t sin t )(sin t + t cos t ) = 0

(9t cos 2 t − 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0 10 + 8 cos 2 t − 8t sin t cos t = 0

or (with t ≠ 0)

7 + 2 cos 2t − 2t sin 2t = 0

or Using “trial and error” or numerical methods, the smallest root is

t = 3.82 s

Note: The next root is t = 4.38 s.


141

142

PROBLEM 11.97 An airplane used to drop water on brushfires is flying horizontally in a straight line at 315 km/h at an altitude of 80 m. Determine the distance d at which the pilot should release the water so that it will hit the fire at B.

SOLUTION First note

v0 = 315 km/h = 87.5 m/s

Vertical motion. (Uniformly accelerated motion) y = 0 + (0)t −

At B: or

1 2 gt 2

1 −80 m = − (9.81 m/s 2 )t 2 2 t B = 4.03855 s

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t

At B:

d = (87.5 m/s)(4.03855 s) d = 353 m ��

or


142

PROBLEM 11.98 Three children are throwing snowballs at each other. Child A throws a snowball with a horizontal velocity v 0 . If the snowball just passes over the head of child B and hits child C, determine (a) the value of v0, (b) the distance d.

SOLUTION (a)


1 2 gt 2

1 −1 m = − (9.81 m/s 2 )t 2 2

At B:

or tB = 0.451 524 s

Horizontal motion (Uniform) x = 0 + (v x ) 0 t

At B: or (b)

Vertical motion: At C: or

7 m = v0 (0.451524 s) v0 = 15.5031 m/s

v0 = 15.50 m/s �

1 −3 m = − (9.81 m/s 2 ) t 2 2 tC = 0.782062 s

Horizontal motion. At C: �

(7 + d ) m = (15.5031 m/s)(0.782062 s) d = 5.12 m �

or


143

144

PROBLEM 11.99 While delivering newspapers, a girl throws a newspaper with a horizontal velocity v0. Determine the range of values of v0 if the newspaper is to land between Points B and C.

SOLUTION Vertical motion. (Uniformly accelerated motion) y = 0 + (0)t −

1 2 gt 2

Horizontal motion. (Uniform) x = 0 + (vx )0 t = v0 t

At B:

y:

x:

y:

or

1 −0.6 m = − (9.81 m/s2)t2 2 tC = 0.35 s

or Then

2.1 m = (v0 ) B (0.452 s) (v0 ) B = 4.65 m/s

or At C:

1 (9.81 m/s2)t2 2

t B = 0.452 s

or Then

−1 m = −

x:

3.7 m = (v0 )C (0.35 s) (v0 )C = 10.57 m/s 4.65 m/s

v0

10.57 m/s


144

PROBLEM 11.100 A baseball pitching machine “throws” baseballs with a horizontal velocity v0. Knowing that height h varies between 0.8 m and 1 m, determine (a) the range of values of v0, (b) the values of a corresponding to h = 0.8 m and h = 1 m.

SOLUTION

(a)


1 2 gt 2

Horizontal motion. (Uniform) x = 0 + (vx )0 t = v0 t

When h = 0.8 m, y = −0.7 m:

1 −0.7 m = − (9.81 m/s2)t2 2 t31 = 0.38 s

or Then

12 m = (v0 )31 (0.38 s)

or

(v0 )31 = 31.58 m/s = 113.69 km/h

When h = 1 m, y = −0.5 m:

1 −0.5 m = − (9.81 m/s2)t2 2 t42 = 0.32 s

or Then

12 m = (v0 ) 42 (0.32 s)

or

(v0 )42 = 37.5 m/s = 135 km/h 113.7 km/h

v0

135 km/h

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. All rights this Manual Manual may may be be displayed, displayed, PROPRIETARY Inc. All rights reserved. reserved. No No part part of of this reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced student using using this this Manual, Manual, distribution to to teachers teachers and and educators educators permitted permitted by by McGraw-Hill McGraw-Hill for distribution for their their individual individual course course preparation. preparation. If If you you are are aa student you are are using using it it without you without permission. permission.

145

146


For the vertical motion v y = (0) − gt

Now When h = 0.8 m:

tan α =

|(v y ) B | (v x ) B

=

gt v0

(9.81 m/s2)(0.38 s) 31.58 m/s = 0.118043

tan α =

α 31 = 6.7°

or When h = 1 m:

(9.81 m/s2)(0.32 s) 37.5 m/s = 0.083712

tan α =

α 42 = 4.8°

or


146

PROBLEM 11.101 A volleyball player serves the ball with an initial velocity v0 of magnitude 13.40 m/s at an angle of 20° with the horizontal. Determine (a) if the ball will clear the top of the net, (b) how far from the net the ball will land.

SOLUTION First note

(vx )0 = (13.40 m/s) cos 20° = 12.5919 m/s (v y )0 = (13.40 m/s) sin 20° = 4.5831 m/s

(a)


At C

9 m = (12.5919 m/s) t or tC = 0.71475 s

Vertical motion. (Uniformly accelerated motion) y = y0 + ( v y ) 0 t −

At C:

1 2 gt 2

yc = 2.1 m + (4.5831 m/s)(0.71475 s) 1 − (9.81 m/s 2 )(0.71475 s)2 2 = 2.87 m yC � 2.43 m (height of net) � ball clears net �

(b)

At B, y = 0:

1 0 = 2.1 m + (4.5831 m/s)t − (9.81 m/s 2 )t 2 2

Solving

t B = 1.271175 s (the other root is negative)

Then

d = (vx )0 t B = (12.5919 m/s)(1.271175 s) = 16.01 m b = (16.01 − 9.00) m = 7.01 m from the net �

The ball lands


147

148

PROBLEM 11.102 Milk is poured into a glass of height 140 mm and inside diameter 66 mm. If the initial velocity of the milk is 1.2 m/s at an angle of 40° with the horizontal, determine the range of values of the height h for which the milk will enter the glass.

SOLUTION First note (vx )0 = (1.2 m/s) cos 40° = 0.91925 m/s (v y )0 = −(1.2 m/s) sin 40° = −0.77135 m/s


Vertical motion. (Uniformly accelerated motion) y = y0 + ( v y ) 0 t −

1 2 gt 2

Milk enters glass at B x:

0.08 m = (0.91925 m/s) t or tB = 0.087028 s

y : 0.140 m = hB + (−0.77135 m/s)(0.087028 s) 1 − (9.81 m/s 2 )(0.087028 s)2 2

or

hB = 0.244 m

Milk enters glass at C x : 0.146 m = (0.91925 m/s) t or tC = 0.158825 s y : 0.140 m = hC + (−0.77135 m/s)(0.158825 s) 1 − (9.81 m/s 2 )(0.158825 s)2 2

or

hC = 0.386 m 0.244 m � h � 0.386 m �

�


148

PROBLEM 11.103 A golfer hits a golf ball with an initial velocity of 48 m/s at an angle of 25° with the horizontal. Knowing that the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and Point B where the ball first lands.

SOLUTION (vx )0 = (48 m/s) cos 25°

First note

(v y )0 = (48 m/s) sin 25° xB = d cos 5°

and at B Now

yB = − d sin 5°

Horizontal motion. (Uniform) x = 0 + (v x ) 0 t d cos 5° = (48 cos 25°)t

At B

or tB =

cos 5° d 48 cos 25°

Vertical motion. (Uniformly accelerated motion) 1 y = 0 + (v y )0 t − gt2 2

(g = 9.81 m/s2)

1 2 gtB 2

At B:

−d sin 5° = (48 sin 25°)tB −

Substituting for t B

cos 5° 1 cos 5° − d sin 5° = (48 sin 25°) d− g 48 cos 25° 2 48 cos 25°

or

2

d2

2 (48 cos 25°)2(tan 5° + tan 25°) 9.81 cos 5° = 214.49 m

d=

d = 215 m

or


149

150

PROBLEM 11.104 Water flows from a drain spout with an initial velocity of 0.75 m/s at an angle of 15° with the horizontal. Determine the range of values of the distance d for which the water will enter the trough BC.

SOLUTION First note

(vx )0 = (0.75 m/s) cos 15° = 0.724 m/s (v y )0 = −(0.75 m/s) sin 15° = −0.194 m/s

Vertical motion. (Uniformly accelerated motion) y = 0 + (v y ) 0 t −

1 2 gt 2

At the top of the trough −2.64 m = (−0.194 m/s) t −

or

1 (9.81 m/s2)t 2 2

t BC = 0.714 s (the other root is negative)


In time t BC

xBC = (0.724 m/s)(0.714 s) = 0.517 m

Thus, the trough must be placed so that xB

0.517 m xC

0.517 m

Since the trough is 0.6 m wide, it then follows that

0

d

0.52 m


150

PROBLEM 11.105 Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B. Knowing that the conveyor belt forms an angle α = 20° with the horizontal, determine the speed v0 of the belt.

SOLUTION (v0 ) x = v0 cos 20° = 0.9397 v0 (v0 ) y = v0 sin 20° = 0.3420 v0

Horizontal motion. At B: Vertical motion. At B:

Using Eq. (1):

x = (v0 ) x t = (0.9397 v0 ) t 9 m = (0.9397 v0 ) t y = (v0 ) y t −

t = 9.578/v0

(1)

1 2 gt 2

1 −5.4 m = v0 (0.3420) t − 9.81t 2 2 −5.4 = (0.3420)(9.578) − (4.905) 9.578 −8.676 = −(4.905) v0

9.578 v0

2

2

v02 = 51.864 m2/s2 v0 = 7.202 m/s

v0 = 7.2 m/s

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Manual may may be be displayed, displayed, PROPRIETARY Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of the publisher, publisher, or or used used beyond beyond the the limited limited reproduced of the student using using this this Manual, Manual, distribution to to teachers teachers and and educators educators permitted permitted by McGraw-Hill for distribution by McGraw-Hill for their their individual individual course course preparation. preparation. IfIf you you are are aa student you are are using using it you it without without permission. permission.

151

152

PROBLEM 11.106 A basketball player shoots when she is 5 m from the backboard. Knowing that the ball has an initial velocity v0 at an angle of 30° with the horizontal, determine the value of v0 when d is equal to (a) 0.2 m, (b) 0.38 m.

SOLUTION First note (vx )0 = v0 cos 30° (v y )0 = v0 sin 30°

Horizontal motion. (Uniform) x = 0 + (vx ) 0 t

At B:

(5 − d) = (v0 cos 30°)t

or tB =

5−d v0 cos 30°

Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t −

1 2 gt 2

At B:

1 = (v0 sin 30°) tB −

Substituting for tB

1 = (v0 sin 30°)

or

v02 =

( g = 9.81 m/s2)

1 2 gtB 2

5−d 1 5−d − g v0 cos 30° v0 cos 30° 2

2g(5 − d)2 1 3 (5 − d) − 1 3

2

d~m


152


(a)

d = 0.2 m:

v02 =

2(9.81)(5 − 0.2)2 3

1 3

(5 − 0.2) − 1 v0 = 9.2 m/s

or (b)

d = 0.38 m:

v02 =

2(9.81)(5 − 0.38)2 3

1 3

(5 − 0.38) − 1 v0 = 9.15 m/s

or

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced distribution you are are aastudent studentusing usingthis thisManual, Manual, distribution to toteachers teachersand andeducators educatorspermitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you you you are are using using itit without without permission. permission.

153

154

PROBLEM 11.107 A group of children are throwing balls through a 0.72-m-inner-diameter tire hanging from a tree. A child throws a ball with an initial velocity v0 at an angle of 3° with the horizontal. Determine the range of values of v0 for which the ball will go through the tire.



When x = 6 m:

6 = (v0 cos 3°)t or t6 =

6 v0 cos 3°


1 2 gt 2

( g = 9.81 m/s 2 )

When the ball reaches the tire, t = t6 � � 6 yB ,C = (v0 sin 3°) � � � v0 cos 3° � � 1 � 6 − g� � 2 � v0 cos 3° �

2

or

v02 =

18(9.81) cos 3°(6 tan 3° − yB , C )

or

v02 =

177.065 0.314447 − yB ,C

At B, y = −0.53 m:

v02 =

177.065 0.314447 − (−0.53)

or

2

(v0 ) B = 14.48 m/s


154


At C, y = −1.25 m:

177.06 s 0.314447 − ( −1.25)

(v0 )C = 10.64 m/s

or �

v02 =

10.64 m/s � v0 � 14.48 m/s �

�


155

156

PROBLEM 11.108 The nozzle at A discharges cooling water with an initial velocity v0 at an angle of 6° with the horizontal onto a grinding wheel 350 mm in diameter. Determine the range of values of the initial velocity for which the water will land on the grinding wheel between Points B and C.

SOLUTION First note (vx )0 = v0 cos 6° (v y )0 = −v0 sin 6°

Horizontal motion. (Uniform) x = x0 + (vx )0 t

Vertical motion. (Uniformly accelerated motion) y = y0 + (v y )0 t −

At Point B:

1 2 gt 2

( g = 9.81 m/s 2 )

x = (0.175 m) sin 10° y = (0.175 m) cos 10° x : 0.175 sin 10° = −0.020 + (v0 cos 6°)t

or

tB =

0.050388 v0 cos 6°

y : 0.175 cos 10° = 0.205 + (−v0 sin 6°)t B −

1 2 gtB 2

Substituting for t B � 0.050388 � 1 � 0.050388 � −0.032659 = (−v0 sin 6°) � � − (9.81) � � � v0 cos 6° � 2 � v0 cos 6° �

2


156


v02

or

=

1 2

(9.81)(0.050388)2

cos 2 6°(0.032659 − 0.050388 tan 6°)

(v0 ) B = 0.678 m/s

or At Point C:

x = (0.175 m) cos 30° y = (0.175 m) sin 30° x : 0.175 cos 30° = −0.020 + (v0 cos 6°)t

or

tC =

0.171554 v0 cos 6°

y : 0.175 sin 30° = 0.205 + (−v0 sin 6°)tC −

1 2 gtC 2

Substituting for tC � 0.171554 � 1 � 0.171554 � −0.117500 = (−v0 sin 6°) � � − (9.81) � � � v0 cos 6° � 2 � v0 cos 6° �

or or

v02 =

1 2

2

(9.81)(0.171554) 2

cos 2 6°(0.117500 − 0.171554 tan 6°)

(v0 )C = 1.211 m/s 0.678 m/s � v0 � 1.211 m/s �

�


157

158

PROBLEM 11.109 While holding one of its ends, a worker lobs a coil of rope over the lowest limb of a tree. If he throws the rope with an initial velocity v0 at an angle of 65° with the horizontal, determine the range of values of v0 for which the rope will go over only the lowest limb.



At either B or C, x = 5 m s = (v0 cos 65°)t B,C

or

t B ,C =

5 (v0 cos 65°)


1 2 gt 2

( g = 9.81 m/s 2 )

At the tree limbs, t = tB ,C � � 1 � � 5 5 yB ,C = (v0 sin 65°) � �− g� � � v0 cos 65° � 2 � v0 cos 65° �

2


158


or

v02 = =

2

1 2

(9.81)(25)

cos 65°(5 tan 65° − yB , C ) 686.566 5 tan 65° − yB, C

At Point B:

v02 =

686.566 5 tan 65° − 5

or

(v0 ) B = 10.95 m/s

At Point C:

v02 =

686.566 5 tan 65° − 5.9

or

(v0 )C = 11.93 m/s 10.95 m/s � v0 � 11.93 m/s ��

�


159

160

PROBLEM 11.110 A ball is dropped onto a step at Point A and rebounds with a velocity v0 at an angle of 15° with the vertical. Determine the value of v0, knowing that just before the ball bounces at Point B, its velocity vB forms an angle of 12° with the vertical.

SOLUTION First note

(vx )0 = v0 sin 15° (v y )0 = v0 cos 15°

Horizontal motion. (Uniform) vx = (vx )0 = v0 sin 15°

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (v0 cos 15°) t − gt 2 2

v y = (v y )0 − gt

y = 0 + (v y ) 0 t −

= v0 cos 15° − gt

At Point B, v y � 0 Then

tan 12° =

(v x ) B v0 sin 15° = |(v y ) B | gt B − v0 cos 15°

� sin 15° � + cos 15° � g = 9.81 m/s 2 � � tan 12° � = 0.22259 v0 v0 g

or

tB =

Noting that

yB = −0.2 m,

We have

1 −0.2 = (v0 cos 15°)(0.22259 v0 ) − (9.81)(0.22259 v0 ) 2 2 v0 = 2.67 m/s ��

or�


160

PROBLEM 11.111 A model rocket is launched from Point A with an initial velocity v 0 of 75 m/s. If the rocket’s descent parachute does not deploy and the rocket lands 120 m from A, determine (a) the angle α that v0 forms with the vertical, (b) the maximum height above Point A reached by the rocket, and (c) the duration of the flight.

SOLUTION Set the origin at Point A.

x0 = 0,

y0 = 0

Horizontal motion:

x = v0 t sin α

Vertical motion:

y = v0t cos α − cos α =

sin 2 α + cos 2 α =

sin α =

x v0t

(1)

1 2 gt 2

1 1 y + gt 2 2 v0 t

(2)

1 1 x 2 + y + gt 2 2 (v0 t )2 x 2 + y 2 + gyt 2 +

(

2

=1

1 24 g t = v02t 2 4

)

1 24 g t − v02 − gy t 2 + ( x 2 + y 2 ) = 0 4

At Point B,

(3)

x 2 + y 2 = 120 m, x = 120 cos 30° m y = −120 sin 30° = −60 m 1 (9.81)2t4 − [752 − (9.81)(−60)]t 2 + 1202 = 0 4 24.06t 4 − 6213.6t 2 − 14400 = 0 t = 16 s and 1.53 s

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161

162

PROBLEM 11.111 (Continued) Restrictions on α :

α

0 tan α =

120° x

y + gt 1 2

2

=

120 cos 30° = 0.086915 −60 + (4.905)(16)2

α = 4.967° 120 cos 30° = −2.141953 −60 + (4.905)(1.53)2 α = 115.03°

and

Use α = 4.967° corresponding to the steeper possible trajectory. (a)

Angle α .

(b)

Maximum height.

α = 5° v y = 0 at

y = ymax

v y = v0 cos α − gt = 0 t=

v0 cos α g

ymax = v0 t cos α − =

(c)

Duration of the flight.

v 2 cos 2 α 1 gt = 0 2 2g

(75)2 cos2 4.967° (2)(9.81)

ymax = 285 m

(time to reach B) t = 16 s


162

PROBLEM 11.112 The initial velocity v0 of a hockey puck is 168 km/h. Determine (a) the largest value (less than 45°) of the angle α for which the puck will enter the net, (b) the corresponding time required for the puck to reach the net.

SOLUTION First note

v0 = 168 km/h = 46.7 m/s (vx )0 = v0 cos α = (46.7 m/s) cos α

and

(v y )0 = v0 sin α = (46.7 m/s) sin α

(a)

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (46.7 cos α )t

At the front of the net, Then or

x = 4.8 m 4.8 = (46.7 cos α ) t tenter =

4.8 46.7 cos α

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (46.7 sin α ) t − gt 2 2

y = 0 + (v y )0 t −

( g = 9.81 m/s2 )

At the front of the net, yfront = (46.7 sin α ) tenter −

1 2 gtenter 2

4.8 = (46.7 sin α ) 46.7 cos α = 4.8 tan α −

Now

1 4.8 g 2 46.7 cos α

2

11.52 g 2180.9 cos2 α

1 = sec2 α = 1 + tan 2 α cos 2 α


163

164


Then or

Then

or or

yfront = 4.8 tan α −

tan2 α −

tan α =

11.52g (1 + tan2 α ) 2180.9

2180.9 2180.9 yfront tan α + 1 + 11.52g 2.4g 2180.9 2.4g

±

(−

2180.9 tan α = ± 4.8 × 9.81

2180.9 2.4g

)

2

(

=0

)

1/2

2180.9 y − 4 1 + 11.52 g front

2 2180.9 − 4.8 × 9.81

2

2180.9 − 1+ yfront 11.52 × 9.81

1/2

tan α = 46.3154 ± [(46.3154)2 − (1 + 19.2981 yfront )] 1/2

Now 0 yfront 1.2 m so that the positive root will yield values of α 45° for all values of yfront. When the negative root is selected, α increases as yfront is increased. Therefore, for α max , set yfront = yC = 1.2 m

(b)

Then

tan α = 46.3154 − [(46.3154)2 − (1 + 19.2981 × 1.2)]1 / 2

or

α max = 14.66°

We had found

tenter =

α max = 14.7°

4.8 46.7 cos α 4.8 = 46.7 cos 14.66° tenter = 0.106 s

or


164

PROBLEM 11.113 The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle α with the horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle α, (b) the angle θ that the velocity of the ball at Point B forms with the horizontal.

SOLUTION First note v0 = 72 km/h = 20 m/s (vx )0 = v0 cos α = (20 m/s) cos α

and

(v y )0 = v0 sin α = (20 m/s) sin α

(a)

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (20 cos α ) t

At Point B:

14 = (20 cos α )t or t B =

7 10 cos α


At Point B:

1 2 1 gt = (20 sin α )t − gt 2 2 2

0.08 = (20 sin α )t B −

( g = 9.81 m/s 2 )

1 2 gt B 2

Substituting for t B � 7 0.08 = (20 sin α ) � α 10 cos �

or Now

8 = 1400 tan α −

� 1 � � 7 �− g� � α 2 10 cos � � �

2

1 49 g 2 cos 2 α

1 = sec2 α = 1 + tan 2 α 2 cos α


165

166


Then or Solving

8 = 1400 tan α − 24.5 g (1 + tan 2 α ) 240.345 tan 2 α − 1400 tan α + 248.345 = 0

α = 10.3786° and α = 79.949°

Rejecting the second root because it is not physically reasonable, we have

α = 10.38° � (b)

We have

vx = (vx )0 = 20 cos α

and

v y = (v y )0 − gt = 20 sin α − gt

At Point B:

(v y ) B = 20 sin α − gt B = 20 sin α −

7g 10 cos α

Noting that at Point B, v y � 0, we have tan θ = = =

|(v y ) B |

vx 7g 10 cos α

− 20 sin α

20 cos α 7 9.81 200 cos 10.3786°

− sin 10.3786°

cos 10.3786°

θ = 9.74° ��

or


166

PROBLEM 11.114* A mountain climber plans to jump from A to B over a crevasse. Determine the smallest value of the climber’s initial velocity v0 and the corresponding value of angle α so that he lands at B.

SOLUTION First note

(vx )0 = v0 cos α (v y )0 = v0 sin α

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (v0 cos α ) t

At Point B: or

1.8 = (v0 cos α )t

tB =

1.8 v0 cos α

Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (v0 sin α ) t − gt 2 2

y = 0 + (v y )0 t −

At Point B:

−1.4 = (v0 sin α ) t B −

( g = 9.81 m/s 2 )

1 2 gt B 2

Substituting for t B � 1.8 � 1 � 1.8 � −1.4 = (v0 sin α ) � �− g� � � v0 cos α � 2 � v0 cos α �

or

v02 =

1.62 g cos α (1.8 tan α + 1.4)

=

1.62 g 0.9 sin 2α + 1.4 cos 2 α

2

2


167

168

PROBLEM 11.114* (Continued) Now minimize v02 with respect to α. We have

dv02 −(1.8 cos 2α − 2.8 cos α sin α ) = 1.62 g =0 dα (0.9 sin 2α + 1.4 cos 2 α ) 2 1.8 cos 2α − 1.4 sin 2α = 0

or

tan 2α =

or

18 14

α = 26.0625° and α = 206.06°

or

Rejecting the second value because it is not physically possible, we have

α = 26.1° � Finally,

v02 =

1.62 × 9.81 cos 26.0625°(1.8 tan 26.0625° + 1.4) 2

(v0 )min = 2.94 m/s �

or


168

PROBLEM 11.115 An oscillating garden sprinkler which discharges water with an initial velocity v0 of 8 m/s is used to water a vegetable garden. Determine the distance d to the farthest Point B that will be watered and the corresponding angle α when (a) the vegetables are just beginning to grow, (b) the height h of the corn is 1.8 m.

SOLUTION First note

(vx )0 = v0 cos α = (8 m/s) cos α (v y )0 = v0 sin α = (8 m/s) sin α

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (8 cos α ) t

At Point B: or

x = d : d = (8 cos α ) t d tB = 8 cos α

Vertical motion. (Uniformly accelerated motion)

1 2 gt 2 1 = (8 sin α ) t − gt 2 ( g = 9.81 m/s 2 ) 2 1 0 = (8 sin α ) t B − gt B2 2 y = 0 + (v y )0 t −

At Point B: Simplifying and substituting for t B

0 = 8 sin α − d=

or (a)

1 � d � g� � 2 � 8 cos α �

64 sin 2α g

(1)

When h = 0, the water can follow any physically possible trajectory. It then follows from Eq. (1) that d is maximum when 2α = 90°

α = 45° �

or Then

d=

64 sin (2 × 45°) 9.81 d max = 6.52 m �

or


169

170


Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as α increases in value from 0 to 45° and then d decreases as α is further increased. Thus, d max occurs for the value of α closest to 45° and for which the water just passes over the first row of corn plants. At this row, xcom = 1.5 m tcorn =

so that

1.5 8 cos α

Also, with ycorn = h, we have h = (8 sin α ) tcorn −

1 2 gtcorn 2

Substituting for tcorn and noting h = 1.8 m, � 1.5 � 1 � 1.5 � 1.8 = (8 sin α ) � �− g� � � 8 cos α � 2 � 8 cos α �

or Now Then or

1.8 = 1.5 tan α −

2

2.25 g 128 cos 2 α

1 = sec2 α = 1 + tan 2 α cos 2 α 1.8 = 1.5 tan α −

2.25(9.81) (1 + tan 2 α ) 128

0.172441 tan 2 α − 1.5 tan α + 1.972441 = 0

α = 58.229° and α = 81.965°

Solving

From the above discussion, it follows that d = d max when

α = 58.2° � Finally, using Eq. (1) d=

64 sin (2 × 58.229°) 9.81 d max = 5.84 m �

or


170

PROBLEM 11.116 A worker uses high-pressure water to clean the inside of a long drainpipe. If the water is discharged with an initial velocity v0 of 11.5 m/s, determine (a) the distance d to the farthest Point B on the top of the pipe that the worker can wash from his position at A, (b) the corresponding angle α.

SOLUTION First note

(vx )0 = v0 cos α = (11.5 m/s) cos α (v y )0 = v0 sin α = (11.5 m/s) sin α

By observation, d max occurs when

ymax = 1.1 m.

Vertical motion. (Uniformly accelerated motion) v y = (v y )0 − gt = (11.5 sin α ) − gt y = ymax

When Then

at

1 2 gt 2 1 = (11.5 sin α ) t − gt 2 2

y = 0 + (v y ) 0 t −

B , (v y ) B = 0

(v y ) B = 0 = (11.5 sin α ) − gt 11.5 sin α g

( g = 9.81 m/s 2 )

or

tB =

and

yB = (11.5 sin α ) t B −

1 2 gt B 2

Substituting for t B and noting yB = 1.1 m � 11.5 sin α � 1 � 11.5 sin α � 1.1 = (11.5 sin α ) � �− g� � g g � � 2 � � 1 = (11.5) 2 sin 2 α 2g

or

sin 2 α =

2.2 × 9.81 11.52

2

α = 23.8265°


171

172

PROBLEM 11.116 (Continued) (a)

Horizontal motion. (Uniform) x = 0 + (vx )0 t = (11.5 cos α ) t

At Point B: where Then

x = d max tB =

and t = t B

11.5 sin 23.8265° = 0.47356 s 9.81

d max = (11.5)(cos 23.8265°)(0.47356) d max = 4.98 m �

or (b)

α = 23.8° ��

From above


172

PROBLEM 11.117 As slider block A moves downward at a speed of 0.5 m/s, the velocity with respect to A of the portion of belt B between idler pulleys C and D is vCD/A = 2 m/s θ. Determine the velocity of portion CD of the belt when (a) θ = 45°, (b) θ = 60°.

SOLUTION We have

vCD = v A + vCD/A v A = (0.5 m/s)(− cos 65°i − sin 65° j)

where

= (0.21131 m/s)i − (0.45315 m/s) j vCD/A = (2 m/s)(cos θ i + sin θ j)

and

vCD = [(−0.21131 + 2 cos θ ) m/s]i + [(−0.45315 + 2 sin θ ) m/s] j

Then (a)

We have

vCD = (−0.21131 + 2 cos 45°)i + (−0.45315 + 2 sin 45°) j

= (1.20290 m/s)i + (0.96106 m/s) j vCD = 1.540 m/s

or (b)

We have

38.6°

vCD = (−0.21131 + 2 cos 60°)i + ( −0.45315 + 2 sin 60°) j

= (0.78869 m/s)i + (1.27890 m/s) j vCD = 1.503 m/s

or

58.3°

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY No part reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced you are are aa student student using using this this Manual, Manual, distribution to to teachers teachers and andeducators educators permitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you distribution you are are using using itit without without permission. permission. you

173

174

PROBLEM 11.118 The velocities of skiers A and B are as shown. Determine the velocity of A with respect to B.

SOLUTION We have

vA = vB + vA/B

The graphical representation of this equation is then as shown. Then

v 2A/B = 102 + 142 − 2(10)(14) cos 15°

or

vA/B = 5.05379 m/s

and or

10 5.05379 = sin α sin 15°

α = 30.8° vA/B = 5.05 m/s

55.8° �

Alternative solution. vA/B = vA − vB

= 10 cos 10i − 10 sin 10 j − (14 cos 25i − 14 sin 25 j) = −2.84i + 4.14 j = 5.05 m/s

55.8°


174

PROBLEM 11.119 Shore-based radar indicates that a ferry leaves its slip with a 70°, while instruments aboard the velocity v = 18 km/h ferry indicate a speed of 19 km/h and a heading of 30° west of south relative to the river. Determine the velocity of the river.

SOLUTION We have

v F = v R + v F/R

or

v F = v F/R + v R

The graphical representation of the second equation is then as shown. We have

vR2 = 182 + 192 − 2(18)(19) cos 10°

or

vR = 3.375 km/h

and or

18 3.375 = sin α sin 10°

α = 67.84°

Noting that v R = 3.4 km/h

7.8°

Alternatively one could use vector algebra.


175

176

PROBLEM 11.120 Airplanes A and B are flying at the same altitude and are tracking the eye of hurricane C. The relative velocity of C with respect to A 75°, and the relative velocity of C with respect is vC/A = 235 km/h to B is vC/B = 260 km/h 40°. Determine (a) the relative velocity of B with respect to A, (b) the velocity of A if ground-based radar indicates that the hurricane is moving at a speed of 24 mi/h due north, (c) the change in position of C with respect to B during a 15-min interval.

SOLUTION (a)

We have

v C = v A + v C/ A

and

v C = v B + v C/ B

Then

v A + vC/A = v B + v C/B

or

v B − v A = v C/A − vC/B

Now

v B − v A = v B/A

so that

v B/A = vC/A − vC/B

or

vC/A = vC/B + v B/A

The graphical representation of the last equation is then as shown. We have

vB2 /A = 2352 + 2602 − 2(235)(260) cos 65°

or

vB/A = 266.798 km/h

and or

260 266.798 = sin α sin 65°

α = 62.03° v B/A = 267 km/h

(b)

We have

v C = v A + v C/ A

or

v A = (24 km/h)j − (235 km/h)(−cos 75°i − sin 75°j)

12.97°

v A = (60.822 km/h)i + (250.99 km/h)j v A = 258 km/h

or

76.4°


176


Noting that the velocities of B and C are constant. We have rB = (rB )0 + v B t

Now

rC = (rC )0 + v C t

rC/B = rC − rB = [(rC )0 − (rB )0 ] + ( vC − v B )t

= [(rC )0 − (rB )0 ] + v C/B t

Then

∆rC/B = (rC/B )t2 − (rC/B )t1 = v C/B (t2 − t1 ) = vC/B ∆t

For ∆t = 15 min:

∆rC/B = (260 km/h)

1 h = 65 km 4 ∆rC/B = 65 km

40°


177

178

PROBLEM 11.121 The velocities of commuter trains A and B are as shown. Knowing that the speed of each train is constant and that B reaches the crossing 10 min after A passed through the same crossing, determine (a) the relative velocity of B with respect to A, (b) the distance between the fronts of the engines 3 min after A passed through the crossing.

SOLUTION (a)

We have

v B = v A + v B/A

The graphical representation of this equation is then as shown. Then

vB2 /A = 662 + 482 − 2(66)(48) cos 155°

or

vB/A = 111.366 km/h

and or

48 111.366 = sin α sin 155°

α = 10.50° v B/A = 111.4 km/h

(b)

10.50° �

First note that at t = 3 min, A is (66 km/h) ( 603 ) = 3.3 km west of the crossing. 7 at t = 3 min, B is (48 km/h) ( 60 ) = 5.6 km southwest of the crossing.

Now

rB = rA + rB/A

Then at t = 3 min, we have rB2/ A = 3.32 + 5.62 − 2(3.3)(5.6) cos 25° rB/A = 2.96 km ��

or


178

PROBLEM 11.122 Knowing that the velocity of block B with respect to block A 70°, determine the velocities of A and B. is v B/A = 5.6 m/s

SOLUTION From the diagram 2 x A + 3xB = constant

Then

2v A + 3vB = 0 2 vA 3

or

|vB | =

Now

v B = v A + v B/A

and noting that vA and v B must be parallel to surfaces A and B, respectively, the graphical representation of this equation is then as shown. Note: Assuming that vA is directed up the incline leads to a velocity diagram that does not “close.” First note

Then or or

α = 180° − (40° + 30° + θ B ) = 110° − θ B 2 v vA 5.6 = 3 A = sin (110° − θ B ) sin 40° sin (30° + θ B )

v A sin 40° =

2 v A sin (110° − θ B ) 3

sin (110° − θ B ) = 0.96418

or

θ B = 35.3817°

and

θ B = 4.6183°

For θ B = 35.3817° :

vB =

or

v A = 5.94 m/s vB = 3.96 m/s

2 5.6 sin 40° vA = 3 sin (30° + 35.3817°)

v A = 5.94 m/s v B = 3.96 m/s

30° � 35.4° �


179

180

PROBLEM 11.122 (Continued) 2 5.6 sin 40° vA = 3 sin (30° + 4.6183°)

For θ B = 4.6183°:

vB =

or

v A = 9.50 m/s vB = 6.34 m/s v A = 9.50 m/s v B = 6.34 m/s

30° � 4.62° ��


180

PROBLEM 11.123 Knowing that at the instant shown block A has a velocity of 8 cm/s and an acceleration of 6 cm/s2 both directed down the incline, determine (a) the velocity of block B, (b) the acceleration of block B.

SOLUTION From the diagram

2 x A + xB/A = constant

Then

2v A + vB/A = 0 | vB/A | = 16 cm/s

or

2a A + aB/A = 0

and

| aB/A | = 12 cm/s2

or

Note that v B/A and a B/A must be parallel to the top surface of block A. (a)

We have

v B = v A + v B/A

The graphical representation of this equation is then as shown. Note that because A is moving downward, B must be moving upward relative to A. We have

vB2 = 82 + 162 − 2(8)(16) cos 15°

or

vB = 8.5278 cm/s

and or

8 8.5278 = sin α sin 15°

α = 14.05° v B = 8.53 cm/s

(b)

54.1°

The same technique that was used to determine v B can be used to determine a B . An alternative method is as follows. We have

a B = a A + a B/A

= (6i ) + 12(− cos 15°i + sin 15° j)* = –(5.5911 cm/s2)i + (3.1058 cm/s2)j a B = 6.40 cm/s2

or

54.1°

* Note the orientation of the coordinate axes on the sketch of the system. PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Manual may may be be displayed, displayed, PROPRIETARY Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of the publisher, publisher, or or used used beyond beyond the the limited limited reproduced of the student using using this this Manual, Manual, distribution to to teachers teachers and and educators educators permitted permitted by by McGraw-Hill McGraw-Hill for distribution for their their individual individual course course preparation. preparation. IfIf you you are are aa student you are are using using it you it without without permission. permission.

181

182

PROBLEM 11.124 Knowing that at the instant shown assembly A has a velocity of 9 cm/s and an acceleration of 15 cm/s2 both directed downward, determine (a) the velocity of block B, (b) the acceleration of block B.

SOLUTION Length of cable = constant L = x A + 2 xB/A = constant v A + 2vB/A = 0

(1)

a A + 2aB/A = 0

(2)

a A = 15 cm/s2 ↓

Data:

v A = 9 cm/s ↓

Eqs. (1) and (2)

a A = −2aB/A

v A = −2vB/A

15 = −2aB/A

9 = −2vB/A

aB/A = −7.5 cm/s2

vB/A = −4.5 cm/s

a B/A = 7.5 cm/s2

(a)

40°

v B/A = −4.5 cm/s

Velocity of B.

v B = v A + v B/A

Law of cosines:

vB2 = (9) 2 + (4.5) 2 − 2(9)(4.5) cos 50°

40°

vB = 7.013 cm/s

Law of sines:

sin β sin 50° = 4.5 7.013

β = 29.44°

α = 90° − β = 90° − 29.44° = 60.56° v B = 7 cm/s

60.6°


182


Acceleration of B. a B may be found by using analysis similar to that used above for vB . An alternate method is a B = a A + a B/A a B = 15 cm/s2 ↓ + 7.5 cm/s2

40°

= −15 j − (7.5 cos 40°)i + (7.5 sin 40°) j = −15 j − 5.745i + 4.821j a B = −5.745i − 10.179 j

a B = 11.7 cm/s2

60.6°


183

184

PROBLEM 11.125 The assembly of rod A and wedge B starts from rest and moves to the right with a constant acceleration of 2 mm/s2. Determine (a) the acceleration of wedge C, (b) the velocity of wedge C when t = 10 s.

SOLUTION (a)

We have

a C = a B + a C/ B

The graphical representation of this equation is then as shown. First note

Then

α = 180° − (20° + 105°) = 55° aC 2 = sin 20° sin 55° aC = 0.83506 mm/s 2

(b)

aC = 0.835 mm/s 2

75° �

vC = 8.35 mm/s

75° �

For uniformly accelerated motion vC = 0 + aC t

At t = 10 s:

vC = (0.83506 mm/s 2 )(10 s) = 8.3506 mm/s

or


184

PROBLEM 11.126 As the truck shown begins to back up with a constant acceleration of 1.2 m/s2, the outer section B of its boom starts to retract with a constant acceleration of 0.5 m/s2 relative to the truck. Determine (a) the acceleration of section B, (b) the velocity of section B when t = 2 s.

SOLUTION (a)

We have

a B = a A + a B/A

The graphical representation of this equation is then as shown. We have

aB2 = 1.22 + 0.52 − 2(1.2)(0.5) cos 50°

or

aB = 0.95846 m/s 2

and or

0.5 0.95846 = sin α sin 50°

α = 23.6° a B = 0.958 m/s 2

(b)

23.6° �

For uniformly accelerated motion vB = 0 + a B t

At t = 2 s:

vB = (0.95846 m/s 2 )(2 s) = 1.91692 m/s

v B = 1.917 m/s

or

23.6° ��


185

186

PROBLEM 11.127 Conveyor belt A, which forms a 20° angle with the horizontal, moves at a constant speed of 1.2 m/s and is used to load an airplane. Knowing that a worker tosses duffel bag B with an initial velocity of 0.7 m/s at an angle of 30° with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt.

SOLUTION First determine the velocity of the bag as it lands on the belt. Now [(vB ) x ]0 = (vB )0 cos 30° = (0.7 m/s) cos 30° [(vB ) y ]0 = (vB )0 sin 30° = (0.7 m/s) sin 30°

Horizontal motion. (Uniform) x = 0 + [(vB ) x ]0 t

(vB ) x = [(vB ) x ]0 = 0.7 cos 30°

= (0.7 cos 30°)t

Vertical motion. (Uniformly accelerated motion) y = y0 + [(vB ) y ]0 t −

1 2 gt 2

= 0.5 + (0.7 sin 30°) t −

(vB ) y = [(vB ) y ]0 − gt 1 2 gt 2

= 0.7 sin 30° − gt

The equation of the line collinear with the top surface of the belt is y = x tan 20°

Thus, when the bag reaches the belt 0.5 + (0.7 sin 30°)t −

1 2 gt = [(0.7 cos 30°)t] tan 20° 2

or

1 (9.81)t 2 + 0.7(cos 30° tan 20° − sin 30°)t − 0.5 = 0 2

or

4.905t 2 − 0.129t − 0.5 = 0

Solving

t = 0.333 s and t = −0.306 s (Reject)


186

PROBLEM 11.127 (Continued) The velocity v B of the bag as it lands on the belt is then v B = (0.7 cos 30°)i + [0.7 sin 30° − 9.81(0.333)]j = (0.606 m/s)i − (2.917 m/s)j

Finally or

v B = v A + v B/A v B/A = (0.606i − 2.917j) − 1.2(cos 20°i + sin 20°j) = −(0.5216 m/s)i − (3.3274 m/s)j

v B/A = 3.37 m/s

or

81.1°


187

188

PROBLEM 11.128 Determine the required velocity of the belt B if the relative velocity with which the sand hits belt B is to be (a) vertical, (b) as small as possible.

SOLUTION A grain of sand will undergo projectile motion. vsx = vsx = constant = −1.5 m/s 0

vs y = 2 gh = (2)(9.81 m/s2)(1 m) = 4.43 m/s ↓

y-direction.

v S/B = v S − v B

Relative velocity. (a)

(1)

If vS/B is vertical, −vS/B j = −1.5i − 4.43j − (−vB cos 15i + vB sin 15j) = −1.5i − 4.43j + vB cos 15i − vB sin 15j

Equate components.

i : 0 = −1.5 + vB cos 15°

vB =

1.5 = 1.55 m/s cos 15°

v B = 1.55 m/s

(b)

15°

vS/C is as small as possible, so make vS/B ⊥ to vB into (1). −vS/B sin 15°i − vS/B cos 15° j = −1.5i − 4.43j + vB cos 15°i − vB sin 15°j

Equate components and transpose terms. (sin 15°) vS/B + (cos 15°) vB = 1.5 (cos 15°) vS/B − (sin 15°) vB = 4.43

Solving,

vS/B = 4.667 m/s

vB = 0.302 m/s

v B = 0.3 m/s

15°


188

PROBLEM 11.129 As observed from a ship moving due east at 9 km/h, the wind appears to blow from the south. After the ship has changed course and speed, and as it is moving due north at 6 km/h, the wind appears to blow from the southwest. Assuming that the wind velocity is constant during the period of observation, determine the magnitude and direction of the true wind velocity.

SOLUTION v wind = v ship + v wind/ship v w = v s + v w/s

Case � v s = 9 km/h →; v w/s ↑

Case � v s = 6 km/h ↑ ; v w/s � 15 = 1.6667 9 α = 59.0°

tan α =

vw = 92 + 152 = 17.49 km/h

v w = 17.49 km/h

59.0° ��


189

190

PROBLEM 11.130 When a small boat travels north at 5 km/h, a flag mounted on its stern forms an angle θ = 50° with the centerline of the boat as shown. A short time later, when the boat travels east at 20 km/h, angle θ is again 50°. Determine the speed and the direction of the wind.

SOLUTION We have

vW = v B + vW/B

Using this equation, the two cases are then graphically represented as shown.

With vW now defined, the above diagram is redrawn for the two cases for clarity.

Noting that

θ = 180° − (50° + 90° + α ) = 40° − α We have

φ = 180° − (50° + α ) = 130° − α

vW vW 5 20 = = sin 50° sin (40° − α ) sin 50° sin (130° − α )


190


Therefore or or

5 20 = sin (40° − α ) sin (130° − α ) sin 130° cos α − cos 130° sin α = 4(sin 40° cos α − cos 40° sin α ) tan α =

sin 130° − 4 sin 40° cos 130° − 4 cos 40°

or

α = 25.964°

Then

vW =

5 sin 50° = 15.79 km/h sin (40° − 25.964°)

vW = 15.79 km/h

26.0° ��


191

192

PROBLEM 11.131 As part of a department store display, a model train D runs on a slight incline between the store’s up and down escalators. When the train and shoppers pass Point A, the train appears to a shopper on the up escalator B to move downward at an angle of 22° with the horizontal, and to a shopper on the down escalator C to move upward at an angle of 23° with the horizontal and to travel to the left. Knowing that the speed of the escalators is 1 m/s, determine the speed and the direction of the train.

SOLUTION v D = v B + v D/B

We have

v D = vC + v D/C

The graphical representations of these equations are then as shown. Then

vD 1 = sin 8° sin (22° + α )

vD 1 = sin 7° sin (23° − α )

Equating the expressions for vD sin 8° sin 7° = sin (22° + α ) sin (23° − α )

or

or or Then

sin 8° (sin 23° cos α − cos 23° sin α ) = sin 7° (sin 22° cos α + cos 22° sin α ) tan α =

sin 8° sin 23° − sin 7° sin 22° sin 8° cos 23° + sin 7° cos 22°

α = 2.0728° vD =

1 sin 8° = 0.341 m/s sin (22° + 2.0728°)

v D = 0.34 m/s

2.07°


192

PROBLEM 11.131 (Continued) Alternate solution using components. v B = (1 m/s)

30° = (0.866 m/s)i + (0.5 m/s)j

vC = (1 m/s)

30° = (0.866 m/s)i − (0.5 m/s)j

v D/B = u1

22° = −(u1 cos 22°)i − (u1 sin 22°) j

v D/C = u2

23° = −(u2 cos 23°)i + (u2 sin 23°) j

v D = vD

α = −(vD cos α )i + (vD sin α ) j

v D = v B + v D/B = vC + v D/C 0.866i + 0.5j − (u1 cos 22°)i − (u1 sin 22°)j = 0.866i − 0.5j − (u2 cos 23°)i + (u2 sin 23°)j

Separate into components, and transpose and change u1 cos 22° − u2 cos 23° = 0 u1 sin 22° + u1 sin 23° = 1

Solving for u1 and u2 ,

u1 = 1.302 m/s

u2 = 1.311 m/s

v D = 0.866i + 0.5j − (1.302 cos 22°)i − (1.302 sin 22°)j = −(0.341 m/s)i + (0.01226 m/s)j

or

v D = 0.866i − 0.5j − (1.311 cos 23°)i + (1.311 sin 23°)j = −(0.341 m/s)i + (0.01226 m/s)j

vD = 0.34 m/s

2.07°


193

194

PROBLEM 11.132 The paths of raindrops during a storm appear to form an angle of 75° with the vertical and to be directed to the left when observed through a left-side window of an automobile traveling north at a speed of 40 km/h. When observed through a right-side window of an automobile traveling south at a speed of 30 km/h, the raindrops appear to form an angle of 60° with the vertical. If the driver of the automobile traveling north were to stop, at what angle and with what speed would she observe the drops to fall?

SOLUTION v R = ( v A )1 + ( v R/A )1

We have

v R = ( v A )2 + ( v R/A ) 2

The graphical representations of these equations are then as shown. Note that the line of action of ( v R/A ) 2 must be directed as shown so that the second velocity diagram “closes.”

From the diagram

(vR ) y = [40 + (vR ) x ]tan 15°

and

(vR ) y = [30 − (vR ) x ]tan 30°

Equating the expressions for (vR ) y [40 + (vR ) x ] tan 15° = [30 − (vR ) x ] tan 30°

or

(vR ) x = 7.8109 km/h

Then

(vR ) y = (40 + 7.8109) tan 15° = 12.8109 km/h

v R = 15.00 km/h

58.6°


194

PROBLEM 11.133 Determine the peripheral speed of the centrifuge test cab A for which the normal component of the acceleration is 10g.

SOLUTION an = 10 g = 10(9.81 m/s 2 ) = 98.1 m/s 2 an =

v2 ρ

v 2 = ρ an = (8 m)(98.1 m/s2 ) = 784.8 m 2 /s 2 v = 28.0 m/s �


195

196

PROBLEM 11.134 To test its performance, an automobile is driven around a circular test track of diameter d. Determine (a) the value of d if when the speed of the automobile is 72 km/h, the normal component of the acceleration is 3.2 m/s 2 , (b) the speed of the automobile if d = 180 m and the normal component of the acceleration is measured to be 0.6 g.

SOLUTION (a)

First note Now or

v = 72 km/h = 20 m/s an =

v2 ρ

d (20 m/s)2 = 2 3.2 m/s 2 d = 250 m �

or (b)

v2 ρ

We have

an =

Then

�1 � v 2 = (0.6 × 9.81 m/s 2 ) � × 180 m � �2 �

or

v = 23.016 m/s v = 82.9 km/h ��

or


196

PROBLEM 11.135 Determine the smallest radius that should be used for a highway if the normal component of the acceleration of a car traveling at 72 km/h is not to exceed 0.7 m/s2.

SOLUTION an =

v2 ρ

an = 0.7 m/s2 v = 72 km/h = 20 m/s

0.7 m/s2 =

(20 m/s)2 ρ

ρ = 571 m


197

198

PROBLEM 11.136 Determine the maximum speed that the cars of the roller-coaster can reach along the circular portion AB of the track if the normal component of their acceleration cannot exceed 3 g.

SOLUTION We have

an =

v2 ρ

Then

(vmax )2AB = (3 × 9.81 m/s2)(24 m)

or

(vmax ) AB = 26.577 m/s (vmax ) AB = 95.7 km/h

or


198

PROBLEM 11.137 Pin A, which is attached to link AB, is constrained to move in the circular slot CD. Knowing that at t = 0 the pin starts from rest and moves so that its speed increases at a constant rate of 20 mm/s 2 , determine the magnitude of its total acceleration when (a) t = 0, (b) t = 2 s.

SOLUTION (a)

At t = 0, v A = 0, which implies (a A ) n = 0 a A = (a A )t a A = 20 mm/s 2 ��

or (b)

We have uniformly accelerated motion v A = 0 + ( a A )t t

At t = 2 s: Now Finally,

v A = (20 mm/s 2 )(2 s) = 40 mm/s (a A ) n =

v A2 (40 mm/s) 2 = = 17.7778 mm/s 2 90 mm ρA

a A2 = ( a A )t2 + (a A ) 2n = (20) 2 + (17.7778) 2 a A = 26.8 mm/s 2 �

or


199

200

PROBLEM 11.138 A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . If the maximum total acceleration of the train must not exceed 1.5 m/s 2 , determine (a) the shortest distance in which the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration at .

SOLUTION When v = 72 km/h = 20 m/s and ρ = 400 m, an =

v 2 (20)2 = = 1.000 m/s 2 ρ 400

a = an2 + at2

But

at = a 2 − an2 = (1.5) 2 − (1.000) 2 = ± 1.11803 m/s 2

Since the train is accelerating, reject the negative value. (a)

Distance to reach the speed. v0 = 0

Let

x0 = 0 v12 = v02 + 2at ( x1 − x0 ) = 2at x1 x1 =

(b)

v12 (20) 2 = 2at (2)(1.11803)

x1 = 178.9 m �

Corresponding tangential acceleration. at = 1.118 m/s 2 �

�


200

PROBLEM 11.139 An outdoor track is 125 m in diameter. A runner increases her speed at a constant rate from 4 to 7 m/s over a distance of 28 m. Determine the total acceleration of the runner 2 s after she begins to increase her speed.

SOLUTION We have uniformly accelerated motion v22 = v12 + 2at ∆s12

Substituting or

(7 m/s)2 = (4 m/s)2 + 2at (28 m) at = 0.59 m/s2

Also

v = v1 + at t

At t = 2 s:

v = 7 m/s + (0.59 m/s2)(2 s) = 8.18 m/s

Now

an =

v2 ρ

At t = 2 s:

an =

(8.18 m/s)2 = 0.535 m/s2 125 m

Finally

a 2 = at2 + an2

At t = 2 s:

a 2 = 0.592 + 0.5352 a = 0.8 m/s2

or


201

202

PROBLEM 11.140 At a given instant in an airplane race, airplane A is flying horizontally in a straight line, and its speed is being increased at the rate of 8 m/s 2 . Airplane B is flying at the same altitude as airplane A and, as it rounds a pylon, is following a circular path of 300-m radius. Knowing that at the given instant the speed of B is being decreased at the rate of 3 m/s 2 , determine, for the positions shown, (a) the velocity of B relative to A, (b) the acceleration of B relative to A.

SOLUTION First note

v A = 450 km/h vB = 540 km/h = 150 m/s

(a)

v B = v A + v B/A

We have


vB2 /A = 4502 + 5402 − 2(450)(540) cos 60°

or

vB/A = 501.10 km/h

and or

540 501.10 = sin α sin 60°

α = 68.9° v B/A = 501 km/h

(b)

First note Now

a A = 8 m/s 2 → (a B )t = 3 m/s 2 ( aB ) n =

v 2B

ρB

=

60°

(150 m/s) 2 300 m (a B ) n = 75 m/s 2

or �

Then

68.9° �

30° ��

a B = (a B )t + (a B )n = 3(− cos 60° i + sin 60° j) + 75(−cos 30° i − sin 30° j) = −(66.452 m/s 2 )i − (34.902 m/s 2 ) j

Finally or

a B = a A + a B/A a B/A = ( −66.452i − 34.902 j) − (8i ) = −(74.452 m/s 2 )i − (34.902 m/s 2 ) j

a B/A = 82.2 m/s 2

or

25.1° �


202

PROBLEM 11.141 A motorist traveling along a straight portion of a highway is decreasing the speed of his automobile at a constant rate before exiting from the highway onto a circular exit ramp with a radius of 168-m. He continues to decelerate at the same constant rate so that 10 s after entering the ramp, his speed has decreased to 32 km/h, a speed which he then maintains. Knowing that at this constant speed the total acceleration of the automobile is equal to one-quarter of its value prior to entering the ramp, determine the maximum value of the total acceleration of the automobile.

SOLUTION First note

v10 = 32 km/h = 8.9 m/s

While the car is on the straight portion of the highway. a = astraight = at

and for the circular exit ramp a = at2 + an2

where

an =

v2 ρ

By observation, amax occurs when v is maximum, which is at t = 0 when the car first enters the ramp. For uniformly decelerated motion v = v0 + at t

and at t = 10 s:

v = constant a=

Then or

a = an =

2 v10 ρ

1 ast. 4

astraight = at

v 2 (8.9 m/s)2 1 at = 10 = ρ 4 168 m

at = −1.9 m/s2

(The car is decelerating; hence the minus sign.)

PROPRIETARY MATERIAL. MATERIAL. ©©2010 2009The TheMcGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be bedisplayed, displayed, PROPRIETARY reproducedor ordistributed distributedininany anyform formor orby byany anymeans, means,without withoutthe theprior priorwritten writtenpermission permission of of the the publisher, publisher,or orused usedbeyond beyondthe thelimited limited reproduced youare areaastudent studentusing usingthis thisManual, Manual, distributiontototeachers teachersand andeducators educatorspermitted permittedby byMcGraw-Hill McGraw-Hillfor fortheir theirindividual individualcourse coursepreparation. preparation.IfIfyou distribution youare areusing usingititwithout withoutpermission. permission. you

203

204


Then at t = 10 s: or Then at t = 0:

8.9 m/s = v0 + (−1.9 m/s2)(10 s) v0 = 27.9 m/s amax =

at2

v2 + 0 ρ

2

(27.9 m/s)2 = (−1.9 m/s ) + 168 m

2

1/2

2 2

amax = 5 m/s2

or


204

PROBLEM 11.142 Racing cars A and B are traveling on circular portions of a race track. At the instant shown, the speed of A is decreasing at the rate of 7 m/s 2 , and the speed of B is increasing at the rate of 2 m/s 2 . For the positions shown, determine (a) the velocity of B relative to A, (b) the acceleration of B relative to A.

SOLUTION First note

v A = 162 km/h = 45 m/s vB = 144 km/h = 40 m/s

(a)

We have

v B = v A + v B/A


vB2 /A = 1622 + 1442 − 2(162)(144)cos165°

or

vB/A = 303.39 km/h

and or

144 303.39 = sin α sin165°

α = 7.056° v B/A = 303 km/h

(b)

First note

(a A )t = 7 m/s 2

60°

(a B )t = 2 m/s 2

45°

52.9° �

v2 ρ

Now

an =

Then

(a A ) n =

or

(a A ) n = 6.75 m/s 2

(45 m/s) 2 300 m

( aB ) n = 30°

(40 m/s) 2 250 m

(a B ) n = 6.40 m/s 2

45°


205

206

PROBLEM 11.142 (Continued) Noting that We have

a = at + a n a A = 7(cos 60°i − sin 60° j) + 6.75(− cos 30°i − sin 30° j) = −(2.3457 m/s 2 )i − (9.4372 m/s 2 ) j

and Finally or

a B = 2(cos 45°i − sin 45° j) + 6.40(cos 45°i + sin 45° j) = (5.9397 m/s 2 )i + (3.1113 m/s 2 ) j

a B = a A + a B/A a B/A = (5.9397i + 3.1113j) − (−2.3457i − 9.4372 j) = (8.2854 m/s 2 )i + (12.5485 m/s 2 ) j

a B/A = 15.04 m/s 2

or

56.6° ��


206

PROBLEM 11.143 A golfer hits a golfball from Point A with an initial velocity of 50 m/s at an angle of 25° with the horizontal. Determine the radius of curvature of the trajectory described by the ball (a) at Point A, (b) at the highest point of the trajectory.

SOLUTION (a)

We have or

(a A ) n =

ρA =

v 2A ρA (50 m/s)2 (9.81 m/s 2 ) cos 25°

ρ A = 281 m �

or (b)

We have

( aB ) n =

vB2 ρB

where Point B is the highest point of the trajectory, so that vB = (v A ) x = v A cos 25°

Then

ρB =

[(50 m/s) cos 25°]2 9.81 m/s 2

ρ B = 209 m �

or


207

208

PROBLEM 11.144 From a photograph of a homeowner using a snowblower, it is determined that the radius of curvature of the trajectory of the snow was 8.5 m as the snow left the discharge chute at A. Determine (a) the discharge velocity v A of the snow, (b) the radius of curvature of the trajectory at its maximum height.

SOLUTION

(a)

We have or

(a A ) n =

v 2A ρA

v A2 = (9.81 cos 40°)(8.5 m) = 63.8766 m 2 /s 2

v A = 7.99 m/s

or (b)

We have

( aB ) n =

40° �

vB2 ρB

where Point B is the highest point of the trajectory, so that vB = (v A ) x = v A cos 40°

Then

ρB =

(63.8766 m 2 /s 2 ) cos 2 40° 9.81 m/s 2

ρ B = 3.82 m ��

or


208

PROBLEM 11.145 A basketball is bounced on the ground at Point A and rebounds with a velocity v A of magnitude 2 m/s as shown. Determine the radius of curvature of the trajectory described by the ball (a) at Point A, (b) at the highest point of the trajectory.

SOLUTION (a)

(a A ) n =

We have

ρA =

or

v 2A ρA (2 m/s)2 (9.81 m/s2) sin 15°

ρ A = 1.6 m

or (b)

( aB ) n =

We have

vB2 ρB

where Point B is the highest point of the trajectory, so that vB = (v A ) x = v A sin 15°

ρB =

Then

[(2 m/s) sin 15°]2 9.81 m/s2

ρ B = 0.027 m

or


209

210

PROBLEM 11.146 Coal is discharged from the tailgate A of a dump truck with an initial velocity vA = 2 m/s 50° . Determine the radius of curvature of the trajectory described by the coal (a) at Point A, (b) at the point of the trajectory 1 m below Point A.

SOLUTION (a)

(b)

(a A ) n =

We have

v 2A ρA (2 m/s)2 (9.81 m/s2) cos 50°

or

ρA =

or

ρ A = 0.634 m

Horizontal motion. (Uniform) (vB ) x = (v A ) x = (2 m/s) cos 50° = 1.286 m/s

Vertical motion. (Uniformly accelerated motion) We have

(

v 2y = (v A ) 2y − 2 g y − y 0A

)

( g = 9.81 m/s2)

where

(v A ) y = (2 m/s) sin 50° = 1.532 m/s

At Point B, y = −3 ft:

(vB ) 2y = (1.532 m/s)2 − 2(9.81 m/s2)(−1 m)

or

(vB ) y = 4.678 m/s

Then and or Now or

vB = (vB ) 2x + (vB ) 2y = (1.286)2 + (4.687)2 = 4.86 m/s tan θ =

vy vx

=

4.687 1.286

θ = 74.66° ( aB ) n =

ρB =

vB2

ρB (4.86 m/s)2 (9.81 m/s2) cos 74.66°

ρ B = 9.1 m

or


210

PROBLEM 11.147 A horizontal pipe discharges at Point A a stream of water into a reservoir. Express the radius of curvature of the stream at Point B in terms of the magnitudes of the velocities v A and v B .

SOLUTION vB2

We have

( aB ) n =

where

(aB ) n = aB cosθ = g cosθ

ρB

Noting that the horizontal motion is uniform, we have ( vB ) x = v A

where

(vB ) x = vB cos θ cos θ =

Then

ρB =

vA vB vB2 g

( ) vA vB

ρB =

or

vB2 � gv A


211

212

PROBLEM 11.148 A child throws a ball from Point A with an initial velocity v A of 20 m/s at an angle of 25° with the horizontal. Determine the velocity of the ball at the points of the trajectory described by the ball where the radius of curvature is equal to three-quarters of its value at A.

SOLUTION Assume that Point B and C are the points of interest, where yB = yC and vB = vC . (a A ) n =

Now

v 2A

ρA

or

ρA =

v 2A g cos 25°

Then

ρB =

v A2 3 3 ρA = 4 4 g cos 25° vB2

We have

( aB ) n =

where

(aB ) n = g cos θ

so that or

ρB

v A2 vB2 3 = 4 g cos 25° g cos θ vB2 =

3 cos θ 2 vA 4 cos 25°

(1)

Noting that the horizontal motion is uniform, we have ( v A ) x = ( vB ) x

where Then or

(v A ) x = v A cos 25° (vB ) x = vB cos θ v A cos 25° = vB cos θ cos θ =

vA cos 25° vB


212

PROBLEM 11.148 (Continued) Substituting for cos θ in Eq. (1), we have

or

vB2 =

� v A2 3 � vA � cos 25° � 4 � vB � cos 25°

vB3 =

3 3 vA 4

3 vB = 3 v A = 18.17 m/s 4 4 cos 25° 3 θ = ± 4.04°

cos θ = 3

v B = 18.17 m/s v B = 18.17 m/s

and�

4.04° � 4.04° �


213

214

PROBLEM 11.149 A projectile is fired from Point A with an initial velocity v 0 . (a) Show that the radius of curvature of the trajectory of the projectile reaches its minimum value at the highest Point B of the trajectory. (b) Denoting by θ the angle formed by the trajectory and the horizontal at a given Point C, show that the radius of curvature of the trajectory at C is ρ = ρ min /cos3 θ .

SOLUTION For the arbitrary Point C, we have (aC ) n =

ρC =

or

vC2

ρC vC2 g cos θ

Noting that the horizontal motion is uniform, we have (v A ) x = (vC ) x (v A ) x = v0 cos α

where

v0 cos α = vC cos θ

Then

cos α v0 cos θ

or

vC =

so that

1 ρC = g cos θ

(a)

2

� cos α � v 2 cos 2 α v0 � = 0 � g cos3 θ � cos θ �

In the expression for ρC , v0 , α , and g are constants, so that ρC is minimum where cos θ is maximum. By observation, this occurs at Point B where θ = 0.

ρ min = ρ B = (b)

(vC ) x = vC cos θ

ρC =

1 cos3 θ

ρC =

ρ min cos3 θ

v02 cos 2 α g

� v02 cos 2 α �� g �

Q.E.D.

�

Q.E.D.

�

� ��


214

PROBLEM 11.150 A projectile is fired from Point A with an initial velocity v 0 which forms an angle α with the horizontal. Express the radius of curvature of the trajectory of the projectile at Point C in terms of x, v0 , α , and g.

SOLUTION We have or

(aC ) n =

ρC =

vC2

ρC vC2 g cos θ

Noting that the horizontal motion is uniform, we have x = 0 + (v0 ) x t = (v0 cos α )t

(v A ) x = (vC ) x

where (v A ) x = v0 cos α Then or so that

(vC ) x = vC cos θ v0 cos α = vC cos θ cos θ =

ρC =

and

(vC ) x = v0 cos α

(1)

v0 cos α vC vC3 g v0 cos α

For the uniformly accelerated vertical motion have (vC ) y = (v0 ) y − gt = v0 sin α − gt

From above Then Now

x = (v0 cos α )t or t = (vC ) y = v0 sin α − g

x v0 cos α

x v0 cos α

(2)

vC2 = (vC ) 2x + (vC ) 2y


215

216

PROBLEM 11.150 (Continued) Substituting for (vC ) x [Eq. (1)] and (vC ) y [Eq. (2)] � x � vC2 = (v0 cos α )2 + � v0 sin α − g � v0 cos α � � � 2 gx tan α g 2 x2 � = v02 �1 − + � � v02 v04 cos 2α ��

or

vC3

=

v03

� 2 gx tan α g 2 x2 + ��1 − v02 v04 cos 2α �

� ��

2

3/ 2

Finally, substituting into the expression for ρC , obtain v02 � 2 gx tan α g 2 x2 ρ= + ��1 − g cos α � v02 v04 cos 2 α

� ��

3/ 2

�


216

PROBLEM 11.151* Determine the radius of curvature of the path described by the particle of Problem 11.95 when t = 0.

SOLUTION We have

v=

dr = R(cos ωn t − ωn t sin ωn t )i + cj + R (sin ωn t + ωn t cos ωn t )k dt

and

a=

dv = R − ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t i dt

(

)

(

)

+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k

or

a = ωn R [−(2 sin ωn t + ωn t cos ωn t )i + (2cos ωn t − ωn t sin ωn t ) k ]

Now

v 2 = R 2 (cos ωn t − ωn t sin ωn t )2 + c 2 + R 2 (sin ωn t + ωn t cos ωn t )2

(

)

= R 2 1 + ωn2 t 2 + c 2

Then and Now

(

1/ 2

R 2ωn2 t dv = 1/ 2 dt � 2 R 1 + ωn2 t 2 + c 2 � � �

(

)

a 2 = at2 + an2 2 2 � dv � � v = � � + �� dt � � ρ

At t = 0:

)

v = � R 2 1 + ωn2 t 2 + c 2 � � �

� ��

2

dv =0 dt a = ωn R(2 k ) or a = 2ωn R v2 = R2 + c2

Then, with we have or

dv = 0, dt a= 2ωn R =

v2

ρ R 2 + c2

ρ ρ=

or

R 2 + c2 � 2ωn R


217

218

PROBLEM 11.152* Determine the radius of curvature of the path described by the particle of Problem 11.96 when t = 0, A = 3, and B = 1.

SOLUTION With

A = 3,

B =1

We have

r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k

Now

v=

and

� � 2 dv a= = 3(− sin t − sin t − t cos t )i + 3 � t + 1 − t �� dt t2 + 1 ��

)

(

� 3t � dr j + (sin t + t cos t )k = 3(cos t − t sin t )i + � 2 � t + 1 �� dt � � t t

2

�� +1 �� j � ��

+ (cos t + cos t − t sin t )k = − 3(2sin t + t cos t )i + 3

1 j (t + 1)1/2 2

+ (2 cos t − t sin t )k v 2 = 9 (cos t − t sin t )2 + 9

Then

t2 + (sin t + t cos t )2 t2 + 1

Expanding and simplifying yields v 2 = t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t v = [t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t ]1/ 2

Then and

Now

dv 4t 3 + 38t + 8(−2 cos t sin t + 4t 3sin 2 t + 2t 4 sin t cos t ) − 8[(3t 2 + 1)sin 2t + 2(t 3 + t ) cos 2t ] = dt 2[t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t ) sin 2t ]1/ 2 2

a =

at2

+

an2

2 2 � dv � � v � = � � + �� dt � � ρ �

2


218

PROBLEM 11.152* (Continued) At t = 0:

a = 3j + 2k

or

a = 13 m/s2 dv =0 dt v 2 = 9 (m/s)2

Then, with

dv = 0, dt

we have

a=

or

ρ=

v2

ρ 9 m2/s2 13 m/s2

ρ = 2.50 m

or


219

220

PROBLEM 11.153 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 160 km above the surface of the planet. Venus: g = 8.53 m/s 2 , R = 6161 km.

SOLUTION an = g

We have Then or

g

R2 r2

and an =

v2 r

R2 v2 = r r2

vcirc. = R

g r

where r = R + h

For the given data vcirc. = 6161 km

8.53 m/s 2 3600 s × 3 1h (6161 + 160) × 10 m vcirc. = 25.8 × 103 km/h �

or


220

PROBLEM 11.154 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 160 km above the surface of the planet. Mars: g = 3.83 m/s 2 , R = 3332 km.

SOLUTION an = g

We have Then or

g

R2 r2

and an =

v2 r

R2 v2 = r r2

vcirc. = R

g r

where r = R + h

For the given data vcirc. = 3332 km

3.83 m/s 2 3600 s × 3 1h (3332 + 160) × 10 m vcirc. = 12.56 × 103 km/h �

or


221

222

PROBLEM 11.155 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 160 km above the surface of the planet. Jupiter: g = 26.0 m/s 2 , R = 69,893 km.

SOLUTION an = g

We have Then or

g

R2 r2

and an =

v2 r

R2 v2 = r r2

vcirc. = R

g r

where r = R + h

For the given data vcirc. = 69,893

26.0 m/s 2 3600 s × 3 1h (69,893 + 160) × 10 m vcirc. = 153.3 × 103 km/h �

or


222

PROBLEM 11.156 Knowing that the diameter of the sun is 1382 × 103 km and that the acceleration of gravity at its surface is 270 m/s2, determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. (See information given in Problems 11.153–11.155.) Earth: (vmean )orbit = 106500 km/h

SOLUTION an = g

We have Then

or

g

R2 r2

and an =

v2 r

R2 v2 = r r2 r=g

R v

2

where R =

1 d 2

For the given data 2

rearth = (270 m/s )

1 2

× 1382 × 103 km 106500 km/h

2

×

1 km 3600 s = 1000 m 1h

2

rearth = 147.3 × 106 km

or

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. Inc. All All rights rights reserved. reserved. No part of of this this Manual Manual may may be be displayed, displayed, PROPRIETARY No part reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced you are are aa student student using using this this Manual, Manual, distribution to to teachers teachers and andeducators educators permitted permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course course preparation. preparation. If If you distribution you are are using using itit without without permission. permission. you

223

224

PROBLEM 11.157 Knowing that the diameter of the sun is 1382 × 103 km and that the acceleration of gravity at its surface is 270 m/s2, determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. (See information given in Problems 11.153–11.155.) Saturn: (vmean )orbit = 34500 km/h

SOLUTION an = g

We have Then

or

g

R2 r2

and an =

v2 r

R2 v2 = r r2 r=g

R v

2

where R =

1 d 2

For the given data rsaturn = (270 m/s2)

1 2

× 1382 × 103 km 34500 km/h

2

×

1 km 3600 s × 1000 m 1h

2

rsaturn = 1403.7 × 106 km

or


224

PROBLEM 11.158 Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth. (See information given in Problems 11.153–11.155.)

SOLUTION an = g

We have Then or

g

R2 r2

and an =

v2 r

R2 v2 = r r2 v=R

g r

where

r =R+h

The circumference s of the circular orbit is equal to s = 2π r

Assuming that the speed of the telescope is constant, we have s = vtorbit

Substituting for s and v 2π r = R

or

torbit = =

g torbit r

2π r 3/2 R g 2π [(6370 + 590) km]3/2 1h × 2 1/2 −3 6370 km [9.81 × 10 km/s ] 3600 s torbit = 1.606 h ��

or


225

226

PROBLEM 11.159 A satellite is traveling in a circular orbit around Mars at an altitude of 300 km. After the altitude of the satellite is adjusted, it is found that the time of one orbit has increased by 10 percent. Knowing that the radius of Mars is 3310 km, determine the new altitude of the satellite. (See information given in Problems 11.153–11.155.)

SOLUTION an = g

We have Then or

g

R2 r2

and an =

v2 r

R2 v2 = r r2 v=R

g r

where

r =R+h

The circumference s of a circular orbit is equal to s = 2π r

Assuming that the speed of the satellite in each orbit is constant, we have s = vtorbit

Substituting for s and v 2π r = R

or

torbit = =

Now or or

g torbit r

2π r 3/2 R g 2π ( R + h)3/2 R g

(torbit ) 2 = 1.1(torbit )1 2π ( R + h2 )3/2 2π ( R + h1 )3/2 = 1.1 R R g g h2 = (1.1)2/3 ( R + h1 ) − R = (1.1)2/3 (3310 + 300) km − (3310 km) h2 = 537 km

or


226

PROBLEM 11.160 Satellites A and B are traveling in the same plane in circular orbits around the earth at altitudes of 190 and 320 km, respectively. If at t = 0 the satellites are aligned as shown and knowing that the radius of the earth is R = 6370 km, determine when the satellites will next be radially aligned. (See information given in Problems 11.153–11.155.)

SOLUTION R2 r2

an = g

We have Then

g

and an =

R2 v2 = r r2

or

v2 r

v=R

g r

r =R+h

where The circumference s of a circular orbit is

s = 2π r

equal to

Assuming that the speeds of the satellites are constant, we have s = vT

Substituting for s and v 2π r = R T=

or Now

hB

g T r

2π r 3/ 2 2π ( R + h)3/ 2 = R g R g hA

(T ) B

(T ) A

Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B completes N orbits, satellite A must complete ( N + 1) orbits. Thus, TC = N (T ) B = ( N + 1)(T ) A

or

N

2π ( R + hB )3/2 2π ( R + hA )3/ 2 = ( N + 1) R R g g


227

228


N=

or

=

Then

( R + hA )3/2 ( R + hB )

(

1 6370 + 320 6370 +190

3/ 2

)

− ( R + hA )

3/2

−1

3/2

=

(

1 R + hB R + hA

)

3/2

−1

= 33.476 orbits

2π ( R + hB )3/2 R g

TC = N (T ) B = N

2π [(6370 + 320) km] 1h = 33.476 × 1/2 6370 km 9.81 m/s2 × 1 km 3600s 3/2

(

1000 m

)

TC = 50.7 h

or Alternative solution

From above, we have (T ) B (T ) A . Thus, when the satellites are next radially aligned, the angles θ A and θ B swept out by radial lines drawn to the satellites must differ by 2π . That is,

θ A = θ B + 2π For a circular orbit

s = rθ

From above

s = vt and v = R

Then

θ=

At time TC : or

R g ( R + hA )

3/2

TC = TC = =

g r

R g R g s vt 1 g = = R t = 3/2 t = t r r r r ( R + h)3/2 r R g ( R + hB )3/ 2

TC + 2π 2π

R g

1 ( R + hA )3/ 2

(

−

1 ( R + hB )3/ 2

2π

1 km (6370 km) 9.81 m/s2 × 1000 m

× ×

1 [(6370 + 190) km] 3/2

1 −

)

1/2

1

[(6370 + 320) km] 3/2

1h 3600 s TC = 50.7 h

or


228

PROBLEM 11.161 The path of a particle P is a limaçon. The motion of the particle is defined by the relations r = b(2 + cos π t ) and θ = π t , where t and θ are expressed in seconds and radians, respectively. Determine (a) the velocity and the acceleration of the particle when t = 2 s, (b) the values of θ for which the magnitude of the velocity is maximum.

SOLUTION We have

r = b(2 + cos π t )

θ = πt

Then

r� = −π b sin π t

θ� = π

and

r�� = −π 2 b cos π t

θ�� = 0

Now

v = r�er + rθ�eθ = −(π b sin π t ) er + π b(2 + cos π t ) eθ

and

a = (�� r − rθ� 2 ) e r + (rθ�� + 2r�θ�) eθ

= [ −π 2 b cos π t − π 2 b(2 + cos π t )]er + (0 − 2π 2 b sin π t )eθ = −2π 2b[(1 + cos π t ) e r + (sin π t ) eθ ]

(a)

At t = 2 s:

v = −(0)er + π b(2 + 1)eθ v = 3π beθ �

or a = −2π 2b[(1 + 1) e r + (0) eθ ]

a = −4π 2 ber �

or (b)

We have

v = π b (− sin π t )2 + (2 + cos π t ) 2 = π b 5 + 4cos π t θ = π t = π b 5 + 4cos θ

By observation,

v = vmax

when cos θ = 1

θ = 2nπ , n = 0, 1, 2, . . . �

or


229

230

PROBLEM 11.162 The two-dimensional motion of a particle is defined by the relation r = 2b cos ω t and θ = ω t where b and ω are constant. Determine (a) the velocity and acceleration of the particle at any instant, (b) the radius of curvature of its path. What conclusion can you draw regarding the path of the particle?

SOLUTION r = 2b cos ω t

θ = ωt

r� = −2bω sin ω t r�� = −2bω 2 cos ω t

(a)

Velocity.

θ� = ω θ�� = 0

vr = �� r = −2bω sin ω t vθ = rθ� = 2bω cos ω t v 2 = vr2 + vθ2 = (2bω )2 [(−sin ω t) 2 + (cos ω t) 2 ] = (2bω )2

Acceleration.

v = 2bω �

ar = �� r − rθ� 2 = −2bω 2 cos ω t − (2b cos ω t)(ω ) 2 = −4bω 2 cos ω t aθ = rθ�� + 2r�θ� = (2b cos ω t)(0) + 2( −2bω sin ω t)(ω ) = −4bω 2 sin ω t a 2 = ar2 + aθ2 = (−4bω 2 ) 2 (cos 2 ω t + sin 2 ω t) = (4bω 2 ) 2

(b)

Since Thus:

a = 4bω 2 �

v = 2bω = constant, at = 0 an = a = 4bω 2 an =

v2

ρ

; ρ=

v 2 (2bω ) 2 = =b ρ =b an 4bω 2

For the path, ρ = constant. Thus, path is a circle.

�


230

PROBLEM 11.163 The rotation of rod OA about O is defined by the relation θ = π (4t 2 − 8t ), where θ and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r = 10 + 6sin π t , where r and t are expressed in meters and seconds, respectively. When t = 1 s, determine (a) the velocity of the collar, (b) the total acceleration of the collar, (c) the acceleration of the collar relative to the rod.

SOLUTION We have

r = 10 + 6sin π t

θ = π (4t 2 − 8t )

Then

r = 6π cos π t

θ = θπ (t − 1)

and

r = −6π 2 sin π t

θ = θπ

At t = 1 s:

r = 10 m

θ = −4π rad

r = −6π m/s

θ =0

r =0

θ = 8π rad/s 2

(a)

We have

v B = re r + rθ eθ v B = −(6π m/s) er

so that

(b)

We have

a B = ( r − rθ 2 )er + ( rθ + 2rθ )eθ

= (10)(8π )eθ

a B = (80π m/s2) eθ

or

(c)

We have

a B/OA = r a B/OA = 0

so that

PROPRIETARY 2009 The The McGraw-Hill McGraw-Hill Companies, displayed, PROPRIETARY MATERIAL. MATERIAL. © © 2010 Companies, Inc. Inc. All All rights rights reserved. reserved. No No part part of of this this Manual Manual may may be be displayed, reproduced or distributed distributed in any form form or or by by any any means, means, without without the the prior prior written written permission permission of or used used beyond beyond the the limited limited reproduced or in any of the the publisher, publisher, or student using Manual, distribution teachers and and educators educators permitted distribution to to teachers permitted by by McGraw-Hill McGraw-Hill for for their their individual individual course coursepreparation. preparation.IfIf you you are are aa student using this this Manual, you you are are using using itit without without permission. permission.

231

232

PROBLEM 11.164 The oscillation of rod OA about O is defined by the relation θ = (2/π )(sin πτ), where θ and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r = 25/(t + 4), where r and t are expressed in meters and seconds, respectively. When t = 1 s, determine (a) the velocity of the collar, (b) the total acceleration of the collar, (c) the acceleration of the collar relative to the rod.

SOLUTION 25 t+4

θ=

2

sin π t

We have

r=

Then

r =−

and

r=

At t = 1 s:

r =5m

θ =0

r = −1 m/s

θ = −2 rad/s

r = 0.4 m/s2

θ =0

(a)

We have

25 (t + 4)2

π

θ = 2 cos π t

50 (t + 4) 2

θ = −2π sin π t

v B = re r + rθ eθ = (−1)e r + (5)(−2)eθ v A = −(1 m/s)er − (10 m/s)eθ

or

(b)

We have

a B = ( r − rθ 2 )er + ( rθ + 2rθ )eθ

= [0.4 − (5)(2) 2 ]e r + [0 + 2(−1)(−2)]eθ

a B = −(19.6 m/s2)er + (4 m/s2)eθ

or (c)

We have

a B/OA = r a B/OA = (0.4 m/s2 ) er

so that

PROPRIETARYMATERIAL. MATERIAL.©©2010 2009The TheMcGraw-Hill McGraw-HillCompanies, Companies,Inc. Inc.All Allrights rightsreserved. reserved. No No part partofofthis thisManual Manualmay maybebedisplayed, displayed, PROPRIETARY anymeans, means,without withoutthe theprior priorwritten writtenpermission permissionofofthe thepublisher, publisher,ororused usedbeyond beyondthe thelimited limited reproducedorordistributed distributedininany anyform formororbybyany reproduced youare areaastudent studentusing usingthis thisManual, Manual, distributiontototeachers teachersand andeducators educatorspermitted permittedby byMcGraw-Hill McGraw-Hillfor fortheir theirindividual individualcourse coursepreparation. preparation.IfIfyou distribution youare areusing usingititwithout withoutpermission. permission. you 233

232

PROBLEM 11.165 The path of particle P is the ellipse defined by the relations r = 2/(2 − cos π t ) and θ = π t , where r is expressed in meters, t is in seconds, and θ is in radians. Determine the velocity and the acceleration of the particle when (a) t = 0, (b) t = 0.5 s.

SOLUTION We have

r=

2 2 − cos π t

θ = πt

Then

r� =

−2π sin π t (2 − cos π t )2

θ� = π

and

r�� = −2π

π cos π t (2 − cos π t ) − sin π t (2π sin π t ) (2 − cos π t )3

= −2π 2

(a)

At t = 0:

Now

2cos π t − 1 − sin 2 π t (2 − cos π t )3

r =2m

θ =0

r� = 0

θ� = π rad/s

r�� = −2π 2 m/s 2

θ�� = 0

v = r�er + rθ�eθ = (2)(π )eθ v = (2π m/s)eθ �

or and

θ�� = 0

a = (r�� − rθ� 2 )e r + (rθ�� + 2r�θ�)eθ

= [ −2π 2 − (2)(π )2 ]er a = −(4π 2 m/s 2 )er �

or (b)

At t = 0.5 s:

θ=

r =1 m r� =

−2π π = − m/s 2 2 (2)

r�� = −2π 2

−1 − 1 π 2 = m/s 2 3 2 (2)

π 2

rad

θ� = π rad/s θ�� = 0


233

234


Now

� π� v = r�er + rθ�eθ = � − � e r + (1)(π )eθ � 2� �π � v = − � m/s � er + (π m/s)eθ � 2 � �

or and

a = (r�� − rθ� 2 )e r + (rθ�� + 2r�θ�)eθ �π 2 � � � π� � =� − (1)(π ) 2 � e r + � 2 � − � (π ) � eθ � � 2� � � 2 �

�π2 � a = − �� m/s 2 �� er − (π 2 m/s 2 )eθ � � 2 �

or


234

PROBLEM 11.166 The two-dimensional motion of a particle is defined by the relations r = 2a cos θ and θ = bt 2 /2, where a and b are constants. Determine (a) the magnitudes of the velocity and acceleration at any instant, (b) the radius of curvature of the path. What conclusion can you draw regarding the path of the particle?

SOLUTION (a)

1 2

We have

r = 2a cos θ

θ = bt 2

Then

r� = −2aθ sin θ

θ� = bt

and

r�� = −2a (θ�� sin θ + θ� 2 + cos θ )

θ�� = b

Substituting for θ� and θ�� r� = −2abt sin θ �� r = −2ab(sin θ + bt 2 cos θ )

Now Then

vθ = rθ� = 2abt cos θ

vr = r� = −2abt sin θ

v = vr2 + vθ2 = 2abt[( − sin θ ) 2 + (cos θ ) 2 ]1/2 v = 2abt �

or Also

ar = �� r − rθ� 2 = −2ab(sin θ + bt 2 cos θ ) − 2ab 2 t 2 cos θ = −2ab(sin θ + 2bt 2 cos θ )

and

aθ = rθ�� + 2r�θ� = 2ab cos θ − 4ab 2 t 2 sin θ = −2ab(cos θ − 2bt 2 sin θ )

Then

a = ar2 + aB2 = 2ab[(sin θ + 2bt 2 cos θ ) 2 + (cos θ − 2bt 2 sin θ ) 2 ]1/2 a = 2ab 1 + 4b 2 t 4 �

or (b)

2

at2

an2

2 2 � dv � � v � = � � + �� dt � � ρ �

Now

a =

Then

dv d = (2abt ) = 2ab dt dt

+

2


235

236


so that or

(

2ab 1 + 4b 2 t 4

)

2

= (2ab) 2 + an2

4a 2b 2 (1 + 4b 2t 4 ) = 4a 2 b 2 + an2

or

an = 4ab 2 t 2

Finally

an =

v2

ρ

�ρ=

(2abt ) 2 4ab 2 t 2

ρ =a �

or Since the radius of curvature is a constant, the path is a circle of radius a.

��


236

PROBLEM 11.167 To study the performance of a race car, a high-speed motionpicture camera is positioned at Point A. The camera is mounted on a mechanism which permits it to record the motion of the car as the car travels on straightway BC. Determine the speed of the car in terms of b, θ , and θ�.

SOLUTION We have

r=

b cos θ

Then

r� =

bθ� sin θ cos 2 θ

We have

v 2 = vr2 + vθ2 = (r�)2 + (rθ�) 2 2

or

� bθ sin θ � � bθ �2 = �� + � � 2 � cos θ � � cos θ � � b2θ� 2 b2θ� 2 � sin 2θ 1 = + � �� = 4 cos 2θ �� cos 2θ � cos θ bθ� v=± cos 2θ

For the position of the car shown, θ is decreasing; thus, the negative root is chosen. v=

bθ� � cos 2θ

v=−

bθ� � cos 2θ

Alternative solution. From the diagram or

r� = −v sin θ bθ� sin θ = −v sin θ cos 2θ

or


237

238

PROBLEM 11.168 Determine the magnitude of the acceleration of the race car of Problem 11.167 in terms of b, θ , θ , and θ .

SOLUTION We have

r=

b cos θ

Then

r=

bθ sin θ cos 2 θ

For rectilinear motion

a=

dv dt

From the solution to Problem 11.167 v=−

Then

a=

bθ cos 2θ

d bθ − dt cos 2θ

= −b

θ cos 2θ − θ (−2θ cos θ sin θ ) cos 4θ a=−

or

b (θ + 2θ 2 tan θ ) cos 2θ

Alternative solution b cos θ

From above

r=

Then

r =b =b

r=

bθ sin θ cos 2 θ

(θ sin θ + θ 2 cos θ )(cos 2θ ) − (θ sin θ )(−2θ cos θ sin θ ) cos 4θ

θ sin θ θ 2 (1 + sin 2 θ ) + cos 2θ cos3θ


238


Now where

a 2 = ar2 + aθ2 � θ�� sin θ θ� 2 (1 + sin 2 θ ) � bθ� 2 ar = �� r − rθ� 2 = b � + �− 2 cos 2θ � cos θ � cos θ 2θ� 2 sin 2 θ � b � �� θ θ sin = + � � cos θ �� cos 2θ �� ar =

and

bθ�� bθ� 2 sin θ aθ = rθ�� + 2r�θ� = +2 cos θ cos 2θ =

Then

b sin θ �� (θ + 2θ� 2 tan θ ) cos 2θ

b cos θ �� (θ + 2θ� tan θ ) cos 2θ

a=±

b (θ�� + 2θ� 2 tan θ )[(sin θ ) 2 + (cos θ )2 ]1/ 2 2 cos θ

For the position of the car shown, θ�� is negative; for a to be positive, the negative root is chosen. a=−

b (θ�� + 2θ� 2 tan θ ) � cos 2 θ


239

240

PROBLEM 11.169 After taking off, a helicopter climbs in a straight line at a constant angle β . Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β , θ , and θ�.

SOLUTION From the diagram r d = sin (180° − β ) sin ( β − θ ) or

d sin β = r (sin β cos θ − cos β sin θ ) tan β tan β cos θ − sin θ

or

r=d

Then

r� = d tan β

−(− tan β sin θ − cos θ ) � θ (tan β cos θ − sin θ )2

= dθ tan β

tan β sin θ + cos θ (tan β cos θ − sin θ ) 2

From the diagram vr = v cos ( β − θ )

where

vr = r�

Then tan β sin θ + cos θ = v(cos β cos θ + sin β sin θ ) dθ� tan β (tan β cos θ − sin θ )2 = v cos β (tan β sin θ + cos θ ) v=

or

dθ� tan β sec β � (tan β cosθ − sin θ )2

Alternative solution. We have

v 2 = vr2 + vθ2 = (r�) 2 + ( rθ�) 2


240

PROBLEM 11.169 (Continued) Using the expressions for r and r� from above � tan β sin θ + cos θ � v = � dθ� tan β � (tan β cos θ − sin θ ) 2 � �

2

1/ 2

or

� (tan β sin θ + cos θ )2 � dθ� tan β v=± + 1� � 2 (tan β cos θ − sin θ ) � (tan β cos θ − sin θ ) � 1/ 2

� � dθ� tan β tan 2 β + 1 =± � 2� (tan β cos θ − sin θ ) � (tan β cos θ − sin θ ) �

Note that as θ increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen. v=

dθ� tan β sec β � (tan β cos θ − sin θ )2


241

242

PROBLEM 11.170* Pin P is attached to BC and slides freely in the slot of OA. Determine the rate of change θ� of the angle θ , knowing that BC moves at a constant speed v0 . Express your answer in terms of v0 , h, β , and θ .

SOLUTION From the diagram r h = sin (90° − β ) sin ( β + θ )

or

r (sin β cos θ + cos β sin θ ) = h cos β r=

or Also, Then

h tan β cos θ + sin θ

vθ = v0 sin (β + θ ) where vθ = rθ� hθ� = v0 (sin β cos θ + cos β sin θ ) tan β cos θ + sin θ

θ� =

or

v0 cos β (tan β cos θ + sin θ ) 2 � h

Alternative solution. h tan β cos θ + sin θ

From above

r=

Then

r� = h

tan β sin θ − cos θ � θ (tan β cos θ + sin θ )2

Now

v02 = vr2 + vθ2 = (r�) 2 + ( rθ�) 2

or

v02

� tan β sin θ − cos θ � = � hθ� 2� � (tan β cos θ + sin θ ) � � � hθ� +� � � tan β cos β + sin θ �

2

2


242


or

v0 = ±

hθ� tan β cos θ + sin θ 1/2

� (tan β sin θ − cos θ ) 2 � +� + 1 � 2 � (tan β cos θ + sin θ ) �

hθ� =± tan β cos θ + sin θ

1/2

� � tan 2 β + 1 � 2� � (tan β cos θ + sin θ ) �

Note that as θ increases, member BC moves in the indicated direction. Thus, the positive root is chosen.

θ� =

v0 cos β (tan β cos θ + sin θ ) 2 � h


243

244

PROBLEM 11.171 For the race car of Problem 11.167, it was found that it took 0.5 s for the car to travel from the position θ = 60° to the position θ = 35°. Knowing that b = 25 m, determine the average speed of the car during the 0.5-s interval.

SOLUTION From the diagram: ∆r12 = 25 tan 60° − 25 tan 35° = 25.796 m

Now

vave = =

∆r12 ∆t12 25.796 m 0.5 s

= 51.592 m/s vave = 185.7 km/h �

or


244

PROBLEM 11.172 For the helicopter of Problem 11.169, it was found that when the helicopter was at B, the distance and the angle of elevation of the helicopter were r = 900 m and θ = 20°, respectively. Four seconds later, the radar station sighted the helicopter at r = 996 m and θ = 23.1°. Determine the average speed and the angle of climb β of the helicopter during the 4-s interval.

SOLUTION We have

θ0 = 20° θ 4 = 23.1°

r0 = 900 m r4 = 996 m

From the diagram: ∆r 2 = 9002 + 9962 −2(900)(996) cos (23.1° − 20°)

or

∆r = 108.8 m

Now

vave =

∆r ∆t 108.8 m = 4s = 27.2 m/s vave = 97.9 km/h

or Also, or

∆r cos β = r4 cos θ 4 − r0 cos θ0 cos β =

996 cos 23.1° − 900 cos 20° 108.8

β = 49.7°

or


245

246

PROBLEM 11.173 A particle moves along the spiral shown; determine the magnitude of the velocity of the particle in terms of b, θ , and θ�.

SOLUTION Hyperbolic spiral.

r=

b

θ

dr b dθ b =− 2 = − 2 θ� dt θ dt θ b b vr = r� = − 2 θ� vθ = rθ� = θ� r� =

θ

θ

2

� 1 � �1� v = vr2 + vθ2 = bθ� � − 2 � + � � � θ � �θ � bθ� = 2 1+θ 2

2

θ

v=

b

θ

2

1 + θ 2 θ� �


246

PROBLEM 11.174 A particle moves along the spiral shown; determine the magnitude of the velocity of the particle in terms of b, θ , and θ�.

SOLUTION Logarithmic spiral.

r = ebθ r� =

dr dθ = bebθ = bebθ θ� dt dt

vr = r� = bebθ θ� vθ = rθ� = ebθ θ� v = vr2 + vθ2 = ebθ θ� b 2 + 1

v = ebθ 1 + b 2 θ� �


247

248

PROBLEM 11.175 A particle moves along the spiral shown. Knowing that θ� is constant and denoting this constant by ω , determine the magnitude of the acceleration of the particle in terms of b, θ , and ω.

SOLUTION b

Hyperbolic spiral.

r=

From Problem 11.173

r� = −

θ

r�� = −

b

θ2

θ�

b �� 2b � 2 θ + 3θ 2

θ

θ

b b 2b ar = r�� − rθ� 2 = − 2 θ�� + 3 θ� 2 − θ� 2

θ

θ

θ

b b b � b � aθ = rθ�� + 2r�θ� = θ�� + 2 � − 2 θ� �θ� = θ�� − 2 2 θ� 2 θ θ θ � θ �

Since θ� = ω = constant, θ�� = 0, and we write: b 2 bω 2 2 − ω ω = 3 (2 − θ 2 ) θ θ3 θ 2 b bω aθ = −2 2 ω 2 = − 3 (2θ ) θ θ ar = +

2b

a = ar2 + aθ2 =

bω 2

θ

3

(2 − θ 2 ) 2 + (2θ ) 2 =

bω 2

θ

3

4 − 4θ 2 + θ 4 + 4θ 2

a=

bω 2

θ

3

4 +θ 4 �


248

PROBLEM 11.176 A particle moves along the spiral shown. Knowing that θ� is constant and denoting this constant by ω , determine the magnitude of the acceleration of the particle in terms of b, θ , and ω.

SOLUTION Logarithmic spiral.

r = ebθ dr = bebθ θ� dt �� r = bebθ θ�� + b 2 ebθ θ� 2 = bebθ (θ�� + bθ� 2 ) r� =

ar = �� r − rθ� 2 = bebθ (θ�� + bθ� 2 ) − ebθ θ� 2 a = rθ�� + 2r�θ� = ebθ θ�� + 2(bebθ θ�)θ� θ

Since θ� = ω = constant, θ�� = θ , and we write ar = bebθ (bω 2 ) − ebθ ω 2 = ebθ (b 2 − 1)ω 2 aθ = 2bebθ w2 a = ar2 + aθ2 = ebθ ω 2 (b2 − 1)2 + (2b) 2 = ebθ ω 2 b4 − 2b 2 + 1 + 4b 2 = ebθ ω 2 b 4 + 2b 2 + 1 = ebθ ω 2 (b2 + 1) 2 = ebθ ω 2 (b2 + 1) a = (1 + b 2 )ω 2 ebθ �


249

250

PROBLEM 11.177 Show that r� = hφ� sin θ knowing that at the instant shown, step AB of the step exerciser is rotating counterclockwise at a constant rate φ�.

SOLUTION From the diagram r 2 = d 2 + h 2 − 2dh cos φ

Then Now or

2rr� = 2dhφ� sin φ r d = sin φ sin θ r=

d sin φ sin θ

Substituting for r in the expression for r� � d sin φ � � � sin θ � r� = dhφ sin φ � �

or

r� = hφ� sin θ

Q.E.D.

�

Alternative solution. First note

α = 180° − (φ + θ )

Now

v = v r + vθ = r�e r + rθ�eθ

With B as the origin vP = dφ�

( d = constant � d� = 0)


250

PROBLEM 11.177 (Continued) With O as the origin

(vP )r = r�

where

(vP )r = vP sin α

Then Now or

r� = dφ� sin α h d = sin α sin θ d sin α = h sin θ


251

252

PROBLEM 11.178 The motion of a particle on the surface of a right circular cylinder is defined by the relations R = A, θ = 2π t , and z = At 2 /4, where A is a constant. Determine the magnitudes of the velocity and acceleration of the particle at any time t.

SOLUTION We have

R=A

θ = 2π t

z=

1 2 At 4

Then

R� = 0

θ� = 2π

z� =

1 At 2

and

�� = 0 R

θ�� = 0

�� z=

1 A 2

Now

v 2 = vR2 + vθ2 + vz2 0 = (� R� ) 2 + ( Rθ�) 2 + ( z� ) 2 �1 � = 0 + ( A + 2π ) 2 + � At � �2 � 1 � � = A2 � 4π 2 + t 2 � 4 � �

2

v=

or and

1 A 16π 2 + t 2 � 2

a 2 = aR2 + aθ2 + az2 0 0 �� 0 − Rθ� 2 ) 2 + ( R� � = (R θ�� + 2� R� θ ) 2 + ( �� z )2

�1 � = [ − A(2π ) 2 ]2 + 0 + � A � �2 � 1� � = A2 �16π 4 + � 4� �

2

a=

or

1 A 64π 4 + 1 � 2


252

PROBLEM 11.179 The three-dimensional motion of a particle is defined by the cylindrical coordinates (see Figure 11.26) R = A/(t + 1), θ = Bt , and z = Ct/(t + 1). Determine the magnitudes of the velocity and acceleration when (a) t = 0, (b) t = ∞.

SOLUTION A , t +1

We have

R=

Then

R� = −

�� = R

and

θ = Bt ,

A , (t + 1) 2

2A (t + 1)3

z=

θ� = B

θ�� = 0

Ct t +1 z� = C

(t + 1) − t (t + 1)2

=

C (t + 1) 2

�� z=−

zC (t + 1)3

Now

v 2 = (vR ) 2 + (vθ )2 + (vz ) 2 = ( R� ) 2 + ( Rθ�) 2 + ( z� ) 2

and

a 2 = ( aR )2 + (aθ ) 2 + (az ) 2 0 �� + 2 R�θ�)2 + ( �� − Rθ� 2 ) 2 + ( Rθ� z )2 = (R

(a)

At t = 0:

R= A R = −A

θ� = B

�� = 2 A R

Then

z� = C �� z = −2C

v 2 = (− A) 2 + ( AB) 2 + (C ) 2 v = (1 + B 2 ) A2 + C 2 �

or and

a 2 = (2 A − AB 2 ) 2 + [ z (− A)( B)]2 + (−2C ) 2 �� 1 � 2 C2 � = 4 A2 ��1 − B 2 � + B 2 + 2 � A �� 2 � �� 1 � � = 4 � � 1 + B 4 � A2 + C 2 � �� 4 � � � 1 � a = 2 � 1 + B 4 � A2 + C 2 � � 4 �

or


253

254


As t → ∞ :

R=0 R� = 0 �� = 0 R

θ� = B

z� = 0 �� z=0

v=0 � a=0 �

and


254

PROBLEM 11.180* For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis.

SOLUTION First note that the vectors v and a lie in the osculating plane. Now

r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k

Then

v=

dr = R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k dt

and

a=

dv dt

(

)

= R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i

(

)

+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k = ωn R[−(2sin ωn t + ωn t cos ωn t )i + (2cos ωn t − ωn t sin ωn t )k ]

It then follows that the vector ( v × a) is perpendicular to the osculating plane.

i j k ( v × a) = ωn R R(cos ωn t − ωn t sin ωn t ) c R(sin ωn t + ωn t cos ωn t ) −(2sin ωn t + ωn t cos ωn t ) 0 (2cos ωn t − ωn t sin ωn t ) = ωn R{c(2 cos ωn t − ωn t sin ωn t )i + R[−(sin ωn t + ωn t cos ωn t )(2sin ωn t + ωn t cos ωn t ) − (cos ωn t − ωn t sin ωn t )(2 cos ωn t − ωn t sin ωn t )] j + c(2sin ωn t + ωn t cos ωn t )k

(

)

= ωn R �c(2 cos ωn t − ωn t sin ωn t )i − R 2 + ωn2 t 2 j + c(2sin ωn t + ωn t cos ωn t )k � � �


255

256


The angle α formed by the vector ( v × a) and the y axis is found from cos α =

Where

( v × a) ⋅ j | ( v × a) || j |

|j| =1

(

( v × a) ⋅ j = −ωn R 2 2 + ωn2 t 2

)

(

|( v × a) | = ωn R ��c 2 (2 cos ωn t − ωn t sin ωn t ) 2 + R 2 2 + ωn2 t 2 �

)

2

1/ 2

+ c 2 (2sin ωn t + ωn t cos ωn t )2 ��

(

)

2 �1/2

(

) ��

= ωn R ��c 2 4 + ωn2 t 2 + R 2 2 + ωn2t 2 �

Then

( ) ω R ��c ( 4 + ω t ) + R ( 2 + ω t ) �� −R ( 2 + ω t ) = �c 4 + ω t + R 2 + ω t � ) ( ) �� ( −ωn R 2 2 + ωn2t 2

cos α =

2

n

2 2 n

2

2 2 n

2

1/2

2 2 n

2

2 2 n

2

2 2 n

2 1/2

The angle β that the osculating plane forms with y axis (see the above diagram) is equal to

β = α − 90° Then

cos α = cos ( β + 90°) = −sin β − sin β =

Then

tan β =

( ) ) + R ( 2 + ω t ) ��

− R 2 + ωn2t 2

(

�c 2 4 + ω 2t 2 n ��

(

R 2 + ωn2 t 2

2

2 2 n

2

1/ 2

)

c 4 + ωn2 t 2

(

� R 2 + ωn2t 2 β = tan � � c 4 + ω 2t 2 n � −1

or

) ��


256

PROBLEM 11.181* Determine the direction of the binormal of the path described by the particle of Problem 11.96 when (a) t = 0, (b) t = π /2 s.

SOLUTION Given:

)

(

r = ( At cos t )i + A t 2 + 1 j + ( Bt sin t )k r − m, t − 5; A = 3, B − 1

First note that eb is given by eb =

v×a |v ×a |

)

(

Now

r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k

Then

v=

dr dt 3t

= 3(cos t − t sin t )i +

j + (sin t + t cos t )k

(

t

)

2 dv t 2 +1 j a= = 3( − sin t − sin t − t cos t )i + 3 t + 12− t dt t +1 + (cos t + cos t − t sin t )k 3 j + (2 cos t − t sin t )k = −3(2sin t + t cos t )i + 2 (t + 1)3/2

and

(a)

t2 +1

At t = 0:

v = (3 m/s)i a = (3 m/s2)j + (2 m/s2)k

Then

and Then

v × a = 3i × (3j + 2k ) = 3(−2 j + 3k ) | v × a | = 3 (−2) 2 + (3) 2 = 3 13

eb =

3( −2 j + 3k ) 3 13

cos θ x = 0

or

=

cos θ y = −

θ x = 90°

1 13 2

(−2 j + 3k )

13

θ y = 123.7°

cos θ z =

3 13

θ z = 33.7°

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257

258


(b)

At t =

π 2

s:

v=−

3π m/s i + 2

a = −(6 m/s2) i +

Then

i 3π v×a = − 2

π2 +4

m/s j + (1 m/s)k

π 24 m/s2 j − m/s2 k 3/2 2 (π + 4) 2

j 3π (π 2 + 4)1/ 2 24 2 (π + 4)3/ 2

−6

=−

3π

k 1 −

π 2

3π 2 24 3π 2 i j + − 6 + 4 2(π 2 + 4)1/2 (π 2 + 4)3/ 2

+ −

36π 18π k + 2 3/ 2 (π + 4) (π + 4)1/ 2 2

= −4.43984i − 13.40220 + 12.99459k

and

Then

or

| v × a | = [(−4.43984) 2 + ( −13.40220)2 + (12.99459)2 ]1/ 2 = 19.18829

1 (−4.43984i − 13.40220 j + 12.99459k ) 19.1829 4.43984 13.40220 12.99459 cos θ x = − cos θ y = − cos θ z = 19.18829 19.18829 19.18829 eb =

θ x = 103.4°

θ y = 134.3°

θ z = 47.4°


258

PROBLEM 11.182 The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION x = 2t 3 − 15t 2 + 24t + 4 dx = 6t 2 − 30t + 24 dt dv = 12t − 30 a= dt v=

so

(a)

0 = 6t 2 − 30t + 24 = 6 (t 2 − 5t + 4)

Times when v = 0.

(t − 4)(t − 1) = 0

(b)

t = 1.00 s, t = 4.00 s

�

Position and distance traveled when a = 0. a = 12t − 30 = 0

t = 2.5 s

x2 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4

so

x = 1.50 m �

Final position For

0 � t � 1 s, v � 0.

For

1 s � t � 2.5 s, v � 0.

At

t = 0,

At

t = 1 s, x1 = (2)(1)3 − (15)(1) 2 + (24)(1) + 4 = 15 m

x0 = 4 m.

Distance traveled over interval: x1 − x0 = 11 m For

1 s � t � 2.5 s,

v�0

Distance traveled over interval | x2 − x1 | = |1.5 − 15 | = 13.5 m

Total distance:

d = 11 + 13.5

d = 24.5 m �


259

260

PROBLEM 11.183 The acceleration of a particle is defined by the relation a = −60 x −1.5, where a and x are expressed in m/s 2 and meters, respectively. Knowing that the particle starts with no initial velocity at x = 4 m, determine the velocity of the particle when (a) x = 2 m, (b) x = 1 m, (c) x = 100 mm.

SOLUTION v

We have When x = 4 m, v = 0:

0

vdv =

�

x

4

(−60 x −1.5 )dx

1 2 v = 120[ x −0.5 ]4x 2

or

� 1 1� v 2 = 240 � − � � x 2�

or (a)

�

v

dv = a = −60 x −1.5 dx

When x = 2 m:

� 1 1� v 2 = 240 � − � � 2 2� v = −7.05 m/s �

or (b)

When x = 1 m:

� 1� v 2 = 240 �1 − � � 2� v = −10.95 m/s �

or (c)

When x = 0.1 m:

� 1 1� v 2 = 240 � − � � 0.1 2 � v = −25.3 m/s �

or


260

PROBLEM 11.184 A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 270 m/s and travels 100 mm before coming to rest. Assuming that the velocity of the projectile is defined by the relation v = v0 − kx, where v is expressed in m/s and x is in meters, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 97.5 mm into the resisting medium.

SOLUTION First note When x = 0.1 m, v = 0:

0 = (270 m/s) − k (0.1 m)

or

k = 2700

(a)

1 s

We have

v = v0 − kx

Then

a=

or

a = −k (v0 − kx)

At t = 0:

1 a = 2700 (270 m/s − 0) s

dv d = (v0 − kx) = −kv dt dt

a0 = −729 × 103 m/s2

or (b)

dx = v = v0 − kx dt

We have At t = 0, x = 0: or

x 0

dx = v0 − kx

t 0

dt

1 − [ln(v0 − kx)]0x = t k

or

t=

v0 1 1 1 = ln ln k v0 − kx k 1 − vk x 0

When x = 0.0975 m:

t=

1 ln 1− 2700 1s

2700 1/s 270 m/s

1 (0.0975 m) t = 1.366 × 10−3 s

or

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261

262

PROBLEM 11.185 A freight elevator moving upward with a constant velocity of 1.8 m/s passes a passenger elevator which is stopped. Four seconds later, the passenger elevator starts upward with a constant acceleration of 0.72 m/s2. Determine (a) when and where the elevators will be at the same height, (b) the speed of the passenger elevator at that time.

SOLUTION (a)

For t t

0:

y F = 0 + vF t

4 s:

yP = 0 + 0(t − 4) +

1 aP (t − 4) 2 2

yF = yP

When

(1.8 m/s) t =

1 (0.72 m/s2)(t − 4)2 2

Expanding and simplifying t 2 − 13t + 16 = 0 t = 1.3765 s and t = 11.6235 s

Solving Most require t At t = 11.6235 s:

t = 11.62 s

4s yF = (1.8 m/s)(11.6235 s)

yF = yP = 20.9 m

or (b)

For t

4 s:

At t = 11.6235 s:

vP = 0 + aP (t − 4) vP = (0.72 m/s2)(11.6235 − 4)s vP = 5.5 m/s

or

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262

PROBLEM 11.186 Block C starts from rest at t = 0 and moves upward with a constant acceleration of 25 mm/s2. Knowing that block A moves downward with a constant velocity of 75 mm/s, determine (a) the time for which the velocity of block B is zero, (b) the corresponding position of block B.

SOLUTION The cable lengths are constant. L1 = 2 yC + 2 yD + constant L2 = y A + yB + ( yB − yD ) + constant

Eliminate yD . L1 + 2 L2 = 2 yC + 2 yD + 2 y A + 2 yB + 2( yB − yD ) + constant 2(yC + y A + 2yB ) = constant

Differentiate to obtain relationships for velocities and accelerations, positive downward. vC + v A + 2vB = 0

(1)

aC + a A + 2aB = 0

(2)

Initial velocities.

(vC )0 = 0, (v A )0 = 75 mm/s

From (1),

1 (vB )0 = − [(vC )0 + (vA )0 ] = −37.5 mm/s 2

Accelerations.

aC = −25 mm/s a A = 0

From (2),

1 aB = − ( aC + a A ) = 12.5 mm/s 2

Motion of block B. vB = (vB )0 + aB t = −37.5 + 12.5t ∆ y B = ( vB ) 0 t +

(a)

Time at vB = 0.

(b)

Corresponding position.

1 aB t 2 = −37.5t + 6.25t 2 2

−37.5 + 12.5t = 0

t = 3.00 s �

∆yB = (−37.5)(3.00) + (6.25)(3.00)2

∆yB = −56.25 mm �


263

264

PROBLEM 11.187 The three blocks shown move with constant velocities. Find the velocity of each block, knowing that the relative velocity of A with respect to C is 300 mm/s upward and that the relative velocity of B with respect to A is 200 mm/s downward.

SOLUTION From the diagram Cable 1:

y A + yD = constant

Then

v A + vD = 0

Cable 2:

(1)

( yB − yD ) + ( yC − yD ) = constant vB + vC − 2vD = 0

Then

(2)

Combining Eqs. (1) and (2) to eliminate vD 2v A + vB + vC = 0

(3)

Now

v A/C = v A − vC = −300 mm/s

(4)

and

vB/A = vB − v A = 200 mm/s

(5)

Then

(3) + (4) − (5) � (2v A + vB + vC ) + (v A − vC ) − (vB − v A ) = (−300) − (200)

v A = 125 mm/s ↑ �

or vB − (−125) = 200

and using Eq. (5)

v B = 75 mm/s ↓ �

or −125 − vC = −300

Eq. (4)

vC = 175 mm/s ↓ �

or


264

PROBLEM 11.188 An oscillating water sprinkler at Point A rests on an incline which forms an angle α with the horizontal. The sprinkler discharges water with an initial velocity v 0 at an angle φ with the vertical which varies from −φ0 to +φ0 . Knowing that v0 = 9 m/s, φ0 = 40°, and α = 10°, determine the horizontal distance between the sprinkler and Points B and C which define the watered area.

SOLUTION First note (v0 ) x = v0 sin φ = (9 m/s) sin φ (v0 ) y = v0 cos φ = (9 m/s) cos φ

Also, along incline Cab y = x tan10°

Horizontal motion. (Uniform) x = 0 + (v0 ) x t = (9 sin φ ) t or t =

x 9 sin φ

Vertical motion. (Uniformly accelerated motion) y = 0 + (v0 ) y t −

1 2 1 gt = (9 cos φ )t− gt 2 2 2

Substituting for t x 1 x − g y = (9 cos φ ) 9 sin φ 2 9 sin φ =

At B:

φ = 40°, x = d B :

d B tan10° =

g x2 x − tan φ 162 sin 2 θ

( g = 9.81 m/s2)

dB 9.81 d B2 − tan 40° 162 sin 2 40° dB = 7 m

or At C:

2

φ = −40°, x = − dC :

− dC tan10° =

− dC 9.81 (− dC ) 2 − tan (−40°) 162 sin 2 (−40°) dC = 9.5 m

or

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265

266

PROBLEM 11.189 As the driver of an automobile travels north at 25 km/h in a parking lot, he observes a truck approaching from the northwest. After he reduces his speed to 15 km/h and turns so that he is traveling in a northwest direction, the truck appears to be approaching from the west. Assuming that the velocity of the truck is constant during the period of observation, determine the magnitude and the direction of the velocity of the truck.

SOLUTION We have

vT = v A + vT/A

Using this equation, the two cases are then graphically represented as shown.

From the diagram (vT ) x = 25 − 15sin 45° = 14.3934 km/h (vT ) y = 15sin 45° = 10.6066 km/h

vT = 17.88 km/h

36.4° �


266

PROBLEM 11.190 The driver of an automobile decreases her speed at a constant rate from 72 to 48 km/h over a distance of 225 m along a curve of 450-m radius. Determine the magnitude of the total acceleration of the automobile after the automobile has traveled 150 m along the curve.

SOLUTION First note

v1 = 72 km/h = 20 m/s v2 = 48 km/h = 13.3 m/s

We have uniformly decelerated motion v 2 = v12 + 2at ( s − s1 )

When v = v2 :

(13.3 m/s)2 = (20 m/s)2 + 2at(225 m)

or

at = −0.496 m/s2

Then when ∆s = 500 ft:

v 2 = (20 m/s)2 + 2(−0.496 m/s2)(150 m) = 251.2 m2/s2

Now

an =

v2

ρ

251.2 m2/s2 450 m = 0.558 m/s2 =

Finally

a 2 = at2 + an2 = (−0.496 m/s2)2 + (0.558 m/s2)2 a = 0.75 m/s2

or

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267

268

PROBLEM 11.191 A homeowner uses a snowblower to clear his driveway. Knowing that the snow is discharged at an average angle of 40° with the horizontal, determine the initial velocity v0 of the snow.

SOLUTION First note

(vx )0 = v0 cos 40° (v y )0 = v0 sin 40°


At B:

4 = (v0 cos 40°) t or t B =

4 v0 cos 40°


At B:

1 2 gt 2

0.4 = (v0 sin 40°) tB −

( g = 9.81 m/s2)

1 2 gtB 2

Substituting for t B 4 1 4 0.4 = (v0 sin 40°) − g 2 v0 cos 40° v0 cos 40°

or

v02 =

1 2

2

(9.81)(16)/ cos 2 40° −0.4 + 4 tan 40° v0 = 6.7 m/s

or

PROPRIETARY MATERIAL. MATERIAL. © © 2010 2009 The The McGraw-Hill McGraw-Hill Companies, Companies, Inc. All rights this Manual Manual may may be be displayed, displayed, PROPRIETARY Inc. All rights reserved. reserved. No No part part of of this reproduced or or distributed distributed in in any any form form or or by by any any means, means, without without the the prior prior written written permission permission of of the the publisher, publisher, or or used used beyond beyond the the limited limited reproduced student using using this this Manual, Manual, distribution to to teachers teachers and and educators educators permitted permitted by by McGraw-Hill McGraw-Hill for distribution for their their individual individual course course preparation. preparation. If If you you are are aa student you are are using using it it without you without permission. permission. 269

268

PROBLEM 11.192 From measurements of a photograph, it has been found that as the stream of water shown left the nozzle at A, it had a radius of curvature of 25 m. Determine (a) the initial velocity vA of the stream, (b) the radius of curvature of the stream as it reaches its maximum height at B.

SOLUTION (a)

We have

(a A ) n =

v A2

ρA

or

�4 � v A2 = � (9.81 m/s 2 ) � (25 m) 5 � �

or

v A = 14.0071 m/s

vA = 14.01 m/s

(b)

We have Where

Then

( aB ) n =

vB2

ρB

vB = ( v A ) x =

ρB

36.9° �

4 vA 5

( 4 × 14.0071 m/s ) = 5

2

9.81 m/s 2

ρ B = 12.80 m �

or


269

270

PROBLEM 11.193 At the bottom of a loop in the vertical plane, an airplane has a horizontal velocity of 150 m/s and is speeding up at a rate of 25 m/s2. The radius of curvature of the loop is 2000 m. The plane is being tracked by radar at O. What are the recorded r , θ� and θ�� for this instant? values of r�, ��

SOLUTION Geometry. The polar coordinates are � 600 � v = (800) 2 + (600) 2 = 1000 m θ = tan −1 � � = 36.87° � 800 �

Velocity Analysis.

v = 150 m/s →

vr = 150 cos θ = 120 m/s

vθ = −150sin θ = −90 m/s r� = 120 m/s �

vr = r� v 90 vθ = rθ� θ� = θ = − 1000 r

θ� = −0.0900 rad/s � Acceleration analysis.

at = 25 m/s 2 an =

v2

ρ

=

(150) 2 = 11.25 m/s 2 2000

a = 25 m/s 2 → +11.25 m/s 2 ↑ = 27.41 m/s 2

24.23°

β = 24.23° θ − β = 12.64°


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PROBLEM 11.193 (Continued) ar = a cos (θ − β ) = 27.41 cos 12.64° = 26.74 m/s 2 aθ = − a sin (θ − β ) = −27.41 sin 12.64° = −6.00 m/s 2 ar = �� r − rθ� 2 �� r = ar + rθ� 2 �� r = 26.74 + (1000)(0.0900) 2

aθ = rθ�� + 2r�θ� a 2r�θ� θ�� = θ − r r −6.00 (2)(120)(−0.0900) = − 1000 1000

r�� = 34.8 m/s 2 �

θ�� = −0.0156 rad/s 2 �


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DYNAMICS solution chapter11

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