CHAPTER 1
1.1 Using Using the definition definition of accelerat acceleration: ion: α=
∆v ∆t
− 0 = 88 ft/ ft/sec − 0 − 609mph = 9.57 ft/ ft/sec 9.2 sec .2 − 0
2
1.2 Differentiate Differentiate x(t) to obtain the velocity: v (t) = x˙ (t) =
−10t 10t + 88 ft/sec. ft/sec.
Differentiating again yields the acceleration: 2
¨(t) = a(t) = x
−10 ft/sec . So v (t) = 0 = −10t 10t + 88.
Solving this for t yields that: v (t) = 0 at t = 8.8 sec. 1.3 Evaluati Evaluating ng x at zero yields x(0) = 5 m. Differen Differentiati tiating ng yields yields x˙ (t) = v (t) = 3t2 so that v (0) = Likewise
−2 m/s.
−2
¨(t) = 6t a(t) = x so that a(0) = 0. Now at t = 3 sec, x(3) = 27
− 2(3) + 5 = 26 m, m/s v(3) = 27 − 2 = 25 m/ and a(3) = 18 m/ m/s2 .
Since the velocity changes sign during this interval, the particle has doubled back and to compute the total distance traveled during the interval you must compute how far it travels before it changes direction and then add this to the distance traveled after the particle has changed direction. The particle changes direction when the velocity is zero, or at the value of t for which v (t) = 3t2 or at time
− 2 = 0,
t = 0.8165. The particle first moves from (0.8165) = 3. 3.9 m or a distance of 1.1 m. x(0) = 5 m to x(0.
1
It then changes direction and moves from 3.9 m to x(3) = 26 m. Thus it travels a total distance of (26
− 3.9) + 1.1.1 = 1.1 + 1.1.1 + 21 = 23. 23.2 m.
1.4 This again is straightforward straightforward differentiation: differentiation: v (t) = x˙ (t) = 2t
− 2.
This is zero when t satisfies: 2t
− 2 = 0 or t = 1 sec.
Next:
¨(t) = v˙ (t) = 2 ft/sec 2 which is constant acceleration. a(t) = x Alternately, the distance traveled can be computed directly from integrating the absolute value of the velocity: d=
3 0
2
|3t − 2|dt = 23.23.178 m
1.5 Solution: Solution: v (t) = x˙ (t) = 6 cos 2t m/s. a(t) = x ¨(t) = v˙ (t) = 12 sin( sin(22t) m/sec2 . Setting 12 sin( sin(22t) = 0, yields 2t 2 t = π, or t = 0, π/2 π/2 s, π, 3π/2 π/ 2...nπ/2, ...nπ/2, for the times for the acceleration to hit zero.
−
−
1.6 From the definition, straightforward differentiation differentiation yields: xA = (3t (3t2 + 6t 6t) ft so the vA = (6t (6t + 6) ft/sec ft/sec and aA = 6 ft/sec2 . Likewise, xB = 3t3 + 2t 2t so that vB = (9t (9t2 + 2) ft/sec ft/sec and aB = 18t 18t ft/sec2 a)Thus at t = 1 sec: xA (1) = 9 ft, vA (1) = 12 ft/sec, and aA (1) = 6 ft/sec2 xB (1) = 5 ft, vB (1) = 11 ft/sec, and aB (1) = 18 ft/sec 2 So that A is ahead of B of B and has the largest velocity, but B is accelerating faster the A. b) Now at t = 2 sec: xA (2) = 24 ft, vA(2) = 18 ft/sec, and aA(2) = 6 ft/sec2 xB (2) = 28 ft, vB (2) = 38 ft/sec, and aB (2) = 36 ft/sec 2 Now B is ahead of A, and has larger velocity and acceleration
2
c) To find where A and B have moved the same distance, let xA (t) = xB (t) and solve for t. This yields 3t2 + 6t 6t = 3t3 + 2t, 2t, or t2 Solving yields t = +1. +1.758s 758s and t =
− t − 4/3 = 0
and and t = 0
−1.758s.
Here we are interested in the positive value of time so that (1.758) = xB (1. (1.758) = 3(1. 3(1.758)2 +6(1. +6(1.758) = 3(1. 3(1.759)3 +2(1. +2(1.758) = 19. 19.81 ft. xA (1.
1.7 Note that this is an inverse inverse problem. Straightf Straightforw orward ard differeniation differeniation yields: yields: sin 10t 10t + e−t 10cos10t 10cos10t) = 3e−t (10cos10t (10cos10t v (t) = 3( e−t sin
− sin sin 10t 10t) sin 10t 10t − 10cos10t 10cos10t) − 3e− (10cos10t (10cos10t − sin sin 10t 10t) a(t) = 3e− (−10 sin = −3e− (99sin10t (99sin10t − 20cos10t 20cos10t) −
t
2
t
t
1.8 From the definition definition x(t) = t3 so that:
2
− 6t − 15t 15t + 40 ft
x˙ = v = 3t2
− 12t 12t − 15 ft/sec, ft/sec, and x¨ = a = 6t − 12 ft/sec. ft/sec. a) v (t) = 0 requires 3t 3 t − 12t 12t − 15 = 0 or t − 4t − 5 = 0. 2
2
Solving for t yields: t=
−1 and 5.
Taking the positive value of t, the velocity is zero at t = 5 sec. b) At t = 5 sec, x(t) = x(5) = (5)3 At rest t = 0,
2
− 6(5) − (15)(5) (15)(5) + 40 = −60 ft.
x(0) = 40. Then the particle has moved from 40 to -60 or 40 + 60 = 100 ft. c) a(5) = (6)(5)
2
− 12 = 18 ft/ ft/sec .
1.9 Note: Note: The purpose purpose of this this proble problem m is to hit home the idea that the distanc distancee trave traveled led and the displaceme displacement nt are differen different. t. This problem problem is easiest easiest to solve using using comput computati ationa onall softw software as it invo involv lves es plots. plots. Studen Students ts could could also also use a symbolic processor to compute the derivatives, although it would be a little over 3
kill and they must be b e reminded that simple derivations should be something they can do in their head, on tests, while complicated derivatives are more accurately done with softwar software. e. The plots are also easy to sketc sketch h by hand, but if they are inexperienced at plotting using software, it is best to start them off with some simple plots. a) v (t) = x˙ = 0.6cos2t 6cos2t ¨= a(t) = x
−1.2sin2t 2sin2t.
b) 3sin2t 3sin2t travels a distance of 0.3m in the time 0 to π4 sec and back another 0.3m returning back to the origin from π4 to π2 s. So the total distance traveled is 0.3 + 0. 0 .3 = 0.6m (maybe look at the plot first). c) The position of the mass at t = π2 s however is x
= 0.3sin = 0. π 2
2π 2
Note the position at any point is not always the distance traveled, which in b) is shown to be 0.6m.
π
t
0 , .01 ..
x t v t a t
2 . 0.3 sin 2 . t 0.6 . cos 2 . t 1.2 . sin 2 . t
1 x t 0
v t a t
0.5
1
1
2 t
FIGURE S1.9
4
1.5
2
1.10 Differentiation yields v(t) = 88.92sin0.26t ft, and a(t) = 23.12cos0.26t ft/sec2 . The plot of each follows.
t
0 , 0.1 .. 12.2
x t
342 . 1
cos 0.26 . t
a t
23.12. cos 0.26. t
v t
88.92. sin 0.26. t
1000
x t
500
v t a t 0
2
4
6
8
10
12
14
500 t
FIGURE S1.10
1.11 The following code in Matlab computes the velocity from the displacement data and plots it: x=[8 9 11 13 14 15 17 18 22 27 32 37 41 44 46 48 49 49 48 47 46]; t=0;,01:02; n=length (x); v=0*x; dt=.01; for i=1:n-1 v(i+1)=(x(i+1)-x(i)/dt; end v plot(t,v),xlabel(‘t*dt or elapsed time’), title (‘velocity versus time’)
5
This produces the following output: v= Columns 1 through 12 0 100 200 200 100 100 200 100 400 500 500 500 Columns 13 through 21 400 300 200 200 100 0 -100 -100 -100 And the following plot:
velocity versus time 500
400
300
200
100
0
-100
0
0.05
0.1 0.15 t*dt or elapsed time
0.2
FIGURE S1.11a
The following produces the corresponding acceleration: EDU>>a=0*v; EDU>>for i=1:n-1 a(i+1)=(v(i+1)-v(i))/dt; end EDU>a a= Columns 1 through 6 0 10000 10000 0 -10000 0 Columns 7 through 12 10000 -10000 30000 10000 0 0 Columns 13 through 18 -10000 -10000 -10000 0 -10000 -10000 6
0.25
Columns 19 through 21 -10000 0 0 EDU>>plot(t,a),xlabel(‘elapsed time’), title (‘acceleration ersus time’)
3
x 10
acceleration versus time
4
2.5
2
1.5
1
0.5
0
-0.5
-1
0
0.05
0.1
0.15 elapsed time
0.2
0.25
FIGURE S1.11b
1.12 Consider the following Matlab code which uses the central difference to compute the velocity: x=[8 9 11 13 14 15 17 18 22 27 32 37 41 44 46 48 49 49 48 47 46]; t=0:,.01:0.2; n=length (x); v=0*x; dt=.01; for i=1:n-1 v(i)=(x(i+1)-x(i-1))/(2*dt); end v plot(t,v),xlabel(’t*dt or elapsed time’),title(velocity versus time’)
7
This results in the following values: v= Columns 1 through 12 0 150 200 150 0 100 150 150 250 0 450 500 500 450 Columns 13 through 21 350 250 200 150 50 -50 -100 -100 0 And the following plot: velocity versus time 500
400
300
200
100
0
-100
0
0.05
0.1 0.15 t*dt or elapsed time
0.2
0.25
The following is the Mathcad code for solving this problem: i
1. . 19
v
xi
i
1
∆ t=0.01
xi
1
1
2. ∆t
600
400 v i
200
0
200 0
0.05
0.1 t
0.15
velocity in mm/s versus time in s
8
0.2
1.13 Solution: 2
y(t) =
−4.905t + 20t (m) ˙ = −9.81t + 20 (m/s) v = y(t) a = v˙ = −9.81 (m/s)
The position, velocity and acceleration at t = 5 are y(5) =
−22.625 m, v(5) = −29.05 m/s, a(5) = −9.81 m/s
2
Note that the ball returns to its initial state of y(5) = 0 when t satisfies 4.905t2 + 20t = 0 or, t = 20/4905 = 2.07 sec.
−
Then to obtain the total distance traveled by the ball, we need to calculate when the ball changes direction, i.e., when v(t) = 0. v(t) = 0 =
−9.81t + 20 or t = 2.0395.
From t = 0 to 2.039 sec. the ball travels a distance of y = 20.39 ft. It then travels back past zero (the top of the building) another 20.39 ft. to y(0) = 0. It then travels on a distance of y(5) =
−22.625, beyond zero.
Thus the total distance traveled by the ball is 20 .39+20.39+22.625 = 63.4 m.
1.14 Solution: x, v and a are given respectively by: x(t) = e−ct sin ωt v(t) =
ct
−ce−
sin ωt + e−ct ω cos ωt
a(t) = c2 e−ct sin(ωt)
ct
2
− 2cωe− cos ωt − ω e− x(0) = 0, v(0) = ω, a(0) = −2cω
ct
sin ωt
1.15 Solution: x(t) = 3t3
2
− 2t
+ 5, x(0) = 5m
˙ = 9t2 v(t) = x(t)
− 4t, v(0) = 0 m/s ˙ = 18t − 4, a(0) = −4 m/s a(t) = v(t)
2
1.16 The velocity and acceleration are respectively: v(t) = 4 t cos(π t)
2
· · · − 2 · t · sin(π · t) · π a(t) = 4 · cos(π · t) − 8 · t · sin(π · t) · π − 2 · t · cos(π · t) · π 2
so that x(0) = v(0) = 0 and a(0) = 4
9
2
1.17 Solution: y(t) = 3t2
− 20, so y(0) = −20 m
˙ = 6t, so v(0) ˙ = 0 m/s v(t) = y(t) a(t) = y¨(t) = 6, so a(0) = 6 m/s2
1.18 Solution: x(t) = exp( 0.1 t) (3cos t cos(2 t) + sin(2 t)), x(0) = 3
− · · · · v(t) = 1.7 · exp(−.1 · t) · cos(2 · t) − 6.1 · exp(−.1 · t) · sin(2 · t), v(0) = 1.7 a(t) = −12.36exp(−0.1t) · cos(2t) − 2.79 · exp(−0.1t) · sin(2t), a(0) = −12.37 1.19 Solution: x(t) = 5 t
· − exp(−t) · 3 · t, x(0) = 0; v(t) = 5 + 3 · exp(−t) · t − 3 · exp(−t), v(0) = 2; a(t) = −3 · exp(−t) · t + 6 · exp(−t), a(0) = 6. 1.20 Solution: There is no derivative at t = 5, however the problem may be split using heaviside functions and differentiated over the two intervals. t
0 , . 01 .. 10
y t
5 . t. Φ 5
t
25
y1 t
5. Φ 5
t
20 . cos
y11 t
20 . π2 . sin
π.t
20 . sin
π
π.t
π.t
.Φ t
π
.Φ t
π
5
.Φ t 5
is the velocity
is the acceleration
200
y11 t y1 t
0
5
200 t
FIGURE S1.20
10
5
10
During the 1 st interv interval al the slope dy/dt is const constan ant. t. From rom the the plot plot y (t) = 20−0 = 5t for t = 0 to 5 sec. In the interval t > 5, y (t) = 25 + A sin(ωt sin(ωt + φ) 4−0 t where ω is the frequency and φ is the plane and A is the amplitude of the sine wave.
−
From the plot A = 20, the period T = 2 sec, so that ω = 2T π = π . To find the phase evaluate y (5) (5) = 25 = 25 + 20(si 20(sin( n(πt πt + φ) so that πt + φ = nπ, nπ, where n is any integer or φ = (n 5)π 5)π, so φ = π will work. Thus:
−
5t 0 5 5 0 5 0 0
2
1.21 Solution: aave =
sin(πt sin(πt
− π)
60 0 mph 5 sec 2
−
=
t>5
60 mph 5 sec
ft mile 1 1 hour = 12 5280 = mile hour sec 3600 sec
(12)(5280) 300
ft/ ft/sec2
= 17. 17.6 ft/ ft/sec
ft 17. 17.6 ft/ ft/sec2 = 17. 17.6 3m = 5.4 m/sec2 .25 sec 2
5.4 m(sec) 9.81 m/sec2
= 0.55 g’s or about 55% of a g.
1.22 One answer: a 4 min mile is near a record speed for trained runners so: v=
1 mile 4 min
=
1 mile 1 min 4 min 60 sec
·
5280 ft mile
= 22 ft/ ft/sec
or about 6.71 meter meter/s, /s, or about 15 mph. On the other hand hand a sprin sprinter ter can cover 100 m dash in 10 seconds, or 10 m/s.
1.23 The Matlab solution follows (see problem 1.11 and 1.12 also): %first assign the data to the vector v v = [0 0.2 0.27 0.3 0.3 0.3 0.3 0.3 0.35 0.4 0.5]; n=length(v); %assign the time step t=0:,0.1:11; dt=0.1; a=0.*v; x=0.*v; % zeros in a and x %using a loop (see statics supplement or student ed. of Matlab)
11
for i=1:n-1 x(i)=(v(i+1)+v(i))*(0.1)/2 end for i=2:n a(i)=(v(i-1)-v(i))/0.1 end %to see the result plotted use the following plot(t,x), xlabel (‘t*dt or elapsed time’), title (‘[position vs time’) plot(t,a), xlabel (‘t*dt or elapsed time’), title (‘acceleration vs time’) Note these plots are not given but appear in the text. 1.24 Solution: Here is the Mathcad solution. soluti on. From the plot estimate the following data: x0 0 x1 0 x2 0 x3 .02 x 4 .04 x 11 0 x5
0.38 x 9 0.5 x 10 These are measured measured every 0.1 sec. Thus the velocity velocity becomes: becomes: n 0 , 1 .. 10 vn
.1 x 6
x n
0.18 x 7
xn
1
0.1
0.25 x 8
0.7
Which is plotted below
5
0 v
n 5
10
0
0. 2
0.4
0. 6
0. 8
1
n. 0.1
Next estimate the acceleration from the velocity: an
vn
vn
1
which is plotted below
0.1 100
a n
0
100
0
0. 2
0. 4
0.6 0.1 . n
12
0.8
1
Next the equivalent Matlab code is given (without the plots as they look the same). This is the same type of code used in problems 1.11, 1.12 and 1.23. x=[0 0 0 0.02 0.04 0.1 0.18 0.25 0.38 0.7]; t=0:0.1:10; dt=0.1; n=length(x); v=0*x; a=0*v; for i=1:n-1 v(i)=(x(i+1)-x(i))/0.1 a(i)=(v(i+1)-v(i))/0.1 end plot (t,v), title (‘velocity versus time’) plot (t,a), title (‘acceleration versus time’)
1.25 This This is of second second order and is linear linear x(t). The term sin(πt sin( πt)) is nonlinear but in t, not x.
1.26 This is first order and linear in v (t).
1.27 This is first order and linear in v (t)
1.28 This is second order in θ and nonlinear because of the term sin θ(t).
1.29 This is second order in θ(t) and nonlinear because of the terms (dθ/dt ( dθ/dt)) dθ/dt and sin θ(t).
|
1.30 This is first order in v and nonlinear because of the term v 2 .
1.31 This is second order and linear in x(t).
1.32 This is still linear of second order in x(t).
13
|
1.33 Solution: a(t) = 3t2
− 4t, x(0) = 0,0, v(0) = 0 (3t − 4t)t = t − 2t ft/ ft/sec v (t) = (3t x(t) = (t − 2t dt = t − t ft t 0 t 0
2
3
2
3
2
1 4 4
2 3 3
The plot from Mathcad is: x(t): = 0.25 t
2 - 2 _ 3 t 3
t
0 , 0. 0 . 01 . . 2
0
x t
1
2 0
1 t
2
FIGURE S1.33
1.34 Solution: 40t cos πt a(tdv) ==40t 40x 40x cos πxdx v(tdx) ==3 3+t + [cos πt(cos+πxtπ sin+ xππtsin− 1]πx − 1)dx 1)dx v 3
t 0
40 π2
x 5
t 0
40 π2
3t + x(t) = 5 + 3t (3t x(t) = 5 + (3t
40 sin πt π3 40 ) + π403 πt
−
−
40 πt
+
40 π3
[2 sin sin πt
[sin πx
− πt cos πx] πx]
14
t 0
− πx cos πx] πx]
The following is typed in Mathcad to produce the desired plot t
0 , .01 .. 10
x t
5
3
40 π
2
.t
40 π
3
. 2 . sin
π
.t
π
. t . cos
π
t
50
x t
0
50
0
5 t
10
The equivalent Matlab code is: sys t x=5+(3-40/pin2)*t+(40/pin3)*(2*sin(pi*t)-pi*t*cos(pi*t)) ezplot(x,[0,10]) (the plot is suppressed here)
1.35 For the car t0 = 0, x0 = 0, vf = 60 mph = 88 ft/sec, tf = 6. For constant acceleration equation (1.28) yields at = vf or acar = 88 = 14.67 ft/sec2 . In 6 terms of g’s, a = 14.67 = 46%g. For the sprinter t0 = 0, x0 = 0, vf = 10 m/s 32.2 and xf = 15 m. From equation (1.30) vf 2 v02 102 = a= 2(xf x0 ) 2(15
− −
In terms of g, a =
3.33 9.81
− 0 = 3.33 m/s − 6)
2
= 34%g.
1.36 a) This is constant acceleration with a0 = 9.81 m/s2 (taking positive x as up), v0 = 10 m/s and x0 = 0. Equation (1.30) relates the displacement, acceleration and velocity. At the top of the motion v = 0, so eq. (1.30) becomes
−
0 = 2( 9.81)(xtop ) + v02 or xtop = 5.1 m
−
15
b) Equation (1.29) relates constant acceleration to velocity, time and position. When the ball returns to its initial position x = 0 and equation (1.29) becomes 0 = a 2t + v0 t + 0. One solution is t = 0 which is the initial state. If t = 0, the relationship becomes 2
0
t = −v2a = 2.038 0
0
∼ 2 sec
1.37 For constant acceleration v = at + v0 and v 2 v2 v02 2x
6 2
4 2
− = (5×10 ) −(10 ) = 6.3 × 1012 m/s2 , (2)(2) 10 −1×10 = 8.0 × 10−7 s. t = v−av = 5×6.3 ×10 a=
6
0
2 0
−v
= 2ax so
4
12
1.38 Here a is a function of x, so consider the development of part 2, eq. (1.17) and (1.18) adx = vdv or
−kxdx = vdv or
2 kx 2 x = v2 x0 2 x0 kx20 = kx2 or
− |
|
v 2 = kx20 v(x) =
− kx k(x20
2
or 2
− x ).
The position time relationship can be found from (1.20):
− √ √ | . = t = dt = − = sin Rearranging and solving for x(t) yields sin− | or √kt = sin− − sin− (1) or √kt + t= t 0
x dx x0 v(x)
1 k
1
x x0
√
x x0
dx k(x20 x2 )
1 k
1
x x0
√
x x0
x(t) = x0 sin( kt + π/2) = x0 cos( kt).
1
x x0
1
x x0
π 2
= sin−1
or x x0
1.39 As the elevator starts from rest with constant acceleration to its operating speed v 2 = 2ax and v = at or t = v/a = (3 m/s)/25 m/s2 = 1.2 sec and travels a distance of x = v 2 /2a = (3 m/s)2 /2(2.5 m/s2 ) = 1.8 m. Which is also the time and distance required to stop the elevator. Hence 2 1.2 s = 2.4 s and 2 1.8 m or 3.6 m are used up in starting up and slowing down, the remaining distance 200 m - 3.6 m or 196.4 m is traveled at a constant velocity of 3 m/s so
×
×
t = x/v = 196.4/3 m/s = 65.5 sec.
16
The total time is then 1.2 + 05.5 + 1.2 = 67.9 sec or a little over one minute. 2
1.40 Since a =
−c x case 2 applies and x − x v −v −c xdx = −c = 2 2 or v = v + c x − c x . Substitution of x = 0, x = 10, v = 30, v = 0, yields 0 = 30 + 0 − c 10 or c = 3. 2
2 0
x
2
2
2
2 0
x0
2
2 0
2
2 2 0 2 2
1.41 Given a(x) = eq. (1.17) x 0
2 2
2
−c x dx = vdv =
1 2 (v 2
or
1 2 (v 2
−
v02 )
=
−
c 3 x 3
0
x0 = 0, t0 = 0, v(0) = v0 we want to determine v(x). From
−cx v v0
2
0
2 0
−v )
or v(x) = v − 2 0
2c 3 x 3
1.42 Since a is given as a function of velocity, case 3 applies: dx = vdv/f (v) upon integrating x
−x
0
=
v v0
vdv/(−v) = −v + v . 0
Since x0 = 0 and v = 0 when it comes to rest, x = 750 mm. 1.43 From the problem statement y0 = 40 km, v0 = 6000 km/s calculate an expression for y. Here acceleration is a function of position, so equations (1.17)-(1.20) apply. Given a(y) =
R2 0 (R+y)2 ,
−g
g0 = 9.81 m/s2 , R = 6370
At t = 0, y0 = y(0) = 40
× 10
3
3
× 10 .
m, v0 = v(0) = 6000 m/s
Note ymax will occur when v = 0. So compute v. a=
dv dt
=
dv dy dy dt
0 v0
R2 0 (R+y)2 .
=
−g
2
ym dy y0 (R+y)2
2
1 R+ym
vdv = −gR − =g R Thus =g R 4.552 × 10− = 02 2
60002 2
v02 2
0
0
2
8
1 (R+y0 )
−
=g R . or . 0
2
1 (R+ym )
1 R+y0
1 (R+ym )
− −
1 R+y0
Integrating yields
1 R+ym
Solving for ym yields ym = 2575.
17
−
1 (R+y0 )
, or
1.44 Solution: 0 vesc
2 vesc 2
vdv =
−g R 0
∞
dy y0 (R+y)2
2
−2g R ∞ = −2g R ∞
=
0
2 esp
−v
dy y0 (R+y)2
2
lim ym
−
2 esp
−2g R
=
1 R+y0
0
= g0 R 2
= −2g R
1 R+ym
→∞ That is −v Then: vesp
dy y0 (R+y)2
2
0
0
1 (R+y0 ) .
1 (R+y0 )
2
= R 2g /(R + y ) = 11.14 km/s = 11.14 × 10 m/s 0
1.45 Given: a(v) =
3
0
−cv = −0.4v,
v0 = 100 km/hr.
Since a = dv/dt we have v dv v0 v
=
−ct or ℓn
v v0
=
Thus v = v0 e−ct .
− = ct.
But v = dx/dt so that
− x=x +v e dt = x + − e− | t 0 0
0
ct
Thus x(t) = x0 +
v0 c
0
v0 (1 c
− e−
ct
ct t 0
).
With x0 = 0, v0 = 100 and c = 0.4 this becomes x(t) = 250(1
1.46 Solution: 2 a(t) = 5 sin(20t) m/s x(0) = 1 m and v(0) = 3 m/s.
Integrating: v 0
t 0
dv = v − v = 5sin20αdα = − 0
5 (cos 20t 20
− 1)
where α is used as the “dummy” variable of integration. Then v(t) = 3
− 0.25cos20t + 0.25 = 3.25 − 0.25cos20t m/s.
Integrating again yields x(t) = x0 + 3.25t
2
− 1.25 × 10 sin20t x(t) = 1 + 3.25t − 0.0125sin20t m
18
0.4t
− c−
).
1.49 This is a free fall problem or uniformly accelerated motion, where the acceleration is given as g = 32.2 ft/s, and the time traveled can be determined by equation (1.29) with tf as the given. Equation (1.29) becomes x(tf ) = 30 ft =
gt2f 2
+ v0 tf + x0
Here v0 = 0, since the ball is dropped, x0 = 0 taking the window as the starting position and hence gt2f 2
= 30 or t = 30/32.2 = 1.365 sec, f
which is the time required to hit the ground. The expression for velocity under uniform or constant acceleration is equation (1.28) or (1.30). From (1.28) v(tf ) = a0 (tf ) + v0 or v(1.365) = (32.2)(1.365) = 43.95 ft/sec.
1.50 This is a case of uniform acceleration a0 = g = 32.2 ft/s2 , with v0 up, and tf = 1.71 s. Using eq. (1.29) again with v0 as the unknown yields v(tf ) = 30
−
(g)t2f 2
+ ( v0 )tf + 0
−
Here v0 is used because v0 is up and we have taken down as positive in writing a plus sign for a0 (= g). This is consistent with the solution to 1.49. Solving for v0 yields
−
v = ( )t − 30 /t = 9.987 ft/sec. 0
g 2
2 f
f
1.51 Given a0 = 0.7g (constant acceleration), vf = 0 (because the car comes to a stop). Convert mph to ft/s (60 mph = 88 ft/s, 45 mph = 66 ft/s, 30 mph = 44 ft/s) and use eq. (1.30) vf 2 = 2a0 (xf
2 0
−x )+v = 0, a = −0.7 g (minus because it decelerates), x 0
where vf 0 distance measurement t = 0) then v02
v02
0
v02 (ft/sec) 2
xf = −2a = 0.4g = (1.4)(32.2) ft/sec so that a) xf = 171.78 ft, b) xf = 96.63 ft, c) xf = 42.95 ft. 2
0
1.52 Following the solution to 1.52m with a0 = 0.4g yields v02
x p = (−2)(−0.4)(32.2) so that a) xf = 300.6 ft, b) xf = 169.1 ft, c) xf = 75.2 ft.
20
= 0 (we start our
1.53 This is a uniform acceleration problem with a0 = 0.6g. Since the car starts from rest v0 = 0 and asume x0 = 0. Let vf = 200 mph = 293.33 ft/sec. Then the time to reach vf can be found from eq. (1.28) vf = a0tf + v0 or tf = 293.33/(0.6)(32.2) = 15.183 s. The distance traveled is found from eq. (1.29) to be (v0 = x0 = 0) xf = a0 t2f /2 = (0.6)(32.2)(15.183)2/2 = 2226.9 ft = 0.422 mile 1.54 Both cars undergo uniform acceleration aA = 0.9g and aB = 0.85g. Let them start at t = 0 in the ame place from rest, i.e., xa (0) = xB (0) = vA (0) = vB (0) = 0. Car A travels 1,000 m or takes the time determine by equation (1.29) xA (tf ) = 1000 =
(0.9g)(t2f ) 2
Then tf = 15.1 sec. During this time car B travels a distance determined by aB (tf )2 2
xB (15.1) =
=
(0.85)(9.81)(15.1)2 2
= 950.6 m
So car A is 1000 - 950.6 = 49.4 m ahead of car B when it crosses the finish line. Note that if tf = 15.05 s is used and not rounded off, then the distance becomes 55.47 mm insteady of 49.4 m.
1.55 This can be solved several ways including graphically by computing the area under the acceleration curve to generate the velocity, and the area under the velocity versus time curve to compute the position: First write the acceleration during each interval. For 0 < t < 50s, a(t) = 2 m/s. For 50 < t < 70s : a(t) = 0, for 70 < t < 100, a(t) = 15(t
− 70). Last for 100 > t > a(t) = 0.
Now calculate the area under the curve in each of these intervals being careful to use the appropriate initial conditions at the beginning of each interval: 0 < t < 50
v(t) = 20t m/s
50 < t < 70 70 < t < 100 100 > t
v(t) = 1000 m/s v(t) = 7.5(t
2
− 70)
+ 1000 m/s
v(t) = 7750 m/s so that v(120) = 7750 m/s
Integrating each of these in the interval yields 0 < t < 50
x(t) = 10t2
50 < t < 70
x(t) = 25000 + 1000(t
70 < t < 100 t > 100
x(t) = 2.5(t
x(t) = 7750(t
3
− 70)
− 50)
+ 1000(t
− 100) + 142, 500 21
− 70) + 45, 000
This last expression yields x(120s) = 297, 500 m = 297.5 km.
1.56 The equation to be solved is of the form dv dt
+ v = et , v(0) = 0, x(0) = 1m
Comparing with equation (1.32) identifies p(t) = 1 and f (t) = et so that the integrating factor becomes λ(t) = e dt = et .
According to equation (1.34) the solution is then v(t) = e−t ( etet dt + C ) = e−t ( 12 e2t + C )
at t = 0, v(0) = 0 ft/s so that C = 0.5 and v(t) = 12 (et
t
− e− ) m/sec
= sin h(t) m/sec Integrating again yields the displacement x(t) = x0 +
t 0
τ
2τ
e− (0.5e
− 0.5)dτ when x
0
= 1 m. Thus x(t) = 12 (et + e−t )m
= cos h(t) m
1.57 The equation to be solved is of the form dv dt
+ v = t, v(0) = 0, x(0) = 1, v(0) = 0
Comparing to equation (1.32): p(t) = 1 and f (t) = t. Thus the integrating factor becomes λ(t) = e dt = et
According to equation (1.34) the solution becomes v(t) = e−t ( ettdt + C ) = e−t [et (t
At t = 0, v(0) = 0 so that 0 =
t
− 1) + C ] = t − 1 + Ce−
t
−1 + C or C = 1. Thus: v(t) = t − 1 + e−
m/s
Integrating again yields (x0 = 1m) 2
x(t) = x0 + 0t (t 1+e−t )dt = 1+ t2
−
t2 2
t
t
−t −e− +1, so that x(t) = 2 − t + − e− m
1.58 a) The equation to be solved is of the form dv dt
t/2
+ tv = e−t
2
Comparing this form to equation (1.3.2), identifies p(t) = t and f (t) = e−t Thus the integrating factor becomes
λ(t) = e
tdt
2
= et
/2
Next, equation (1.34) yields that the solution is 2
v(t) = e−t
/2
( et
2
/2
2
e−t
/2
2
dt + c) = e−t
/2
2
t + ce−t
/2
At 0, v(0) = 10 ft/s so that c = 10 and v(t) = 10e−t 22
2
/2
2
+ e−t
/2
t
/2
.
Integrating again (x(0) = 0) yields x(t) = 0t (10 + τ )e−τ /2 dτ which yields the error function when integrated, i.e., x(t) = 12.5 erf(0.71t) et /2 + 1. 2
−
2
b) This does not have an integrating factor, or other closed formed solution, so the solution must be found numerically by writing the equation in first order for and applying an Euler or Runge-Kutta solution. A Mathcad solution is shown. i
∆t
0 .. 4000
x0
1
v0
xi
1
vi
1
ti
i . ∆t
5
2 t .v
a v, t
0.001
3. t
1
vi . ∆t
xi vi
a vi , ti . ∆t
15
x v
10 i i 5
0
1
2 t i
3
4
FIGURE S1.58
To solve this problem with Matlab, create and run the following code: x(1)=1; v(1)=5; t(1)=0; dt=0.001; for n=1:4000; x(n+1)=x(n)+v(n)*dt; v(n+1)=(-t(n)ˆ.2*v(n)+1+3*t(n))*dt+v(n); t(n+1)=t(n)+dt; end plot(t,x),plot(t,v)
23
1.59 First solve the homogeneous equation x ¨ + 5x˙ + 4x = 0 by following eq. (1.41), assume a solution of the form x(t) = Aeλt where λ must satisfy λ2 + 5λ + 4 = (λ + 4)(λ + 1) = 0 so that λ1,2 = 4, 1. Thus the homogeneous solution is of the form xh (t) = A1 e−4t + A2 e−t . The particular solution is guessed to be x p = a + bt, of the form of the forcing function where a and b are to be determined. Substitution of the assumed form for x p (t) into the equation of motion yields
− −
5b + 4(a + bt) = 3t + 0to Comparing coefficients of t and to yields 5b + 4a = 0 and 4b = 3 so that b = 3/4 and a = 54 ( 34 ) = 15 . Thus x p = 15 + 34 t. This is called the 16 16 method of undetermined coefficients. The total solution is the sum ( x = xh +x p ) so that
−
x(t) = A1 e−4t + A2 e−t + 34 t
−
−
−
15 16
To determine the coefficients A1 and A2 apply the initial conditions 15 16 A2 + 34
x(0) = 0.5 = A1 + A2 v(0) = 0 =
−4A − 1
−
which represents two equations in the two unkowns A1 and A2 . Solving yields A1 =
−0.229 and A
2
= 1.667 and hence: x(t) =
4t
−0.229e−
+ 1.667e−t + 34 t
−
15 16
k c 1.60 Define ω 2 = m = 41 = 4 and ζ = 2mω = 2.795 > 0 so the system is over damped. Then the problem in standard form is
x¨ + 2ζω x˙ + ω 2 x = x¨ + 5x˙ + 4x = 0 Assume solutions of the form x = Aeλt . The characteristic equation becomes λ2 + 5λ + 4 = 0 which has roots λ1 = 1, λ2 = 4. Thus the general solution is of the form
−
−
x(t) = A1 e−t + A2 e−4t Applying the initial condition yields x(0) = 5 = A1 + A2 v(0) = 0 =
−A − 4A 1
2
which is a system of two linear equations in the two unknowns A1 and A2 . −t 5 e−4t Solving yields A1 = 20 and A2 = 53 . Thus the solution is x(t) = 20 3 3 e 3 20 t 4t − − and v(t) = 3 ( e + e )
−
−
−
24
1.61 The equation of motion has the form (after dividing by 1000) a=
dv dt
=
−cv − 400x, v(0) = 0, and x(0) = 0.01m
Following along with equation (1.51), the Euler method of integration yields 0 vi+1 v cvi ∆t 400xi ∆t v0 = i = , 0.01 xi+1 xi + vi ∆t x0 Using a high level language (Matlab, Mathcad or Mathematica) yields (some students may know the analytical solution for this equation. Others will know how to use the more sophisticated higher-order Runge-Kutta integration the following plot. Values of c are varied until the plot produces only two oscillations.)
−
−
∆t
i
.01 1.5 0 .. ∆t
c
v0
0
x0
.01
vi
1
xi
1
vi
15
c . v i . ∆t xi
400. x i . ∆t vi . ∆t
0.01
0.005 x
i 0
0.005
0
0.5
1
1.5
i . ∆t
FIGURE S1.61
Note here that the oscillation dies out at about t = 1 second, for a value of c = 15, or a damping value of 15, 000 kg/s. The Matlab code for doing this is given below using an Euler method. This can also be done using ODE which involves a Runge-Kutta routine. Create the 25
following Matlab code then run it with different values of c until the desired response results: c=15 x(1)=0.01;v(1)=0.0;t(1)=0; dt=0.01; for n=1:150; x(n+1)=x(n)+v(n)*dt; v(n+1)=v(n)-c*v(n)*dt-400*x(n)*dt end plot(t,x) Run this Matlab code with various values of c until the response decays within two cycles as desired.
1.62 Following the development of the numerical integration section equation (1.51) becomes 2
v = v − 900v ∆t − 4000(x ) ∆t i+1
i
i
i
xi + vi ∆
xi+1
with initial condition v0 = 0 and x0 = 20 mm. The Mathcad code is: i ∆t
0 .. 1000
v0
0
.001
x0
20
vi
1
xi
1
vi
900. vi . ∆t xi
4000. xi . x i . ∆t vi . ∆t
20
x
i
10
0
0.2
0.4
i . ∆t
FIGURE S1.62
26
0.6
0.8
1
The equivalent Matlab code can be either Euler (see previous problem) or Runge-Kutta Method. To use RK, first save the following Matlab code under Onept62.m: function xdot=onept62(t,x) xdot=[x(2);-900*x(2)-4000*x(2)-4000*x(2)*x(2)]; % the equation of motion Then the following commands will compute and plot the solution EDU>tspan=[0 1] % defines the time interval of interest EDU>x0=[20;0]; %enters the initial conditions, displacement first EDU>ode45(*onept62’,tspan,x0); % calls the RK routine and applies % it to 1.62.
1.63 Following the development of the numerical integration section (1.51) becomes 3 i
v = v − 90v ∆t − 100x ∆t i+1
i
i
xi + vi ∆t
xi+1
with initial condition v0 = 0 and x0 = 10 mm. The Mathcad code is: ∆t
0.001
v0
0
xi
1
vi
1
N x0
500 10
0 .. N a v, x
ti
i . ∆t
c. v
k.x
c
90
k
100
3
vi . ∆t
xi vi
i
a vi , xi . ∆t
10
5 x i 0
5
0
0.1
0.2
0.3 t
FIGURE S1.63
The catch gets near zero within 1/20 second.
27
i
0.4
0.5
The Matlab code is to prepare the following file named onept63.m: Function xdot=onept63(t,x) c=20;k=100; xdot=[x(2);-c*x(2)-k*x(1)ˆ3]; Then type the following in the command window: EDU>tspan-[0 0.5]; EDU>x0=[10;0]; EDU>ode45(‘onept63’,tspan,x0);
1.64 Following the solution to 1.63 equation (1.51) becomes: 3
v = v − cv ∆t − 100(x ) ∆t i+1
i
i
i
xi + vi ∆t
xi+1
Repeat the numerical solution to problem 1.63 with successively smaller values of damping (c) each time until the solution oscillates twice before coming to rest. A value of about c = 20 1/s comes close as illustrated. The Mathcade code is: ∆t
0.001
v0
0
xi
1
vi
1
N x0
500 10 xi
vi
i
0 .. N a v, x
ti
i . ∆t
c. v
k.x
c
20
k
100
3
vi . ∆t a vi , xi . ∆t
10
x i
0
10
0
0.1
0.2
0.3 t
FIGURE S1.64
28
i
0.4
0.5
The Matlab code requires the following file saved as onept64.m: function xdot=onept64(t,x) c=20;k=100; xdot=[x(2);-c*x(2)-k*x(1)ˆ3]; The type the following in the command window EDU>tspan=[0 0.5]; EDU>x0=[10;0]; EDU>ode45(‘onept64’,tspan,x0);
1.65 Solution: First set up the Euler form of the equation for numerical integration:
v = v − cv |v |∆t − 4kx |x |∆t i+1
i
i
i
i
i
xi + vi ∆t
xi+1
Then resolve for various values of c, k and x0 until a response that dies out in one oscillation results. There are many answers, the plot shows this is achieved for x0 = 0.01 m, k = 400 1/ms2 , and c = 1000 m−1 . Another solution is x0 = 2, c = 6 and k = 40. The Mathcad solution is: c
1000 ∆t
i
0 .. 3000
0.01
x0
vi
1
xi
1
.01
vi
v0
0
c . vi . vi . ∆t xi
400 . xi . xi . ∆t v . ∆t i
0.01
0.005 x
i 0
5
10
15
0.005 i. ∆t
FIGURE S1.65
29
20
25
30
Save the following Matlab code as a file named “onept65.m”: function xdot=onept65(t,x) c=1000;k=400; xdot=[x(2);-c*x(2)*abs(x(2))-k*x(1)*abs(x(1))] In the command window: EDU>tspan=[0 30]; EDU>x0=[0.01; 0]; EDU>ode45(‘onept65’,tspan,x0); ˆ so that m¨r = m¨ ˆ . Then m¨r = g jˆ 1.66 Assuming r = xˆi + y jˆ + z k xˆi + m¨ y jˆ + m¨ zk yields x ¨ = 0, y¨ = g/m and z¨ = 0. These are linear, decoupled equations.
−
−
1.67 Yields the 3 scalar equations x ¨ + cx˙ = 0, y¨ + cy˙ + g = 0 and z¨ + cz˙ = 0 which are decoupled, linear equations. ˆ and v = vxˆi + vy jˆ + vz k ˆ yields 1.68 Assuming r = xˆi + u jˆ + zk
x¨ = −c (v + v + v ) v , y¨ = −c v + v + v (v ) and z¨ = −c v + v + v 2 x
2 y
2 z
x
2 x
2 y
2 z
y
2 x
2 y
2 z
(vz ). These are coupled, nonlinear equations.
ˆ and v = vxˆi + vy jˆ + vz k ˆ j 1.69 Assuming r = xˆi + y jˆ + zk x¨ = 3t2 , y¨ =
− sin(πt) and z¨ = xz
The x and y equations are linear and decouple. The z equation is nonlinear and coupled to x.
1.70 Consider the plane trajectory equations given by eq. (1.74). In this case we know xf = 450 ft, zf = 12 ft, g = 32.2 ft/s2 , x0 = 0, z0 = 0 and v0 = 130 ft/s. Thus equation 1.73 becomes 450 = (130 cosθ)t + 0 12 =
2
−16.2t
+ (130 sinθ)t
which is two nonlinear algebraic equations in two unknowns t and θ. Solving yields t = 6.86 sec, θ = 1.042 rad (59.7◦ ) and t = 4.088, θ = 0.561 rad (32.143◦ ).
30
These solutions are found using Mathcad. Note that there are two solutions found by taking different initial guesses to the iterative solution of these nonlinear algebraic equations. One solution has a low angle which would drive into the bunker and one (correct) that has a larger angle which will loft onto the green.
1.71 Choose (0,0) in the x z plane to be on the ground so that vz (0) = 2 m/s, zf = 0, x0 = 0, z0 = 3m, xf = d, xf = 0. Then equation (1.73) becomes d = 2t, 0 = 9.81 t2 + v0 (0) + 3. Combining 4.905 t2 = 3 and t = d2 yields 2
−
d2 =
12 4.905
−
or d = 1.56 m.
1.72 Looing at the top half spray, Eq. 1.73 becomes d2 cos10◦ = 20 cos70◦ t + 0 d2 sin 10◦ =
2
−16.1t
(1)
+ 20sin70◦ t + 0
(2)
which is a system of two equations in the two unknowns: d2 and t. From (1) t =
d2 cos10◦ 20cos70◦
= 0.1459d2 d2 =
20sin 20◦ (0.1439) sin10◦ (16.1)(0.1439)2
−
or d2 = 7.588 ft.
Next consider the spray to the left: d1 cos10◦ = 20 cos50◦ t + 0 0=
2
−16.1t
+ 20sin50◦ t + d1 sin10◦
(1) (2)
From (1) t = .0766d1 or t2 = .005268d21 and eq. (2) becomes d1 =
(20)2 cos2 50◦ (cos10 (16.1)(cos2 10◦ )
◦ tan50◦ + sin 10◦ ) = 14.26 ft.
1.73 Working with equation 1.73 for projectile motion, let the hose be at x0 = z0 = 0 and assume it hits at x(tf ) = x and z(tf ) = 0, then eq. (1.73) becomes t2
x = v0 cos θtf and 0 = g 2f + v0 sin θtf . Solving this last expression for tf yields tf = 2v gsin θ , the time to hit the ground. Then from the expression for x 0
x(tf ) =
2v02 g
sin θ cos θ 2v02
The max value of x occurs at dx/dθ = 0 or g ( sin2 θ + cos2 θ) = 0. This requires sin θ = cos θ or θ = 45◦ , the value at which xf will be maximum.
31
−
1.74 From the problem statement, taking the batters “foot” as the origin, the value of x0 = 0, z0 = 4 ft, v0 = 140 ft/s, θ = 20◦ and (since it hits the ground) zf = 0. Equation (1.73) then becomes x(tf ) = 14
− cos20◦ t
f
and 0 =
2 f +
−16.1t
140 sin 20◦ tf + 4
Solving the last expression for tf yields tf = 3.055 sec. and -0.081 sec. Obviously the physical value is tf = 3.055, which from the first equation yields xf = 140(cos20◦ )(3.055) = 402 ft.
1.75 From the projectile equation for z: z = 16.1t2 +140sin20t+4. The maximum value of the parabolic trajectory would occur at tf /2 except the value of tf calculated in 1.74 assumes the trajectory is 4 ft off the ground. The equation for time of flight is 16.1t2f +140sin20◦ tf = 0 or tf = 2.97 sec, and tf /2 = 1.487
−
sec. Then zmax
− = −16.1
2.92 2
2
+ 140 sin 20 + 4 = 39.6 ft. 2.92 2
1.76 Consider the projectile equation 1.73 and first solve for v0 so the ball just clears the bottom window. Picking a coordinate system 1m off the ground yields x0 = z0 = 0, xf = v0 cos30◦ tf , zf = 2m =
−
9.81 2 + 2 tf
v0 sin θtf
where xf = 6.5m. This yields two equations in two unknowns: 6.5 = v0 (.886)tf or tf =
7.5 v0
Thus v0 = 12.56 m/s. With yf = 3m, this becomes v0 = 19.18 m/s so that he must kick through with a speed: 12.56 < v0 < 19.18 m/s.
1.77 The initial velocity is given as v0 = 10 m/s, x0 = z0 = 0, zf = 2m (for smallest and 3m for largest). xf = 6.5m so the first equation of (1.73) becomes (let t denote the time at which the ball reaches the window). 6.5 = 1
− cos θt or t = 0.65/ cos θ
The second projectile equation yields 2
2=− + 10 sin θ 9.81 2
0.65 cos θ
0.65 cos θ
which can be solved numerically for θ = 40.865◦. Changing the value of zf = 3m and repeating yields θ = 55.58◦ . Thus he needs to kick through at an angle between 40.9◦ < θ < 55.6◦ to make it through the window with an initial velocity of 10 m/s. Each of these two equations have 2 solutions so it will also make it in for 59.44◦ < θ < 66.28◦ on the high lofty solution.
32
√
1.78 The given values are z0 = d/ 2, v0 = 25 m/s, θ = 0. Equation 1.73 becomes xf = 25t = √d2 or t = d = 180.2 m.
d , 25 2
√
0 =
−
9.81 2 t 2
√2
+ √d2 , or d =
2
9.81
d2 (625)(2)
, so
1.79 Sample 1.14 gives the equation for a particle in projectile motion with wind resistances. The equations are nonlinear and coupled and must be solved numerically. The initial conditions of x(0) = 0, vx (0) = 25, y(0) = 0, vy (0) = 0 will allow the solution computed numerically following sample 1.14. The tra jectory can then be plotted along with a line at 45 ◦ representing the hill. The intersection will yield the value of d. Since we do not know d, it is best to put the coordinate system at the end of the ski run and let z (or y) evolve in the negative direction. Such a line passing through the origin has slope -1 and can be written as d = x, or di = xi in incremental form. The Mathcad code is
−
i
0 .. 1100 vx0
25
vxi
∆t
0.005
x0
0
vyi
vyi
1
yi
1
0.04
g
9.81
0
y0
0
c . vxi . ∆t vxi . ∆t
xi
1
c
vy0
vxi
1
xi
di
−
g
c . vyi . ∆t
yi
vyi . ∆t
0
20
xi
d i
40
60
80
100
120
140
50
y i 100
150 x i
FIGURE S1.79
Form the plot, then cross about 111 m out or d = 111/ cos45◦ = 157 m down the incline. 33
The Matlab code for solving and plotting is given next with the plots suppressed as they are the same as the above: function xdot=onept78 (t,x): c=0.04; g=9.81; xdot=[x(2);-(c*x(2);x(4);-g-c*x(3)]; In the command window: EDU>tspan=[0 140] EDU>x0-[0;25;0;0]; EDU>[t,x]=ode45(‘onept78’,tspan,x0); EDU>d=-x(:1); EDU>plot(x(:,1),x(:,3),‘t‘,x(:,1),d,‘*’)
1.80 This is just a repeat of the previous problem with a more accurate nonlinear damping term. The Mathcad code is: i
0 .. 1000 vx0
25
vxi
0.005
x0
0
c
0.002
c . vxi . xi
1
vyi
1
yi
g
vy0
vxi
1
xi
di
∆t
vyi
g
c . vyi .
1
yi
9.81
0
vxi
y0
2
vyi
0
2 . ∆t
vxi . ∆t vxi
2
vyi
2
. ∆t
vyi . ∆t
xi
0
d y
20
40
60
50 i i 100
150 x
FIGURE S1.80
34
i
80
100
120
In this case the skier makes it about 108 meters out or so which is 108 = d cos(45◦ ) so that d = 152.7 m. The Matlab code is the same as the previous problem except the “x dot” becomes: xdot=[x(2);-c*x(2)*sqrt(x(2)ˆ2+x(4)ˆ2);x(4);-g-c*x(2)*sqrt(x(2)ˆ2 + x(4)ˆ2)];
1.81 Solution: 300 yards = 900 ft so that xf = x(tf ) = 900. Given that the ball is at zero to start with x0 = y0 = 0, and hits the ground at yf = 0. With θ = 90◦ given, the trajectory equations are 900 = v0 cos9◦ t + 0 0=
2
−16.1t
+ v0 sin9◦ t + 0
(1) (2)
Solve (1) for t and (2) for v0 to get v0 = 306 ft/s (about 209 mph). 1.82 Solution: 200 mph = 293.3 ft/s. Here x0 = y0 = 0, θ = 9◦ . Then eq. (1.71) becomes (1) x = (293.3)cos9◦ t = 289.7t 0=
2
−16.1t
+ 45.8t
From (2) t = 2.849 sec so from (1) x = 825.6 ft
35
(2)
1.83 This follows directly from the solution of sample 1.14 with the following values and equations: x0 = y0 = 0, c = 0.05, vx0 = 293.3cos9◦ ft/sec and vy0 = 293.3sin9◦ ft/s. i
∆t
0 .. 600 x0
0
vyi yi
vyi
1
32.2
vx0 y0
293.3. cos 9 .deg
0
c . vyi . ∆t
g
vyi . ∆t
yi
1
g
vxi . ∆t
xi
1
0.05
c . vxi . ∆t
vxi
1
c
293.3. sin 9 . deg
vy0
vxi xi
0.005
40
20 y
i 0
200
400
600
800
1000
20 x
i
FIGURE S1.83
From the figure x = 758 ft (found by using the trace funciton in Mathcad). The Matlab code is given in the Matlab supplement.
1.84 Again use the projectile equations of eq. (1.73). Here: x0 = 0, y0 = 0 (so yf = 3 ft), xf = 20 ft, θ = 45◦ and hence 20 = v0 √12 t or t = 3=
2
−16.1t
√
20 2 v0
+ v0 √12 t
·
Solving yields v = 0
(16.1)800 17
= v0 = 27.53 ft/s.
36
1.85 Using the projectile motion equations with x0 = y0 , xf = 20 ft, yf = 3 ft and v0 = 30 ft/sec, equations (1.73) becomes 20 = (30) cosθt 3= Substitution of t = (3) =
−16.1
4 9cos2 θ
2 3cos tθ
2
−16.1t
(1)
+ (30) sinθt
(2)
from (1) into (2) yields the transindental equation
+ 20 tan θ
Solving for θ yields θ = 33.7◦ and 64.8◦ . Either angle will “work”, however the lower angle gives a trajectory up through the bottom of basket whereas the 64.8◦ solution gives the lofty shot and then goes through the top of the hoop. The following Mathcad code solves the problem: Specify the known parameters. v0 30 g 32.2
x 0
0
y0
7
x f
20
y f
10
Initial guess for time and angle 3 θ 30. deg t f Given x f v 0. t f . cos y f Find
g. θ
t f 2
x 0
θ
v 0. t f . sin
2
, t f =
y0
θ
1.132
1.132
1.568
deg
= 64.859 Angle in degrees
A second solution can be found with a lower angle.
v0 30 g 32.2
x 0
0
y0
7
x f
20
y f
10
Initial guess for time and angle These values are assumed less. 1 θ 10. deg t f Given x f v 0. t f . cos y f Find
g. θ
t f 2 2
, t f =
θ
x 0
v 0. t f . sin
θ
y0
0.588
0.588
0.801
deg
= 33.69 Angle in degrees
The second solution is not valid as the ball would hit the net from below.
37
1.86 Using the projectile motion equations with x0 = 0, y0 = 10, θ = 0, v0 = 120 mph = 176 ft/s and yf = 0 (i.e., hits the ground) yields xf = 176t 0=
2
−16.1t
(1)
+ 10
(2)
From (2) t = 0.7885 sec and from (1) xf = 138.7 ft
1.87 This again uses the projectile motion equation. a)Let x0 = 0, xf = 6ft, y0 = 0, so yf = 15 4 = 11 ft, θ = 80◦ and the unknown is v0 . Equation (1.73) becomes
−
xf = 6 = (v0 cos 80◦ )t + 0 yf = 11 = (v0 sin80◦ )t Substitute t =
6 v0 cos 80
2
− 16.1t
(1) +0
(2)
from (1) into to (2) to get 11 = 6 tan80◦ − 16.1
62 v02 cos2 80◦
(3)
Solving yields v0 = 28.9 f /s. b) Repeating (a) with xf = 16 eq. (3) becomes 11 = 16 tan 80◦ − 16.1
162 v02 cos2 80◦
or v0 = 41.4 ft/s. 1.88 This is circular motion with R = 150 ft, at = 12 ft/s2 . Compute the time t at which an = 24 ft/s2 . From Eq. (1.84), at = αr so that α = art = 12140ft/sft = 0.08 rad/s2 a constant. dw = 0.08 so that w w0 = 0.08t or w = 0.08t + w0 . dt 2 From (1.84) an = rw = 25 = 150(w0 + 0.08ts )2 (1), where ts = time to slip. Also at ts , a = a2t + a2r = 252 + 122 = 27.73 ft/s2 at slip. Solving (1), with 2
−
√
ω0 = 0 for ts yields ts =
1 0.08
25 150
1/2
= 5.104 s.
1.89 This is a circular motion with r = 2 m and at (t) = 6sin πt(m/s2 ). The particle starts at rest so that θ(0) = ω(0) = 0. For circular motion v(t) = 0t at dt = t 6 cos πt). Also ar = vr = 12 ( π36 )(1 cosπt)2 = π18 (1 cos π6)2 . 0 6sin πt = π (1 From equation (1.81) taking the magnitude of a(t) yields
2
−
(t) = a + a = 6 sin πt + [
a
2 t
2 n
−
2
2
2
18 (1 π2
38
2 2
− cos πt) ] .
2
−
The plots follow: t
0 , 0.001 .. 2 6
v t
π
a t
. 1
cos π . t
at t
2 ^ s / m l e c c a s / m l e v
2
an t
6 .sin π . t
at t
18
an t
π2
. 1
cos π . t
2
2
10
a t v t
5
0
0.5
1 t time s
1.5
2
FIGURE S1.89
The Matlab code for producing the plots is given in the following file: syms t % declares t symbolic v=(6/pi)*(1-cos(pi*t)); an=0.5vˆ2; at=6*sin(pi*t); a=sqrt(atˆ2+anˆ2) ezplot(a,[0,2]), ezplot(v,[0,2]) 3
1.90 From the solution to 1.89 a(t) = (36 sin3 πt + 18 (cos πt 1)4 )1/2 . Find ts when π a(t) = 5. The answer can be seen from the plot given in figure S1.89 or from solving
−
4
52 = 36 sin2 πts +
182 (cos πts π4
− 1)
4
for ts which has 2 solutions in the interval of interest. From Mathcad they are: ts = 0.312 s, and 1.688s. 1.91 Given r = 200 m, v = 30 km/hrs = motion, equation (1.81) yields an =
v2 4
=
8.33 200
− 0.3469 m/s
2
30 1
×
103 m hr 1 3600 sec hr
or an = 0.35 m/s2 .
39
· = 8.33 m/s. For circular
1.92 This is circular motion starting from rest so that θ(0) = ω(0) = 0, with r = 4 m. a) α(t) = 2t2 r/s2 so that dω = 2t2 and ω ω0 = 3t2 or w(t) = 23 t3 . Thus from dt eq. 1.84 an = r(ω)2 = 4 49 t6 and at = 4(2t2 ) = 8t2 . At t = 2s, an (2) = 113.8, at (2) = 32 so that a(t) = 11382 + 322 = 118.2 m/s2 .
−
·√
3
2
b) From eq. (1.82) ddt s = ra = 8t2 so that dv = 8t2 dt or v 0 = 83 t3 . Thus ds = 83 t3 dt and s(2) = 83 02 t3 dt and the total distance traveled is s = 10.67 m.
−
2
1.93 Solution: α(t) = 3t2 2t rad/s2 with ω0 = θ0 = 0. Integrating yields dω = dt 2 6 2 2 3 2 3 2 3t 2t or 0 dω = 0 3t 2t or ω(t) = t t rad/s. Likewise dθ = (t t )dt t so that θ(t) = 0t (t3 t2 )dt = t4 or θ(t) = 14 t4 13 t3 rad. 3
−
− −
−
4
−
−
3
−
−
1.94 Solution: α(t) = t cos(πt) rad/s2 , ω0 = 2 rad/s, θ0 = 30◦ = π6 rad. Thus dω(t) = t cos πtdt or upon integrating ω 2 = 0t x cos πxdx = π1 [cos πt + πt sin πt 1] so that ω(t) = (2 π1 ) + π1 (cos πt + πt sin πt) rad/s. Integrating again yields
−
− [2sin πt − πt − πt cos πt + 2π t] 2
θ
−
π 6
=
1 π3
and π 6
θ(t) =
−
2
2
3
1 (2 sin πt π3
+ 2t +
− πt − πt cos πt) rad
1.95 Solution: ω0 = 3 rad/s and α(ω) = 2ω 2 rad/s2 . From eq. (1.87) α = ω dω = dt ω θ 2ω 2 . Solving yields 3 dω = 2 0 dθ, or ℓn ω3 = 2θ. ω = 3e−2θ so that ω dθ = 3e−2θ or 0θ e2θ dθ = 0t 3dt = 3t. Evaluating the other integral yields dt
−
1 2θ θ 0 2e
|
= 3t or e2θ
− −
−
− 1 = 6t
and 2θ = ℓn(6t + 1) or θ(t) = 12 ℓn(6t + 1) and θ(10) = 12 ℓn(61) = 2.055 rad. 1.96 Solution: ω0 = 2 rad/s and α(ω, t) = 0.01ω+4t rad/s2 . Then dω = 0.01ω+4t dt which is a first order differential equation of the form: ω˙ + 0.01ω = 4t, ω0 = 2. Using the integrating factor x(t) = e0.01t , equation (1.34) yields the solution ω(t) = e−0.01t [C + 40, 000e0.01t(0.0t 1)]
−
Since ω(0) = 2, C = 0.01t
−3998 and
−
−
−3998e− + 4000(0.01t − 1) + 40t rad/s = [0.01t − 1 + e− ω(t) = −4000 − 3998e− ω(t) =
0.01t
Since ω =
dθ , dt
0.01t
integrating this ω(t) yields θ(t).
θ(t) = 400 + [0.005t2
− t − 0.01e−
0.01t
](4
where 3998 has to be rounded to 4000.
40
4
× 10 ) rad
](4
4
× 10 ) rad/s
1.97 The equation of motion can be written as dω = 2sin θ 0.4ω. Since ω = dt this can be written as a second order nonlinear equation in θ: ˙ =0 θ¨ + 0.4θ˙ 2sin θ = 0 θ(0) = π and θ(0)
−
−
dθ , dt
6
which can be solved by numerical integration as suggested in equation (1.93). That is
ω = ω + (2 sin θ − 0.4ω )∆t , ω = 0 n+1
n
n
n
0
θn + ωn ∆t
θn+1
π 6
θ0
which is plotted in the following Mathcad file:
i
0 .. 500
ωi
1
θi
1
∆t
ω0
0.01
ωi
0
2 .sin θi
π
θ0
6
0.4 . ωi . ∆t
ωi . ∆t
θi
6
4
θ
i 2
0
1
2
3 i. ∆t
FIGURE S1.97
The Matlab code is: function xdot=onept97(t,x); xdot=[x(2);2*sin(x(1))-0.4*x(2)]; command window: EDU>tspan=[0 5]; EDU>x0-[pi/6;0]; EDU>ode45(‘onept97’, tspan, x0);
41
4
5
1.98 This is a second order, nonlinear differential equation in θ which can be solved numerically by using the first order form suggested in equation (1.93). From ˙ the given form of θ¨ + 0.02θ˙ θ˙ 3cos θ = 0 θ(0) = π6 and θ(0) = 0. Equation (1.93) becomes
| |−
ω = ω + (3 cos θ − 0.02ω |ω |]∆t n+1
n
n
n
n
θn + ωn ∆t
θn+1
which is plotted below in Mathcad:
i
0 .. 500
ωi
1
θi
1
∆t
ω0
0.01
ωi
0
3 .cos θi
θi
θ0
π 6
0.02 . ωi . ωi . ∆t
ωi . ∆t
3
2
θ
i 1
0
1
2
3 i. ∆t
FIGURE S1.98
In Matlab the code is: function xdot=onept98(t,x); xdot=[x(2);3*cos(x(1))-0.02*x(2)*abs(x/2))]; command window EDU>tspan=[0 5]; EDU>x0=[pi/6;0]; EDU>ode(‘onept98’,tspan,x0);
42
4
5
ˆ m, differentiation 1.99 Follow sample 1.18. Given r(t) = 3cos tˆi + 3 sin t jˆ + 4tk ˆ and a(t) = 3cos tˆi 3sin t jˆ so that yields v(t) = 3sin tˆi + 3 cost jˆ + 4k v(3) = ˆi + jˆ + 4k and a(3) = ˆi jˆ. The unit tangent vector ˆet is calculated from eq. (1.97) to be ˆ eˆt (3) = v(3) = 0.085ˆi 0.594 jˆ + 0.8k
−
−
|v(3)|
−
−
−
−
at (3) = (a et )ˆet = 0, an = a at = a so that eˆn = an = a = 0.99ˆi 0.141 jˆ
·
|an |
eˆb = eˆt
−
|a|
× eˆ
n
−
ˆ = 0.133ˆi + 0.792 jˆ + 0.6k
1.100 From problem 1.99 ˆ and a(t) = v(t) = 3sin tˆi + 3 cos t jˆ + 4k
−
−3cos tˆi − 3sin t jˆ m/s
2
Note magnitude of both v(t) and a(t) is constant. 2
1.101 From equation 1.100, ρ(t) = |avn(t) where v 2 (t) is v(t) v(t) = 9 sin2 t +9cos2 t + (t)| 16 = 9 + 16 = 25. Compute an (t) . From eq. (1.97) ˆet = |vv| = 15 ( 3sin tˆi + ˆ . Then from eq. (1.98) 3cos t jˆ + 4k
|
·
|
−
ˆt)ˆet = 15 [( 3cos t)( 3sin t) + ( 3sin t)(3 cos t) + (0)(4)]ˆet = 0. at (t) = (a e
·
Now an Thus
− − − √ = a − a = a − 0 = a. Thus |a (t)| = |a(t)| = 3 t
2 cos2 t
n
+ 32 sin t = 3.
2
= 13 [25] = 8.33 m, a constant so that the motion is circular, moving ρ(t) = |avn(t) (t)| at a constant angular velocity. ˆ m, successive differentiation yields the velocity 1.102 Given r(t) = t2ˆi + 3t jˆ +10sin tk and acceleration: ˆ m/s v(t) = r˙ (t) = 2tˆi + 3 jˆ + 10 cos tk a(t) = ¨ r(t) = 2ˆi
− 10sin tkˆ m/s
2 2
From eq. (1.100) the radius of curvature is ρ(t) = v|a(t) n| Now v 2 = v v = 4t2 +9+100 cos2 t. Following eq. (1.97) eˆt (t) = |vv| , at = (a eˆt)ˆet , an = a at and ρ(t) = |avn| . Programming these formulations and evaluating at each value of t yields
−
·
·
2
a) t = 1s, ρ(1) = 7.23 m b) t = 3s, ρ(3) = 126.645 m c) t = 5s, ρ(5) = 13.346 m
43
The solution in Mathcad is given in the following: t
5
2.t v
2
3
a
0
10. cos t
10. sin t
v et
a . et . et
at
an
v
a
at
v.v
ρ
an
ρ = 13.346
The Matlab code is t=5;v=[2*t;3;10*cos(t)];a=[2;0;-10*sin(t)]; et=v/norm(v);at=dot(a,et)*et;an=a-at; pro=dot(v,v)/norm(an) 1.103 Solution: r(t) = 12 (1 + t2 ) and θ = πt2 so that r(t) ˙ = t and r¨(t) = 1, θ˙ = 2πt and θ¨ = 2π. From equation (1.103) ˙ θ = tˆ v = rˆ ˙ er + r θe er + πt(1 + t2 )ˆeθ a = (¨ r
− rθ˙ )ˆe 2
r
˙ eθ = (1 + (r θ¨ + 2r˙ θ)ˆ
2 2
2
− 2π t (1 + t ))ˆe
To plot the motion define t : 0, 0.1...2, define r = x = r cos θ and y = r sin θ which is plotted below.
44
r
+ (π(1 + t2 ) + 4t2 π)ˆeθ
1+t2 2
and θ = πt2 . Then let
The Mathcad code is:
t
0 , 0 .01 .. 2 2
1
r t
t
θ
2
t
2
π .t
x t
θ
r t 8 cos
t
r t . sin
y t
θ
t
2
20
y t
10
0
10
20
2
4 x t
FIGURE S1.103
The Matlab code is: EDU>t=linspace(0,2); EDU>r=(1+tˆ2)/2;th=pi*t.ˆ2; EDU>x=r.*8.*cos(th); y=r.*sin(th); EDU>plot(x,y)
1.104 The value of v(t) is always positive. Note that v(t) =
t 1 + π (1 + t ) = ds/dt so that s−s = t + π t (1 + t ) dt = 18.977 m. 2
0
2 2
2 0
2
2 2
| |
t + π t (1 + t ) = 2
2 2
2 2
2 2
1.105 Solution: r(t) = 2 so that r˙ = r¨ = 0 and there is circular motion. θ(t) = sin πt so that θ˙ = π cos πt and θ¨ = π 2 sin πt. From equations (1.103): ˙eθ = 2π cos πtˆ v(t) = r θˆ eθ and
−
a(t) =
−2π
2
cos2 πtˆ er
− 2π
2
sin2 πtˆ eθ
Let x = r cos θ and y = r sin θ to plot the motion as illustrated below.
45
The Mathcad solution is:
t
0 , 0.01 . . 2
sin π. t
θ t
r t
2 90 2
120
60
1.5
150
30
1 0.5
r t
180
0
0
210
330 240
300 270 θ t
FIGURE S1.105
The Matlab code is: EDU>linspace(0,2); EDU>r=2;th=sin(pi*t); EDU>x=r*cos(th);y=5*sin(th); EDU>plot(x,y)
1.106 Solution: ds dt
s
= v(t) =
−s
| |
0
=
√4π
2 cos2
πt = 2π cos πt
|
2 0
|2π cos πt|dt = 8 m
|
˙ 1.107 Since the particle starts from rest, r(0) ˙ = 0, θ(0) = 0 r(0) = 1, and θ(0) = 0. ˙ To determine v(t) and a(t) from eqs.(1.103) we need r(t), θ(t), r(t) ˙ and θ(t) which we can calculate by integrating ¨r and θ¨ t 0
t 0
x
t
− r(0) ˙ = r¨(x)dx = 2e− dx = 2(1 − e− ) r(t) − 1 = 2 (1 − e− )dx so that r(t) = 1 + 2(t + e− − 1) = −1 + 2t + 2e− ˙ r(t)
t 0
Likewise ˙ 0=π θ(t)
−
x
t
t 0
t 0
dx = πt and θ(t) − 0 = π xdx = 46
πt2 2
t
Now from eqs. (1.103) t
t
− e− )ˆe + [(2e− + 2t − 1)πt]ˆe a(t) = [2e− − (2e− + 2t − 1)π t ]ˆ e + [(2e− + 2t − 1)π + 2(2 − 2e− )πt] v(t) = 2(1
t
r
θ
t
2 2
t
r
t
2
1.108 Let r(t) = 1 + 2(t + e−t 1) and θ(t) = πt2 . Then define x(t) = r(t)cos θ(t), y(t) = r(t)sin θ(t) and plot (Mathcad solution)
−
t
0 , 0 .01 .. 2
r t
1
2. t
r t . cos θ t
x t
e
t
2
1
θ t
π .t 2
r t . sin θ t
y t
2
3
2
1
0
1
2
3
4
y t 2
4 x t
FIGURE S1.108
The distance traveled is The Matlab code is:
2 0
|v(t)|dt = 14.438 m
EDU>linspace(0,2); EDU>r=1+2*(t-exp(-t)-1);th=(pi*t.ˆ2)/2; EDU>x=r.*cos(th);y=r.*sin(th); EDU>plot(x,y) 1.109 Given ω = 2π rad/s and r(t) = r0 + ra sin2πt, to determine the acceleration ˙ Since ω = 2π, dθ = 2πdt and θ = θ0 + 2πt, requires expressions for r, θ, r˙ and θ. θ˙ = 2π, and θ¨ = 0. Likewise r˙ = 2πra cos2πt and r¨ = 4π2 ra sin2πt. From eq. 1.103, ar = r¨ rθ˙2 = 4π2 ra sin2πt (r0 + ra sin2πt)4π2 = 4π2 (2ra sin2πt + r0 ) m/s2 aθ = r θ¨ 2r˙ θ˙ = (2)(2πr a cos2πt)(2π) = 8π2 ra cos2πt m/sec2
−
−
−
−
−
47
−
1.110 Solution: a) Let r0 = 2, ra = 1.5 < r0 and θ0 = 0, r(t) = 2 + 1.5sin2πt, θ(0) = 0 so θ(t) = 2πt. For one revolution let t = 0, 0.01...1 sec, then let x(t) = r(t)cos θ(t) and y(t) = r(t)sin θ(t) which is plotted below using Mathcad t
0 , 0 .01 .. 2 r t
1.5 . sin 2 . π . t
2
2.π .t
θ t
90 120
60
150 r t
30
180
0
1
2
210
330 240
300 270 θ t
FIGURE S1.110
The Matlab code for this plot is EDU>linspace(0,2); EDU>r=2+1.5*sin(2*pi*t);th=2*pi*t; EDU>x=r.*cos(th);y=r.*sin(th), EDU>plot(x,y) ˙eθ so that: b) The velocity is v(t) = rˆ ˙ er + r θˆ
˙ |v(t)| = | r˙ + r θ | = π 1.5cos 2πt + (2 + 1.5sin2πt) so s = |v(t)|dt = 14.407 m 2
2 2
0
3
2
2
1 0
48
1.111 Solution: ar = 2t, aθ = cos(πz), r(0) = .5 m, θ(0) = 0. Since it starts from ˙ rest θ(0) = r(0) ˙ = 0. From eq. (1.103) ¨r r θ˙2 = 2t, r θ¨ + 2r˙ θ˙ = cos(πt) which is a system of 2 coupled 2nd order equations which are nonlinear and inhomogeneous. The initial conditions (4) are given above. To solve numerically ˙ follow sample 1.22 (use x = r, ˙ y = θ)
−
r¨ = rθ˙ 2 + 2t θ¨ =
−2
r˙ θ˙ r
+
x˙ = r y 2 + 2t r˙ = x : y˙ = 2xy/r + cosr πt θ˙ = y
·
cos πt r
−
The Euler formula becomes (tn+1 = tn + ∆t) xn+1 = (rn yn2 + 2tn )∆t + xn rn+1 = rn + xn ∆t yn+1 = ( 2xn yn /rn + cosrnπtr )∆t + yn θn+1 = yn ∆t + θn
·
−
·
x 0 r = .5 y 0 0
0
0
θ0
0
Once these are solved the polar coordinate r(t) and θ(t) are given by the digital record for rn and θn .
49
1.112 The trajectory is obtained by plotting rn cos θn vs. rn sin θn from above. The Mathcad solution is: i
0 . . 2000
t
i . ∆t
α
v, r,
ω, θ
∆t
1
,t
r
0
r0
0.5
ω
0
θ0
xi
ri . cos θ i
yi
2. v. ω
2
vi
1
ri
1
θ
π. t
r. ω
ω
0
0
. cos
2. t
a v, r, ω, θ , t
v0
0.001
i
1
i
1
a v ,r , i
i
ω
i
,
α
i
, t . ∆t i
vi , ri ,
ω , θ i i
ω . ∆t
θ i
i
ri . sin θ i
Trajectory in meters 0.8
0.6
y
i
0.4
0.2
0 0.5
1
1.5
2 x i
2.5
FIGURE S1.112
50
v
i
vi . ∆ t
ri ω i
θ
3
3.5
, t i . ∆t
The equivalent Matlab code is: function xdot=onept112(t,x); xdot=[x(2);x(2)*x(3)ˆ2+2*t;x(4);-2*x(1)*x(3)/x(1)+(cos(pi*t))/x(2)]; In the command window: EDU>tspan=[0 2]; EDU>x0=[0;5;0;0] EDU>[t,x]=ode45(‘onept112’,tspan,x0); EDU>xc=x(:,2).*cos(x(:,4));ys=x(:,2).*sin(x(:,4)); EDU>plot(xc,ys)
1.113 The acceleration components in polar coordinates are given in eq. (1.04) to be ˙ ar = r¨ rθ˙ 2 and aθ = rθ¨ + 2r˙ θ.
−
˙ First We are given ¨r and θ¨ which we need to integrate to get r˙ , r, θ and θ. consider θ¨ = 0 so that θ˙ = constant = 1.5 rad/s. Integrating again yields θ(t) = 1.5t. Then the above becomes simply 2
ar = r¨
− (1.5) r, a = 2(1.5)r˙ = 3r˙ Integrating r¨ = 3 − 0.01r˙ requires the solution of θ
r¨ + 0.01r˙ = 3
3t Which is a second order differential equation with particular solution rP = 0.01 . The homogeneous equation is r¨ + 0.01r˙ = 0 which has solution r1 = A and r2 = Beλt . Substitution yields
λ2 + 0.01λ = 0 or λ = rH (t) = A + Be −0.01t
−0.01 so the homogeneous solution is
and the general solution is r = rH + rP = A + Be−0.01t
− 300t
To get A and B apply the initial condition r(0) = 0.4 and r(0) ˙ =0 ˙ = 0.01B + 300 = 0 or B = r(0) = A + B = 0.4, r(0) A = 30, 000.4 Thus 0.01t
−30, 000.4 − 30, 000e− + 300t m Thus r˙ = 300(1 − e− ), r¨ = 3e− so that − 2.25(30000.4 − 30000e− + 300t) a = 3e− + 900t a = (3)(30000.4) − 90, 000e− r(t) =
0.01t
r
θ
0.01t
0.01t
0.01t
0.01t
51
−30, 000
1.114 The trajectory is a part of x(t) = r cos θ versus y = r sin θ. The form of r(t) and θ(t) are given in the previous problem. Let t = 0, 0.1...4 and plot y vs. x(t) as given below from Mathcad t
0 , 0.1 .. 4
r t
30000.4
θ t
1.5. t
30000 . e
0.01. t
300. t
90 120
60
150 r t
30
180
0 0
1000 2000
210
330 240
300 270 θ t
FIGURE S1.114
The Matlab code is: EDU>t=linspace(0,4); EDU>th=1.5*t;r=3000.4-3000*exp(-0.01*t)+300*t; EDU>x=r.*cos(th);y=r.*sin(th); EDU>plot(x,y)
1.115 In order to determine the velocity and acceleration from eq. (1.103), r(t), r(t), ˙ r¨, ˙ ¨ ˙ ¨ θ, θ and θ are needed. Since θ = π/4 so that θ = 0 and θ(t) = (π/4)t (assuming so that r(t) ˙ = 15π sin πt and θ(0) = 0). Then r(t) = r(θ(t)) = 100 + 60 cos πt 4 4 15π πt cos 4 . r¨ = 4
−
−
2
52
Thus from eq. (103) v(t) = ( 15π sin πt )ˆ + ( π4 )(100 + 60 cos πt )ˆ 4 er 4 eθ
and
−
(t) = −
a
a(t) =
15 2 cos πt 4 π 4
2
− 100 + 60 cos eˆ + − π 4
−(100 + 120 cos
πt π 2 ) ˆ 4 16 er
πt 4
−[
15π 2 2
r
15 2 sin πt4 2 π
sin πt4 ]ˆeθ
eˆ
θ
1.116 Here we need to find r(t) from the drawing and knowledge of θ(t). θ(t) = π sin πt, so that θ˙ = π4 cos πt and θ¨ = π4 3 sin πt = π2 θ(t). From the drawing 4 300 = r(t)cos θ(t) so that 2
−
−
r(t) = 300 sec θ(t) = 300 sec ( π4 sin πt) ˙ = 75π2 tan( π4 sin πt) sec ( π4 sin πt)cos πt r(t)
r¨(t) = 300
π2 16
2
sin
πt tan( π4
sin πt) +
π4 16
Then
cos
2
πt(tan2 ( π4
sin πt) − 1) sec(
π 4
sin πt)
v = [75π 2 tan( π4 sin πt)sec( π4 sin πt)cos πt]ˆ er + [75π2sec( π4 sin πt)cos πt]ˆeθ
¨ The aceleration becomes (in terms of θ, θ˙ and θ) ¨ er + 300 sec θ[θ¨ + 2 tan θθ˙2 ]ˆeθ a = 300 sec θ[(2 tan2 θ)θ˙2 + tan θ θ]ˆ
53
1.117 Let θ be the angle between r(t) and the 60 mm line. Let β (t) be the angle between the 100 mm radius and the end of r(t). From this triangle r(t) = 60 cos θ + 100 cos β and from the law of sines: (1) 60 sin θ = 100sin β . Differentiation of the law ˙ . These last two expressions can be of sines yields (2) 60 cos θθ˙ = 100cos β β used to remove the β dependence. Differentiating r(t) and using (1) and (2) to ˙ term yields remove the β ˙ = 60 cos θ tan β θ˙ r˙ = 60 sin θθ˙ 100 sin β β
−
−
−
where we note that eq. (1) can be used to remove the β dependence in favor of θ. Now θ˙ = constant = π is given, so r˙ becomes 1
−60π sin θ − 60π cos θ tan β, where β = sin− (0.6sin θ). From (1) r¨ = −60π cos θ + 60π sin θ tan β − r˙ =
2
602 π 2 cos2 θ 100cos3 β
2
which can be verified symbolically using one of the codes. With r, r˙ and r¨ given, θ˙ = π, so that θ = πt and θ¨ = 0, then a can be determined from equation (104): ˙ ar = r¨ rθ˙ 2 and aθ = 2r˙ θ˙ = 2πr.
−
The Mathcad code for plotting this follows: 2 ar t ddr t r t .π aθ t 2. dr t . π Accelerations vs time 2000
2
s / m n a r t i n o i t a θ t a r e l e c c A
1000
0
1000
2000
3000 0
0.5
1 t time s
FIGURE S1.117
54
1.5
2
1.118 From the problem statement α(t) = 0.5cos t. Integrating yields ω(t) = 0.5sin t+ ω0 = 0.5sin t since it starts from rest. Integrating again (θ0 = 0) yields θ(t) = 0.5cos t + C = 0.5cos t + 0.5. Now recall the formulation of the previous problem β (t) = a sin(0.6sin(θ(t)). Then using the formulas developed in problem 1.117, ar and aθ can be determined. These are illustrated in the figure where the derivatives are confirmed by symbolic computations.
−
t
−
β t
0.5 . cos t
θ t
0 , 0.01 .. 7
asin 0.6 . sin θ t
0.5 . sin t
ω t
0.5
60 . cos θ t
r t
α t
0.5 . cos t
100 cos β t
First symbolicaly calculate the time derivative of r(t) and define it by rd(t) d 60 . cos θ t dt
100 cos β t
d 60 . sin θ t . θ t dt
100. sin β t
.d β t dt
d 60. sin θ t . θ t dt
rd t
d 100. sin β t . β t dt
Next symbolically calculate the second deriviative and denote it by rdd(t) d dt
d 60 . sin θ t . θ t dt
d θ t 60. cos θ t . dt rdd t
2
d 100. sin β t . β t dt d2 θ t 60. sin θ t . d t2
d 60. cos θ t . θ t dt
2
d β t 100. cos β t . dt
d2 60. sin θ t . θ t d t2
2
d2 β t 100. sin β t . d t2
d 100. cos β t . β t dt
Now define the acceleration components in terms of r and its derivatives, theta and omega: ar t
rdd t
2 r t . ω t
aθ t
2 . rd t . ω t
r t .α t
100
50 ar t aθ t
0
2
4
6
50
100 t
FIGURE S1.118
55
8
2
d2 100. sin β t . β t d t2
1.119 The acceleration and velocity in cylindrical coordinates are given in eq. (1.109) ¨ Here r = 1.5, θ˙ = π and z = and (1.111) and require r, r, ˙ r¨, z, ˙ z¨, θ˙ and θ. 0.5cos2πt so that r˙ = r¨ = 0, θ¨ = 0, θ¨ = 0, z˙ = π sin2πt and z¨ = 2π2 cos2πt. Thus
−
v = 1.5πˆ eθ
−
− π sin(2πt)ˆe m/s a = −1.5π eˆ − 2π cos(2πt)ˆ e m/s 2
z
r
2
2
z
The Mathcad code for generating this plot is: i
0 . . 40
xi
t
1.5. cos π. ti
0.1. i
1. 5 sin π . ti
yi
0.4 0.2 0 0.2
1
0. 5. cos 2. π . ti
zi
0
1
1
0 1
x, y, z FIGURE S1.119
The Matlab code for forming this plot is given in the following file: i=(0:1:40); x=1.5*cos(pi*i*0.1);y=1.5sin(pi*i*0.1);z=0.5*cos(pi*i*0.1); plot3(x,y,z)
56
1.120 The distance traveled is calculated from ds = v or s = s0 + 0t v dt. Assuming dt the particle starts out s0 = 0, s = 02 (1.5π)2 + (π sin2πt)dt = 10.398 m.
||
| |
The total distance traveled can also be calculated using software. For example, the Mathcad code for computing this distance follows: t
0 , 0.01 . . 2 0 2
1.5. π
v t
v t
π . sin 2. π . t
d t = 10.398
0
FIGURE S1.120
1.121 Solution: r(t) = R hz R, = R h−0.1ht R = R R + 0.1Rt = 0.1Rt so that h R r˙ = 10 and r¨ = 0. θ(t) = 2πt so that θ˙ = 2π and θ¨ = 0.
−
z(t) = h(1 Thus
−
−
− 0.1t) so that z˙ = −0.1h and z¨ = 0.
v = 0.1Rˆ er + (0.1Rt)(2π)ˆeθ
− 0.1hˆe = 0.1Rˆe + 0.2πRtˆe − 0.1hˆe , and a = (−0.1Rt)4π eˆ + ( · 2π)ˆ e = −0.4Rπ tˆ e + 0.4πRˆ e. The particle reaches the bottom of the cone when z(t) = 0 or h(1 − 0.1t) = 0 2
r
R 5
z
r
2
θ
θ
r
z
θ
or at t = 10 sec. Since θ = 2πt, as 10 sec have pased, then θ has gone around 10 times before it reaches the bottom (2π10 = 20π or 10 complete cycles).
1.122 Solution: for R = 3m and h = 5 m, v(t) = 0.3ˆer + 0.6tˆ eθ 0.5ˆez so that v = (0.3)2 + (0.6t)2 + (0.5)2 as it takes 10 sec to travel to the bottom
|| |v|dt = (0.3) + (0.6t) + (1.5) dt = 94.67 m s= 10 0
10 0
2
2
−
2
1.123 From eq. (1.111) the given form of the differential equations are ˆer : 3 = r¨ r θ˙2 (1), eˆθ : 2r˙ θ¨ = 0.1 (2) and from eˆz : z¨ = 2. Since the particle starts from rest at zero, all the initial conditions are zero and since the z coordinate is decoupled it can be directly integrated to yield z˙ = 2t and z¨ = t2 . Equations (1) and (2) for r and θ on the other hand are two coupled nonlinear equations which must be solved numerically. They are r¨ = rθ˙ 2 + 3
−
θ¨ =
− −
−2
r˙θ˙ r
+
0.1 r
57
−
Putting these in 1st order form or Euler integration yields
r˙ 0 r = 1 , θ˙ 0 0
θ
r˙ rθy˙ + 3 r = dr θ˙ 0.1/r − 2drdθ/r 2
θ
dθ
The Euler equations then becomes (follow sample 1.22) as illustrated in
r dr θ
n+1 n+1
n+1
dθn+1
r + dr · ∆t r 0 = (r · dθ + 3) · ∆t + dr , dr = 0 dθ + θ +−dθ2 · ∆t· · ∆t θ 0 n
n
n 2 n
n
n 0.1 rn
n drn dθn rn
0
n
0
0
0
dθ0
Figure S1.123 shows the integration along with a plot of the first 3 s using Mathcad. i
0 .. 3000
∆t
0.001 ∆t. i
ti
ar r , dr ,
θ,
dθ
3
r . dθ2
aθ r , dr ,
θ,
dθ
1.
0.1
r0
dr0
1
dri ri dθi θ
i
r
0
0
θ
0
1
i
0
i
aθ ri , dri , θi , dθi . ∆t θ dθ . ∆t
dθi
1
dθ0
ar ri , dri , θi , dθi . ∆t r dr . ∆t
dri
1
2 . dr . dθ
1
i
i
90 120
60
150 r i
30
180
0
5
0
10
210
330 240
300 270 θ
i
FIGURE S1.123
58
The required Matlab code is as follows: function xdot=onept123(t,x); xdot=[x(2);3+x(1)*x(4)ˆ2;x(4);(0.1-2*x(2)*x(4))/x(1)], In the command window: EDU>tspan[0 3]; EDU>0=[1;0;0;0]; EDU>[t,x]=ode(‘onept123’,tspan,x0); xc=x(:,1).*cos(x(:,3));yc=x(:,1).*sin(x(:,3)); plot(xc,yc) ˙ ¨ 1.124 Since R(t) = 0.3 + 0.1t2 , R(t) = 0.2t and R(t) = 0.2. Since θ(t) = π sin(πt), 2 3 ˙ = π cos πt and θ(t) ¨ = π sin πt. Since φ(t) = π te−t , φ(t) ˙ = π (e−t te−t ) θ(t) 2 2 π π −t t t t − − − ¨ and φ(t) = 2 ( e + ( e + te )) = 2 e (t 2). Now from equation (1.116)
−
−
−
v(t) = 0.2tˆ eR + π2 (0.3 + 0.1t2 )e−t (1
−
−
2
− t)ˆe +[0.3 + 0.1t ][sin( φ
πt t )](π2 cos πt)ˆeθ 2 e
−
From Eq. (1.118) 2
π 2
t 2
2
πt 2
t
2
− (0.3 + 0.1t )( (1 − t)e− ) − (0.3 + 0.1t )[sin( e− )] (π +[(0.3 + 0.1t )( (t − 2)e− ) + 2(0.2t)( e− (1 − t)) −(0.3 + 0.1t )(π cos πt) sin( e− )cos( e− )] eˆ +[(0.3 + 0.1t )(−π sin πt) sin( e− ) + 0.4t(π cos πt) sin( e− ) +2(0.3 + 0.1t )( e− (1 − t))(π cos πt) cos( e− ]ˆe
a(t) = [0.2
2
2
π 2
2
2
2
π 2
t
2
t
t
t
φ
πt 2
2
πt 2
2
cos2 πt)]ˆeR
t
πt 2
t
πt 2
3
π 2
πt 2
4
t
t
θ
1.125 From Sample Problem 1.24 we have a = g sin φˆ eφ , and we are given that R = 200 ˙ mm, (Rθ)0 = 600 mm/s and that φ0 = π/2, θ0 = 0, g = 9810 mm/s2 . Since h is constant (R = 200 θ˙0 = 3), h = θ˙0 sin2 φ0 = 3 rad/s. Now from the geometry of spherical coordinates the relation to rectangular coordinates is (from fig. 1.21)
⇒
x(t) = R sin φ cos θ, y = R sin φ sin θ, z = R cos φ Now to solve the problem we need to integrate (via Eulerian) the two coupled equations θ˙ = h and φ¨ = h cos φ + g sin φ 2
sin2 φ
sin2 φ
R
subject to the initial condition φ0 = π2 , θ0 = 0, φ˙ 0 = 0.
59
The Euler formulation and a plot of the motion is given in the following Mathcad code: i
0 .. 800
∆t
0.001
2 h . cos φ
aφ φ
sin φ
3
R
200
g
9810
. sin φ
2
sin φ dφ0
xi
R
h
3
dθ φ
dφi
g
2
∆t . i
ti
0
1
θi
1
θ0
2
dφi . ∆t
φi θi
0
aφ φi . ∆t
dφi
1
φi
π
φ0
dθ φi . ∆t
R .sin φi . cos θi
yi
R . sin φi . sin θi
zi
0
100 0 100 0 100 200
x, y, z
FIGURE S1.125
60
100
R .cos φi
The Matlab code is (using an Euler method): x(1)=0;p(1)=pi/2,th(3)=0;h=3;R=200;g=9810 dt=0.001; for n=1:800; x(n+1)=x(n)+(hˆ2*cos(p(n))/(sin(p(n))ˆ2 + (g/R)*sin(p(n))*dt; p(n+1)=p(n)+x(n)*dt; th(n+1)=th(n)+(3/sin(p(n))ˆ2)*dt; end xc=R*sin(p).*cos(p);yc=R*sin(p).*sin(p);zc=R*cos(p); plot3(xc,yc,zc) 1.126 This is a repeat of problem 1.125 for the case that θ˙0 = 0. Since θ˙0 and the constant of motion θ˙ sin2 φ = h must hold for all θ, h must be zero and the kinematic equations become φ¨ = Rg sin φ and θ˙ = 0 which are numerically integrated in the following figure (Mathcad): 0 .. 800 ∆t
i
dθ φ
0
dφ0
0
dφi
1
φi
1
θi
1
xi
0.001 g
aφ φ
R π
φ0
3
R
200
g
9810
0
aφ φi . ∆t
dφi
dφi . ∆t
φi θi
h
. sin φ θ0
2
∆t. i
ti
dθ φi . ∆t
R . sin φi . cos θi
yi
R .sin φi . sin θi
zi
R .cos φi
0
100 00.4 0.6 0.8 1.2 100 0 100 200
x, y,z
FIGURE S1.126
61
The Matlab code is contained in the following file which when run will produce an identical plot: x(1)=0;p(1)=pi/2;th(1)=0;h=3;R=200;g=9810; dt=0.001 for n=1:800; x(n+1)=x(n)+(g/R)*sin(p(n))*dt; p(n+1)=p(n)+x(n)*dt; th(n+1)=th(n); end xc=R*sin(p).*cos(p);yc=R*sin(p).*sin(p);ac=R*cos(p); plot3(xc,yc,zc) 1.127 From the solution to problem 1.126, φ¨ = Rg sin φ for the case that the initial velocity in the circumferential direction is zero ( h = 0). We are given that φ = π + β , β small so that using the trig identity for sin(π + β ) we have sin φ = sin(π + β ) = sin π cos β + cos π sin β = sin β . But since β is assumed small, we can use the small angle approximation: sin β = β , and our differential d ¨ ¨ equation becomes (φ = dt (π + β ) = β )
−
2
−
−
2
¨ g β = 0 β + R which was solved analytically in Sample Problem 1.9. The solution is ˙0 sin( g t) β (t) = β 0 cos( g t) + g β
R
R
R
˙ 0 are the initial angle and angular velocity given to the particle. where β 0 and β
1.128 From the value of a in spherical coordinates, the following 3 equations result ¨ Rφ˙ 2 Rθ˙2 sin2 φ 12 = R
−
− 4 = Rφ¨ + 2R˙ φ˙ − Rθ˙
2
sin φ cos φ 5 = Rθ¨ sin φ + 2R˙ θ˙ sin φ + 2Rφ˙ θ˙ cos φ which can be numerically integrated subject to the initial conditions ˙ =2 R(0) = 2 R(0) ˙ ˙ = 2 (since vθ = 4 = R(0) sin φ(0)θ(0)) θ(0) = 0 θ(0) ˙ ˙ + 2θ(0) ˙ + 1) = 1/2 (since vφ = R(0)θ(0) φ(0) = π/2 φ(0) by direct comparison with v(0) = 2ˆ er + ˆeφ 4ˆeθ .
−
−
−
62
Once one numerically integrated, the expressions x = R sin φ cos θ, y = R sin φ sin θ and z = R cos φ can be used to construct a 3-D plot of the motion as described in the figure that follows using Mathcad. 0 .. 1000 ∆t
i
0.001
aR R , vφ , vθ , φ
1 R
2
vRi Ri
1
vφi
1
φi vθi
θi
xi
1
R0
1
2 . vR . v φ
Ri
π 2
vθ0
2 . R . vφ . vθ . cos φ
θ0
2
0
vRi . ∆t
aφ Ri , vRi , vφi , vθi , φi . ∆t φ vφ . ∆t i
vθi
2 . vR . v φ . sin φ
φ0
0.5
2 R . vθ . sin φ . cos φ
aR Ri , vφi , vθi , φi . ∆t
vRi
1 1
. 4
vφ0
2
vφi
2 2 R . vθ . sin φ
1 . 5 . R sin φ
aθ R , v R , vφ , vθ , φ
vR0
2
R . vφ
12
aφ R , v R , vφ , vθ , φ
∆t . i
ti
i
aθ Ri , vRi , vφi , vθi , φi . ∆t θ vθ . ∆t i
Ri . sin φi . cos θi
i
Ri . sin φi . sin θi
yi
zi
Ri . cos φi
0 1 2 3 15
5
10 5
0
0
x, y, z
FIGURE S1.128
63
The required Matlab code is: R(1)=2;vR(1)=2;p(1)=pi/2);vp(1)=0.5;th(1)=0;vth(1)=-2; dt=0.001 for n=1:1000; R(n+1)=R(n)+vR(n)*dt vp(nt)=vp(n)+((1/R(n))*4-2*vR(n)*vp(n)+R*vth(n)ˆ2*sin(p(n)*cps)xdt; p(n+1)=p(n)+vp(n)*dt; vth(n+1)=vth(n+1)+(1/(R(n)*sin(p(n)))*(5-2*vR(n)*vp(n)*sin(p(n)) -2*R(n)*vp(n)*vth(n)*cos(p(n))*dt; th(n+1)=th(n)+vth(n)*dt; vR(n+1)=vR(n)+(12+R(n)*vp(n)ˆ2+R(n)*vth(n)ˆ 2 * sin *p(n))+ dt; end x=R.*cos(p).*sin(th);y=R.*sin(p).*sin(th);z=R.*cos (th); plot3(x,t,z) 1.129 We are given xA = 2t + 6 and xB/A = 3t2 + 2t + 3. From equation (1.120) xB = xB/A + xA = 3t2 + 4t + 9 so that x˙ B = vB = (6t + 4) m/s, and aB = 6 m/s2 . Likewise vB/A = 6t + 2 and aB/A = 6 m/s2 from straight forward differentiation.
1.130 Since the particles start from rest vA (0) = vB (0) = 0 and since vB/A(0) = vB vA , vB/A(0) = 0. Since xA = 3t + 2 and xB = 6t2 + 2, vA = 3 and aA = 0, vB = 12t and aB = 12. Since aB and aA are both constant and different in value: the two particles never have the same acceleration. The two particles reach the same position when xB/A = 0. xB/A = 6t2 + 2 (3t + 2) = 6t2 3t = 0 when t = 0 and when t = 1/2s. Since xB/A = 6t2 3t, vB/A = 12t 3 so that they have the same velocity when t = 3/12 or t = 1/4s.
−
−
−
−
−
1.131 Here you must pick t0 carefully as the cars do not start moving at the same time. Car B moves with a constant velocity of vB = 15 m/s so that xB = 15t + xB (0) for all time. Car A on the other hand, starts moving when car B crosses a position 100 meters out. Starting the clock at 0 when car B crosses the 100 m mark and taking the position x = 0 m as the reference point we can write xB = 15t+100, since vB = 15 m/s implies that xB (t) = 15t+xB (0) = 15t+100. Now consider car A. We have that xA (t) = t2 until it reaches the speed of 20 m/s. This happens when 2t = 20 or at t = 10 sec. Note that xA (10) = 100 m,
64
so that the cars cannot meet until after t > 10 s (i.e., car B started 100 m out). For t > 10 s, we are given that vA = 20 m/s. Integrating yields xA xA (10) dx
=
t 10
20dt or x (t) = 20t − 200 + x (10) A
A
But xA (10) = 100. Then for t > 10, xA (t) = 20t 100. Now for t > 10, xB/A = 15t + 100 (20t 100) = 5t + 200. Then xB/A = 0 yields tm = 40s, the time at which the cars meet. They have traveled at distance of xA (40) = (20)(40) 100 = 700 m which can be checked by calculating xB (40) = (15)(40)+100 = 700 m.
−
−
−
−
−
1.132 Particle B reverses direction when the velocity is zero. Since xB = 12 + 18t 18 4.9t2 , vB = 18 9.8t = 0 at tr = 9.8 = 1.83s, at this time particle B changes directions. The two particles will meet when xA = xB or 5+2t = 12+18t 4.9t2 , which has solution tc = 0.391, and 3.656s. So they collide at tc = 3.656s. With relative velocity vB/A (tc ) = 16 9.8t = 19.83 m/s. since tc > tr , the particles collide after B reverses direction as illustrated in the figure.
−
−
−
−
t
−
−
0 , 0.1 .. 4 xB t
18 . t
12
2
4.9 . t
xA t
5
2.t
30
xA t xB t
20
10
0
1
2 t
3
4
FIGURE S1.132
1.133 Let car A be moving to the right with vA (0) = 60 mph = 88 ft/s, xA (0) = 0, and aA = 16 ft/s2 . Let car B be moving to the left with xB (0) = 300, vB (0) = 88 ft/s and aB = 15 ft/s2 . Integrating each yields
−
vA = vA (0) xA vB xB
−
t 0
− 16dt = 88 − 16t and = x (0) + 88t − 8t = 88t − 8t = v (0) + 16 dt = −88 + 16t = x (0) − 88t + 8t = 300 − 88t + 8t 2
A
B
B
2
t 0
2
2
65
If they collide xB = xA or 300 88t + 8t2 = 8t2 + 88t or t = 2.109 and 8.891. Now check to see if the cars are still moving at that time. The cars come to rest at vA = vB = 0 or 88 16t = 0 or t = 88 = 5.5 sec. Thus the cars hit at 16 t = 2 sec. At that time
−
−
−
vB/A(t) = vB (2.109)
A (2.109)
−v
= ( 54.256)
−
− (54.256) = 108.5 ft/s.
1.134 Solution: vA (0) = 45 mph = 66 ft/s, vB (0) =58 mph = 85 ft/s, xB/A(0) = 500 ft, taking xA (0) = 0 then xB (0) = 500 ft, aA = 4 ft/s and aB = 2 ft/s. Integrating each yields
−
vA (t) = vA (0) + 2t = 66 + 2t, xA (t) = xA (0) + 66t + t2 = 66t + t2 , vB (t) = vB (0) 2t = 85 2t, xB (t) = xB (0) + 85t t2 = 500 + 85t t2 . The cars meet when xB/A = 0 or when 500+85t 2t2 = 0. This yields two roots for t, one is negative and the other is t = 33.795s. Using the above formulas, the cars meet at
−
−
xA (33.795) = 3, 373 ft
−
−
−
∼ 0.64 miles and their velocities are
vA (33.795) = 133.59 ft/s and vB (35.795) = 17.41 ft/s
1.135 Given: aB/A = 3 m/s2 , vB/A (0) = 0, xB/A (0) = 10 m, and xA = 3t2 + 1 compute aB . Differentiation of xA yields vA = 6t and aA = 6. So that aB/A = 3 = aB aA = aB 6, and solving yields aB = 9 m/s2 . Integrating aB/A = 3 yields vB/A(t) = vB/A(0) + 3t = 3t. Integrating again yields xB/A = vB/A(0) + 3t22 = 10 + 3t2 . Since xB/A = xB xA we have 10 + 3t2 = xB 3t2 1 or xB (t) = 4.5t2 + 11 m.
·
−
−
2
−
−
−
1.136 Since the time is so large in seconds, leave this in mph, miles and hours. For the 747, vB = 575 mph and xB = 575t miles. At t = 3 hr., xB (3) = 575(3). At t = 3 the Concorde takes off and vA = 1336 mph so that xA = 1336t miles. For t 3, xB can be rewritten as xB = (575)(3) + 575t, t 3. The Concorde catches the 747 at xA = xB or (575)(3)+575t = 1336t or t = (575)(3) = 2.27 hrs. (76)
≥
≥
1.137 The pendulum falls as θ (t) = θ cos t, with θ˙ (0) = 0. Consider the vertical (θ = 0) position as the reference point at which we would like to know the time = 1.121 s, t , i.e., solving θ cos t = 0 for t yields t = or t = B
c
0
g ℓ
0
g ℓ
B
g ℓ c
π 2
c
π 2
ℓ g
the time it takes the pendulum to cover the angular distance from 20 ◦ to zero (note this does not depend on the value of θ0 ). Next consider particle A. It 66
starts at rest (xA (0) = 0) 10 meters to the left, so that xA (0) = 10 m. The particle’s acceleration can be written as an = a m/s2 where a is an unknown constant. Integrating yields vA = vA (0) + at = at. Integrating again yields: xA (t) = xa (0) + a2 t2 = a2 t2 10. The value at collision for t will be tc = π2 gℓ
−
so that a2 t2
− −
− 10 = 0 or a =
20 t2c
=
= 15.9 m/s2 .
80g π2 ℓ
1.138 From eq. (1.77) vB = ℓθ˙B = 5θ˙B since the pendulum is in uniform circular motion or radius ℓ = 5 m. Differentiation of the expression for θB (t) yields g g so that vB = ℓθ˙ θ0 gℓ sin . Let tc be the θ˙B = θ0 gℓ sin ℓt ℓt time of impact calculated in problem 1.37. Then from 1.37, vA = atc so that g 20 vB/A = θ0 gℓ sin t t . Substitute tc = π2 gℓ and θ0 = π9 radians (20◦ ) ℓ t c
√ − − √ − − √gℓ + = 20.279 m/s. yields v = 2
c
B/A
π 9
40 π
g ℓ
1.139 The acceleration of each particle is given by aA =
−9.81 − 0.1v
(1) aB =
A
−9.81 − 0.1v
(2)
B
with initial conditions vA (0) = 30 m/s, xA (0) = 100 m vB (0) = xB (0) = 0. Equation (1) becomes x ¨ A + 0.1x˙ A = 9.81 and (2) becomes x ¨ B + 0.1x˙ B = 9.81
−
−
These are second order, uncoupled linear differential equations with analytical solutions. The homogeneous solutions are xA = A, a constant and xA = Be 0.1t or xA (t) = A+Be −0.1t . The particular solution is Ct. Substitution of (xA ) p = Ct into (1) yields C = 98.1. Thus (xA ) p = 98.1t and the total solution is xA = A + Be −0.1t 98.1t. Applying the initial condition yields
−
−
−
100 = A + B from the initial position 30 =
−0.1B − 98.1 from the initial velocity Solving yields B = −681 and A = 781. Thus x (t) = 781 − 681e− − 98.1t and v (t) = 68.1e− − 98.1. 0.1t
A
0.1t
A
Next consider particle B which has the same form but different initial conditions, i.e., xB = A′ + B ′ e−0.1t 98.1t. The initial conditions yield
−
0 = A′ + B ′ 0=
−0.1B′ − 98.1
Solving yields A′ = 981 and B ′ = xB (t) = 981(1
0.1t
− e−
)
−981 so that
− 98.1t and v
B (t)
= 98.1e−0.1t
− 98.1
Computing xB/A (t) and setting this equal to zero yields 0 = 981
− 781 + (−981 + 681)e−
0.1tc
Solving yields tc = ℓn(2/3) −0.1 = 4.0555, the time of collision. The relative velocity at tc = 4.055 is vB (4.055) vA (4.055) = 20 m/s
−
67
The two differential equations for the acceleration of the particles are separable and can be solved by hand. However, to gain a better conceptual understanding of the particle motion, the equations are solved by numerical integration using the following Mathcad code: 0.1 . v
a v 9.81 i 0 . . 4060 0.001 ∆t t i . ∆t vA 0
0
yA 0
100
vB 0
30
yB 0
0
vA i
1
yA i
1
vB i
1
yB i
1
vA i yA i vB i yB i
a vA i . ∆t vA . ∆t i
a vB i . ∆t vB . ∆t i
Position- time of A and B 100
yA yB
50
0 0
1
2
3
4
5
ti time s
yA = 29.241 vB
yB = 29.252
vA 4055 = 19.999
The impact occurs 29.2 m above the original position of B (taken as the origin).
The relative velocity at impact is 20 m/s
68
The Matlab code for this is: function xdot=onept139(t,x); xdot=[x(2);-9.81-0.1*x(2);x(4);-9.81-0.1*x(4)]; In the command window: EDU>tspan=[0 4]; EDU>x0=[100;0;0;30]; EDU>[t,x]=ode45(‘onept139’,tspan,x0); EDU>plot(t,x(:,1),t,x(:,3))
1.140 Let vs denote the velocity of the swimmer starting from rest and vc = 0.1+0.1xs denote the velocity of the river (opposite of vs ). Then vs/c = 1.5 m/s = vs vc s so that vs = vc + 1.5 = 0.1 0.01xs + 1.5 or dx + 0.01xs = 1.4 which can dt be solved by use of an integrating factor for xs . Here p(t) = 0.01, f (t) = 1.4, λ(t) = e .01dt = e0.01t so that xs (t) = e0.01t [ e0.01t (1.4) + C 0 ] = 140 + C 0 e−0.01t . Since xs (0) = 0, C 0 = 140 and xs (t) = 140(1 e−0.01t ), xs (60) = 140(1 e−0.6t ) = 63.2 m.
−
− −
i
−
0 . . 14000
v x, t
x
1.4
x0
0
i+1
xi
∆t
−
−
0.01
0.01. x
v xi , t . ∆t
150
100 x i
50
0 0
50
100 t
FIGURE S1.140
69
i
150
The Matlab code is: x(1)=0;dt=0.001,t(1)=0 for n=1:14000 x(n+1)=x(n)+(1.4-0.01*x(n))*dt t(n+1)=t(n)+dt end plot(t,x) 1.141 Since xs (t) = 140(1 e−0.6t ) from problem 1.40, the swimmer will reach 100 m at time t that satisfies 100 = 140(1 e−0.6t ) or t = −ℓn(1/7) = 125 s. 0.1
−
−
1.142 Given vA (t) = jˆ m/s (constant), vB (0) = 0, aB (t) = 2[cos 30◦ˆi + sin 30◦ j] n/s a constant. Also rA (0) = 0, rB (0) = 4ˆi + jˆ m. From va (t) = 3 jˆ. rA (t) = rA (0) + 3t jˆ = 3t jˆ. Integration of aB (t) yields: vB (t) = vB (0) + 1.73tˆi + t jˆ m/s = 1.73tˆi + t jˆ. Integrating again yields rB (t) = 1.73 t2ˆi + 1 t2 jˆ + rB (0) = (0.866t2 4)ˆi + (0.5t2 + 1) jˆ
−
2
−
2
Thus rB/A (t) = (3t + 4 0.866t2 )ˆi (0.5t2 + 1) jˆ. For this to be zero, each component must be zero, which cannot be because the jˆ component is always positive, hence t has no real roots. The distance between the particles is rB/A = (3t + 4 0.866t2 )2 + (0.5t2 + 1)2 . The minimum can be found from
−
−
− [(3 + 4 − 0.866t ) + (0.5t + 1) ] = 0 2(3t + 4 − 0.866t )(3 − 1.73t) + 2(0.5t + 1)(t) = 0 d dt
2 2
2
|
|
2
2
2
which is cubic in t and has positive solutions. Then t = 0 or rB/A = 4i = jˆ is when the two particles are closest. As time evolves they move apart (roots found using Mathcad). 1.143 Using y up along north and x to the right along east, the airplane has relative speed of vA/W = 200(cos θˆi + sin θ jˆ) mph where θ is the unknown direction. The wind velocity is vw = 50(cos 45◦ˆi + sin 45◦ i) = √502ˆi + √502 jˆ = constant. We would like vA = 200ˆi. From the definition then vA = vW + vA/W = (50)[(0.707)ˆi + (0.707 jˆ)] + 200(cos θˆi + sin θ) jˆ
which results in the two scalar equations vA = 35.4 + 200 cos θ 0 = 35.4 + 200 sinθ Solving yields: θ =
−10.2◦ and v
A
= 232.2 mph.
70
1.144 Note that the flanker travels 40 ft (20 ft/s 2 s = 40 ft) before the quarter back throws the ball at t = 0. Let xf (0) = 40ˆi denote the initial position of the quarter back. For the ball ˆ ab = g k
×
−
so that ˆ vb = v cos θ cos β ˆi + ( gt + v sin θ) jˆ + v cos θ sin β k
−
and
ˆ xb = [v cos θ cos βt + xb (0) ˆi + [ −gt + v sin θt + yb (0)] jˆ + [v cos θ sin βt]k 2 2
where xb (0) = 30 ft (i.e. 30 yards behind the line of scrimmage) and yb (0) = 6 ft. For the flanker x˙ f = 20ˆi, and integrating yields ˆ xf (t) = (20t + xf (0))ˆi + 8 jˆ + 60k
−
Setting xb = xf yields the three equations from ˆi: 20t + 40 = v cos θ cos βt 30 from jˆ: 8 =
−
gt2 2
+ v sin θt + 6
−
ˆ : 60 = v cos θ sin β . from k Solving these three nonlinear equations in the three unknowns t, β , v for θ = 30◦ using a numerical procedure yields v = 69.719 ft/s, β = 28.152◦ and t = 2.106. From the expression for xf the yards gained on the play are (20)(2.106)+40 3
= 27.375 yd
1.145 Following sample 1.28, let vs be the absolute velocity of the swimmer. With ˆi perpendicular to the shore and jˆ up stream, vs = vˆi, where v is the unknown absolute speed of the swimmer. Let θ be the angle that vs/w (swimmer relative to the water) so that vs/w = 6 cosθˆi + 6 sin θ jˆ (ft/s) vw =
−5 jˆ ft/s, so that
vs = vw + vs/w becomes
vˆi =
−5 jˆ + 6 cos θˆi + 6 sin θ jˆ
from ˆi: v = 6 cos θ from jˆ: 0 = 6 sin θ
−5
which is 2 equations in two unknowns: θ and v. Solving yields θ = 56.4◦ , = 60.3 sec. v = 3.3 ft/s. The time to travel 200 ft is t = dv = 200 3.3
71
1.146 Solution: We can form a block model of the United States as shown. Seattle
21.80 45 0
450
1200 mi
1200 mi
600 mi
1200 mi
Miami
The direct flight from Seattle to Miami would be along a pathSouth21.8 0 of East. The wind directionfor the 1200 miles was estimatedto be South 450 East, during the next 600 miles the wind direction is to the East and during thelast 1200 miles, the wind is North 450 East. The wind speedis estimatedto be 50 mph. The generalrelative motionequationfor any segmentof the flight is: v P = v w + v p / w For the first leg of the flight, thesevectors are:
[ ] = 50[cos(45) ˆi − sin( 45) ˆj ] = 300[cos(θ ) ˆi + sin(θ ) ˆj ]
v p = vp cos(21.8) ˆi − sin( 21.8) ˆj vw
v p / w
The scalar equationsof relative motionare solved using the Given-Find functionin Mathcad: Make an initial guessfor the velocity of the plane andthe bearingof the plane vp 300 10. deg θ Given v p. cos 21.8. deg 50. cos 45. deg 300. cos θ v p. sin 21.8. deg 50. sin 45. deg 300. sin θ
72
1.147 Solution: Seattle
21.80
1200 mi
Wind Vel
750 mi
250 mi
E
S 85 0 E
500 mi E
1000 mi
500 mi
N 30 0 E
Miami
S 30 0 E
The solution will be made using the Given - Find function of Mathcad: Leg 1.
Make an initial guess for the velocity of the plane and the bearingof the plane vp 300 10. deg θ Given v p. cos 21.8. deg 100. cos 0. deg 300. cos θ v p. sin 21.8. deg 100. sin 0. deg 300. sin θ 390.541 0.505 Find v p , θ = = 28.934 0.505 deg vp = 391 mph
θ =
-30 0
Leg 2.
Make an initial guess for the velocity of the plane and the bearingof the plane vp 300 θ 10. deg Given v p. cos 21.8. deg 100. cos 85. deg 300. cos θ v p. sin 21.8. deg 100. sin 85. deg 300. sin θ Find v p ,
θ
=
331.502 0.078
0.078 deg
73
= 4.469
1.148 Solution: S 10 0 W
Seattle
San Francisco
45 0
450
New York
1200 mi
1200 mi
600 mi
1200 mi
Miami
The flight is now to theWest againstthejet stream. The flight will be divided into three legs. Leg 1.
Make an initial guess for the velocity of the plane andthe bearingof the plane vp 200 θ 10. deg Given v p. cos 10. deg
v p. sin 10. deg Find v p , θ =
50. cos 45. deg 200. cos θ 50. sin 45. deg 200. sin θ 167.082
0.032
0.032
deg
= 1.833
The velocity of the planeandthecourse bearingare: vp = 167 mph
θ = S 2
0
W
Leg 2.
Make an initial guess for the velocity of the plane andthe bearingof the plane vp 200 θ 10. deg Given v p. cos 10. deg 50. cos 0. deg 200. cos θ v p. sin 10. deg 50. sin 0. deg 200. sin θ 150.571 0.131 Find v p , θ = = 7.506 0.131 deg
74
1.149 Solution:
21.80
Chicago
1200 mi
Memphis
Wind Vel
750 mi
250 mi
500 mi
1000 mi
E
S 85 0 E
E
N 30 0 E
500 mi S 30 0 E
The desired flight path from Chicago to Memphis is assumedto assumedto be S 800 E under a 100mph wind N 300 E. Make an initial guess for the velocity velocity of the planeand the bearingof the plane vp 300 θ 80. deg Given v p. cos 80. deg 100 100. cos cos 30. deg 300. cos cos θ v p. sin 80. deg 100. sin 30. deg 300. sin θ 250.701 1.715 Find v p , θ = = 98.262 1.715 deg The velocity of the plane and the course bearingare: vp = 251 mph
θ = S
1.150 From sample 1.29 the time to reach the opposite shore is tf = sec. The expression for y (t) at y (tf ) = 500 becomes
· 500 = 5 10 cos θ 200
2
403 3 cos3 θ
·
−
θ 2 10cos 200
·
402 2cos 2 θ
+ 1010 sinsin θ
82 0 W
2d vB/w cos θ
=
40 cos θ
40 cos θ
which which is transendental transendental in θ and has solution (see numerical root finder) θ = 56. 56.43◦.
75
1.151 From sample 1.29 the angle is determined θ = sin−1
= sin = 7.6◦ (2)(2) (3)(10)
2vw 3vB/w
where vw = 2 ft/s is the velocity of the river.
1.152 From sample 1.30, the 3 equations determining vs , θ and β with vRT = 0.7 become θ + β = 90◦
−0.3cos θ sin β = 0.7 − 0.3sin θ
−v
s
vs
cos β =
Solving using Mathcad yields: θ = 24. 24.8◦, β = 64. 64.3◦ and vs = 0.632 m/ m/s.
1.153 We are given: xB = 2 sin πtˆi + 5t 5t jˆ, xG = 4t cos cos 30◦ˆi + 4t 4t sin30◦ jˆ so that differentiation yields vB = 2π cos πtˆi + 5 jˆ, vG = 4 cos 30◦ˆi + 4 sin30 jˆ and aB =
−2π
2
sin πtˆi, aG = 0
From eq. 1.126, vG/B = vG and aG/B = aG
−a
B
−v
B
= (4 cos30◦
= 2π 2 sin πtˆi
76
− 2π cos πt) (4 sin sin 30◦ − 5) jˆ πt)ˆi + (4
1.154 From the previous problem: xG/A = (4t (4t cos30◦
− 2sin πt) (4t sin sin 30◦ − 5t) jˆ πt)ˆi + (4t
so that the distance between the two as a function of time is ((4t cos30◦ d(t) = x(t) = ((4t
| |
2
− 2sin πt) πt)
+ (4t (4t sin30◦
which is plotted below using Mathcad: θ
30 . deg
t
2 .sin π . t 5.t
xB t
2 1/2
− 5t) )
0 , 0. 0.1 .. 3 4 .t .cos θ 4 .t .sin θ
xG t
0
d t
xB t
xG t
0
15
m e c n a t s i D
10 d t 5
0
1
2
3
t Time s
FIGURE S1.154
1.155 Given xB = 2 sin sin πtˆi + 5t 5t jˆ and xG = 4t cos cos 30◦ˆi + 4t 4t sin30sin
aB/G
πt 2
j
− 4t cos cos 30◦ )ˆi + (5t (5t − 4t sin30◦ sin ) jˆ = (2π (2π cos πt − 4cos30◦ )ˆi + (5 − 2πt sin30◦ cos − 4sin30◦ sin = −2π sin πtˆi + (−4π sin sin 30◦ cos + π t sin30 sin30◦ sin ) jˆ
xB/G = (2 sin sin πt vB/G
ˆ,
πt 2
πt 2
πt 2
2
77
2
πt 2
πt ˆ ) j 2
1.156 From xB/0 B/ 0 of the previous problem: [4π4 sin2 πt + (1 d(t) = xB/G (t) = [4π
|
|
− 4π sin30◦ cos
πt 2
+ πt sin sin 30◦ sin πt2 )2 ]1/2
which is plotted in the figure using Mathcad. 30 . deg
θ
xB t
t
0 , 0. 0.1 .. 3 4 .t .cos θ
2 .sin π . t 5.t
xG t
4 .t .sin θ . sin
0
π .t 2
0
d t
xB t
30
m e c n a t s i D
20 d t 10
0
1
2
3
t Time s
FIGURE S1.156
The Matlab code is: syms t th=pi/6; xb=[2*sin(pi*t);5*t;0];xG=[4*t*cos(th);4*t*sin(th)*sin(pi*t/2);0]; d=sqrt(dot(xb,xG)); ezplot(d,[0,3])
78
xG xG t
1.157 From the given function aA = d (vA ) = ˆi + 1.224 jˆ dt
− 1.224kˆ − 0.2v
A
which is a vector differential equation of the form ˆ=b v˙ A + 0.2vA = ˆi + 1.224 jˆ 1.224k
−
The solution of the homogeneous equation vAH = Ae−0.2t plus a particular 1 ˆ ˆ ) or solution: vp = 0.2 (i + 1.224 jˆ 1.224k vA = Ae0.2t +
− (ˆi + 1.224 jˆ − 1.224k)
1 0.2
where A is evaluated from the listed condition vA (0) = 0 which yields ˆ ) or A = 1 (1.224 jˆ 1.224k ˆ) 0 = A + 1 (ˆi + 1.224 jˆ 1.224k
−
0.2
Then vA = 5(1
− e−
0.2t
−
)(ˆi + 1.224 jˆ
Integration again yields xA
or
A (0)
−x
= 5(t + 5e−0.2t
−
0.2
− 1.224k) m/s
− 5)(ˆi + 1.224 jˆ − 1.224kˆ)
− 1))(ˆi + 1.224 jˆ − 1.224kˆ) ˆ m/s. Integrating yields = 3tˆi − 2cos πt k ˆi − ˆ where v (0) = 0 k
xA = 5(t + 5(e−0.2t
Next consider aB vB
B (0)
−v
=
3t2 2
2sin πt π
Integrating again yields 2 xB xB (0) = t ˆi (1 3
−
2
Now form
π2
− cos πt)kˆ where x
B (0)
− 5(1 − e ) ˆi − 6.12(1 − e ˆ + 6.12(1 − e− ) − sin πt k
vB/A =
and
−
B
3t2 2
0.2t
0.2t
0.2t
=0
) jˆ
2 π
− − 5(t + 5(e − 1)) ˆ − [6.12(t + 5(e− = (cos πt − 1) + 6.12(t + 5(e− − 1)) ˆ + t3 2
xB/A
1 π2
0.2t
i
0.2t
79
k
0.2t
− 1))] jˆ
The figure represents a numerical solution and plots of the magnitude of xB/A in Mathcad: i
∆t
0 .. 3000 vx0
0
x0
0.001
0
vy0
0
0.2 . vx
y0
1
ay vy , t
2 .0.0612
0.2 . vy
az vz , t
0.621. 2
0.2 . vz
xi vyi
1
1
yi
1
vzi
1
zi
ax vxi , ti . ∆t xi vx i . ∆t
vyi
ay vyi , ti . ∆t yi vy i . ∆t
vzi
az vzi , ti . ∆t
vi
vxi
2
0
z0
2
yi
vyi
2
vzi
2
xi
xi
10
x
i
0
vzi . ∆t
zi
1
vz0
2 . cos π . t
vxi
1
0
3.t
ax vx , t
vxi
i . ∆t
ti
2
zi
2
10
v
5
0
1
2 t
3
i
5
0
1
2 t
i
3
i
FIGURE S1.157
The Matlab code ode45 can be used to solve this numerically. The plots of x(t) and r(t) can be obtained symbollically from: sys t x=[tˆ3/2-5*(t-5*(exp(-0.2*b)-1));6.12*(t+5*(exp((-0.2*t)-1));(1/piˆ2)*(cos(pi*t)1)]; Nx=SVD(x); ezplot(Nx(3),[0,3])
80
1.158 Define d(t) = xB/A where xB/A is defined as derived in Problem 1.157. Using Mathcad to plot yields figure S1.157.
|
|
1.159 Since xA starts to move 1.5s after xB , shift the time of xB ahead by 1.5s by replacing t with t + 1.5 in the expression for xB derived in Problem 1.157. This yields 1 ˆ xB = (t+1.5) ˆi (1 cos(π(t + 1.5))k 3
−
3
Then for t
−
π2
≥ 4.5s, − − 5(t + 5(e − 1) ˆ − 6.12 [t + 5(e − 1)] ˆ = + 6.12 [t + 5(e− − 1)] − (1 − cos(πt + 1.5)) ˆ
xB/H
(t+1.5)3 2
0.2t
2 π2
0.2t
and for 0 < t < 1.5 (t+1.5) ˆ 2 xB/A = i (1 3
2
−
π2
0.2t
i
j
k
− cos(π(t + 1.5))kˆ.
The Mathcad solution: i
0 .. 3000
vx 0
0
∆t
x0
ax vx , t
0.001
0
vy0
i . ∆t
ti 0
y0
0.2 . vx . Φ t
1
0
vz0
1.5
3 t
ay vy , t
2 . 0.0612
0.2 . vy . Φ t
1.5
az vz , t
0.621. 2
0.2 . vz . Φ t
1.5
vx i xi vy i
1
ax vx i , ti . ∆t xi vx i . ∆t
vy i
ay vy i , ti . ∆t yi vy i . ∆t
vzi
az vzi , ti . ∆t
1 1
yi
1
vzi
1
zi
vx i
vi
vx i
2
vy i
2
vzi
2
xi
20
x
i
2 . cos
z0
0
π.t
vzi . ∆t
zi
1
0
xi
2
yi
2
zi
2
40
v
10
0
1
2
i
3
t i
20
0
1
2 t i
FIGURE S1.159
81
3
1.160 This is similar to sample problem 1.32. Using the coordinate system suggested in Sample 1.32, yB denotes the distance to the bracket and xA to the man. Then xA + 2yB = ℓ 2vB =
−2v = 1 ft/s, v = −2 ft/s (down)
−v
A
Since vB
or vA =
B
A
1.161 Use the pulley as a reference point, and let xA be the distance from the pulley to A and xB the distance from the pulley to B. Then xA + xB = constant and differentiating yields xA =
B,
−v
and aA =
−a
B.
1.162 Let vc be the cable velocity: vc = 0.5 m/s. Let xT denote the distance from the tree to the end of the truck. The length to the cable is xc + 2xT = ℓ so that vc =
−2v
T
or vT =
−
1 2 vc
=
−0.25 m/s (right)
1.163 Let xA be the distance for the pulley to the man at A, x p be the distance between the top pulley and the bottom pulley. Let xB be the distance from the top pulley to B and let the bracket for the bottom pulley be a height of c. Then xA + 2x p + xB
− c, but x
p
= (xB
− c)
so that xA + 3xB =constant and vA =
−3v
B
with xB , xA pointed down, B traveling upwards yields vB =
−10 ft/s and v
A
=
−3 · (−10) = 30 ft/s (down)
1.164 Let xA denote the distance from the top of the pulley to block A, yB denote the distance from the top of the pulley down to the pulley holding B and y denote the distance from the top of the pulley down to mass c (down position, to the right position). Then the length of the rope is yc + 2yB + xA = constant differentiation yields vA + 2vB + vC = 0 or vc =
−2v − 3 m/s and a = −2a − 1 m/s B
c
A
82
1.165 Let h be the hypothesis of the triangle. Then from the right triangle h2 = 152 + x2
(1)
for any value of x. Also from length of rope ℓ = h + (15
− y) = 30
(2)
Solve this for h = y + 15
(3)
Now substitute (3) into (1) to get: (15 + a)2 = 152 + x2
(4)
for any x and y. Taking a time derivative yields 2(15 + y)y˙ = 2xx˙ Solving for y˙ yields y˙ =
xx˙ 15 + y
At y = 10 (15)2 = 152 + x2 or, x = 20 m Then y˙ =
0.5 m/s = 0.4 m/s, at y = 10 m. 2
20 25
1.166 Let xA extend from the fixed pulley above B to the left of pulley B to mass A and let yB extend (positive) from the same pulley down to the mass at B. Let d be the distance between the two fixed pulleys. Then the length of the rope is ℓ = 2yB + xA + (xA or
− d) + (x − d) A
const. = 2yB + 3xA . Differentiate to get: y˙ B = −23 x˙ A and y¨B =
−
3 x¨ . 2 A
1.167 Use the top fixed pulleys as a reference and let xA denote the distance to mass A, xB the distance to mass B, xc the distance to mass C and denote x p the distance to tohe pulley that is free to move. All distances are vertical with positive downward. There are two separate ropes. Let ℓ1 denote the from A to the movable pulley so that ℓ1 = xA + xB + (xB
83
−x ) p
(1)
Let ℓ2 denote the length of the rope from mass C up to ground so that ℓ2 = xc + 2x p
(2)
Differentiation of (2) yields v p = vc /2 which is substituted into the derivative of (1) to yield 4vB + vc = 2vA m/s
−
−
1.168 Let the top pulley be the fixed frame of reference, and x denote the distance from the reference down (t) to the mass, and let xm be the distance from the reference down (t) to a point on the rope connecting to the motor. Them ℓ = xm + 2x and differentiation yields that v = 12 vm From section 1.7 vm = rw = (0.2)(9.55 rad/s = 1.91 m/s and v = 12 vm = 0.955 m/s
1.169 Let yB denote the distance from the pulley to the mass at A (+ down), let xB denote the distance from the pulley to the collar along the shaft. The length of the rope is ℓ = da + yB . From the right triangle made by “d”, the bar the collar rides on, and the rope: ℓ21. Thus ℓ = d + yB = yB + d2 + x2B . Differentiating yields 0 = y˙ B + 12 2xB x˙ B .
√
d2 +x2B
1.170 Pick a fixed point to left of block A as the reference and define xA and xc (both positive to the right) as the distance from the reference to block A and C respectively. Then define yc to be the distance down from the first pulley to B. The length of the rope ℓ is then ℓ = xA + xc + 2yB anad differentiation yields vA = vc 2vB .
− −
1.171 Solution: ds dt
= 20 m/s = constant so s = 20t and
ds2 dt2
s ) so that θ(t) = 4 sin(0.5t) θ(s) = 4 sin( 2000
From equation 1.147 with x0 = y0 = 0 x(s) =
s 0
s 0
cos(θ(η))dη y(s) = sin θ(η)dη 84
=0
The Mathcad solution: s
0 , 1 0 .. 6 000
s
4 .sin
θ s
2000
s cos θ u
x s
s
du
sin θ u
y s
0
du
0
2 d 20 . θ s ds
an s
Road as viewed from above 2000
1000
y s
m k
4000
3000
2000
1000
0
1000
1000 x s km
2 ^ s / m n o i t a r e l e c c a l a m r o n
1
an s
0.5
0
2000
4000 s distance along road in km
6000
FIGURE S1.171
From (1.146): an =
ds dt
2 dθ(s) ds
4 = (20)2 200 cos
y(s) and an (s) are plotted in the figure.
|
|
= 800 cos s 200
s 200
√
1.172 We are given that a(s) = 5 so that integrating vdv = 5ds yields v(s) = 10s s 2s for zero initial condition. Also since θ = 1 + 2000 = 1000 . But an (s) = , dθ ds 2s 2s = (10s)( 2000 ). Thus a = a2n + a2t = (10s)2 ( 2000 )2 + 25. When v(s)2 dθ ds a = 8 the car spins out so that s must satisfy 2
2
||
2
||
2s 82 = (10s)2 ( 2000 )2 + 25.
2
2
Solving for s yields: s =
√39(2000)
2
20
or s = 1, 118 m.
85
2
2
m hour 5280 ft 1.173 The given constant speed means that ds = 60 hour = 88 ft/s. To dt 3600s m determine the normal acceleration from equation (1.146) we need to calculate d s (ds/dt)2 = 882 and dθ/ds. From the given value of θ: ds + e−5/1000 = 2000
·
1 2000
882 1000
1 s/1000 1000 e
−
Thus an = The Mathcad solution:
−
v
88
0.5 − e−
s
s/1000
= 7.744 − e− ft/s . 1 2
s/1000
2
0 , 10 .. 5000 s
s
s
θ
e
2000
1000
2 d v . θ s ds
an s
Magnitude of Normal Acceleratio 4
2 ^ s / t f
an s 2
0
1000
2000
3000 s distance traveled in f
4000
5000
FIGURE S1.173
1.174 This requires the use of the 3-D formulation of equation 1.150. The component of a normal to the above can be found by taking the dot product of the acceleration with the unit normal vector. To define the unit normal vector can be found from an = a = at = a a t. First we need dθ and dβ . From the given ds ds 1 π πs form: dθ = 1000 (0.5 e−s/1000 ), dβ = 5000 cos 5000 . The Mathcad solution: ds ds
− ·
−
v
88
s
2
0 , 10 .. 5000
s
s
θ s
2 ^ s / t f n i n o i t a r e an s l e c c a l a m r o n
1000
e
2000
an s
v
β s
2.
d θ s ds
2
π . sin
. cos β s 2
π. s 5000
d β s ds
2
20
10
0
1000
2000
3000 s distance traveled in f
FIGURE S1.174
86
4000
5000
1.175 Solution: s
v
88
s
0 , 1 0 .. 5000
s
θ s
e
2000
1000
β s
π . sin
π .s 5000
2 . 1.5 . s
v s
an s
v s
2 ^ s / t f n i n o i t a r e l e c c a l a m r o n
2
d θ s ds
2.
d β s d s
2
. cos β s
2
40
an s
20
0
1000
2000
3000 s distance traveled in ft
4000
5000
FIGURE S1.175
1.176 Solution: s
v
88
s
0 , 1 0 .. 5000
s
θ s
e
2000
1000
β s
π . sin
π .s 5000
1
ρ s d θ s ds t f n i e r u t a v r u c f o s u i d a r
2
. cos β s
2
2
d β s ds
3000
2000
ρ s 1000
0
1000
2000
3000 s distance traveled in ft
FIGURE S1.176
87
4000
5000
1.177 Solution: s
0 , 1 .. 50
v
s
θ s
π
β s
100
2
. sin
π.s 25
10
2 v .
an s
2 ^ s / m n i n o i t a r e l e c c a l a m r o n
2
d d s
2
. cos β s
θ s
2
d d s
β s
20
an s
10
0
10
20
30 s distance traveled in m
40
50
FIGURE S1.177
1.178 Solution: s
0 , 1 .. 100
v s
an s
v s
a s 2 ^ s / m n i n o i t a r e l e c c a l a t o t
2.
d θ s ds
at s
2
π
β s
100
2
π .s
5 . cos
10
s
θ s
. sin
π .s 25
25
2
. cos β s
an s
d β s ds
2
2
at s
d v s ds
2
60
40 a s 20
0
20
40
60 s distance traveled in m
FIGURE S1.178
88
80
100
.v s
CHAPTER 2
2.1 Given: µk = 0.1, W = 200 lb and P = 50 lb. The free body diagram yields:
− mg − P sin 30 = 0 (sum of forces in y-direction) P cos 30 − f = ma (sum of forces in x-direction)
(1)
N
(2)
x
From (1) N = mg + P sin 30 = 200 + 50( 12 ) = 225 lb Friction force f = µk N = 0.1(225) = 22.5 lb Therefore max = 50(
√3 2
)
20.80 200/32.2
− 22.5 = 20.80 lb 2
⇒a = = 3.34 ft/s ⇒ v = 3.34t (assuming initially at rest) ⇒ x = 1.67t (assuming x(0) = 0) x
x
2
Therefore t =
100 1.67
= 7.73 sec
_ F
y o 30
x
f N
FIGURE S2.1
2.2 Solution: µs = 0.4, m = 150 kg, µk = 0.3
− f − mg sin θ = ma N − mg cos θ = ma = 0 ⇒ N = mg cos θ = 150(9.81) cos θ = 150(9.81) √ T
x
y
100 1002 +12
1
100
θ
mg T
x
f N
θ = tan -1 (1/100) = .573o ~ ~ 0.01 rad FIGURE S2.2
90
= 1471 N
f max = µs N = 0.4(1471) = 588.4 N Therefore T max = f max + mg sin θ = 588.4 + (150)(9.81) √10012 +1 = 603 N f = µk N = 0.3(1471) = 441 N ax =
1 (T m
− f − mg sin θ) =
1 150
ax = 0.98 m/s2
603 − 441 − 150(9.81) √
1 1002 +1
Therefore vx = 0.98t m/s assuming zero initial velocity. Therefore t =
5 0.98
= 5.1 sec
2.3 Examining the maximum friction µs = 0.2. This can produce an acceleration of ac = 0.2(32.2) = 6.44 ft/s2 . Therefore the box slips because the force of static friction mac is less than that applied by the truck ac = 0.15(32.2) = 4.83 ft/s2 ac/T = 4.83
2
− 10 = −5.17 ft/s
(We could now determine where the crate falls off if we knew length of tuck bed.) mg
f N FIGURE S2.3
2.4 Solution: a)
F = ma
100
− 50 =
100 32.2 a 2
a = 16.1 ft/s b) 100 T
− T =
− 50 =
100 32.2 a
50 32.2 a 150 a 32.2 150 32.2 a
Therefore 50 = c) 150
− T =
2
⇒ a = 10.73 ft/s 91
100 32.2
− 100 = a ⇒ 50 = a ⇒ a = 6.44 ft/s T
2
250 32.2
50
50
100
100
150
50
T
T
T
T
100
100
50
150
100
FIGURE S2.4
2.5 Solution: N
− mg = ma, or
N = mg + ma for: a = +2 : for: a =
−2.5 :
100
N ↑ = m(9.81 + 2) = 1181 N N ↓ = 100(9.81
− 2.5) = 731 N mg
+
N
FIGURE S2.5
2.6 Solution:
−mg sin30◦ + µmg cos 30◦ = ma along the conveyor. Or: −0.5g + 0.2598g = a, thus: a = −.24g. 92
v=
−0.24gt + v x = −0.12gt + 0.2t 0.12(9.81)t − 0.2t − 10 = 0 0
2
2
t = 3s
Note that this slips immediately as µs g cos(30◦ ) < 0.5g .
10 9.8
30o
f
30o N
FIGURE S2.6
2.7 From S2.7a the constraint equation is 2(xA
B ) + (xc
−x
Thus 2xA
− 3x
B
B)
−x
= constant
= constant and differentiation yields
2aA = 3aB
(1)
Now from S2.7b, Newton’s law in the horizontal direction yields 3T = (40)aB
(2)
and from S2.7c, Newton’s law yields 100
− 2T = (20)a
(3)
A
This is 3 equations in 3 unknowns which yields aB = 1.765 m/s2 , aA = 2.647 m/s2 and T = 23.533 N 40g
20g
XB
A
B
B
3T
2T
A
xC x
(a)
N (b)
A
FIGURE S2.7
93
N (c)
100
2.8 Constraints to motion are from figure S2.8a are : xc + x p = ℓ1 , therefore: a p = And xA
c
− x + x − x = ℓ , hence: − 2a = 0, and thus a + a p
aA + aB
−a .
B
p
2
p
A
+ 2ac = 0
B
C
P
x
B A
FIGURE S2.8a
From the FBD’s of figure S2.8b: 25g
− T = 25a 2
A
T 1 = 2T 2 20g
− T = 20a 50g − 2T = 50a . But a = − (a + a 2
B
2
c
1 2
c
B)
A
Subtract 2x the first equation from third to get 50(ac
A)
−a
= 0, or ac = aA .
From the constraint aB = 5g = 85aA
→a
A
−3a
A.
Subtract 2nd from 1st
= 0.577 m/s2
T 2 = 231 N aB =
2
−1.731 m/s
T 1 = 462 N ac = 0.577 m/s2 T2
T2
25g
20g
T 1
2T 2
FIGURE S2.8b
94
T1
50g
2.9 From figure S2.9a, the constraints are xA + 2x p = ℓ1 , and (xB
−x )+x p
B
= ℓ2 . x P
x A P
xP A
50
1 00
B
FIGURE S2.9a
Differentiating yields aA =
−2a
p
and 2aB = a p so that aA =
Next the FBD’s of Figure S2.9b yields: 100 9.81
−4a
B
· − T = 100a from A T − 2T = 0 (assumed massless) from P (50)(9.81) − 2T = 50a from B 2
1
A
1
2
B
A
B
P
T 1
T 1
100(9.81)
T 1
T 2
2T
2
50(9.81) T2 =2T 1
FIGURE S2.9b
These three equations, plus the constraints aA = 4aB form a system of four equations and four unknowns which can be solved for aA . This yields
−
aA = 8.324 m/s2 Integrating vdv = aA dx from rest conditions yields v2 2
= (9.324)(0.5) or v = 2.885 m/s
2.10 From S2.10, (100)a = 100x or x¨ =
−50x vdv = −50dx = −25(x − 0.3 ) v = 50(0.09 − x ) √ = 7.07 0.09 − x v2 2
2
2
2
dx dt
2
95
t 0.3
t dx √0.09 −x2 = 0 7.07dt sin−1 x |x = 7.07t
sin− − sin
0.3
1
x 0.3
1
x 0.3
− sin− (1) = 7.07t = (7.07t + ) 0.3
1
π 2
x = 0.3 sin(7.07t + π2 ) x = 0.3 cos(7.07t)
x
Fs FIGURE S2.10
Alternately from S2.10 ˙ = 0, x(0) = 0.3m −kx = m¨x where x(0) m = 2 kg,
k = 100 N/m,
x(0) = 0.3
x¨ + 50x = 0 Integrating Factor x = Aeat ,
a2 + 50 = 0,
a=
x = A sin(7.07t) + B cos(7.07t)
±7.07i
x(0) = 0.3 = B
→ B = 0.3 ˙ = 0 = 7.07A → A = 0 x(0) x = 0.3 cos(7.07t)
2.11 From the FBD of the system as a whole: P a=
1 [P 3m
− 3mg sin α]
− 3mg sin α = 3ma, so that
From the FBD for car C: P
− T − mg sin α = ma = BC
Solving yields
1 [P 3
− 3mg sin α].
T BC = 2/3P . Likewise a FBD of the system consisting of B and C taken together yields T AB = 13 P
96
mg
P
mg
mg
C mg α
B
P
C
A
T BC
α
FIGURE S2.11
2.12 From a FBD of the entire system as given in the top of Figure S2.12: 70sin20◦
− 70cos20◦(0.2) =
70 32.2 a
a = 4.96 ft/s2
Now from the bottom of S2.12, the FBD yields 50sin20◦
− T − 0.2(50) cos 20◦ =
50 (4.96) 32.2
So T = 0!
This result is expected, as the motion is independent of mass and both blocks slides independently. o 2 0
o 2 0
2 0 5 0
o 2 0
T 5 0
f
FIGURE S2.12
2.13 Sled will start to move when a) 500(t2 + 2t) = 0.5(200)(9.81) t=
−1 ±
√
4+4(1.962) 2
t = 0.721 s or t =
=
t2 + 2t
−1 ± √2.962
−2.721
97
− 1.962 = 0
F(t)
f FIGURE S2.13
b) From t = 0.721 to 6s
− 0.4(200)(9.81) = 200a [500(t + 2t) − 0.4(200)9.81]dt v= = [0.833t + 2.5t − 3.924t]| F (t)
6 1 200 0.721 3
2
2
6 0.721
= 248 m/s (555 mi/hr)
v(t) = 0.833t3 + 2.5t2 x(6) = [0.208t4 x(6) = 386 m
− 3.924t + 1.217 + 0.833t − 1.962t + 1.217t| 8
2
6 0.721
c) From 6s on:
−0.4g = a a = −3.924 v = −3.924t + 248 time where v = 0 t = 63.2s x=
−
3.924 2 2 t
+ 248t
= 7837 m Total distance = 386 + 7837 = 8223 m or 8.2 km (5.1 mi)
98
2.14 Solution: f max = 0.3(90)(9.81) = 265 N. Therefore system slips. a) Assume whole system slips as a whole. Then summing forces yields: 1200 (.25)(990)(9.91) = (90)(a). Solving yields a = 10.88 m/s2 .
−
20 40 30
1200N f
90(9.81) A+B+C
a=10.9 m/s
2
1200 220.7 A+B
f=60a a=2.9 m/s
60
2
f=0.3(60)(9.81) = 176.58 A+B slips
FIGURE S2.14a
b) Assume slip between each surface 20g
f 1
N
f 2
1
N
30g
40g f 2 N1
2
f 2
1200
f 3
N2
N3
FIGURE S2.14b
0.25(20)(9.81) = 20aA from FBD of the 20kg block. Thus aA = 2.45 m/s2 0.25(60)(9.81)
− 0.25(20)9.81) = 40a
B
from the FBD of the 40kg block thus
aB = 2.45 m/s2
Therefore A and B the top two blocks move together (they have the same acceleration) 99
c) Next C slips out from under A and B (the FBD is S2.14c). Then 0.25(60)(9. 25(60)(9.81) = 60a 60aAB . Thus aAB = 2.45 m/s2 . From the FBD on the bottom of Figure S2.14c: 1200
− 0.25(60)(9. 25(60)(9.81)
−0.25(90)(9. 25(90)(9.81) = 30a 30a
c
Thus the acceleration of the bottom block is
27.74 m/s2 ac = 27. N 60g
AB f
AB 120
30g f AB
fC
NAB NC
FIGURE S2.14c
2.15 Will the system slip? 20 40 400
30
FIGURE S2.15a
Is 400
3(90)(9.81) = 265 N? ≥ 0.3(90)(9.
Yes, so the system slips.
mg A B
N
f max =176.6 a=2.9 m/s 2
FIGURE S2.15b
100
From the FBD of A and B : 400
− 0.25(90)9. 25(90)9.81 = 90a 90a
T
aT = 1.99 m/s2
176.6 which implies a = 2.9 m/s2 f max max = 176. So A and B must move with C .
2.16 Solution: 50g
N1
f
2 10g
f 1 o 30
N
2
N
2
f 2
f 3 N3
FIGURE S2.16a
Assume no slip and determine the required value of µs . From FBD of A: cos 30◦ N 1 + µN 2 cos
− N sin30◦ = 0 cos 30◦ − 50g 50g = 0 µN + µN sin30◦ + N cos
(1)
−µN cos30◦ + N sin30◦ − µN = 0 −µN sin30◦ − N cos cos 30◦ + N − 10g 10g = 0
(3)
1
2
2
(2)
2
From the FBD of B : 2
2
2
2
3
(4)
3
This is a system of four nonlinear equations in the four unknowns µs , N 1 , N 2 , and N 3 . Solved by Mathcad Eqs. 1-4 required µs = 0.242.
Since the µ required for equilibrium is larger than the 0.2 given the system will slip. Knowing it will slip, the equations of motion are (with zero acceleration into the walls) cos 30◦ N 1 + µk N 2 cos
− N sin30◦ = 0 cos 30◦ − 50g 50g = 50a 50a µ N + µ N sin30◦ + N cos 10a −µ N cos30◦ + N sin30◦ − µ N = 10a −µ N sin30◦ − N cos cos 30◦ + N − 10g 10g = 0 k
1
k
2
2
2
k
2
2
k
k
2
2
3
3
A
B
µk = 0.15, this is a system of 5 unknowns N 1 , N 2 , N 3, aA , aB and only 4 equations equations.. Hence Hence we need a constraint constraint of the motion. motion. The accelerat acceleration ion of the 101
blocks in the normal direction at the contacting surface is the same (see S2.16b). Thus aA cos cos 30◦ = aB sin sin 30◦ which provides the 5 th equation. Solving these 5 nonlinear equations in the 5 unknowns yields the desired acceleration (using a computer code such as the MATLAB file at the end of this solution):
−
aA =
2
−3.401 m/ m/s
m/s2 aB = 5.891 m/
From d = a2 t2 , the time for box A to move 0.1 m is t=
(2)(0. (2)(0.1) 3.401
(0.242) = xB (0.
= 0.242 s. During this time box B moves 5.891 (0. (0.242)2 2
= 0.173 m a A θ θ
a B o -a cos 30 = a sin 30 o A B
FIGURE S2.16b
This must be added to the distance the block B slides to the right due to its separation velocity. The velocity of separation is: 1.426 m/s vB = 5.891 0.242 = 1.
·
The deceleration of the block is block stops is: t=
1.426 1.472
−0.15 · 9.81 = −1.472 and the time when the
= 0.969 s
The total distance traveled is: xB =
2
81(0.969) −0.075 · 9.81(0.
+ 1. 1.426 0.969 + 0. 0.173 = 0. 0.864 m
·
102
The MATLAB solution: Main program to solve the first first nonlinear equation(‘non1.m’)
% Main m-file to solve the nonlinear equation % Initial guess values for the solution % x(1)=N1; x(2)=N2; x(3)=N3; x(4)=Mu_s; g=9.81; x0=[ 10*g 6*g 10*g 0.4]; % solve the nonlinear equation with given initial guess F=fsolve('neqn1',x0); % display solutions disp(' ') disp(sprintf(' disp(sprintf(' disp(sprintf(' disp(sprintf('
N1= %6.4f',F(1))) N2= %6.4f',F(2))) N3= %6.4f',F(3))) Mu_s= %6.4f',F(4)))
Function program to solve the first nonlinear equation(‘neqn1.m’)
function F=neqn1(x) % Function m-file to define the nonlinear equation deg2rad=pi./180; g=9.81; % constants required for the calculation N1=x(1); N2=x(2); N3=x(3); Mu_s=x(4); % assign the values F1= N1 + Mu_s * N2 * cos(30*deg2rad) - N2 * sin(30*deg2rad) ; F2= Mu_s * N1 + Mu_s * N2 * sin(30*deg2rad) + N2 * cos(30*deg2rad) - 50*g ; F3= -Mu_s * N2 * cos(30*deg2rad) + N2 * sin(30*deg2rad) - N3 * Mu_s ; F4= -Mu_s * N2 * sin(30*deg2rad) - N2 * cos(30*deg2rad) + N3 - 10*g ; F=[F1 F2 F3 F4]'; Main program to solve the second second nonlinear equation(‘non2.m’)
% Main m-file to solve the nonlinear equation % Initial guess values for the solution % x(1)=a_A; x(2)=a_B; x(3)=N1; x(4)=N2; x(5)=N3; g=9.81; x0=[ -0.3 0.4 10*g 6*g 10*g ]; % solve the nonlinear equation with given initial guess F=fsolve('neqn2',x0); % display solutions disp(' ') disp(sprintf(' a_A= %6.4f',F(1))) disp(sprintf(' a_B= %6.4f',F(2))) disp(sprintf(' N1= %6.4f',F(3))) disp(sprintf(' N2= %6.4f',F(4))) disp(sprintf(' N3= %6.4f',F(5))) Function program to solve the first nonlinear equation(‘neqn2.m’)
function F=neqn2(x) % Function m-file to define the nonlinear equation deg2rad=pi./180; g=9.81; Mu_k=0.15; % constants constants required required for the calculation calculation a_A=x(1); a_B=x(2); N1=x(3); N2=x(4); N3=x(5); % assign the values F1= N1 + Mu_k * N2 * cos(30*deg2rad) - N2 * sin(30*deg2rad) ; F2= Mu_k * N1 + Mu_k * N2 * sin(30*deg2rad) + N2 * cos(30*deg2rad) - 50*g - 50 * a_A; F3= -Mu_k * N2 * cos(30*deg2rad) + N2 * sin(30*deg2rad) - N3 * Mu_k - 10 * a_B; F4= -Mu_k * N2 * sin(30*deg2rad) - N2 * cos(30*deg2rad) + N3 - 10*g ; F5= a_A * cos(30*deg2rad) + a_B*sin(30*deg2rad); F=[F1 F2 F3 F4 F5]';
103
2.17 The geometry of the sketched spring in Fig. S2.17 yields ℓ2 = ℓ20 + x2 (ℓ0 + ∆)2 = ℓ20 + x2 ℓ20 + 2ℓ0 ∆ + ∆2 = ℓ20 + x2 where ∆ is the spring elongation. l
o
m x l
Fs N mg
FIGURE S2.17
Thus ∆2 + 2ℓ0 ∆
2
−x √ ±
= 0, or
4ℓ2 +4x
−2ℓ
0 2 0 ∆= = 2 of the spring force is
|F | = k s
ℓ20
+
x2
−ℓ ± 0
−ℓ
0
ℓ20 + x2 . Using the positive root, the magnitude
From the FBD: mg
−k
ℓ20
+
x2
−ℓ √ 0
x
= ma
ℓ20 +x2
Thus d2 x dt2
=g− ℓ k m
a(x) = g
−
kx m
2 0
1
+
x2
−ℓ √
ℓ0 ℓ20 +x2
−√
x ℓ20 +x2
0
.
2.18 Solution: From the figure (ℓ + ∆)2 = x2 + ℓ2 ℓ2 + 2ℓ∆ + ∆2 = x2 + ℓ2 ∆2 + 2ℓ∆ ∆ = −2ℓ±
2
−x √
=0
4ℓ2 +4x2 2
=
√ℓ
2
+ x2
− F sin θ − µ N = m¨x, N − F cos θ = 0, cos θ = √ mg
s
s
k
− ℓ (could write this directly) sin θ = √ℓ2x+x2
ℓ ℓ2 +x2
104
l
θ
Fs
x
θ Ν mg f
FIGURE S2.18
√ N = k ℓ + x − ℓ √ , F = k∆ N = kℓ 1 − √ x 1 − √ + µ ℓ 1 − √ x¨ = g − 2
2
ℓ ℓ2 +x2
s
ℓ ℓ2 +x2
k m
x¨ = g
−
k (x m
ℓ ℓ2 +x2
+ µk |xx˙˙ | ℓ)(1
x˙ k x˙
ℓ ℓ2 +x2
||
ℓ ) ℓ2 +x2
−√
which would require a numerical solution.
105
2.19 Solution: l
0.3
k
300
m
2
g
9.81
i 0 . . 2000 ∆t 0.001
v0
0
x0
0
a x
k
g
vi
1
xi
1
l2
.
m
x2
x
l . l2
x2
a x i . ∆t v . ∆t
vi xi
i
Position vs time 0.6 0.4 x
i
0.2 0 0.2 0
2.20 First note that 30g
0.5
1 . i ∆t Time s
1.5
2
≥ 0.5(20)g, so it does in fact slip. The constraint is that
L = xA + xB + const
⇒ 0 = x¨ ⇒0=a
A
+ x¨θ
A
+ aB so that aA =
From the FBD of (B): 30(9.8)
− T = 30a
B.
(1)
−a
(2)
B
From the FBD of A:
F = ma x
A
becomes:
− µ N = 20a T − µ mg = 20a T − 0.4(20)(9.81) = (20)a T
k k
A
A
(3)
A
Adding (2) and (3), using (1) yields (30)(9.81)
− (0.4)(20)(9.81) = 50a
A
106
So that aA = 4.316 m/s2 . Then d =
t=
1 4.326
aB 2 t 2
or
= 0.481 s. 20g T +
T
A
B
F
+
N
30g (b)
(a)
FIGURE S2.20
2.21 First determine the unit normal vector to the surface of the ice, n. T
t z
n
100 ft
45 30 x
y
FIGURE S2.21
Then compute the component of the weight vector along n, denoted N . The force parallel to the ice will then be S = W N , where W is the weight. Once the unit normal along S is determined, the motion can be treated as rectilinear along the face of the ice. The following Mathcad code completes the solution.
−
107
cos 45. deg T
n
0
0 sin 45. deg T
t
T
t
t
cos 30. deg sin 30. deg
0 W
0 130
0.655 n=
0.378 0.655
Let us find the component componentof the weight that acts normal to the face of the ice. N
W. n . n
55.714 N =
32.167 55.714
The force parallel to the ice face causing her to slip is S S
W
S =
N
55.714 32.167 74.286
We can first see if she slips by comparing|S| comparing|S| to µk |N| S
= 98. 271
1 . 0. N = 8 5 . 1 0 5
Therefore she slips. The unit vector in the direction of the slip is s s
S S
We can now treat the the motionas rectilinear rec tilinear along the ice face in the directions. a
32.2
.
S
130 a = 7.477 ft/s 2
0 . 8. N
The distances she slides in the s direction is: d=
66.144 ft −100(t · s) = 66. Now her velocity after sliding 66.144 ft can be determined. √ = = 2 , = 31. 31 45ft/ 45ft s Yes, she will be injured.
dv v dx
a, v
ad, ad v
.
/.
108
2.22 Constraint: xA + xB = ℓ therefore aA =
−a
B
x A
Constraint xA+x = l B ... aA = -a B
2 0 6 0 x B
30o
FIGURE S2.22
From the FBD of A: 20g 20g sin sin 30 + 0.2(20g 2(20g )cos30◦ From the FBD of B : 60g 60g sin sin 30◦ Thus
20a − T = 20a
A
2(20g )cos30◦ − 0.1(80g 1(80g )cos30◦ − T = 60a 60a − 0.2(20g
B
132. 132.1
− T = 20a 20a 192. 192.4 − T = 60a 60a
A B
Which along with the constraint equation is a system of 3 equations in 3 unknowns. Solving yields m/s2 aB = 0.753 m/ aA = aA/B
2
−0.753 m/ m/s = a − a = −1.506 m/ m/s A
B
2
NA
f A A
T f A
mAg
T
B f B mA g NB
NA (a)
(b)
FIGURE S2.22a,b
109
2.23 The rope length yields: 2x 2 xA + yB = ℓ, so that 2a 2aA + aB = 0 x
8 kg
150
2x + y = l A B y
A
2aA + aB = 0
3 kg
FIGURE S2.23
A FBD of A yields: 150
− 2T = 8a
(1)
A
A FBD of B yields: 3 (9. (9.81)
− T = 3a
·
(2)
B
Using the constraint, the 1 st equation becomes 150
− 2T = −4a
B
Multiply (2) by 2 yields 58. 58.86
− 2T = 6a
B
Subtracting these last two equations yields 10a −10a
m/s, and a −9.114 m/ Also from (2): T = 3(9. 3(9.81) − 3a = 56. 56.8 N B
= 91. 91.14, so aB =
A
= 4.557 m/ m/s2
B
Solving
vB = aB t for t yields: t=
2.5 4.557
= 0.549 s
2.24 The constraint here is that xA + xB = constant, so that aA = that the system moves to the left.
B.
−a
Assu Assume me
From FBD of mass B for y direction: N B µk mB g cos β from x direction:
B g cos β
−m
−T + m
Bg
Thus we have: aB = g (sin β
= 0, so N B = mB g cos β and f B = µk N B =
sin β
−µ
k
− f
B
cos β )
110
= mB aB
−
T mB
(1)
a+ B
a T
m g B
T f A = µ k N A
f B = µ k N B
β
+ A
α
NB
NA mA g
(a)
(b)
FIGURE S2.24a,b
From the FBD of mass A from the y -direction: N A from the x-direction: Thus aA = x¨A
A g cos α
= 0 or N A = mA g cos α
−m
−T + m g sin α − µ m = g (sin α − µ cos α) − . A
Ag
k
cos α = mA aA
T mA
k
(2)
Now use the constraint aA + aB = 0 along with equations (1) and (2) to eliminate the tension T . T . This yields g (sin α
−µ
k
cos α)
−
T mA
Solving yields T = g
+ g (sin β
mA mB mA +mB
k
cos β )
T mB
− =0 (sin α + sin β − µ (cos α + cos β )))) −µ
k
Thus the value of constant acceleration from (2) is:
x¨A
= g (sin α − µ cos α) − g k
mA mB mA +mB
[sin α + sin β − µ (cos α + cos β )])] k
2.25 Solution: mg
a
y
θ x m
µs N
N
FIGURE S2.25
From the FBD, the sum of forces in the x direction yields: N
− mg = ma sin θ 111
The force sin in the y direction is: µs N = ma cos θ Thus µs (mg + ma sin θ) = ma cos θ or µs g = a(cos θ or
−µ
s
sin θ)
µs g . cos θ µs sin θ
a=
−
2.26 Solution: mg
a
θ m
x f y
N
FIGURE S2.26
From the FBD, the force sum in the x direction is: µs N = ma cos θ The force sum in the y direction is: mg
− N = ma sin θ
Thus:
N = m(g and µs m(g
− a sin θ)
− a sin θ) = ma cos θ
µs g = a(cos θ + µs sin θ) a=
µs g cos θ+µs sin θ
112
2.27 From the top drawing, the constraints are x
B
y
A
3T 2 2T1 T1
2T
2
20g
FIGURE S2.27
2yA + x p = ℓ1 and xB + 2(xB
− x ) = ℓ . Thus 3a p
2
B
+ 4aA = 0
(1)
From the FBD of the pulley, T 1 = 2T 2 . Thus the FBD of A yields: 20(9.81)
− 4T = 20a 2
(2)
A
The FBD of B yields:
−3T = 10a → −3T = −10 2
B
2
4 3 aA
so T 2 =
40 9 aA
(3)
Substitution of (3) into (2) yields:
20(9.81) − 4 a 40 9
A
= 20aA
or
(180 + 160)aA = 9(20)9.81 and 2 aA = 5.194 m/s . aA 2 t 2 5.194 2 t, 2
xA (t) =
thus
1.5 =
T 2 = 23.1 N, t = 0.76 s, T 1 = 46.2 N
2.28 From the drawing, the constraints are: xA + xB + (xB xA + 2xB +
xc 2
−x )= ℓ p
2
and 2x p + xc = ℓ1 combining yields
= const
Thus: 2aA + 4aB + ac = 0. Let T 1 be the tension in cable 1 and T 2 = T 1 /2 be the tension in cable 2.
113
x B C
A
A
B
T 1
2T
10g
80g
1
P
C
2T 2
T 2
T1
20g
FIGURE S2.28
Now since a = 12 at2 , t =
=
(2)(4) 4.5
2d a
From the FBD of A: 10g
= 1.33 s.
− T = 10a From the FBD of B: 80g − 2T = 80a From the FBD of C : 20g − = 20a 1
A
1
B
T 1 2
c
Rearranging these 4 equations in 4 unknowns yields: 10aA + T 1 = 10g, 80aB + 2T 1 = 80g 20ac + 12 T 1 = 20g 2aA + 4aB + ac = 0 In matrix form these become: 10 0 0 1 −1 10 9.81 aA 0 80 0 2 80 9.81 aB = 0 0 20 0.5 20 9.81 ac 2 4 1 0 0 T 1 which can be solved using a code to yield:
aA =
· · ·
−11.319,
aB = 4.528, aC = 4.528,
T 1 = 211.292 and T 2 =
T 1 2
Now since a = 12 at2 , t =
= 105.646 2d a
=
(2)(4) 4.5
= 1.33 s.
2.29 Solution: Does system slip? 100 > 30(9.81)0.2 = 58.9 so it slips. From the FBD of the system as a whole: 100
− 0.15(30)(9.81) = 30a
T
Thus
114
20 10
100
f
N FIGURE S2.29
aT = 1.862 m/s2 From the FBD of B 0.2(20)9.81 = 20aB or aB = 1.962 which is larger than aT so there is enough friction for the system to move as one.
2.30 Solution: 20g
0.15 N B N
0.15N B
B
NB 100
10g
0.15 N
A
NA
FIGURE S2.30
From B (0.15)(20)g = 20aB or aB = 1.47 m/s2 From A 100 So
− (0.15)(20)g − (0.15)(30)g = 10a
A
aA = 2.643 m/s2 Block B slips.
2.31 Solution: Note that aA = a cos θˆi + a sin θ jˆ and aB = x¨Bˆi + y¨B jˆ so that
115
30g B a A
30
N
y B
NB
o
x
50g
30o N A
FIGURE S2.31 aB/A = aB
= (¨xB
−a
A
− a cos θ)˙ ˆi + (¨y − a sin θ)ˆy B
Since aB/A cannot have a component in the jˆ direction we have the constraint that y¨B = a sin θ
(1)
From the FBD of B we have (x direction) 0 = mB x¨B so that x¨B = 0 (y direction) N B
= mB y¨B
−m
Bg
(2)
From the FBD of A we have (x direction)
−N sin θ = m x¨ = m a cos θ (y direction) −N − m g + N cos θ = m a sin θ A
A A
B
A
(3)
A
A
(4)
A
This yields a system of 4 equations in the 4 unknowns yB , a, N A and N B which can be written in matrix form as 1 sin θ 0 0 0 y¨B 30 0 0 1 (30)(9.81) a = 0 50 cos θ sin θ 0 0 N A 0 50 sin θ cos θ 1 (50)(9.81) N B This has solution
−
a=
−
− 2
−6.824 m/s , y¨
B
=
2
−
which has magnitude 0 + (6.824)(.866) = 5.91 m/s2 . Thus t=
2d a
=
2(.1) 5.91
−
−3.411 m/s , N
The relative acceleration is aB/A = (¨ xB a cos θ)ˆi
−
= 0.184 s
116
A
= 591.0 N, N B = 191.9 N.
2.32 Solution: y
x l
Fs
α
N
α mg
FIGURE S2.32
The spring force is F s = k∆. From the geometry ∆=
√ℓ
2
+ x2
− ℓ. Summing forces in the x direction (see S2.32) yields √ − µ N = m¨x mg cos α − k( ℓ + x − ℓ) √ 2
x ℓ2 +x2
2
x˙
k
Summing forces in the y direction yields: N
− mg sin α − k(√ℓ
Therefore x¨ =
1 m
or x¨ =
1 m
2
+ x2
ℓ ℓ2 +x2
− ℓ) √
|x˙ |
=0
mg cos α − kx 1 − √ − µ mg sin α + kℓ 1 − √ || ∗ mg cos α − kx + kℓµ 1 − √ − µ mg sin α ℓ ℓ2 x2
x˙ k x˙
x˙ k x˙
ℓ ℓ2 +x2
||
117
ℓ ℓ2 +x2
x˙ k x˙
||
2.33 Solution: k := 40
µ k := 0.2
L := 0.2
m := 5
g := 9.81
α := 30⋅ deg
i := 0 .. 4000
∆t := 0.001
v0 0.00001 := x 0 0 a( x, v )
:=
1
⋅ m⋅g ⋅ cos (α ) − k⋅ x + µk ⋅ L⋅
vi+ 1 vi + a(xi , vi)⋅ ∆t m
:=
x
i+ 1
x
i
+ v i⋅ ∆t
v v
⋅ 1 −
− µ ⋅ m⋅ g⋅ sin (α )⋅ k 2 2 L + x L
3
2 xi 1
0
0
1
2
3
i⋅ ∆t
118
4
v v
2.34 Solution: 50x 50 ft
x x
110
(a)
(b) FIGURE S2.34a,b
First determine her velocity when the bungee cord becomes taut
√2 · g · ℓ = 62.16 ft/s Now the FBD is illustrated in S2.34b and yields −kx + mg = m¨ x, with v = v . Thus a = v = g − x x¨ = +g, but vdv = +gdx, so that v0 =
0
dv dx
k m
Integrating both sides yields:
vdv = g − x dx 0 v0
d 0
−gℓ = gd − √ −± d=
k m
k 2 2m d
g 2 + 2kgℓ m
g
k m
−
Therefore 150
= 77.893 ft (positive root)
− 60 − 77.9 = 12.1 ft above the ground.
Also, note from the quadratic formula above (using k = c/ℓ) that d is proportional to ℓ. Thus, F x = kd = cdℓ is independent of ℓ. For the numbers given, F x = 5 77.9 = 389.5 lb.
·
2.35 Solution: cv y x mg FIGURE S2.35
From the FBD, the sum of the forces in the y-direction when the skydiver has reached her terminal velocity (i.e. zero acceleration) yields: cv
− mg = 0 119
Solving for c = 4.0A = Thus A =
1 4.0
·
(60)(9.81) 9
mg . v
= 16.4 m2
2.36 From the FBD of the man there are 3 forces acting on him: gravity, the drag force, which shuts “off” after 10 meters, and the buoyancy force which shuts off when he stops. The force summation yields
−F − F + mg = m¨x b
c
0
Now F c = cv from 10 m or F c = cv < x Likewise F B Thus x¨ =
−
cv m
− 10 > = −mg from 13m or < x − 13 > 0
0
0
− 10 > −g < x − 13 > F c
+g F b
+ mg FIGURE S2.36
√
The force at impact is m¨ x = mg cv, where v = 2ad and x = 10, thus F impact = 70 9.81 500 (2)(10)(9.81) = 6313N. The depth beneath the surface that the diver travels is d = 13.195-10 =3.195 m (see Mathcad soln).
·
|
−
m := 70 a( v , x)
c := 500
:= g −
c⋅ v m
−
−
∆t := 0.001
⋅ Φ ( x − 10) ⋅
v v
g := 9.81
i := 0 .. 5000
− g⋅ Φ ( x − 13)
vi+ 1 vi + a(vi , xi)⋅ ∆t
v0 0 := x 0 0
x
:=
i+ 1
x
i
+ v i⋅ ∆t
20
x
= 13.194
x
= 13.195
x
= 13.195
3057
vi
3058
10
5000
0
0
5
10 xi
120
15
2.37 This can be solved by “hand” or by using the singularity function and numerically integrating. By hand: for the first 5 seconds: a(t) = 0.7t, v(t) = s(t) =
0.7 2 2 t 0.7 3 6 t
At t = 5, we have v5 = 8.75 m/s, s5 = 14.58 m. After the first 5 seconds a = over at t = 0) v(t) =
−2 m/s and integrating yields (starting the clock
−2t + v = −2t + 8.75 s(t) = − + 8.75t 5
2t2 2
Solving v(t) = 0 for t yields t =
8.75 2
= 4.38s.
Thus in the second interval s(4.38) = 19.14 m is reached. The total distance traveled is 14.58 + 19.14 = 33.72 m. This is solved by simple numerical integration in the following Mathcad code. i
0 . . 940
∆t ti
g
9.81 m
1200
0.01 i. ∆t
a t
Φ 5
v0
0
x0
0
vi
1
xi
1
t . 0 .7. t
vi xi
Φ t
5 .2
a ti . ∆t vi . ∆t
40
v x
i
20
i 0
5
10
20 t i
The total distance traveled is 33.8 meters.
121
2.38 The free body diagram given in S2.38: F d
F
k
+ mg
FIGURE S2.38
Here F d = cx˙ < x 35 >0 < v > is the damping force applied by the mat and F k = k(x 35)2 < x 35 >0 is the spring force applied by the mat. The equation of motion obtained by summing forces in the vertical direction is just m¨ x = mg F d F k . Two design criteria must be met: total force must be less than 600 lb and the total mat displacement must be less than 6 ft. The following Mathcad code computes the displacement by numerical integration, and can be evaluated for various values of c and k simply by changing the definition statements for these constants. The solution is shown for one possible combination of c and k that satisfy the design criteria.
−
−
−
− −
120
g := 32.2
m :=
a( v , x) := g
− Φ ( x − 20) ⋅ ⋅ ( x − 20) − Φ ( x − 20) ⋅ ⋅ v ⋅ Φ ( v )
:= m⋅a
c := 20
k
2
c
m
i := 0 .. 8000
v0 0 := x 0 0 i
k := 32
m
∆t := 0.001
F
g
(v , x )
t
i
:= i⋅ ∆t
vi+ 1 vi + a(vi , xi)⋅ ∆t :=
x
i+ 1
x
i
+ vi⋅ ∆t
i i
20
0 Fi
xi
500 0
0
5 i⋅ ∆t
0
5 ti
122
10
2.39 The sum of forces in the vertical direction ignoring gravity is m¨ x = 50000[< 3
4 n=1
0
− t > +
− 6n >< 6n + 3 − t >]
The following Mathcad code integrates the equation of motion and display the displacement and velocity: m := 500 a( t) :=
F := 50000
F m
4
⋅ Φ ( 3 − t) +
∆t := 0.01
∑ n
=1
i := 0 .. 2700
v0 100 := x 0 0
(Φ (t − 6⋅ n)⋅ Φ ( 6⋅ n + 3 − t)) t
i
:= i⋅ ∆t
vi+ 1 vi + a(ti)⋅ ∆t :=
xi+ 1
x
i
+ v i⋅ ∆t
v
3
= 1.604 × 2700
x
2700
= 2.3 ×
10 4
10
2000 2 .10
4
vi
xi 0 0
10
20
0
30
0
10
ti
20
30
ti
2.40 A FBD of the ball is given in S2.40a. y
mg
x 20o
F FIGURE S2.40
The problem is solved by setting up the computational solution and then performing a trial and error procedure to find the required force. The Mathcad solution is shown for one possible value of F (roughly the minimum possible value). The initial conditions are vx (0) = 120 ft/s, x(0) = 0, y(0) = 4 ft (this is assumed to be the level of the ball when it is struck, based on an average batter size and general location of the strike zone) and vy (0) = 0. The force required is found to be about 128 lb.
−
123
W
:=
5
g
16
F := 13200⋅ m ax( t) :=
F m
i
F
m :=
W g
= 128.106
⋅ Φ ( 0.015 − t) ⋅ cos (20⋅ deg )
∆t := 0.001 t
:= 32.2
ay ( t) := −g
+
F m
⋅ Φ ( 0.015 − t) ⋅ sin(20⋅ deg )
i := 0 .. 4550
:= i⋅ ∆t
vxi+ 1 vxi + ax(ti)⋅ ∆t x i+ 1 xi + vxi⋅ ∆t vy := i+ 1 vy i + ay (ti)⋅ ∆t y i+ 1 yi + vy i⋅ ∆t
vx0 −120 x0 0 vy := 0 0 y 4 0
x
= 354.065
y
= 7.137 × 4534
4534
100
yi
50
0
0
100
200
300
xi
2.41 A free body diagram of the box is given in the figure. The forces are F k = kx < x
0
− 4 > , f = µ
k
cos θ |vv|
The FBD yields the following two equations
F : N − mg cos45◦ = 0 F : m¨x = −f − F + mg sin45◦ y
x
k
124
400
−4
10
y
f
N
F k mg FIGURE S2.41
The following Mathcad code integrates to find the solution. 0 . . 5000 ∆t
i
µ:= 0.6
0.001 t i
45. deg
θ
m
1
. m. g. sin θ m
a v, x
2
i . ∆t g
9.81 k
µ. cos θ .
v0
0
vi
1
x0
0
xi
1
500
v
Φ x
v
4 . k. x
4
a vi , x i . ∆ t
vi xi
vi . ∆t
5
v x
i 0
i
1
2
3
5 ti
125
4
5
The equivalent MATLAB code is: function yprime = ds2pt41(t,y) yprime(1,:) = y(2,:); % friction force if y(2) = = 0, Ef = 0; else Ef = 4.162*y(2,:)/abs(y(2,:)): end % spring force if y(1)<=0, Fk = 0; else Fk = 250*y(1,:); end yprime(2,:) = -EF-Fk+6.947; return
2.42 Working with the free body diagram of the previous problem replace the forces F k with F k + F c where F c = cv(x 4). Thus the equation of motion can be written as: ma = mg(sin θ µk cos θ |vv| ) < x 4 >◦ (k(x 4) + cv)
−
−
−
−
−
The following MATLAB code can be used to solve for the velocity and displacement using ODE for plotting the response. function yprime = ds2pt41(t,y) yprime(1,:) = y(2,:); %friction force if y(2) = = 0, Ff = 0; else Ef = 4.162*y(2,:)/abs(y(2,:)); end % spring and damper force if y(1) < =0, Fc = 0; Fk = 0; else Fc = 12*y(2,:); Fk = 250*y(1,:); end yprime (2,:) = -Ef-Fc-Fk+6.947; return
126
Next the Mathcad solution is presented, complete with the plots: k := 500
µk := 0.6
i := 0 .. 5000 a( v , x, c)
:=
m := 2
∆t := 0.001 1 m
g := 9.81 t
i
θ := 45⋅ deg
:= i⋅ ∆t
⋅ m⋅ g⋅ sin(θ ) − µk⋅ cos (θ )⋅
v v
− Φ ( x − 4)⋅ [ k⋅ ( x − 4) + c⋅ v]
vi+ 1 vi + a(vi , xi , 20)⋅ ∆t
v0 0 := x 0 0
:=
x
x
i+ 1
i
+ vi⋅ ∆t
6
4 xi 2
0
0
2
4
6
ti
vi+ 1 vi + a(vi , xi , 200)⋅ ∆t
v0 0 := x 0 0
:=
x
x
i+ 1
i
+ v i⋅ ∆t
6
4 xi 2
0
0
2
4
6
ti
These codes are run several times for different values of c until one is found that keeps the box from losing contact with the spring. A value of c = 20 very nearly keeps it in contact, but a value as high as c = 200 is needed to truly keep the box in contact with the spring.
127
2.43 Assuming positive in the direction of the expanding bag: 30 m/s
v B
FIGURE S2.43
F d = 5 Ns/m vH/B m/s
·
vH/B = 30 + vB 5(30 + vB ) = 500 vB = 70 m/s 2.44 Solution: mg
Fa
Fs
FIGURE S2.44
From the free body diagram, Newton’s law in the vertical direction yields F = mg 10H (x 5)(x 5) 0.5 v v Where the Heaviside function indicates that the spring forces does not act until x = 5.
−
−
− − ||
The MATLAB code for solving this consists of the following m-file, then using ODE and PLOT to compute and plot the solution x(t). The m-file is function yprime = ds2pt41(t,y) yprime(1,:) = y(2,:); % viscous force Fd = 0.1*y(2,:)*abs(y(2,:)); % spring force if y(1) < =0, Fk = 0; else Fk = 2*y(1,:); end yprime (2,:) = -Fd-Fk+9.81; return 128
The Mathcad equivalent is given next along with the plot of x(s) vs. t: m
5
k
10
c
0.5
k . Φ x
5 . x
g
9.81
i 0 . . 10000 ∆t 0.001 ti i. ∆t
m. g
a v, x v0
0
x0
0
vi
1
xi
1
vi
5
c. v. v
a vi , xi . ∆t xi vi . ∆t Position vs time 15
10 xi
5
0 0
2
4
i. ∆ t time s
6
8
2.45 Solution: ρ
130
N
FIGURE S2.45
ρ = 100 ft v = 60 mph = 88 ft/s
F = ma n
N
− 130 =
n
=
mv2 ρ
130 v2 32.2 100
129
10
Therefore N − 130 = 130 32.2
882 100
N = 443 lb up
2.46 Solution: mg
500 ft
FIGURE S2.46
To fly the hill N = 0 2
v , v= mg = m 500
√32.2 × 500 = 126.9 ft/s = 86.5 mph
2.47 Solution: mg
ρ f
β
β
N
FIGURE S2.47
N cos β
− mg − f sin β = 0,
f = µs N N = mv2 ρ
N sin β + f cos β =
mv2 p
mg cos β µ sin β
−
(sin β +µ cos β ) = mg (cos β −µ sin β )
v = ρg
sin β +µ cos β cos β µ sin β
−
2.48 Solution: 25 mph = 36.7 ft/s. Therefore 0.418(cos β β = 6.02◦
36.72 100
− 0.3sin β ) = sin β + 0.3cos β
130
= 32.2
sin β +0.3cos β cos β 0.3sin β
−
2.49 Solution: mg
P
L v
FIGURE S2.49
amax = 5g = ρ=
(1027)2 5(32.2)
v2 p
v = 700 mph = 1027 ft/s
= 6551 ft
2.50 Solution:
mg
β
β
FIGURE S2.50
mg sin β = mRβ¨ wdw = w2 w = Rg ( a w0 (1 w2 = 2g R
|
g R
sin βdβ
β 0
− cos θ) − cos β ) + w
2 0
At separation:
2
mg cos β = m vR g cos β = 2g(1 β = cos−1
2 3
v2 R2
=
2g (1 R
− cos β ) +
+
v02 3gR
v02 R
− cos β ) +
v02 R2
3g cos β = 2g +
131
v02 R
2.51 Solution: θ
N
θ mg
FIGURE S2.51
The following equations can be written: mg cos θ = mRθ¨ N
− mg sin θ = mRθ˙
(1)
2
(2)
Eq (1) can be rewritten as ω dω = dθ
g R
˙ cos θ (where ω = θ).
Integrating this gives: ω2 2
=
g R
(sin θ
Thus: ω=
2g R
− sin θ )
(sin θ
0
− sin θ ) 0
and
v = 2gR (sin θ − sin θ ) 0
2.52 Solution: mRθ¨ = mg cos θ
θ˙ θ˙
− f | |
−mg sin θ + N = mRθ˙
2
Therefore mRθ¨ = mg cos θ
θ˙ k θ˙
− µ | | [mg sin θ + mRθ˙ ] 2
Canceling the mass ˙ θ¨ = g cos θ µk θ [g sin θ + Rθ˙ 2 ] R
−
R θ˙
||
132
2.53 Solution: The mass cancels out of the equation of motion and values have been assumed for the radius and the coefficient of kinetic friction. The Mathcad code is: i
0 . . 2000
∆t ti
0.001
g
µ k
9.81
0.15
R
0.3
i . ∆t g
α ω,θ ω 0
R
. cos θ
µ k ω . . g. sin θ R ω
2
R . ω
0 10. deg
θ0 ω i
1
θi
1
α ω i , θi . ∆t
ω i
ω i . ∆t
θi
Angular Position vs time 3
θi
2 1 0 0
0.5
1 ti Time s
1.5
2
The equivalent code in MATLAB is: function yprime = ds2pt53(t,y) g = 9.81; R = 1; gg = g/R; mu = 0.07; yprime(1) = y(2); if y(2)==0 yprime(2) = -mu*(y(2)ˆ2+gg*sin(y(1)))+gg*cos(y(1)); else yprime(2) = -mu*(y(2)ˆ2+gg*sin(y(1)))*y(2)/abs(y(2))+gg*cos(y(1)); end return
133
2.54 See the free body diagram given in Sample Problem 2.12. It is given that: F s = kR(2 cos θ2 ˙
− 1)
F f = µk N |θθ˙| From Sample 2.12, the FBD yields upon summing forces in the normal and tangential directions F n = N + mg sin θ + F s cos θ = mRθ˙2
| | −
2
N = mg sin θ + kR(2cos
θ 2
− 1) cos − mRθ˙ θ 2
2
F = −F + F sin − mg cos θ = mRθ¨ The equation of motion is thus: ¨ ˙ mRθ = −µ mg sin θ + kR(2 cos − 1)cos − mRθ | | +kR(2cos − 1) sin − mg cos θ ¨ = − sin + (2 cos − 1)cos − ˙ t
f
θ 2
s
θ 2
k
θ 2
θ
µk
g R
k +m (2cos θ2
θ 2
2
θ˙ θ˙
θ 2
k m
θ
θ 2
θ 2
− 1) sin −
g R
θ 2
˙
θ2 |θθ˙|
cos θ
When the slider is in equilibrium, θ¨ = 0 and θ˙ = 0. The Mathcad solution follows. The solution shown here is for a coefficient of kinetic friction of 0.18, a value that was chosen for a typical response. The spring would be unloaded when θ = 120◦ but there would still be the gravitational acceleration acting on the slider. It may be observed that the slider is oscillating about a slighter higher angle (roughly 123◦ ), which is the equilibrium angle. Note the slow start of the slider as it overcomes the friction force. The spring force is high when θ = 30◦ but this contributes to a high normal force and therefore a high friction force. If the coefficient of kinetic friction is too high, could the problem be such that friction drives the motion? No, this is an impossibility. In this formulation the friction term in the equation of motion is ˙ θ˙ to make sure the friction always opposes motion; this term is multiplied by θ/ zero when the angular velocity is zero so friction will never initiate motion. To determine whether motion would occur in a particular case, determine whether the spring force minus the gravitational force in the initial configuration is greater than the coefficient of static friction times the normal force. That would be a separate calculation as the friction force in this equation of motion is zero until motion starts.
||
134
m := 1
k := 600
R := 0.2
g := 9.81
µ k := 0.18
(ω , θ ) := m⋅g ⋅ sin (θ ) + k⋅ R ⋅ 2⋅cos θ − 1 ⋅ cos θ − m⋅R ⋅ ω2
N
α (ω , θ ) :=
−g R
⋅ cos (θ ) +
i := 0 .. 15000
ω0 0 := π θ 0 6
k m
2
⋅ 2⋅ cos
∆t := 0.0001
θ
2
t
i
2
− 1 ⋅ sin
θ
2
−
µk
⋅
m⋅ R
:= i⋅ ∆t
ωi+ 1 ωi + α (ωi , θ i)⋅ ∆t := t θ θ + ω ⋅ ∆ i 1 i i +
200
θi deg
100
0
0
0.5
1
1.5
ti
( ) := −g ⋅ cos (θ ) +
f θ
R
x := 120⋅ deg
k m
⋅ 2⋅ cos
root( f( x) , x) deg
θ
2
= 123.35
135
− 1 ⋅ sin
θ
2
(ω , θ ) ⋅
N
ω ω
2.55 We are given α = constant and so ω = αt. r ω2
rα FIGURE S2.55 f = m(rαˆ et + rω 2 eˆn ) is the total force. The total acceleration is
µs mg = m (rα)2 + (rω 2 )2 µs =
1 g
(rα)
2
+ (r(αt)2 )2 =
rα g
√1 + α t
2 4
2.56 Solution: _ ω
B
A
RA RB
FIGURE S2.56
RA + RB = ℓ T = mB RB ω 2 , T = mA RA ω 2 Therefore mB RB = mA RA . RA = ℓ RB = RB = RA =
−R ℓ−
B
mA mA mB mB RB mA mA +mB ℓ mB mA +mB ℓ
136
2.57 Solution: mg
θ R
<
en
<
en N <
et
FIGURE S2.57
Sum forces along tˆ: N cos θ mg = 0 sum forces along n ˆ : N sin θ = mRω2 sin θ
−
Therefore N = mRω2 , mRω2 cos θ = mg, θ = cos−1
2.58 First-determine the required velocity at top. N <
n
mg
FIGURE S2.58a
From the figure S2.58a: 2
N + mg = m vR , N = 12 mg
v= gR or ω = 3 2
T
3 g 2 R.
θ
R
θ
mg
ω 0
N
FIGURE S2.58b
137
. g Rω2
From S2.58b: mRθ¨ = mg sin θ (sum of tangential forces) ωT ω0
−
ωdω =
π g 0 R
sin θdθ
2 ωT ω02 = + Rg cos θ π0 = 2 g + 32 Rg = 11 ω02 = 4g R 2 R
v0
−
|
−
2g R
= gR 11 2
2.59 From Problem 2.58 mg <
n
N
FIGURE S2.59
N
− mg =
N =
13 2
mg
mv02 R
=m
·
11 g 2
or N = mg +
11 mg 2
6.5 times body weight could realistically cause spinal injuries.
138
2.60 See free body diagram shown in the solution to problem 2.61 (S2.61). From the solution to problem 2.61: y¨A =
ℓθ˙ 2 sin θ+g cos θ sin θ , γ +sin2 θ
where γ = mA /mB ˙ For ℓ = 0.2 m, θ = 30◦ , θ(0) = 0, mA = 3 kg and mB = 2 kg: y¨A = 2.427 m/s (acceleration of block A). The acceleration of block B is: ˆ ℓθ¨ sin θˆi aB = aA jˆ + ℓθ¨ cos θ j
−
From the solution to problem 2.61: (mA + mB sin2 θ)θ¨ + mB θ˙2 cos θ sin θ + (mA + mB ) gℓ sin θ = 0 For the numbers given: θ¨ = 35.036 rad/s2 .
−
And:
aB = 2.427 jˆ
− 6.068 jˆ + 3.504ˆi = 3.504ˆi − 3.641 jˆ m/s.
139
2.61 The FBD’s are given in the figure: m Ag
y
θ(t)
x
mAg
T θ
m g B
θ m g B
T
N
FIGURE S2.61
In the y-direction for block A: (1) T sin θ = mA y¨A and in the radial and tangential directions for block B: (2) T + mB g cos θ = mB ( ℓθ˙2 + y¨A sin θ) (3)
− −m
B g sin θ
−
= mB ℓθ¨ + mB y¨A cos θ
Substitution of (2) into (1) yields sin θ mB ℓθ˙2 mB y¨A sin θ + mB g cos θ = mA y¨A
{
−
}
or mA 2 + sin ℓθ˙2 sin θ + g cos θ sin θ = ( m θ)¨ yA B Thus we have (4) y¨A =
ℓθ˙2 sin θ+g cos θ sin θ . γ +sin2 θ
where γ = mA /mB Substitute 4 into 3: ˙2 cos θ sin θ cos θ = ℓθ¨ + ℓθ sinγ θ+g +sin2 θ
·
−g sin θ.
(γ + sin2 θ)θ¨ + θ˙2 cos θ sin θ + gℓ cos2 θ sin θ + γ gℓ sin θ + gℓ sin3 θ = 0 which can also be written as: 2 (1 + sin θ )θ¨ + 1 θ˙2 cos θ sin θ + γ
γ
g γℓ
cos2 θ sin θ + gℓ sin θ +
g γℓ
sin3 θ = 0
When mA >> m B (γ very large) this becomes the pendulum equation: θ¨ + g sin θ = 0 ℓ
Resubstituting for γ into the original equation and simplifying: (mA + mB sin2 θ)θ¨ + mB θ˙2 cos θ sin θ + (mA + mB ) g sin θ = 0 ℓ
140
2.62 The free body diagrams are given in S2.62: T
m g A y
B:
A:
x
β N
mB g
T
FIGURE S2.62
¨ with θ = 0 and θ˙ = 0: From the solution to problem 2.63 for θ, θ¨ = −g sin β cos β (mA2+mB ) ℓ[mA +mB sin (β ]
Writing the equation of motion in the x-direction for block A in the coordinate system shown in S2.62: mA aA = mA g sin β + T sin β Writing the equation of motion in the y-direction for block B in the coordinate system shown in S2.62: T cos β mB g cos β = mB ℓθ¨ sin β
−
Solving the previous equation for T : T = mB g + mB ℓθ¨ tan β Substituting this gives the acceleration of block A (which is in the x-direction): aA = mA +mB g sin β + mB ℓθ¨ sin β tan β mA
mA
And the acceleration of B is: ˆ aB = aAˆi + ℓθ¨ cos β ˆi + ℓθ¨ sin β j
141
2.63 Solution: m g A
T y
B:
A:
θ
x
θ β
N
mB g
T
FIGURE S2.63
T sin(β + θ) + mA g sin β = mA aA
−T sin θ = m a T cos θ − m g = M a B Bx
B
B By
aB = aA + aB/A aA = aA (cos β ˆi
− sin β jˆ) + ℓθ˙ (− sin θˆi + cos θ jˆ) 2
(1) T sin(β + θ) + mA g sin β = mA aA (2) T sin θ = mB [aA cos β + ℓθ¨ cos θ
−
(3) T cos θ
−m
B gg T mA
Therefore aA =
− ℓθ˙
2
sin θ]
= M B [ aA sin β + ℓθ¨ sin θ + ℓθ˙2 cos θ]
−
sin(β + θ) + g sin β
Substituting into (2) and (3) 2)
−T sin θ = m
B
T sin θ +
mB mA
sin(β + θ)cos β + g sin β cos β + ℓθ¨ cos θ
T mA
sin(β + θ)cos β = −m
B gsinβ cos β
β +ℓθ¨ cos θ −ℓθ˙2 sin θ] T = −mB [gsinsinθ+β cos mB sin(β +θ)cos β
¨
−
B ℓθ cos θ
−m
ℓθ˙2 sin θ
+ mB ℓθ˙2 sin θ
mA
− ℓθ¨ sin θ − ℓθ˙ cos θ] sin(β + θ)sin β ] = m [q(1 − sin β ) + ℓθ¨ sin θ + ℓθ˙
3) T cos θ = mB g T [cos θ + T =
mB mA
B [aA
−m
sin β
2
B
2
mB [g(1 sin2 β )+ℓθ¨ sin θ+ℓθ˙ 2 cos θ] m cos θ+ mB sin(β +θ)sin β
−
A
Equating two expressions for T
−[cos θ + = [sin θ +
mB mA mb mA
sin(β + θ)sin β ] [g sin β cos β + ℓθ¨ cos θ
· sin(β + θ)cos β ] · [g cos
2
− ℓθ˙
2
sin θ]
β + ℓθ¨ sin θ + ℓθ˙2 cos θ]
Therefore g sin β cos β cos θ + ℓθ¨ cos2 θ ℓθ˙2 sin θ cos θ +g cos2 β sin θ + ℓθ¨ sin2 θ + ℓθ˙ sin θ cos θ
−
B +m sin(β + θ)[g sin2 β cos β + ℓθ¨ cos θ sin β mA
− ℓθ˙
2
sin θ sin β ]
B +m sin(β + θ)[g cos2 β cos β + ℓθ¨ sin θ cos β + ℓθ˙ 2 cos θ cos β ] mA
142
cos θ]
B sin(β + θ)[g cos β + ℓθ¨ sin(β + θ) + ℓθ˙2 cos(β + θ)] = 0 g sin(β + θ)cos β + ℓθ¨ + m mA B ¨ + ) + ℓθ[1 g sin(β + θ)cos β ( mAm+m A
mB mA
sin2 (β + θ)]
B +ℓθ˙2 m sin(β + θ)cos(β + θ) = 0 mA
(mA +mB )+ℓθ˙2 mB sin(β +θ) cos(β +θ) θ¨ = −g sin(β +θ) cos β ℓ[m 2 A +mB sin (β +θ)]
2.64 Solution: (in Mathcad) i 0 . . 2000 ∆t 0.001 ti i. ∆ t mA
3
mB
2
g β L
9.81 0 0.5 mA
ddθ θ , dθ
m B . g . sin β
θ . cos β
L. m A dθ0
θi
m B . sin β
θ
2
0 . 20 deg
θ0 dθi
m B . L . dθ 2. sin β
dθ i
1
ddθ θ i , dθi . ∆t θ dθ . ∆ t i
1
i
Angle- time relationship 40
θi
20
deg
0 20 40 0
0.5
1 t i time s
143
1.5
2
θ . cos β
θ
2.65 Solution: (in Mathcad) i 0 . . 2000 ∆t 0.001 . ti i ∆t
mA
3
mB
2
g β L
9.81 30. deg 0.5
mA
ddθ θ , dθ
m B . g. sin β
θ . cos β
L. m A dθ0
2 m B . L . dθ . sin β
m B . sin β
θ
2
0 20. deg
θ0 dθi
dθ i
1
θi
ddθ θ i , dθi . ∆t dθ . ∆ t θ i
1
i
Angle- time relationship 50
θ i
0
deg 50
100 0
0.5
1 t i time s
2.66 Solution: an = 3 g =
·
p =
(293)2 96.6
v2 p
v = 200 mph = 293 ft/s
= 889 ft
From the solution of 2.47: an =
g(sin β +µ cos β ) cos β µ sin β .
3(cos β
−
− µ sin β ) − (sin β + µ cos β ) = 0 −(3µ + 1) sin β + (3 − µ)cos β = 0 3−µ Therefore tan β = 3µ+1
µ = 0.85 β = 31.2◦ .
144
1.5
2
θ . cos β
θ
2.67 The Mathcad code to solve this problem is: N := 917
n := 0 .. N
v0 0 s0 0 := −70⋅ deg θ0 0 dθ0
∆t := 0.001
t
n
:= n⋅ ∆t
L := 3
g
v − g⋅sin(θ n) + 0.3⋅ g⋅ cos (θ n) + dθn⋅ (v )2 ⋅ ∆t n n s + v ⋅ ∆t vn+ 1 n n s n+ 1 s 3 n := −70⋅ deg ⋅ 1 − θ n+ 1 L d θ 2 n+ 1 s ) ( n 3⋅ 70⋅ deg ⋅ 3
s
N
= 3.001
N := 817
v
N
= 4.227
n := 0 .. N
v0 0 s0 0 := −70⋅ deg θ0 0 dθ0
t
N
L
N
= 2.999
v
N
= 0.917
∆t := 0.001
t
n
:= n⋅ ∆t
L := 3
g
:= 9.81
v − g⋅sin(θ n) + 0.0⋅ g⋅ cos (θ n) + dθn⋅ (v )2 ⋅ ∆t n n s + v ⋅ ∆t vn+ 1 n n s n+ 1 s 3 n := 70⋅ deg ⋅ 1 − − θ n+ 1 L 2 dθn+ 1 s ) ( n 3⋅ 70⋅ deg ⋅ 3
s
:= 9.81
= 6.668
t
N
L
= 0.817
The child’s velocity is 4.227 m/s at the bottom of the slide when the coefficient of kinetic friction is 0.3, and 6.668 m/s when there is no friction. It takes the child 0.917 seconds to reach the bottom of the slide when the coefficient of kinetic friction is 0.3, and 0.817 seconds when there is no friction.
145
The MATLAB code to solve the problem is as follows: function yprime = ds2pt67(t,y) s = y(1); %position s = y(2); %speed L = 3; %length of slide mu = 0.3; %kinetic coefficient of friction g = 9.81;
%acceleration due to gravity
theta = -1.222*(1-(s/L)ˆ3); %angle theta = 1.222*3 *(s/L)ˆ2/L; %dtheta/ds % differential equations yprime(1,:) = s ; yprime(2,:) = - mu*(s ˆ2*theta + g*cos(theta))*sign(s )-g*sin(theta); % v x and v y yprime(3,:) = s *cos(theta); yprime(4,:) = s *sin(theta); return
146
2.68 The Mathcad code to solve the problem: The heightof the slide must first be determined numerically as shownbelow. The height was originally choosento be 3 m, and was reducedto 2.854 m so that the bottom would be 0.6 m from the ground. x0
0
y0
2.854
s
0 , 0.1 . . 3 s 3
70. deg. 1
θ s
3
s x s
x0
cos θ ζ
dζ
sin θ ζ
dζ
0 s
y s
y 0
0
y 3 = 0.6
Slide Contour 3
2.5 m e c n a t s i d l a c i t r e v
2 y s 1.5
1
0.5 0
0.5
1 x s
1.5
2
horizontal distance m
The solution for the child falling from the top is straight forward as the only force acting on the child is gravitational acceleration. :
y(t) = -g m/s
2
.
y(t) = - gt m/s 2
- gt y(t) = + 2.854 m 2 The time when the child hits the ground is t = 0.763s and the velocity is 7.485 m/s.
147
2.69 Solution: N g
6263 9.81 0.5 47. deg 46 0.. N 0.001 .i ∆t
µ α L i ∆t ti
vi 1
si
1
θi
1
dθi
0
s0
0
θ0
α
dθ0
0
g. sin θ
a θ , dθ , v
vi
v0
µ . g. cos θ
v 2. dθ
a θ i , dθ i , vi . ∆t si vi . ∆ t α. 1
si 2 L
si . . 2 α L2
1
t 6263 = 6.263 v6263 = 3.453 s6263 = 46.005 x0
0
xi
1
xi
y
20
yi
1
yi
0
cos θi . si sin θ i . si
1
si
1
si
Slide Profile 20
yi
10 0 10 0
10
20 30 xi Horizontal Position m
40
The slide provides a very small velocity at the bottomand the time on the slide is 6.263 s. A longer time on the slide may be obtained by changing the initial angle and the coefficient of friction as well as the lengthof the slide.
148
2.70 Solution: θ(s) = (s/25000) cos(s/500), for 0 < s < 3000 ds dt
= 100 km/hr =
dθ ds
=
100 3.6
1 [cos(s/500) 25000
s 500
− · sin(
∆s := 1 θ ( s ) :=
= 27.8 m/s s )], N (s) 500
i := 0 .. 3000 s 25000
⋅ cos
s
i
= cos(θ(s)) +
v2 dθ g ds
:= i⋅ ∆s
500 s
( ) ( )
x0 0 xi+ 1 xi + cos θ (s i) ⋅ ∆s := := y 0 y 0 i+ 1 yi + sin θ (s i) ⋅ ∆s 50 0 yi 50 100
0
500
1000
1500
2000
2500
3000
xi
1
dθ( s ) :=
25000
v := 100⋅
N
i
⋅ cos
− 500
1000
v
3600
( ( ))+
:= cos θ
s
s
i
v
s 25000
⋅ sin
⋅ 1 500 500 s
= 27.778
2
9.81
⋅ dθ
(s ) i
1.02
Ni 1
0
500
1000
1500 xi
149
2000
2500
3000
2.71 Solution: s
0 , 10 . . 7000 s s . cos θ s 4000 400 1 s . cos dθ s 4000 400 v 66 g 32.2
1 s 400
. sin
s
1
400
4000
0.5. g
a max
v . dθ s
an s
20
18.941213 an s
0 a max
15.547668
20 0 0
1000
2000
3000
Yes, the occupant slips near the end of the ride.
2.72 Solution: 0 < s < 20 θ = s sin−1 [0.2 s˙ = 1
− 0.5 cos(
s )] 10
F −= m/s F = ms˙ θ′ 2
n
F max = µsN = µs mg > ms˙ 2 θ′ µs g > s˙ 2 θ′ no slip condition µs >
s˙ 2 θ g
′
=
102 0.2571 9.81
θ = 5 sin−1 [0.2 1 θ′ = 5 √1− u′ u2
where u = 0.2 s u′ = 0.5 sin( 10
4000 s
= 2.62
− 0.5cos(
− 0.5cos( )·
s )] 10
s ) 10
1 10
max θ′ (s) = 0.2571 r/m @ s = 17.17 m
150
5000
6000
7000 3 7. 10
The following is a Mathcad analysis: v := 10
s
:= 0 .. 20
θ ( s ) := 5⋅ a sin 0.2 − 0.5⋅ co s
dθ( s ) :=
d ds
s
10
θ (s )
2
an( s ) := v ⋅ dθ( s ) 40
an( s )20
0
µ( s ) :=
0
10
an ( s )
20
s
9.81
µ( 17) = 2.62
2.73 Differentiate the given r and θ: r(t) = 0.300 + 0.100 cos(πt) ˙ = r(t)
−0.100π sin(πt) r¨(t) = −0.100π cos(πt) 2
θ(t) =
π 6
˙ = θ(t)
π2 6
sin(πt)
¨ = θ(t)
cos(πt) π3 6
− sin(πt) v(t) = −0.100π sin(πt)ˆ e + (0.300 + 0.100 cos πt) cos (πt)ˆe a(t) = [−0.100π cos(πt) − (0.3 + 0.1 cos(πt)) cos πt]ˆ e +[(0.3 + 0.1cos πt)(− sin πt) + 2(−0.1π sin(πt))( cos πt)]ˆe π2 6
r
π4 36
2
π3 6
2
2
θ
r
π2 6
θ
F = mar eˆr + maθ eˆθ
=m
{−0.100
−0.3m
π3 6
π4 36
cos3 (πt)
− 0.300 sin(πt){1 + cos(πt)}eˆ
π4 36
cos2 (πt)
θ
151
− 0.100π
2
cos(πt) eˆr
}
2.74 It is given that: F(t) = (3t2
− 1)ˆe
r
+ cos( πt6 )ˆeθ , m = 1
Therefore upon comparison with eq. (2.39) in the book: m(¨r rθ˙ 2 ) = 3t2 1
−
−
˙ = cos( πt ). m(r θ¨ + 2r˙ θ) 6 This gives the following differential equation to solve: r¨ = rθ˙ 2 + 1 (3t2 1) m
θ¨ =
−2
r˙ ˙ rθ
+
1 mr
−
cos( πt6 )
With initial conditions ˙ ˙ = 0 θ(0) =0 r(0) = 2 θ(0) = 0 r(0) The solution in MATLAB is given in the following file: function yprime = ds2pt74(t,y) m = 1; Fr = 3*tˆ2-1; Fth = cos(t*pi.6); r = y(1); th = y(2); v = y(3); w = y(4);
% radial position % angular position % radial velocity % angular velocity
yprime(1) = v; yprime(2) = w; yprime(3) = Fr/m + r*wˆ2; yprime(4) = (1/r)*(Fth/m-2*v*w); return
152
The Mathcad solution follows: m := 1
(
ddr r, d θ , t
) := r⋅ dθ2 +
(
dd θ dr , r , dθ , t
r
∆t := 0.001
dr 0 0 r 0 2 := 0 d θ0 0 θ 0 i
m
(
()
:= r i⋅ cos θ i
)
2
⋅ 3⋅ t − 1
) := −2⋅ dr ⋅ dθ −
i := 0 .. 2000
x
1
⋅ cos
1
6
m⋅r t
i
π⋅ t
:= i⋅ ∆t
(
)
dr i+ 1 dr i + ddr r i , dθi , ti ⋅ ∆t r + dr ⋅ ∆t r i+ 1 i i := dθi+ 1 dθi + dd θ dr i , r i , dθi , ti ⋅ ∆t θ i + d θi⋅ ∆t θ i+ 1
(
y
i
)
()
:= r i⋅ sin θ i
0
yi 2
1.5
2
2.5
3
xi
2.75 The force acting on m along r is: Differentiating the given value of θ yields: F r =
−k(r − 0.2)ˆe . r
θ = 0.1t2 θ˙ = 0.2t θ¨ = 0.2
Using equation 2.39 yields: m(¨r rθ˙ 2 ) = k(r 0.2) ˙ =0 m(r θ¨ + 2r˙ θ)
−
− −
The governing differential equations and initial conditions are: θ¨ = 2 r˙ θ˙
− θ¨ = −
4 2 ˙ r˙ θ r
k = 400N/m m = 4kg ˙ = 0 r(0) ˙ = 0 r(0) = 0.2 θ(0) = 0 θ(0) 153
3.5
The MATLAB code is: function yprime = ds2pt75(t,y) m = 4; k = 40; L = 0.2; th = 0.1*tˆ2; thdot = 0.2*t; yprime(1) = y(2); yprime(2) = y(1)*thdotˆ2-(k/m)*(y(1)-L); return The Mathcad solution is: i 0 . . 10000 ∆t 0.001 k 40 m 4 ti i . ∆t
θi dθ i
0.1. ti 2 0.2. t i
dr0
0
r0
0.2
dri ri
dri
1 1
k
2
r. dθ
ddr r , dθ
. r m
0.2
ddr ri , dθ i . ∆t r dr . ∆ t i
i
Trajectory Plot
0.33073
90 120 150
0.3
60
0.2
ri
Starting point 30
0.1 180
0.2
0
0
210
330 240
300 270 θ i
154
2.76 Solution:
f=-µ
Fs
k
v mg(k ___ ) | v|
θ
FIGURE S2.76
˙eθ ). The unit vector along v for computing f is: v = (rˆ ˙ ex + r θˆ v | v| =
˙eθ rˆ ˙ ex +rθˆ r˙ 2 +r2 θ˙2
√
Equation (2.39) from the book yields: m(¨r rθ˙ 2 ) = k(r 0.2) µr mg
−
− −
√
−
r˙
(r˙ 2 +r2 θ˙ 2 )
or r¨ = rθ˙ 2
−
k (r m
− 0.2) − µ · g · √ k
r˙
r˙ 2 +r2 θ˙2
The MATLAB code is: function yprimem = d s2pt75(t,y) m = 4; k = 20; l = 0.2; g = 9.81; mu = 0.2; th = 0.1*tˆ2; thdot = 0.2*t; v = sqrt(y(2)ˆ2+(y(1)*thdot)ˆ2); if v==0, yprime(1) = y(2); yprime(2) = y(1)*thdotˆ2-(k/m)*(y(1)-1); else yprime(1) = y(2); yprime(2) = y(1)*thdotˆ2-(k/m)*(y(1)-1)-mu*g*y(2)/v; end return
155
The Mathcad solution is: k 20 m 4 µ 0.2 g 9.81 i 0 . . 10000 0.001 ∆t . ti i ∆t θ t 0.1. t2 dθ t 0.2. t r. dθ 2
ddr r , d r , dθ , t
dr0
0
r0
0.2
dri
1
ri
1
k m
. r
0.2
dr
µ . g. dr 2
2 r. dθ
ddr ri , dri , dθ ti , ti . ∆t r dr . ∆ t
dri
i
i
Particle movement 0-10 s
0.515584
90 120
60 0.4
150
Starting point of the particle t=0 s
30 0.2
r i
0.199999 180
0
0
210
330
240
300 270 θ t i
2.77 We are given m = 4 kg, k = 650 N/m and θ˙ = 1.5 rad/s Differentiate: r(θ) = 0.200(2
− cos θ)
˙ = 0.200(sin θ)θ˙ r(θ)
¨ r¨(θ) = 0.200[(cos θ)θ˙2 + (sin θ)θ] 156
The equation of motion in the radial direction is: N r k(r 0.100) = m(¨r rθ˙2 ), where θ˙ = 1.5. Thus:
−
−
− N = k[0.200(2 − cos θ) − 0.100] + m[0.200 cos θ(1.5) − 0.200(2 − cos θ)(1.5) ] = 650(0.300 − 0.200 cos θ) + 4[0.400 cos θ(1.5) − 0.200(2)(1.5) ] 2
r
2
Thus
N r = 191.4
− 126.4cos θ.
2.78 We are given θ˙ = 2 θ¨ = sin θθ˙ = sin θ(2 Thus
− cos θ, so that − cos θ)
− 0.100) + m(¨r − rθ˙ ) r¨(t) = 0.200[cos θ(2 − cos θ) + sin rθ˙ = 0.200(2 − cos θ)(2 − cos θ) 2
N r = k(r
2
2
2
θ(2
2
− cos θ)]
The following MATLAB code computes the solution function Nr = ds2pt78(theta) thdot = 2-cos(theta); thddot = sin(theta).*thdot; r = 0.2*(2-cos(theta)); rdot = 0.2*sin(theta).*thdot; rddot = 0.2*sin(theta).*thddot + 0.2*cos(theta).*thdot.ˆ2; k = 650; m = 4; l = 0.1; Nr = m*(rddot-r.*thdot. ˆ2)+k*(r-l); return The following computes the solution in Mathcad:
157
2
2
θ := 0 , Fs
π 48
.. 4⋅ π
(θ ) := 650⋅ 0 .2⋅ (2 − co s(θ )) − 0.1 ( ) := 0.2⋅ cos(θ )⋅ (2 − cos (θ ))2 + (sin (θ ))2⋅ (2 − cos (θ ))
ddr θ
( ) := 0.2⋅ (2 − cos (θ ))3
r ωω θ
( ) := Fs (θ ) + 4⋅ (d dr( θ ) − r ωω(θ ))
Nr θ
300
200 Nr ( θ ) 100
0
0
200
400
600
800
θ deg
2.79 Following the solution of sample Problem 2.17 the following codes are used to numerically adjust the mass until the desired response results. In MATLAB, the code is: function yprime = ds2pt79(t,y) m = 10.6; k = 500; r0 = 0.3; g = 9.81; r = y(1); theta = y(2); rdot = y(3); thdot = y(4); % velocities yprime(1,:) = rdot; yprime(2,:) = thdot; % accelerations yprime(3,:) = r*thdotˆ2-(k/m)*(r-r0)+g*cos(theta); yprime(4,:) = (-2*rdot*thdot-g*sin(theta))/r; return In Mathcad the code is:
158
k g N ∆t i v0
1600 9.81 2400 0.0005 0.. N 0
r0
0.3
ω 0
0 30. deg
θ0
m:= 3.6 kg The mass was adjustedby trial and error to producethe required path of motion. k 2 . r a v, r, ω , θ r. ω g. cos θ 0.3 m 1 . g. sin θ α v, r, ω, θ 2. v. ω r vi 1 vi a vi , ri , ω i , θ i . ∆t
ri
ω i
1
θi
1
xi yi
ri
1
ω i
vi . ∆ t
α vi , r , ω i , θ . ∆t i i θi ω i . ∆t
ri . sin θ i 0.3 ri . cos θ i
Pendulum Trajectory 0.05
y
i
0
-0.05 -0.15
-0.1
-0.05
0.05 i horizontal position (m)
2.80 Solution:
159
x
0.1
0.15
k(r-L)
θ
θ mg
FIGURE S2.80
From the free body diagram, the equations of motion can be written: k(r L) + mg cos θ = m¨ r mrθ˙2
− −
r¨ = rθ˙ 2 + g cos θ and:
−
k m
(r
−
− L)
−mg sin θ = 2mr˙ θ˙ + mrθ¨ θ¨ = − sin θ − 2 θ˙ g r
r˙ r
The MATLAB code to solve the problem numerically is: function yprime = ds2pt80(t,y) m = 5; r = y(1); th = y(2); rdot = y(3); thdot = y(4); Fr = 1-sin(th); Fth = -(2+t-tˆ2); yprime(1) = rdot; yprime(2) = thdot; yprime(3) = r*thdotˆ2+Fr/m; yprime(4) = (-2*rdot*thdot+Fth/m)/r; return The numerical solution in Mathcad is given on the following page. Note that the solution (pendulum angle versus time) to Sample Problem 2.13 is also computed and plotted (in blue). The two solutions differ when k = 60 but are in good agreement when k = 600. The length of the pendulum is also plotted as a function of time showing the degree to which the mass is oscillating on the spring (note that the oscillations are very small when k = 600, i.e. in that case the pendulum is behaving approximately as if the spring were a rigid bar or string).
160
m := 2
L := 2
(
g := 9.81
) := r⋅ dθ2 + g⋅cos (θ ) −
ddr r, d θ , θ , k
(
dd θ dr , r , dθ , θ i := 0 .. 60000
k m
⋅ ( r − L)
) := −g ⋅ sin(θ ) − 2⋅ dr⋅ dθ r
r
∆t := 0.0001
t
i
:= i⋅ ∆t
(
)
(
)
dr i+ 1 dr i + ddr r i, dθi , θ i , 60 ⋅ ∆t r + dr ⋅ ∆t r i+ 1 i i := dθi+ 1 dθi + dd θ dr i , r i , dθi , θ i ⋅ ∆t θ i + dθi⋅ ∆t θ i+ 1 g d βi − ⋅ sin (β i)⋅ ∆t dβ0 0 dβi+ 1 L := 30⋅ deg := β i+ 1 β 0 β i + dβi⋅ ∆t dr 0 0 r 0 L := 0 d θ0 30⋅ deg θ 0
50
3
θi deg
βi
0
r i
2.5
deg 50
0
2
4
2
6
0
2
ti
dr 0 0 r 0 L := 0 d θ0 30⋅ deg θ 0
4
6
4
6
ti
(
)
(
)
dr i+ 1 dr i + ddr r i, dθi , θ i , 600 ⋅ ∆t r + dr ⋅ ∆t r i+ 1 i i := d dd dr r d t θ + θ , , θ , θ ⋅ ∆ d θ i i i i+ 1 i i θ i + d θi⋅ ∆t θ i+ 1
50
3
θi 2.5
deg
βi
0
r i 2
deg 50
0
2
4
6
ti
1.5
0
2 ti
161
2.81 The free-body diagram is:
R
θ
θ
N
mg
FIGURE S2.81
Using equation (2.39) in the book the equations of motion are: N + mg cos θ = m(¨r r θ˙2 )
− − −mg sin θ = m(rθ¨ + 2r˙θ)˙ ˙ Subject to: θ(0) = vR0 we have that: θ¨ = ωdω = dθ 1 (ω 2 2
−
v02 ) R2
=
r (cos θ R
−
r R
sin θ, or:
− 1), thus: ω
2
=
ω v0 /R
ωdω =
2g (cos θ R
− 1) +
The normal force will be a minimum at θ = 180◦ 2
N =
−mg + mRω Therefore ω ≥ = (−2) + √ v = 5gR 2
2g R
r R
v02 R2
0
2.82 The free-body diagram is:
mg
θ
f
N
FIGURE S2.82
From the diagram and equation (2.39) in the book: N + mg cos θ = mRθ˙2
−
−
162
θ 0
−
g R
v02 R2
sin θdθ
θ˙ θ˙
−mg sin θ − µ |N | | | = mRθ¨ k
Solving for the normal force: N = mg cos θ + mRθ˙2
And substituting this into the equation for angular acceleration: ˙ θ¨ = g sin θ µk g cos θ + Rθ˙2 θ
−
R
−
|| θ˙
R
The computer solution in Mathcad is (independent of mass): R := 2
µ k := 0.2
α (ω , θ ) :=
−g R
v0 := 5
g := 9.81
µk 2 ω ⋅ g ⋅cos (θ ) + R ⋅ ω ⋅ R ω t := i⋅ ∆t ∆t := 0.001 i
⋅ sin(θ ) −
i := 0.. 5000 v ω0 0 := R θ 0 0
ωi+ 1 ωi + α (ωi , θ i)⋅ ∆t θ i+ 1
:=
θ i + ωi⋅ ∆t
100
θi
50
deg
0
50
0
2
4
6
ti
2.83 From the geometry in Figure S2.83a:
R
β L
θ R/2
FIGURE S2.83a
From the law of cosines: L2 = R2 + L=
R 2
R2 4
−R
2
√5 − 4cos θ
cos(θ)
163
And: R2 4
= R2 + L2
R2 4
= R2 +
− 2RL cos β √ (5 − 4cos θ) − R 5 − 4cos θ cos β
R2 4 2 cos θ 5 4cos θ
2
cos β = √ −− From the law of sines: sin θ L
=
sin β R/2
θ sin β = √5−sin4cos θ The spring force is:
F s = k(L
−L ) = 0
kR 2
√5 − 4cos θ − 1 4
Components of the spring force are: F sn F sθ
√ = −F cos β = − 1− (2 − cos θ) − 1 − √ sin θ = −F sin β = − s
kR 2
1 4 5 4cos θ
s
kR 2
1 4 5 4cos θ
−
>
eθ
>
en
β N Fs
FIGURE S2.83b
From the free-body diagram in S2.83b, the equation of motion is: 2
mR ddt2θ = d2 θ dt2
−
kR 2
1 −
=− 1− k 2m
√ 1 sin θ 4 5−4cos θ
√ 1 sin θ 4 5−4cos θ
164
2.84 Solution: >
>
eθ
β
en
θ
N Fs
mg
FIGURE S2.84
With the addition of the weight as shown in S2.84, the equation of motion becomes: 2
mR ddt2θ = d2 θ dt2
−
kR 2
1 −
=− 1− k 2m
√ 1 sin θ − mg cos θ 4 5−4cos θ
√ 1 sin θ − Rg cos θ 4 5−4cos θ
2.85 Solution: >
>
eθ
β
en
f
N Fs
FIGURE S2.85
The friction force is: f =
θ˙ θ˙
−µ N | | k
With the addition of the friction force as shown in S2.85, the equation of motion becomes:
=− 1− 1 − √ =−
2 mR ddt2θ
d2 θ dt2
kR 2
k 2m
where: N =
kR 2
1 −
˙ √ 1 sin θ − µk N |θθ˙| 4 5−4cos θ
1 4 5 4cos θ
−
sin θ −
µk N mR
θ˙ θ˙
||
√ 1 (2 − cos θ) − mRθ˙2 4 5−4cos θ
165
2.86 Solution: >
>
eθ
en
θ
β
f
N Fs
mg
FIGURE S2.86
With the addition of both the weight and the friction force as shown in S2.86, the equation of motion becomes:
d2 θ dt2
θ˙ θ˙
√ = −mg cos θ − 1− sin θ − µ N | | − √ = − cos θ − 1− sin θ −
2 mR ddt2θ
kR 2
g R
k 2m
1 4 5 4 cos θ
N = mg sin θ +
kR 2
1 −
µk N mR
1 4 5 4cos θ
−
where:
1 4 5 4cos θ
√−
k
θ˙ θ˙
||
(2 − cos θ) − mRθ˙
2
A Mathcad code for integrating the equation of motion is as follows with some choice of m, k, R, µk and v0 such the the motion damps out after two oscillations. There are infinitely many other combinations that would satisfy this criterion.
166
m := 2
k := 70
( ) := 1 −
f θ
(
N d θ , θ
µ := 0.3
g := 9.81
R := 1
1 4⋅ 5 − 4⋅ cos
(θ )
) := k ⋅ R ⋅ f (θ )⋅ (2 − cos (θ )) − m⋅R ⋅ dθ2 + m⋅g ⋅ sin (θ ) 2
(
dd θ d θ , θ
) :=
i := 0 .. 40000
−k ⋅ f (θ )⋅ sin (θ ) − 2⋅ m ∆t := 0.0001
dθ0 0 := θ 0 30⋅ deg
g
⋅ cos (θ ) −
R t
i
µ ⋅ N(dθ , θ )⋅ m⋅ R
dθi+ 1 dθi + dd θ(dθi , θ i)⋅ ∆t θ i+ 1
:=
θ i + d θi⋅ ∆t
0
deg
50
100
0
2
4
ti
2.87 We are given: 24
M = 5.98
× 10 kg r = 6380 × 10 m G = 66.7 × 10− m /kg · s 3
12
3
2
m
F
FIGURE S2.87 GmM r2
v=
=
mv2 r
GM r
=
5.98 66.7 500+6380
×
× 10
3
dθ
:= i⋅ ∆t
50
θi
dθ
= 7614 m/s
167
2.88 Solution: Rω2 m =
GM m R2
GM R3
ω=
The period τ is 3
2π ω
τ =
= √2π R2 GM
2.89 Solution: τ = 23 hrs 56 min. = 86.16 3/2 τ = √2π r 0 GM 3/2
r0
=
√
τ GM 2π
= 273.87
r0 = 42210 km. r = r0
× 10
×10
9
3
s.
m
− R = 42210 − 6380
r = 35830 km v=
GM r0
= 3074 km/s.
2.90 Solution: First, let us determine the velocity at final circular orbit of 700 km altitude: R4 = 6380 + 700 = 7080 km
= 7.506
=
= 7.670
v4 =
MG RA
× 10
3
× 10
3
m/s
R1 = 6380 + 400 = 6780 km v1
MG R1
m/s
From Eq. (2.67) for a conic section: 1 r
= A(1 + e cos θ)
For elliptic orbit r0 Therefore
rapogee r0
=
7080 6780
rapogee
= (1+e) (1−e)
Solving for e: e = ra r0
⇒ θ = 0,
⇒ θ = 180
(ra /r0 ) 1 (ra /r0 )+1
−
= 1.044
e = 0.022 h is the angular momentum/unit mass = r0 v0 . From eq. 2.69 e=
h2 GM r0
−1=
r0 v02 GM
−1
The velocity to go into elliptical orbit is v2 =
GM (1 + R1
e) = 7.754
3
× 10
m/s 168
Since angular momentum is concerned in elliptical orbit = 7.411 v3 ra = v2 r0 , or v3 = v2 6780 7080 Therefore the first boost v1 The second boost v3 7.411
is
3
× 10
3
m/s.
3
m/s
2
is 7.670
3
m/s
→v
4
→v → 7.506 × 10
× 10
× 10
3
m/s or 7.754
3
× 10
2.91 From Problem 2.89 R4 = 42, 170 km v4 = 3.075
× 10
3
m/s
From Problem 2.90 R1 = 6780 km, v1 = 7.670 e=
× 10
3
m/s
R4 /R1 1 R4 /R1 +1
v2 = v3 =
− = 0.723
GM (1 + R1
R1 R4
e) = 10.070
v2 = 1.619
Therefore v1
× 10
3
× 10
m/s 3
3
→ v : 7.670 × 10 → 10.070 × 10 m/s v → v : 1.619 × 10 → 3.075 × 10 m/s 2
3
3
4
3
2.92 For an elliptical orbit e = 0.12 v p = 4000 m/s 1 r0
=
GM [1 + h2
e], h = r0 v0
r0 =
GM (1 + v02
e), M = 5.98
r0 =
5.98 66.7 1012 (1 16 106
×
×
×
9
× 10
1 ra
=
GM (1 h2
1 ra
=
5.98 66.7 10!2 (0.88) 111.683 1018
×
kg, G = 66.7
+ 0.12) = 27.92
Therefore k = 111.68 At the apogee
24
× 10
m2 /s
6
× 10
× 10−
r
m
− e) × ×
ra = 35.7
6
= 2.8 k ra
× 10−
8
× 10 m v = = 3128 m/s Altitude = r − r = (35.7 − 6.378) × 10 a
a
e
169
6
= 29, 332 km
m2 /kg s2
·
m/s
2.93 Solution: For a parabolic arc: e = 1 1 r
GM (1 + k2
=
e cos θ) θ = 140◦
r = 200, 000, 000, cos 140◦ =
−0.766 Therefore h = 2 × 10 × 5.98 × 10 × 66.7 × 10− h = 136.6 × 10 2
8
24
12
9
h v
d=
=
(1
− 0.766)
136.6 109 2.222 103
× ×
v
d
FIGURE S2.93
d = 23, 395 km Re = 6380 km Altitude = d
−R
e
= 17, 015 km
2.94 Solution: 3
ra = 6380 1 rp
+ 392.62
3
3
m
3
3
m
× 10 = 6772.62 × 10 + 385.42 × 10 = 6765.42 × 10 r = [1 + e], e = −1 τ = 92.34 · 60 = 5.54 × 10 seconds √ h = (r + r ) r r = 5.196 × 10 − 1 = 5.108 × 10− e= p
× 10 = 6380 × 10
3
h2 GM r0
GM h2
3
π τ
a
p
10
a p
h2 GM rp
4
2.95 Final velocity (Mir’s velocity at the perigee), using values from 2.94: v4 =
GM (1 + rp
e) = 7.687
× 10
3
m/s
R1 = 6380 + 100 = 6480 km v1
=
MG R1
=
5.98 66.7 6.480
×
× 10
3
= 7.846
× 10
170
3
m/s
The Hohmann transfer-elliptic intercept orbit R1 = 6480 km, r p = 6765 km e2 =
rp 1 R1 rp +1 R1
−
= 0.022
The velocity to go into this orbit is v3 =
GM (1 + R1
e2 ) = 7932
× 10
3
m/s
Conservation of momentum: v3 = 7932 v1 v3
→v →v
2 4
R1 rp
3
× 10 × = 7598 × 10 7.846 → 7.932 km/s 7.598 → 7.687 km/s
2.96 Solution: v4
=
MG , R4
3
m/s
R4 = 6380 + 150 = 6530 km
v4 = 7816 m/s
R1 = 6765 km v1 = 7687 m/s
ra
rp FIGURE S2.96
e=
4a rp ra rp
−1
ra = 6765, r p = 6530
+1 ,
ra rp
= 1.036,
e = 0.018, v2 to go into elliptic transfer orbit v2
=
GM (1 R
v3 = 7618
×
− e) = 7618 m/s
ra rp
= 7892 m/s
Retrofiring v1 v3
→v →v
2 4
7687
→ 7618 m/s 7892 → 7816 m/s
171
2.97 Solution: mg
f
fw b β N
b Nw
FIGURE S2.97
−mgkˆ = −N e
w= Nω
ω R
ˆ sin β ) f ω = +µN ω ( eˆeθ cos β + k ˆ) N b = N b (ˆeθ sin β + cos β k
−
ˆ sin β ) f b = µN b ( eˆθ cos β + k
−
The equations of motion are N ω = mrθ˙2
− − −µN cos β + N sin β − µN cos β = mrθ¨ −mg + µN sin β + N cos β + µN sin β = m¨z ˆ ) − rθ˙ eˆ Constraint: a = a(+cos β eˆ − sin β k −µmrθ˙ cos β + N (sin β − mu cos β ) = ma cos β −mg + µmkθ˙ sin β + N (cos β + µ sin β ) = −ma sin β ω
b
b
ω
b
b
2
θ
2
r
b
2
b
Mulp (1) by sin β and (2) by cos β and add. 2
2
−mg cos β + N (cos β − µ sin β cos β + sin β − µ sin β cos β ) = 0 b
mg cos β Therefore N b = 1−2µ sin β cos β a = 1 ( µmrθ˙ 2 + mg(sin β −µ cos β ) ) m
−
1 2µ sin β cos β
Therefore θ¨ = α = 1
mr
z¨ =
1 m
−
−µmrθ˙ cos β + 2
−mg + µmrθ˙ sin β +
z¨ = +µrθ˙ 2 sin β
2
−
mg cos β (sin β µ cos β ) 1 2µ sin β cos β
−
−
mg cos β (cos β +µ sin β ) 1 2µ sin β cos β
−
g sin β (sin β µ cos β ) 1 2µ sin β cos β
−
−
The Mathcad solution:
172
(1) (2)
i 0 . . 5000 ∆t 0.001 ti i . ∆t g β µ r
9.81 30. deg 0.1 0.5 1 . α ω r
µ . r. ω . ω . sin β
a ω v0
sin β µ . cos β 1 2. µ . sin β . cos β sin β µ . cos β g. sin β . 1 2. µ . sin β . cos β
µ . r. ω . ω . cos β
0
vi
1
z0
0
zi
1
ω 0
0
ω i
1
θ0
0
θi
1
g. cos β .
a ω i . ∆t
vi
v i . ∆t
zi
ω i
α ω i . ∆t ω i . ∆t
θi
Velocity in z direction 5 vi r. ω i
0
5 0
1
2
3 ti time s
40
θ i
20
0 0
2
4
6
ti
173
4
5
2.98 We are given: ˙ r = 1.5, h = 50, vθ = vθ = r θ, 0 a = (¨ r
− rθ˙ )ˆe 2
20 1.5
¨ eθ + z¨k ˆ + (2r˙ θ˙ + r θ)ˆ
r
mg
N
FIGURE S2.98 F=
−N ˆe − mgkˆ x
r˙ = r¨ = 0 N = mrθ˙2 0 = rθ¨
−mg = m¨z θ¨ = 0, z¨ =
−g,
θ˙ = θ˙0 , z˙ =
θ = θ˙0 t
−gt,
z=
−
1 f t2 2
+ z0
The following MATLAB code plots this: clear all format short e format compact t = linspace(0,2); g = 9.81; v0 = 20; R = 1.5; w0 = v0/R; theta = w0*t; z = -(g/2)*t.ˆ2; r = R*ones(size(t)); x = r.*cos(theta); y = r.*sin(theta);
174
plot3(x,y,z) The Mathcad code is: i
0 . . 200 i
t
100
ω
20 1. 5
x
1.5. cos ω . ti
y
1.5. sin ω . ti
zi
1 2
. 9.81. t 2
50
3-D trajectory
50
45
40
35
1 1
1
0
x, y, z
175
0
1
2.99 Solution: 0 v= r ω vz
·
mrθ¨ =
f = f =
µmrθ˙2 (rωˆeθ r2 ω 2 +vg2
√−
µmrθ˙2 ˙ rθ, r2 ω 2 +v32
−√
−µN v/|v| z¨ =
ˆ) + vz k µr θ˙ 2 vz r2 ω 2 +vz2
−g − √
These are solved numerically by the following MATLAB code: function yprime = ds2pt99(t,y) R = 1.5; g = 9.81; mu = 0.2; th = y(1); z = y(2); thdot = y(3); zdot = y(4); v = sqrt((R*thdot)ˆ2+zdotˆ2); vprime(1) = thdot; yprime(2) = zdot; yprime(3) = -mu*R*thdotˆ3/v; yprime(4) = -mu*R*thdotˆ2*zdot/v-g; return
176
The Mathcad code follows: i
∆t t
0 . . 390 0.01 .i ∆t
g
µ
9.81
0.2
3
µ . r. ω
α vz , ω
2 r. ω
vz2 2
a vz , ω
µ . r. ω . vz
g
2 r. ω
vz0 ω 0
0 50 20 1.5
θ
0
z0
vzi
1
zi
1
ω i
1
θi
1
n x y
vz2
vzi
a vzi , ω i . ∆t zi vzi . ∆ t
ω i
α vz , ω i . ∆t ω ii . ∆ t
θ
0 . . 391 r. cos θ n r. sin θ n
zn
zn 3-D Trajectory
40
20
0 1 1 0
1
x, y, z
177
0
1
r
1.5
2.100 This follows 2.99 and the codes are the same with different initial condition. The Mathcad version follows:
i ∆t t
0 . . 390 0.01 i. ∆t
g
µ
9.81
0.2
3
µ . r. ω
α vz , ω
2 r. ω
vz2 2
a vz , ω
µ . r. ω . vz
g
2 r. ω
vz0
20 30 20
z0 ω 0
1.5
θ0
0
vzi
1
zi
1
ω i
1
θi
1
n xn yn
zn
vz2
vzi
a vzi , ω i . ∆t zi vzi . ∆ t
ω i
α vz i , ω i . ∆t θ ω . ∆ t i
i
0 . . 391 r. cos θ n r. sin θ n
zn
178
r
1.5
2.101 We have vθ = v0 cos θ and vz = v0 sin β
β
v o
y x
z
θ
mg
N
FIGURE S2.101 N=
−N ˆe
R
W = mg(cos θˆ er
m¨ z=0 mrθ˙2 =
−
mrθ¨ =
− sin θˆe ) θ
(1)
−N + mg cos θ
(2)
−mg sin θ
(3)
Eq. (3) may be written = ω dω dθ
−g/r sin θ
Integrating yields: ω2 ω 2 ω0 2
θ 0
| = g/r cos θ| ω − ω = (cos θ − 1) 2g r
2 0
(4)
From Eq. 2, minimum velocity can be determined when θ = 180◦ and N = 0. 2 The angular velocity ω180 = g/r. From (4) g/r vθ0
7g r
2 0
− ω = − Therefore ω = √ √ = 4gr = v cos β, v = 0
0
0
5g r
5gR cos β .
2.102 From the previous problem the forces are N=
−N ˆe
r
− mg(cos θˆe − sin θˆe ) f = √ − (rωˆe + v eˆ ) W
r
µN (rω)2 +vz 2
θ
θ
z z
rω0 = v0 cos β vz0 = v0 sin β
−N + mg cos θ = −mrω
2
179
m¨ z=
(rω) + 2
2
−µNv / vz mrθ¨ = −mg sin θ − µNrω/ (rω) z
From 2.101 ω
2
+ vz 2
=
g/r when θ = 180◦
N = mg cos θ + mrω 2 2 )v z Therefore z¨ = −µ(g cos θ+rω 2 2
√ θ¨ = − sin θ − √
(rω) +vz
r r
µ(g cos θ+rω 2 )ω (rω)2 +(vz)2
The minimum initial velocity of 10.6 m/s was determined by trial and error using the following codes. The motion damps out after 4 seconds and the particle comes to rest at 360 degrees. The MATLAB code is: function yprime = ds2pt102(t,y) m = 3; R = 0.6; g = 9.81; mu = 0.2; th = y(1); z = y(2); thdot = y(3); zdot = y(4); v = sqrt((R*thdot)ˆ2+zdotˆ2); N = m*g*cos(th)+m*R*thdotˆ2; yprime(1) = thdot; yprime(2) = zdot; yprime(3) = (-m*g*sin(th)-mu*N*R*thdot/v)/m/R; yprime(4) = (-mu*N*zdot/v)/m; return
180
The Mathcad code is: I:= 0 ..5000 ∆t 0.001 . ti i ∆t µ
0.2
50. deg
β
r. ω
V vz , ω
g
α vz , ω , θ
r
v0. sin β
z0
0 v0.
θ0 1
zi
1
ω i
1
θi
1
ω min
2 r. ω .
a vzi , ω i , θ i . ∆t z vz . ∆t i
ω i
i
α vz i , ω i , θi . ∆t θi
r
0.6
vz V vz , ω
µ . g. cos θ
cos β r 0 vzi
10.6
vz 2
. sin θ
vz0
vzi
9.81 v0
µ . g. cos θ
a vz , ω , θ
ω 0
2
g
ω i . ∆t
g r
ω min = 4.044
181
2 r. ω .
ω V vz , ω
20
ω
10
i 0
10 0
100
200
300
400
500
θi
deg
angular position degrees
Angle vs time 600 θ i
400
deg 200 0 0
1
2
3
4
5
ti time s
The minimum initial velocity of 10.6 m/s was determined by trial and error. The motion damps out after 4 seconds and the particle comes to rest at 360 degrees.
182
2.103 The forces are: x mg y
θ N
3
β
FIGURE S2.103 N=
−N ˆe
r
ˆ) W = mg(cos β sin θˆ er + cos β cos θˆ eθ + sin β k The equations of motion are: N + mg cos β sin θ = mrθ˙2
−
mg cos β cos θ = mrθ¨
−
mg sin β = m¨ z The data is: m = 2 kg θ(0) = 0 r=1m β = 20◦ The z equation can be quickly integrated: z¨ = g sin β z˙ = g(sin β )t z = g(sin β )t2 /2 The MATLAB code is: function yprime = ds2pt103(t,y) g = 9.81; R = 1; beta = 20*pi/180; yprime(1) = y(2); yprime(2) = (g/R)*cos(beta)*cos(y(1)); return
183
The Mathcad code is: i 0 . . 2500 ∆t 0.001 . ti i ∆t 20. deg β r g
1 9.81 g
α θ
r
ω 0
0 0
θ0 ω i
1
θi
1
n
zn
. cos β . cos θ
α θ i . ∆t
ω i
ω i . ∆ t
θi
0 . . 2500
g. sin β .
tn 2 2
Angular position vs time 200 θ i
100
deg 0 100 0
0.5
1
1.5
2
2.5
t i time s
Position down sluiceway 15
zi
10
5
0 0
0.5
1
1.5 t
i
time s
184
2
2.5
2.104 We are given: µk = 0.3 Friction acts in a direction opposite to the velocity ˆ v = rωˆ eθ + vz k N=
−N ˆe
r
ˆ) W = mg(cos β sin θˆ er + cos β cos θˆ eθ + sin β k f =
v v
−Nµ | |
The equations of motion are: 2
−N + mg cos β sin θ = −mrω˙ = mrθ¨ mg cos β cos θ − µN √ = m¨ mg sin β − µN √ z rω (rω)2 +vz 2
vz (rω)2 +vz 2
Therefore N = m(g cos β sin θ + rω 2) and α(vz,ω,θ) =
1 r
g cos β cos θ
a(vz,ω,θ) = g sin β
µN m
µN m
rω (rω)2 +vz 2
− √
vz . (rω)2 +vz 2
− √
The MATLAB code is: function yprime = ds2pt103(t,y) m = 2; g = 9.81; R = 1; mu = 0.3; beta = 20*pi/180; th = y(1); z = y(2); thdot = y(3); zdot = y(4); v = sqrt((R*thdot)ˆ2 + zdot ˆ2); N = m*g*cos(beta)*sin(th)+m*R*thdotˆ2; yprime(1) = thdot; yprime(2) = zdot; yprime(3) = (m*g*cos(beta)*cos(th)-mu*N*R*thdot/v)/m/R; yprime(4) = (m*g*sin(beta)-mu*n*zdot/v)/m; return
185
The Mathcad code is: i ∆t ti
β
0 . . 6000 0.001 .i ∆t 20.deg r
1
g
9.81
2 g. cos β . sin θ r. ω 1 . g. cos β . cos θ α vz , ω , θ r
N ω,θ
g. sin β
a vz , ω , θ
2 r. ω
vz
µ. N ω , θ .
2 r. ω
vz0
0 0 0 0
z0 ω 0 θ0
vzi
1
zi
1
ω i
1
θi
1
vzi
vz2
a vzi , ω i , θ i . ∆t z vz . ∆t i
ω i
r. ω
µ. N ω , θ .
i
α vz i , ω i , θ i . ∆t θ ω . ∆t i
i
200 θi deg
100
0 0
1
2
3 ti time s
4
5
6
Position along Sluiceway 20
zi
10
0 0
1
2
186
3 t i time s
4
5
6
vz2
2.105 We have m = 2 kg, R = 1.5 m, v(0) = 10ˆeθ , φ(0) = 90 mg φ
θ
N
FIGURE S2.105
The forces are: N=
−N ˆe
R
W = mg(sin φˆ eφ
− cos φˆe
R)
The coordinate accelerations are ¨ Rφ˙ 2 R sin2 φθ˙2 aR = R aφ
−
− = Rφ¨ + 2R˙ φ˙ − R sin φ cos φθ˙
2
aθ = R sin φθ¨ + 2R˙ θ˙ sin φ2 Rφ˙ θ˙ cos φ
−N − mg cos φ = −mR(φ˙ mg sin φ = mR(φ¨ − sin φ cos φθ˙ )
2
Therefore
+ sin2 φθ˙ 2 )
2
0 = R sin θθ¨ + 2Rφ˙ θ˙ cos φ
From sample problem 2.22, the last equation θ˙ = sin φ02vθ0 = 102 φ¨ =
R sin φ
R sin φ
g R
2 sin2 φ cos φvθ0 0 2 R sin3 φ
sin φ +
The MATLAB code is: function yprime = ds2pt105(t,y) g = 9.81; R = 1.5; ph = y(1); th = y(2); phdot = y(3); yprime(1) = y(3); yprime(2) = 10/(R*sin(ph)ˆ2); yprime(3) = -(g/R)*sin(ph) + 100*cos(ph)/(Rˆ2*sin(ph)ˆ3); return
187
The Mathcad code is (first define R and g in the code): ∆t ti
0.001 i . ∆t g.
ddφ φ dφ 0
sin φ
100. cos φ
R
R 2. sin φ
0 90. deg
φ0 dφ i
dφ i
1
φi
φi
1
R . sin φ
θ0 xi yi zi
i
10
dθ φ
θi
ddφ φ i . ∆t dφ . ∆ t
2
0 dθ φ i . ∆t R . sin φ i . cos θ i R . sin φ i . sin θi R . cos φ
θi
1
i
φ 0 = 1.571
0 0.1 0.2 0.3 0.4 1
1 0 1
0 1
x, y, z
3-D Trajectory
188
3
2.106 We are given: R = 1.5, θ˙0 = ˙ eφ + R sin φθˆ ˙eθ v = Rφˆ N=
−N ˆe
10 , 1.5
θ0 = 0, φ0 = 90◦
R
W = mg(sin φˆ eφ v
f =
− cos φˆe ) r
−µN | | −N − mg cos φ = −mR(φ˙ + sin φθ˙ ) mg sin φ − µN | | = mR(φ¨ − sin φ cos φθ˙ ) −µN | | = mR(sin φθ¨ + 2φ˙ θ˙ cos φ) v
2
2
2
Rφ˙
2
v
R sin φθ˙ v
The mass drops out. N/m = R(φ˙ 2 + sin2 φθ˙ 2 ) Therefore φ¨ = r sin φ R
θ¨ =
1 sin φ
− µR(φ˙
−µ
2
N sin φθ˙ m v
− g cos φ
˙ + sin2 φθ˙2 ) |φv| + sin φ cos φθ˙2
−
2φ˙ θ˙ cos φ
The MATLAB code is: function yprime = ds2pt105(t,y) g = 9.81; R = 1.5; mu = 0.3; ph = y(1); th = y(2); phdot = y(3); thdot = y(4); v = R*sqrt(phdotˆ2+sin(ph)ˆ2*thdotˆ2); N = g*cos(ph)+vˆ2/R; yprime(1) = y(3); yprime(2) = y(4); yprime(3) = - (g/R)*sin(ph) -mu*Nphdot/v+sin(ph)*cos(ph)*thdotˆ2; yprime(4) = -mu*N*thdot/v-2*phdot*thdot/tan(ph); return
189
The Mathcad code is: R 1.5 g i 0 . . 4000 0.001 ∆t . ti i ∆t
R . dφ
N dφ , φ , dθ
ddφ dφ , φ , dθ ddθ dφ , φ , dθ
1
dθi
1
zi
0 dφ i
1
φi
yi
g. cos φ
dθ
1.5
θ0
xi
2.
10
dθ0
θi
sin φ
0 90. deg
φ0
dφ i
2
0.3
2 2 R . dφ R . sin φ . dθ g dφ 2 . sin φ sin φ . cos φ . dθ µ . N dφ , φ , dθ . R v dφ , φ , dθ 1 dθ . µ . N dφ , φ , dθ . sin φ . 2. dφ . dθ . cos φ sin φ v dφ , φ , dθ
v dφ , φ , dθ
dφ0
µ
9.81
ddφ dφ i , φ i , dθ i . ∆t dφ . ∆t φ i
dθi
i
ddθ dφ i , φ i , dθ i . ∆t dθ . ∆t θ i
1
R . sin φ i R . sin φ i R . cos φ
i
. cos θ i . sin θ i
i
2.107 Solution: z
mg
ψ
y
x
N ψ _ R _ w
z x
y
eϕ ϕ
FIGURE S2.107
190
N=
−N ˆe W = mg(− cos φˆ e + sin φˆ e ) f = −µN | | ¨ − Rφ˙ − R sin φθ˙ a =R ¨ + 2R˙ φ˙ − R sin φ cos φθ˙ a = Rφ φ
e
φ
v v
2
2
R
2
φ
2
aθ = R sin φθ¨ + 2R˙ θ˙ sin φ + 2Rφ˙ θ˙ cos φ
φ = const ˙ eθ ˙ eR + R sin φθˆ v = Rˆ R˙ v
−mg cos φ − µN = m(R¨ − R sin φθ˙ ) −N + mg sin φ = −mR sin φ cos φθ˙ = m(R sin φθ¨ + 2R˙ θ˙ sin φ) −µN 2
2
2
R sin φθ˙ v
N/m(dθ,R) = g sin φ + R sin φ cos φθ˙2 v(dR,R,dθ) ¨= R
=
dR2 + (R sin φdθ)2 2
−g cos φ + R sin − θ¨ = −µ v θ˙ mv
φθ˙2
2R˙ θ˙ R
−µ
N ˙ m R/v
The MATLAB code is: function yprime = ds2pt107(t,y) g = 9.81; mu = 0; phi = pi/6; r = y(1); th = y(2); rdot = y(3); thdot = y(4); v = sqrt(rdotˆ2+(r*sin(phi)*thdot)ˆ2); N = g*sin(phi)+r*sin(phi)*cos(phi)*thdotˆ2; yprime(1) = rdot; yprime(2) = thdot; yprime(3) = -mu*N*rdot /v - g*cos(phi)+r*sin(phi)ˆ2*thdotˆ2; yprime(4) = -mu*N*thdot/v - 2*rdot*thdot/r; return
191
The Mathcad code is: i 0 . . 6000 ∆t 0.001 ti i . ∆t 9.81 µ 30. deg
g
φ
0.0
v dθ , dR , R
dR
dR 0
0
R0
R0
1
Ri
1
dθi
1
xi yi zi
0 dR i
ddR dθi , dR , R i . ∆t i R dR . ∆t i
dθ i
i
ddθ dθ , dR , R i . ∆t i i θ diθ . ∆ti i
1
i
R i . sin φi R . sin φ
. cos θ i . sin θ i i i . R i cos θ i
Path of Motion 0
5 2
2
dθ
v dθ , dR , R
R0 . sin φ
θ0
2.
dθ
10
dθ0
θi
R . sin φ
µ . N dθ , R .
ddθ dθ , dR , R
2
R0
cos φ
2 R . sin φ . dθ
2
g. cos φ
ddR dθ , dR , R
dR i
2 R . sin φ . cos φ . dθ
g. sin φ
N dθ , R
0
0
x, y, z
192
µ . N dθ , R . 2.
dR . dθ R
dR v dθ , dR , R
2.108 Use the analysis and codes of 2.107 with a new friction coefficient. The codes are essentially the same, so only the Mathcad plot is shown. Path of Motion
100 0 100 0
50 0
50
50
x, y, z
2.109 Solution: mg
N
FIGURE S2.109 N = N ˆ eR W = mg(
− cos φˆe
+ sin φˆeφ ) ˙ eφ + R sin φθˆ ˙eθ V = Rφˆ f =
v
R
−µN | | N − mg cos φ = −m(Rφ˙ + R sin φθ˙ ) mg sin φ − µN = m(Rφ¨ − R sin φ cos φθ˙ ) = m(R sin φθ¨ + 2Rφ˙ θ˙ cos φ) −µN v
2
2
2
Rφ˙ v
2
R sin φθ˙ v
193
v(dφ,φ,dθ) = R N/m = g cos φ φ¨ =
g R
θ¨ =
−µ
sin φ N θ˙ mv
·
dφ2 + (sin(φ)dθ)2
− R(φ˙
2
N φ˙ mv
−µ − 2φ˙ θ˙
+ sin2 φθ˙2 )
+ sin φ cos φθ˙2
cos φ sin φ
The MATLAB code is: function yprime = ds2pt107(t,y) g = 9.81; mu = 0; r = 6; ph = y(1); th = y(2); phdot = y(3); thdot = y(4); v = r*sqrt(phdotˆ2+(sin(ph)*thdot)ˆ2); N = g*cos(ph) -vˆ2/r; yprime(1) = phdot; yprime(2) = thdot; yprime(3) = -mu*N*phdot/v + (g/4)*sin(ph) +sin(ph)*cos(ph)*thdotˆ2; yprime(4) = -mu*N*thdot/v -2*phdot*thdot/tan(ph); return
194
The Mathcad code is: i 0 . . 900 0.001 ∆t . ti i ∆t R g µ
6 9.81 0.0 2
N dφ , φ , dθ
g
ddφ dφ , φ , dθ
R
. sin φ
µ . N dφ , φ , dθ
µ . N dφ , φ , dθ .
ddθ dφ , φ , dθ
dθ v dφ , φ , dθ
.
dθ v dφ , φ , dθ 2. dφ . dθ .
2 sin φ . cos φ . dθ
cos φ sin φ
0. 5
dφ 0
φ0
6 10.deg
dθ 0
4 6. sin 10.deg
θ0
0
dφ i
1
φi
1
dθ i
1
θi
2 sin φ . dθ 2 2 R . dφ sin φ . dθ
R . dφ g. cos φ
v dφ , φ , dθ
ddφ dφ i , φ i , dθ i . ∆t dφ . ∆t φ
dφ i
i
i
ddθ dφ i , φ i , dθ i . ∆t dθ . ∆t θ
dθ i
i
1
i
10
5 N dφ , φ , dθ i i i 0
5
n
866
0
200
400
N dφ n , φ n , dθ = 3.95 10 3 n φ 866 deg θ866 deg
600
800
1000
i
= 41.541
= 75.735
The particle will fall off the surface when the normal force goes to zero and the position coordinates are shown.
195
2.110 Use the same codes and analysis as in 2.109. The MATLAB code becomes: clear all format short 3 format compact g = 9.81; r = 6; ph0 = pi/18; th0 = 0; phdot0 = -0.5/r; thdot0 = 4/r/sin(ph0); [t,Y] = ode45(’ds2pt109’,0,1,[ph0 ; th0; phdot0; thdot0]); ph = Y(:.1); th = Y(:.2); phdot = Y(:.3); thdot = Y(:.4); v = r*sqrt(phdot.ˆ2+(sin(ph).*thdot).ˆ2); N = g*cos(ph( - v.ˆ2/r; figure(1),plot(t,N)
196
The Mathcad code is: i 0 . . 2000 ∆t 0.001 . t i ∆t R g µ
6 9.81 0.2 2
v dφ , φ , dθ
R.
N dφ , φ , dθ
g. cos φ g . sin φ R
ddφ dφ , φ , dθ
dφ
2
2 sin φ . dθ
µ . N dφ , φ , dθ
dθ
.
v dφ , φ , dθ
dθ v dφ , φ , dθ
2. dφ . dθ .
2 sin φ . cos φ . dθ
cos φ sin φ
0.5
dφ 0
φ0
6 . 10 deg
dθ 0
4 6. sin 10. deg
θ0
0 1
φi
1
dθi
1
θi
R . dφ
µ . N dφ , φ , dθ .
ddθ dφ , φ , dθ
dφ i
2 sin φ . dθ
ddφ dφ i , φ i , dθ i . ∆t dφ . ∆t φ
dφ i
i
i
ddθ dφ i , φ i , dθ i . ∆t dθ . ∆t θ
dθ i
i
1
i
10
N dφ , φ , dθ i i i
0
-10
n 1816 N dφ n , φ n , dθn
φn deg
θn deg
0
500
1000 i
= 0.011
1500
2000
= 51.168 = 113.121
The mass stays on the surface for a greater lengthof time when friction is present. If the friction is too great, the particle will not leave the surface before it stops.
197
2.111 The data is: y
x
θ
ϕ
z
mg
FIGURE S2.111
φ(0) = 30◦
θ(0) = 0
Fs =
m = 3 kg
R(0) = 2
−k(R − R )ˆe W = mg(cos φˆ e − sin φˆ e ) 0
F
R
φ
The equation of motion become: ¨ Rφ˙ 2 R sin2 φθ˙2 ] = mg cos φ m[R
− − k(R − 2) m[Rφ¨ + 2R˙ φ˙ − sin φ cos φθ˙ ] = −mg sin φ −
2
m[R sin φθ¨ + 2R˙ θ˙ sin φ + 2Rφ˙ θ˙ cos φ] = 0, or: ¨ = g cos k (R 2) + Rφ˙ 2 + R sin2 φθ˙ 2 R m ˙ ˙ g φ¨ = R sin φ 2RRφ + sin φ cos φθ˙2 ˙ θ˙ φ 2φ˙ θ˙ cos θ¨ = 2R R sin φ
−
− − − −
−
The last equation can be written 1 d ˙ =0 [R2 sin2 φθ] R sin φ dt
Therefore R2 = sin2 φθ˙ = h h = R2 sin2 φ0 θ˙0 = R0 sin φ0 vθ0 0
For this case h = 2sin30◦ (0.5) = 0.5 θ˙ = 2 h 2 R sin φ
The differential equations may be written as 2 φh2 ¨ = g cos φ k (R 2) + Rφ˙ 2 + R sin R 4 4
− − ¨ = g cos φ − (R − 2) + Rφ˙ R + φ¨ = − sin φ − m k m
g R
2
2R˙ φ˙ R
R sin φ
+
R3
h2 sin2 φ
sin φ cos φh2 R4 sin4 φ
198
˙ = 0.5. vθ = 0.5, therefore θ(0)
The MATLAB code is: function yprime = ds2pt111(t,y) m = 3; k = 200; g = 9.81; R = 2; v0 = 0.5; h = v0*R*sin(pi/6); r = y(1); ph = y(2); th = y(3); rdot = y(4); phdot = y(5); thdot = h/r*sin(ph))ˆ2; yprime(1) = rdot; yprime(2) = phdot; yprime(3) = thdot; yprime(4) = -(k/m)*(r-R)+g*cos(ph) + r*phdot + 4*sin(ph)ˆ2*thdotˆ2; yprime(5) = -(g/r)*sin(ph) -2*(rdot/r)*phdot + sin(ph)*cos(ph)*thdotˆ2; return
199
The Mathcad code is: 9.81 h 0.5 g i 0 . . 5000 ∆t 0.001 ti i . ∆t
k
g
ddφ dR , R , dφ , φ
2
dφ 0
0 30.deg
φ0 dR i
dR i
1
Ri
1
dφ i
k
. R m
. sin φ
2.
2
h2
2 R . dφ
R 3. sin φ
dR . dφ R
2
cos φ . h2 R 4. sin φ
3
ddR dR i , R i , dφ i , φ i . ∆t R dR . ∆t i
i
dφ i
1
φi
1
ddφ dR i , R i , dφ i , φ i . ∆t dφ . ∆t φ i
θ0
0 . . 5000 0
θi
1
xi
R
3
0
R0
i
m
g. cos φ
ddR dR , R , dφ , φ
dR 0
200
h 2.
sin φ i R i . sin φ i . cos θi Ri
2
i
. ∆t
θi yi
R i . sin φ i . sin θ i
200
zi
R i . cos φ i
1 0 1 2 1
1 0
0.5 0
1
0.5 1
y, x, z
Path of motion
0.5
y
i
0
0.5 2
1
0 x i
1
2
2
1
0 xi
1
2
1.6 1.8 zi 2 2.2
201
2.112 The MATLAB code is: clear all format short e format compact N = 100; W = 3000; g = 32.2; s = linspace(0,6000,N); v = 88; th = s/2000 +exp(-s/1000); be = (pi/12)*sin(pi*s/3000); th = 1/2000 -(1/100)*exp(-s/1000); be = (piˆ2/36000)*cos(pi*s/3000); Gvec = [ -sin(th).*cos(be).*th -cos(th).*sin(be.*be ;... cos(th).*cos(be).*th -sin(th).*sin(be.*be ;... cos(be).*be ]; for i = 1:N, mag Fn(i) = (W/g)*vˆ2*norm(Gvec(:,i)); end Figure (1),plot(s,mag Fn) xlabel(’distance along road (ft)’) ylabel(’normal force (lbf)’) grid
202
The Mathcad code is:
s W g
0 , 10 . . 6000 6000 32.2 s
θ s β s
s 2000
e1000
π. π. s sin 12 3000 s
dθ s dβ s
Γ s
1
1
2000
1000
π
. cos
36000 sin θ s
cos θ s
e1000
π. s 3000 . cos β s . dθ s . cos β s . dθ s
cos β s v s F n s
cos θ s sin θ s . dβ s
. sin β s
. dβ s
. sin β s
. dβ s
4000
5000
88 W g
. v s 2. Γ s 1000
F n s
500
0 0
1000
2000
203
3000 s Position ft
6000
2.113 Solution: a = 2,
0 ≤ s ≤ 3000 √ v = 2as v(3000) = 109.5 ft/s (75 mph) a=
1 3001 ≤ s ≤ 6000 −vdv ads = v 109.5
v2 2
−
v=
s 3000
(109.5)2 2
=
109.5
2
−s + 3000 − 2(s − 3000)
The Mathcad code follows: s W g
0 , 10 . . 6000 6000 32.2 s
θ s β s
s
e1000 2000 π π. s . sin 12 3000 s
dθ s dβ s
1
1
2000
1000
π 36000
. cos
sin θ s
Γ s
cos θ s
e1000
π. s 3000
. cos β s . dθ s . cos β s . dθ s
cos θ s
cos β s v s F n s
Φ 3000 W g
s .
4. s
Φ s
sin θ s . dβ s
. sin β s . sin β s
3001 . 109.5 2
2. s
. dβ s . dβ s
3000
. v s 2. Γ s 1500
1000 F n s 500
0 0
1000
2000
204
3000 s Position ft
4000
5000
6000
2.114 Using the analysis of 2.113 the MATLAB code is: clear all format short e format compact N = 100; W = 3000; g = 32.2; s = linspace(0,6000,N)’; v = sqrt( 4*s.(s>=0) -6*(s-3000).*(s>=3000)); a = 2*(s>=0) -3*(s>=3000); t = s/2000 +exp(-s/1000); b = (pi/12)*sin(pi*s/3000); t = 1/2000 - (1/1000)*exp(-s/1000); b = (piˆ2/36000)*cos(pi*s/3000); avec = [... a.*cos(t).*cos(b)+v.ˆ2.*(-sin(t).*cos(b).*t -cos(t).*sin(b).*b ),... a.*sin(t).*cos(b)+v.ˆ2.*( cos(t).*cos(b).*t -sin(t).*sin(b).*b ),... a.*sin(b)+v.ˆ2.*cos(b).*b ]; mag a = sqrt(sum((avec.*avec)’)’); figure(1),plot(s,mag a) xlabel(’distance along road (ft)’) ylabel(’acceleration (ft/sˆ 2)’) grid
205
The Mathcad code is: s W g
0 , 10 . . 6000 6000 32.2 s
θ s
β s
s
e 1000
2000
π. π. s sin
12
3000 s
dθ s dβ s
Γ s
1
1
2000
1000
π
. cos
36000 sin θ s
cos θ s
e 1000
π. s 3000 . cos β s . dθ s . cos β s . dθ s
cos β s
v s
an s at s
s . 4. s
Φ 3000
sin θ s . dβ s
3001 .
Φ s
. sin β s
cos θ s
. sin β s
109.52
. dβ s . dβ s
2. s
3000
1. v s 2. Γ s s .2
Φ 3000
a s
2
an s
Φ s
3001
at s 2 Acceleration Magnitude vs position
8
a s
6
4
2 0
1000
2000
206
3000 s Position ft
4000
5000
6000
2.115 Solution: m g
f N FIGURE S2.115
−mgkˆ + N ˆn − µˆt = ma n f = −µN | | ˆ=| | F=
Γ Γ
v v
ˆ] ˆt = cos θ(s)cos β (s)ˆi + sin θ(s)cos β (s) jˆ + sin β (s)k ˆ n ˆ +m N = mgk
·
ˆˆ a(s) = −mgkm·t−µN
·
v2 p
The MATLAB code is: function yprime = ds2pt115(t,y) m = 30; g = 9.81; mu = 0.0; s = y(1); v = y(2); th = pi*s/20; be = -(pi/3)*(1-(s/10).ˆ2); th = pi/20; be = 2*pi*s/300; kvec = [ 0 ; 0 ; 1 ]; tvec = [... cos(th).*cos(be);... sin(th).*cos(be);... sin(be) ]; Gvec = [... -sin(th).*cos(be).*th -cos(th).*sin(be).*be ;... cos(th).*cos(be).*th -sin(th).*sin(be).*be ;... cos(be).*be ]; 207
nvec = Gvec/norm(Gvec); N = m*g*(kvec’*nvec) + vˆ2*norm(Gvec); yprime(1) = y(2); yprime(2) = -g*(kvec’*tvec) - mu*N/m; return The Mathcad code is: m µ g
30 0 9.81
π
dθ s
20 π. s
θ s
20 π
dβ s
30
π
β s
3
. 2.
s 10 s 2
. 1
10 sin θ s . cos β s . dθ s cos θ s . cos β s . dθ s
Γ s
cos θ s t s
sin θ s
. cos β s . cos β s
cos β s
sin β s Γ s
n s
Γ s 0
k
0 1
N v, s
m. g. k. n s
a v, s
g. k . t s
m. v . Γ s
µ. m
N v, s .
208
v v
cos θ s sin θ s . dβ s
. sin β s . sin β s
. dβ s . dβ s
i 0 . . 1580 ∆t 0.001 ti i . ∆t
v0
0
s0
0
vi
1
si
1
vi
a vi , si . ∆t si vi . ∆t
s1580 = 10.008 Position on slide vs time 15
10
s i
5
0 0
0.5
1 t i time s
1.5
2
v1580 = 10.987 m/s
This is the velocity at the bottom of the slide. Note the time at the bottom was found by trial and error and for this case was 1.58 s.
209
2.116 Use the analysis of 2.115. The MATLAB code is: function yprime = ds2pt116(t,y) m = 30; g = 9.81; mu = 0.2; s = y(1); v = y(2); th = pi*s/20; be = - (pi/3)*(1-(s/10).ˆ2); th = pi/20; be = 2*pi*s/300; kvec = [ 0 ; 0 ; 1 ]; tvec = [... cos(th).*cos(be);... sin(t).*cos(be);... sin(be) ]; Gvec = [... -sin(th).*cos(be).*th -cos(th).*sin(be).*be ;... cos(th).*cos(be).*th -sin(th).*sin(be).*be ;... cos(be).*be [; nvec = Gvec/norm(Gvec); N = m*g*(kvec’*nvec) + vˆ2*norm(Gvec); yprime(1) = y(2); yprime(2) = -g*(kvec’*tvec) - mu*N/m; return
210
The Mathcad code is: m
µ g
30 0.2 9.81
π
dθ s
20 π. s
θ s
20
π
dβ s
30
π
β s
3
. 2.
s 10
. 1
10
. cos β s . dθ s . cos β s . dθ s
sin θ s
Γ s
2
s
cos θ s
cos β s
t s
cos
θ
s
. cos
β
s
sin
θ
s
. cos
β
s
sin
β
s
Γ s
n s
Γ s 0
k
0 1
N v, s
m. g. k. n s
a v, s
g. k. t s
m. v . Γ s
µ m
.N v, s .
211
v v
cos θ s sin θ s . dβ s
. sin β s . sin β s
. dβ s . dβ s
