CHAPTER 1
1.1 Using the definition of acceleration:
α =∆v
∆t− 60 mph − 0
9.2 − 0=
88 ft/sec − 0
9.2 sec= 9.57 ft/sec2
1.2 Differentiate x(t) to obtain the velocity:
v(t) = x(t) = −10t + 88 ft/sec.
Differentiating again yields the acceleration:
a(t) = x(t) = −10 ft/sec2.
So v(t) = 0 = −10t + 88.
Solving this for t yields that:
v(t) = 0 at t = 8.8 sec.
1.3 Evaluating x at zero yields x(0) = 5 m.
Differentiating yields x(t) = v(t) = 3t2 − 2
so that v(0) = −2 m/s.
Likewise
a(t) = x(t) = 6t
so that a(0) = 0.
Now at t = 3 sec,
x(3) = 27 − 2(3) + 5 = 26 m,
v(3) = 27 − 2 = 25 m/s
and a(3) = 18 m/s2.
Since the velocity changes sign during this interval, the particle has doubledback and to compute the total distance traveled during the interval you mustcompute how far it travels before it changes direction and then add this to thedistance traveled after the particle has changed direction. The particle changesdirection when the velocity is zero, or at the value of t for which
v(t) = 3t2 − 2 = 0,
or at time
t = 0.8165.
The particle first moves from
x(0) = 5 m to x(0.8165) = 3.9 m or a distance of 1.1 m.
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 + 21 = 23.2 m.
1.4 This again is straightforward differentiation:
v(t) = x(t) = 2t − 2.
This is zero when t satisfies:
2t − 2 = 0 or t = 1 sec.
Next:
a(t) = x(t) = v(t) = 2 ft/sec2 which is constant acceleration.
Alternately, the distance traveled can be computed directly from integratingthe absolute value of the velocity:
d =∫ 30 |3t2 − 2|dt = 23.178 m
1.5 Solution: v(t) = x(t) = 6 cos 2t m/s. a(t) = x(t) = v(t) = −12 sin(2t) m/sec2.Setting −12 sin(2t) = 0, yields 2t = π, or t = 0, π/2 s, π, 3π/2...nπ/2, for thetimes for the acceleration to hit zero.
1.6 From the definition, straightforward differentiation yields: xA = (3t2 + 6t) ft
so the vA = (6t + 6) ft/sec
and aA = 6 ft/sec2. Likewise, xB = 3t3 + 2t
so that vB = (9t2 + 2) ft/sec
and aB = 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/sec2
So that A is ahead of B and has the largest velocity, but B is accelerating fasterthe 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/sec2
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) andsolve for t. This yields
3t2 + 6t = 3t3 + 2t, or t2 − t − 4/3 = 0 and t = 0
Solving yields t = +1.758s and t = −1.758s.
Here we are interested in the positive value of time so that
xA(1.758) = xB(1.758) = 3(1.758)2+6(1.758) = 3(1.759)3+2(1.758) = 19.81 ft.
1.7 Note that this is an inverse problem. Straightforward differeniation yields:
v(t) = 3(−e−t sin 10t + e−t10 cos 10t) = 3e−t(10 cos 10t − sin 10t)
a(t) = 3e−t(−102 sin 10t − 10 cos 10t) − 3e−t(10 cos 10t − sin 10t)
= −3e−t(99 sin 10t − 20 cos 10t)
1.8 From the definition
x(t) = t3 − 6t2 − 15t + 40 ft
so that:
x = v = 3t2 − 12t − 15 ft/sec,
and x = a = 6t − 12 ft/sec.
a) v(t) = 0 requires 3t2 − 12t − 15 = 0
or t2 − 4t − 5 = 0.
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 − 6(5)2 − (15)(5) + 40 = −60 ft.
At rest t = 0,
x(0) = 40.
Then the particle has moved from 40 to -60 or 40 + 60 = 100 ft.
c) a(5) = (6)(5) − 12 = 18 ft/sec2.
1.9 Note: The purpose of this problem is to hit home the idea that the distancetraveled and the displacement are different. This problem is easiest to solveusing computational software as it involves plots. Students could also use asymbolic processor to compute the derivatives, although it would be a little over
3
kill and they must be reminded that simple derivations should be something theycan do in their head, on tests, while complicated derivatives are more accuratelydone with software. The plots are also easy to sketch by hand, but if they areinexperienced at plotting using software, it is best to start them off with somesimple plots.
a) v(t) = x = 0.6 cos 2t
a(t) = x = −1.2 sin 2t.
b) 3 sin 2t travels a distance of 0.3m in the time 0 to π4
sec and back another0.3m returning back to the origin from π
4to π
2s. So the total distance traveled
is 0.3 + 0.3 = 0.6m (maybe look at the plot first).
c) The position of the mass at t = π2s however is x
(
π2
)
= 0.3 sin(
2π2
)
= 0.
Note the position at any point is not always the distance traveled, which in b)is shown to be 0.6m.
t ..,0 .01π
2
x t .0.3 sin .2 tv t .0.6 cos .2 t
a t .1.2 sin .2 t
0 0.5 1 1.5 2
2
1
1
x t
v t
a t
t
FIGURE S1.9
4
1.10 Differentiation yields v(t) = 88.92 sin 0.26t ft, and
a(t) = 23.12 cos 0.26t ft/sec2.
The plot of each follows.
t ..,0 0.1 12.2
x t .342 1 cos .0.26 t v t .88.92 sin .0.26 t
a t .23.12 cos .0.26 t
0 2 4 6 8 10 12 14
500
500
1000
x t
v t
a t
t
FIGURE S1.10
1.11 The following code in Matlab computes the velocity from the displacement dataand 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:
0 0.05 0.1 0.15 0.2 0.25-100
0
100
200
300
400
500
t*dt or elapsed time
velocity versus time
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
Columns 19 through 21
-10000 0 0
EDU>>plot(t,a),xlabel(‘elapsed time’), title (‘acceleration ersus time’)
0 0.05 0.1 0.15 0.2 0.25-1
-0.5
0
0.5
1
1.5
2
2.5
3x 10
4
elapsed time
acceleration versus time
FIGURE S1.11b
1.12 Consider the following Matlab code which uses the central difference to computethe 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:
0 0.05 0.1 0.15 0.2 0.25-100
0
100
200
300
400
500
t*dt or elapsed time
velocity versus time
The following is the Mathcad code for solving this problem:
velocity in mm/s versus time in s
. .1 19
i 1
xi 1 xi 1
.2 ∆t
0 0.05 0.1 0.15 0.2200
0
200
400
600
vi
t
i
v
t=0.01∆
8
1.13 Solution:
y(t) = −4.905t2 + 20t (m)
v = y(t) = −9.81t + 20 (m/s)
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/s2
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 calculatewhen 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. toy(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) = −ce−ct sin ωt + e−ctω cos ωt
a(t) = c2e−ct sin(ωt) − 2cωe−ct cos ωt − ω2e−ct sin ωt
x(0) = 0, v(0) = ω, a(0) = −2cω
1.15 Solution:
x(t) = 3t3 − 2t2 + 5, x(0) = 5m
v(t) = x(t) = 9t2 − 4t, v(0) = 0 m/s
a(t) = v(t) = 18t − 4, a(0) = −4 m/s2
1.16 The velocity and acceleration are respectively:
v(t) = 4 · t · cos(π · t) − 2 · t2 · sin(π · t) · πa(t) = 4 · cos(π · t) − 8 · t · sin(π · t) · π − 2 · t2 · cos(π · t) · π2
so that x(0) = v(0) = 0 and a(0) = 4
9
1.17 Solution:
y(t) = 3t2 − 20, so y(0) = −20 m
v(t) = y(t) = 6t, so v(0) = 0 m/s
a(t) = y(t) = 6, so a(0) = 6 m/s2
1.18 Solution:
x(t) = exp(−0.1 · t) · (3 cos 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 splitusing heaviside functions and differentiated over the two intervals.
t ..,0 .01 10
y t ..5 t Φ 5 t .25 .20 sin .π t π Φ t 5
y1 t .5 Φ 5 t ..20 cos .π t π Φ t 5 is the velocity
y11 t ...20 π2 sin .π t π Φ t 5 is the acceleration
0 5 10
200
200
y11 t
y1 t
t
FIGURE S1.20
10
During the 1st interval the slope dy/dt is constant. From the plot y(t) =20−04−0
t− = 5t for t = 0 to 5 sec. In the interval t > 5, y(t) = 25 + A sin(ωt + φ)where ω is the frequency and φ is the plane and A is the amplitude of the sinewave.
From the plot A = 20, the period T = 2 sec, so that ω = 2πT
= π. To find thephase evaluate y(5) = 25 = 25 + 20(sin(πt + φ) so that πt + φ = nπ, where n isany integer or φ = (n − 5)π, so φ = π will work. Thus:
y(t) ={
5t 0 < t < 520 sin(πt − π) t > 5
y′(t) ={
5 0 < t < 520π cos(πt − π) t > 5
y′′(t) ={
0 0 < t < 5−20π2 sin(πt − π) t > 5
1.21 Solution: aave = 60−0 mph5 sec
= 605
mphsec
= 125280 ftmile
milehour
1sec
1 hour3600 sec
= (12)(5280)300
ft/sec2
= 17.6 ft/sec2
17.6 ft/sec2 = 17.6 m3.25
ftsec2 = 5.4 m/sec2
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 mile4 min
= 14
milemin
1 min60 sec
· 5280 ftmile
= 22 ft/sec
or about 6.71 meter/s, or about 15 mph. On the other hand a sprinter cancover 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. From the plot estimate the following data:x0 0 x1 0 x2 0 x3 .02 x4 .04
x9 0.5 x10 0.7x5 .1 x6 0.18 x7 0.25 x8 0.38
These are measured every 0.1 sec. Thus the velocity becomes:n ..,0 1 10
x11 0
vn
x n 1 xn
0.1Which is plotted below
0 0.2 0.4 0.6 0.8 110
5
0
5
vn
.n 0.1
Next estimate the acceleration from the velocity:
which is plotted belowan
vn 1 vn
0.1
0 0.2 0.4 0.6 0.8 1100
0
100
an
.0.1 n
12
Next the equivalent Matlab code is given (without the plots as they look thesame). 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 is of second order and is linear x(t). The term sin(πt) is nonlinear but int, 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|and sin θ(t).
1.30 This is first order in v and nonlinear because of the term v2.
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, v(0) = 0
v(t) =∫ t0(3t
2 − 4t)t = t3 − 2t2 ft/sec
x(t) =∫ t0(t
3 − 2t2dt = 14t4 − 2
3t3 ft
The plot from Mathcad is:
t . .,0 0.01 2
0 1 22
1
0
x t
t
x(t): = 0.25 t2 -
2_3
t3
FIGURE S1.33
1.34 Solution:
a(t) = 40t cos πt∫ v3 dv =
∫ t0 40x cos πxdx
v(t) = 3 + 40π2 [cos πt + tπ sin πt − 1]
∫ x5 dx = 3t + 40
π2
∫ t0(cos πx + xπ sin πx − 1)dx
x(t) = 5 + 3t + 40π3 sin πt − 40
πt + 40
π3 [sin πx − πx cos πx]t0
x(t) = 5 + (3t − 40π
t) + 40π3 [2 sin πt − πt cos πx]
14
The following is typed in Mathcad to produce the desired plot
t ..,0 .01 10
x t 5 .340
π2
t .40
π3
.2 sin .π t ..π t cos π t
0 5 1050
0
50
x t
t
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 constantacceleration equation (1.28) yields at = vf or acar = 88
6= 14.67 ft/sec2. In
terms of g’s, a = 14.6732.2
= 46%g. For the sprinter t0 = 0, x0 = 0, vf = 10 m/sand xf = 15 m. From equation (1.30)
a =v2
f − v20
2(xf − x0)=
102 − 0
2(15 − 6)= 3.33 m/s2
In terms of g, a = 3.339.81
= 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, accelerationand velocity. At the top of the motion v = 0, so eq. (1.30) becomes
0 = 2(−9.81)(xtop) + v20 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) becomes0 = a0t2
2+ v0t + 0. One solution is t = 0 which is the initial state. If t 6= 0, the
relationship becomes
t = v0
−2a0= 2.038 ∼ 2 sec
1.37 For constant acceleration v = at + v0 and v2 − v20 = 2ax so
a =v2−v2
0
2x= (5×106)2−(104)2
(2)(2)= 6.3 × 1012 m/s2,
t = v−v0
a= 5×106−1×104
6.3×1012 = 8.0 × 10−7 s.
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
−kx2
2|xx0
= v2
2|x0
kx20 = kx2 or
v2 = kx20 − kx2 or
v(x) =√
k(x20 − x2).
The position time relationship can be found from (1.20):
t =∫ t0 dt =
∫ xx0
dxv(x)
=∫ xx0
dx√k(x2
0−x2)
= 1√k
sin−1(
xx0
)
|xx0.
Rearranging and solving for x(t) yields
t =√
1k
sin−1(
xx0
)
|xx0or
√kt = sin−1
(
xx0
)
− sin−1(1) or√
kt+ π2
= sin−1(
xx0
)
or
x(t) = x0 sin(√
kt + π/2) = x0 cos(√
kt).
1.39 As the elevator starts from rest with constant acceleration to its operating speed
v2 = 2ax and v = at or
t = v/a = (3 m/s)/25 m/s2 = 1.2 sec
and travels a distance of
x = v2/2a = (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 slowingdown, the remaining distance 200 m - 3.6 m or 196.4 m is traveled at a constantvelocity 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.
1.40 Since a = −c2x case 2 applies and
v2 − v20
2=∫ x
x0
−c2xdx = −c2
(
x2 − x20
2
)
or v2 = v20 + c2x2
0 − c2x2. Substitution of x0 = 0, x = 10, v0 = 30, v = 0, yields0 = 302 + 0 − c2102 or c = 3.
1.41 Given a(x) = −cx2 x0 = 0, t0 = 0, v(0) = v0 we want to determine v(x). Fromeq. (1.17)
−c∫ x0 x2dx =
∫ vv0
vdv = 12(v2 − v2
0)
or12(v2 − v2
0) = − c3x3 or v(x) =
√
v20 − 2c
3x3
1.42 Since a is given as a function of velocity, case 3 applies: dx = vdv/f(v)
upon integrating
x − x0 =∫ vv0
vdv/(−v) = −v + v0.
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 expres-sion for y. Here acceleration is a function of position, so equations (1.17)-(1.20)apply. Given
a(y) = −g0R2
(R+y)2, g0 = 9.81 m/s2, R = 6370 × 103.
At t = 0, y0 = y(0) = 40 × 103 m, v0 = v(0) = 6000 m/s
Note ymax will occur when v = 0. So compute v.
a = dvdt
= dvdy
dydt
= −g0R2
(R+y)2. Integrating yields
∫ 0v0
vdv = −gR2∫ ymy0
dy(R+y)2
= g0R2[
1(R+ym)
− 1(R+y0)
]
, or
02
2− v2
0
2= g0R
2[
1R+ym
− 1R+y0
]
.
Thus60002
2= g0R
2[
1(R+y0)
− 1(R+ym)
]
or
4.552 × 10−8 =[
1R+y0
− 1R+ym
]
.
Solving for ym yields ym = 2575.
17
1.44 Solution:∫ 0vesc
vdv = −g0R2∫∞y0
dy(R+y)2
v2esc
2= −2g0R
2∫∞y0
dy(R+y)2
−v2esp = −2g0R
2∫∞y0
dy(R+y)2
= g0R2
limym → ∞
[
1R+ym
− 1R+y0
]
= −2g0R1
(R+y0).
That is −v2esp = −2g0R
2 1(R+y0)
Then: vesp = R√
2g0/(R + y0) = 11.14 km/s = 11.14 × 103 m/s
1.45 Given: a(v) = −cv = −0.4v, v0 = 100 km/hr.
Since a = dv/dt we have∫ vv0
dvv
= −ct or ℓn vv0
= − = ct.
Thus v = v0e−ct.
But v = dx/dt so that
x = x0 + v0
∫ t0 e−ctdt = x0 +
(
−v0
c
)
e−ct|t0Thus x(t) = x0 + v0
c(1 − e−ct).
With x0 = 0, v0 = 100 and c = 0.4 this becomes x(t) = 250(1 − c−0.4t).
1.46 Solution:
a(t) = 5 sin(20t) m/s2 x(0) = 1 m and v(0) = 3 m/s.
Integrating:∫ v0 dv = v − v0 =
∫ t0 5 sin 20αdα = − 5
20(cos 20t − 1)
where α is used as the “dummy” variable of integration. Then
v(t) = 3 − 0.25 cos 20t + 0.25 = 3.25 − 0.25 cos 20t m/s.
Integrating again yields
x(t) = x0 + 3.25t− 1.25 × 102 sin 20t
x(t) = 1 + 3.25t− 0.0125 sin 20t m
18
1.47 Solution:
v0 = 0.6 ft/s, a = −v3 ft/s2
Thus this is case 3 on page 19. However a straightforward integration of
a = dv/dt yields a(v) = −v3 = dvdt
.
Then dt = −dvv3 and integrating yields
t = − 12v2 + 1.389.
Rearrange to get v(t) =(
12t+2.778
)1/2ft/s.
At t = 4 sec, v(4) = 0.305 ft/s.
1.48 Follow example 1, because the acceleration is a function of velocity so case 3 isused.
Note that a = dvdt
= g − cv2 or dvg−cv2 = dt. Integrating both sides using the
stated initial conditions yields
t =∫ v
0
dv
g − cv2=
1
c
∫ v
v0
dvgc− v2
=(
1
c
)
1
2√
g/c
ln
√
g/c + v√
g/c − v
for gc
> v2 and 4(1c) 1
2√
g/cln
(
v−√
g/c
v+√
g/c
)
for v2 > g/c, from using a table of
integrals. Thus there are two possibilities. For g/c > v2; t = 1
2√
g/clm
[√g/c+v√g/c−v
]
.
Solving for v yields v(t) =√
gc
(
e2√
gct−1e2
√gct+1
)
; g/c > v2. For g/c < v2,
t = 12√
gcln
v−√
g/c
v+√
g/c
and solving for v(t) yields
v(t) =√
gc
(
1+e2√
gct
1−e2√
gct
)
; v2 > g/c.
Note from this second expression the v2 = g/c results in the expression −e2√
2ct =
e2√
2ct which has no solution. Thus v cannot reach the value g/c, i.e., v = g/c isthe driver’s terminal velocity. Next consider integrating again to calculate x(t),i.e., dx = vdt or
∫ x
0dx =
√
g
c
∫ t
0
(
e2√
gct − 1
e2√
gct + 1
)
dt
=
√
g
c
[
∫ t
0
e2√
gct
e2√
gct + 1dt −
∫ t
0
dt
e2√
gct + 1
]
x(t) =1
c
[
ℓn(
e2√
gct + 1)
−√gct − 0.693
]
19
1.49 This is a free fall problem or uniformly accelerated motion, where the accel-eration is given as g = 32.2 ft/s, and the time traveled can be determined byequation (1.29) with tf as the given. Equation (1.29) becomes
x(tf ) = 30 ft =gt2f2
+ v0tf + x0
Here v0 = 0, since the ball is dropped, x0 = 0 taking the window as the startingposition and hence
gt2f2
= 30 or tf =√
30/32.2 = 1.365 sec,
which is the time required to hit the ground. The expression for velocity underuniform 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, andtf = 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 writinga plus sign for a0(= g). This is consistent with the solution to 1.49. Solving forv0 yields
v0 =[
(g2)t2f − 30
]
/tf = 9.987 ft/sec.
1.51 Given a0 = 0.7g (constant acceleration), vf = 0 (because the car comes to astop). Convert mph to ft/s (60 mph = 88 ft/s, 45 mph = 66 ft/s, 30 mph = 44ft/s) and use eq. (1.30)
v2f = 2a0(xf − x0) + v2
0
where vf = 0, a0 = −0.7 g (minus because it decelerates), x0 = 0 (we start ourdistance measurement t = 0) then
xf =v2
0
−2a0=
v2
0
0.4g=
v2
0(ft/sec)2
(1.4)(32.2) ft/sec2
so that a) xf = 171.78 ft, b) xf = 96.63 ft, c) xf = 42.95 ft.
1.52 Following the solution to 1.52m with a0 = 0.4g yields
xp =v2
0
(−2)(−0.4)(32.2)
so that a) xf = 300.6 ft, b) xf = 169.1 ft, c) xf = 75.2 ft.
20
1.53 This is a uniform acceleration problem with a0 = 0.6g. Since the car starts fromrest v0 = 0 and asume x0 = 0. Let vf = 200 mph = 293.33 ft/sec. Then thetime 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 = a0t2f/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 themstart 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)(t2
f)
2
Then tf = 15.1 sec. During this time car B travels a distance determined by
xB(15.1) =aB(tf )2
2= (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 becomes55.47 mm insteady of 49.4 m.
1.55 This can be solved several ways including graphically by computing the areaunder the acceleration curve to generate the velocity, and the area under thevelocity 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 carefulto use the appropriate initial conditions at the beginning of each interval:
0 < t < 50 v(t) = 20t m/s
50 < t < 70 v(t) = 1000 m/s
70 < t < 100 v(t) = 7.5(t − 70)2 + 1000 m/s
100 > t 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 − 50)
70 < t < 100 x(t) = 2.5(t − 70)3 + 1000(t − 70) + 45, 000
t > 100 x(t) = 7750(t− 100) + 142, 500
21
This last expression yields
x(120s) = 297, 500 m = 297.5 km.
1.56 The equation to be solved is of the formdvdt
+ 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(∫
etetdt + C) = e−t(12e2t + C)
at t = 0, v(0) = 0 ft/s so that C = 0.5 and v(t) = 12(et − e−t) m/sec
= sin h(t) m/sec
Integrating again yields the displacement
x(t) = x0 +∫ t0 e−τ (0.5e2τ − 0.5)dτ when x0 = 1 m. Thus x(t) = 1
2(et + e−t)m
= cos h(t) m
1.57 The equation to be solved is of the formdvdt
+ 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 − 1) + C] = t − 1 + Ce−t
At t = 0, v(0) = 0 so that 0 = −1 + C or C = 1. Thus: v(t) = t − 1 + e−t m/s
Integrating again yields (x0 = 1m)
x(t) = x0+∫ t0(t−1+e−t)dt = 1+ t2
2−t−e−t+1, so that x(t) = 2 − t + t2
2− e−tm
1.58 a) The equation to be solved is of the form
dvdt
+ tv = e−tt/2
Comparing this form to equation (1.3.2), identifies p(t) = t and f(t) = e−t2/2.Thus the integrating factor becomes
λ(t) = e∫
tdt = et2/2
Next, equation (1.34) yields that the solution is
v(t) = e−t2/2(∫
et2/2e−t2/2dt + c) = e−t2/2t + ce−t2/2
At 0, v(0) = 10 ft/s so that c = 10 and v(t) = 10e−t2/2 + e−t2/2t
22
Integrating again (x(0) = 0) yields x(t) =∫ t0(10 + τ)e−τ2/2dτ which yields the
error function when integrated, i.e., x(t) = 12.5 erf(0.71t) − et2/2 + 1.
b) This does not have an integrating factor, or other closed formed solution, sothe solution must be found numerically by writing the equation in first order forand applying an Euler or Runge-Kutta solution. A Mathcad solution is shown.
i ..0 4000 ∆t 0.001 ti.i ∆t
x0 1 v0 5
a ,v t .t2
v 1 .3 t
xi 1
vi 1
xi.vi ∆t
vi.a ,vi ti ∆t
0 1 2 3 4
5
10
15
xi
vi
ti
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) .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) =A1e
−4t +A2e−t. The particular solution is guessed to be xp = a+bt, of the form
of the forcing function where a and b are to be determined. Substitution of theassumed form for xp(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(3
4) = −15
16. Thus xp = −15
16+ 3
4t. This is called the
method of undetermined coefficients. The total solution is the sum (x = xh+xp)so that
x(t) = A1e−4t + A2e
−t + 34t − 15
16
To determine the coefficients A1 and A2 apply the initial conditions
x(0) = 0.5 = A1 + A2 − 1516
v(0) = 0 = −4A1 − A2 + 34
which represents two equations in the two unkowns A1 and A2. Solving yields
A1 = −0.229 and A2 = 1.667 and hence: x(t) = −0.229e−4t + 1.667e−t + 34t − 15
16
1.60 Define ω2 = km
= 41
= 4 and ζ = c2mω
= 2.795 > 0 so the system is over damped.Then the problem in standard form is
x + 2ζωx + ω2x = x + 5x + 4x = 0
Assume solutions of the form x = Aeλt. The characteristic equation becomesλ2 + 5λ + 4 = 0 which has roots λ1 = −1, λ2 = −4. Thus the general solutionis of the form
x(t) = A1e−t + A2e
−4t
Applying the initial condition yields
x(0) = 5 = A1 + A2
v(0) = 0 = −A1 − 4A2
which is a system of two linear equations in the two unknowns A1 and A2.Solving yields A1 = 20
3and A2 = −5
3. Thus the solution is x(t) = 20
3e−t − 5
3e−4t
and v(t) = 203(−e−t + e−4t)
24
1.61 The equation of motion has the form (after dividing by 1000)
a = dvdt
= −cv − 400x, v(0) = 0, and x(0) = 0.01m
Following along with equation (1.51), the Euler method of integration yields[
vi+1
xi+1
]
=[
vi − cvi∆t − 400xi∆txi + vi∆t
]
,[
v0
x0
]
=[
00.01
]
Using a high level language (Matlab, Mathcad or Mathematica) yields (somestudents may know the analytical solution for this equation. Others will knowhow to use the more sophisticated higher-order Runge-Kutta integration thefollowing plot. Values of c are varied until the plot produces only two oscilla-tions.)
c 15∆t .01
i ..01.5
∆t
v0
x0
0
.01
vi 1
xi 1
vi..c vi ∆t ..400 xi ∆t
xi.vi ∆t
0 0.5 1 1.50.005
0
0.005
0.01
xi
.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. Thiscan 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 desiredresponse 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 withintwo cycles as desired.
1.62 Following the development of the numerical integration section equation (1.51)becomes[
vi+1
xi+1
]
=[
vi − 900vi∆t − 4000(xi)2∆t
xi + vi∆
]
with initial condition v0 = 0 and x0 = 20 mm. The Mathcad code is:
i ..0 1000
∆t .001
v0
x0
0
20
vi 1
xi 1
vi
..900 vi
∆t ...4000 xi
xi
∆t
xi
.vi
∆t
0 0.2 0.4 0.6 0.8 1
10
20
xi
.i ∆t
FIGURE S1.62
26
The equivalent Matlab code can be either Euler (see previous problem) orRunge-Kutta Method. To use RK, first save the following Matlab code un-der 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
[
vi+1
xi+1
]
=[
vi − 90vi∆t − 100x3i ∆t
xi + vi∆t
]
with initial condition v0 = 0 and x0 = 10 mm.
The Mathcad code is:
∆t 0.001 N 500 i ..0 N ti.i ∆t c 90 k 100
v0 0 x0 10 a ,v x .c v .k x3
xi 1
vi 1
xi.vi ∆t
vi.a ,vi xi ∆t
0 0.1 0.2 0.3 0.4 0.55
0
5
10
xi
ti
FIGURE S1.63
The catch gets near zero within 1/20 second.
27
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) 3];
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:
[
vi+1
xi+1
]
=[
vi − cvi∆t − 100(xi)3∆t
xi + vi∆t
]
Repeat the numerical solution to problem 1.63 with successively smaller valuesof damping (c) each time until the solution oscillates twice before coming torest. A value of about c = 20 1/s comes close as illustrated.
The Mathcade code is:
∆t 0.001 N 500 i ..0 N ti.i ∆t c 20 k 100
v0 0 x0 10 a ,v x .c v .k x3
xi 1
vi 1
xi.vi ∆t
vi.a ,vi xi ∆t
0 0.1 0.2 0.3 0.4 0.510
0
10
xi
ti
FIGURE S1.64
28
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) 3];
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:
[
vi+1
xi+1
]
=[
vi − cvi|vi|∆t − 4kxi|xi|∆txi + vi∆t
]
Then resolve for various values of c, k and x0 until a response that dies out inone oscillation results. There are many answers, the plot shows this is achievedfor 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 0.01 x0 .01v0 0
i ..0 3000
vi 1
xi 1
vi...c vi vi ∆t ...400 xi xi ∆t
xi.vi ∆t
0 5 10 15 20 25 30
0.005
0.005
0.01
xi
.i ∆t
FIGURE S1.65
29
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);
1.66 Assuming r = xi + yj + zk so that mr = mxi + myj + mzk. Then mr = −gjyields 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 whichare decoupled, linear equations.
1.68 Assuming r = xi + uj + zk and v = vxi + vy j + vzk yields
x = −c√
(v2x + v2
y + v2z) vx, y = −c
√
v2x + v2
y + v2z (vy) and z = −c
√
v2x + v2
y + v2z
(vz). These are coupled, nonlinear equations.
1.69 Assuming r = xi + yj + zk and v = vxi + vy j + vzkj
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 weknow 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 cos θ)t + 0
12 = −16.2t2 + (130 sin θ)t
which is two nonlinear algebraic equations in two unknowns t and θ. Solvingyields 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 solutionsfound by taking different initial guesses to the iterative solution of these non-linear algebraic equations. One solution has a low angle which would drive intothe bunker and one (correct) that has a larger angle which will loft onto thegreen.
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) becomesd = 2t, 0 = −9.81
2t2 + v0(0) + 3. Combining 4.905 t2 = 3 and t = d
2yields
d2 = 124.905
or d = 1.56 m.
1.72 Looing at the top half spray, Eq. 1.73 becomes
d2 cos 10◦ = 20 cos 70◦t + 0 (1)
d2 sin 10◦ = −16.1t2 + 20 sin 70◦t + 0 (2)
which is a system of two equations in the two unknowns: d2 and t.
From (1) t = d2 cos 10◦
20 cos 70◦= 0.1459d2 d2 = 20 sin 20◦(0.1439)−sin 10◦
(16.1)(0.1439)2or d2 = 7.588 ft.
Next consider the spray to the left:
d1 cos 10◦ = 20 cos 50◦t + 0 (1)
0 = −16.1t2 + 20 sin 50◦t + d1 sin 10◦ (2)
From (1) t = .0766d1 or t2 = .005268d21 and eq. (2) becomes
d1 = (20)2 cos2 50◦
(16.1)(cos210◦)(cos 10◦ tan 50◦ + 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
x = v0 cos θtf and 0 = gt2f2
+ v0 sin θtf . Solving this last expression for tf yieldstf = 2v0 sin θ
g, the time to hit the ground. Then from the expression for x
x(tf ) =2v2
0
gsin θ cos θ
The max value of x occurs at dx/dθ = 0 or2v2
0
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 valueof 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 − cos 20◦ tf and 0 = −16.1t2f + 140 sin 20◦tf + 4
Solving the last expression for tf yields tf = 3.055 sec. and -0.081 sec. Obviouslythe physical value is tf = 3.055, which from the first equation yields xf =140(cos 20◦)(3.055) = 402 ft.
1.75 From the projectile equation for z: z = −16.1t2 +140 sin 20t+4. The maximumvalue of the parabolic trajectory would occur at tf/2 except the value of tfcalculated in 1.74 assumes the trajectory is 4 ft off the ground. The equationfor time of flight is −16.1t2f +140 sin 20◦tf = 0 or tf = 2.97 sec, and tf/2 = 1.487
sec. Then zmax = −16.1(
2.922
)2+ 140 sin 20
(
2.922
)
+ 4 = 39.6 ft.
1.76 Consider the projectile equation 1.73 and first solve for v0 so the ball just clearsthe bottom window. Picking a coordinate system 1m off the ground yields
x0 = z0 = 0, xf = v0 cos 30◦ tf , zf = 2m = −9.812
t2f + v0 sin θtf
where xf = 6.5m. This yields two equations in two unknowns:
6.5 = v0(.886)tf or tf = 7.5v0
Thus v0 = 12.56 m/s. With yf = 3m, this becomes v0 = 19.18 m/s so that hemust 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 smallestand 3m for largest). xf = 6.5m so the first equation of (1.73) becomes (let tdenote 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 = −(
9.812
) (
0.65cos θ
)2+ 10 sin θ
(
0.65cos θ
)
which can be solved numerically for θ = 40.865◦. Changing the value of zf = 3mand repeating yields θ = 55.58◦. Thus he needs to kick through at an angle be-tween 40.9◦ < θ < 55.6◦ to make it through the window with an initial velocityof 10 m/s. Each of these two equations have 2 solutions so it will also make itin 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 = d√2
or t = d25
√2, 0 = −9.81
2t2 + d√
2, or d =
√2
29.81
(
d2
(625)(2)
)
, sod = 180.2 m.
1.79 Sample 1.14 gives the equation for a particle in projectile motion with windresistances. The equations are nonlinear and coupled and must be solved nu-merically. The initial conditions of x(0) = 0, vx(0) = 25, y(0) = 0, vy(0) = 0will 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. Theintersection will yield the value of d. Since we do not know d, it is best to putthe coordinate system at the end of the ski run and let z (or y) evolve in thenegative direction. Such a line passing through the origin has slope -1 and canbe written as d = −x, or di = −xi in incremental form. The Mathcad code is
i ..0 1100 ∆t 0.005 c 0.04 g 9.81
vx0 25 x0 0 vy0 0 y0 0
vxi 1
xi 1
vyi 1
yi 1
vxi..c vxi ∆t
xi.vxi ∆t
vyi.g .c vyi ∆t
yi.vyi ∆t
di xi
0 20 40 60 80 100 120 140
150
100
50di
yi
xi
FIGURE S1.79
Form the plot, then cross about 111 m out or d = 111/ cos 45◦ = 157 m downthe incline.
33
The Matlab code for solving and plotting is given next with the plots suppressedas 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 nonlineardamping term. The Mathcad code is:
i ..0 1000 ∆t 0.005 c 0.002 g 9.81
vx0 25 x0 0 vy0 0 y0 0
vxi 1
xi 1
vyi 1
yi 1
vxi..c .vxi vxi
2 vyi2 ∆t
xi.vxi ∆t
vyi.g .c .vyi vxi
2 vyi2 ∆t
yi.vyi ∆t
di xi
0 20 40 60 80 100 120
150
100
50di
yi
xi
FIGURE S1.80
34
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” be-comes:
xdot=[x(2);-c*x(2)*sqrt(x(2) 2+x(4) 2);x(4);-g-c*x(2)*sqrt(x(2) 2 + x(4) 2)];
1.81 Solution: 300 yards = 900 ft so that xf = x(tf ) = 900. Given that the ball isat zero to start with x0 = y0 = 0, and hits the ground at yf = 0. With θ = 90◦
given, the trajectory equations are
900 = v0 cos 9◦t + 0 (1)
0 = −16.1t2 + v0 sin 9◦t + 0 (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
x = (293.3) cos 9◦t = 289.7t (1)
0 = −16.1t2 + 45.8t (2)
From (2) t = 2.849 sec so from (1) x = 825.6 ft
35
1.83 This follows directly from the solution of sample 1.14 with the following valuesand equations: x0 = y0 = 0, c = 0.05, vx0 = 293.3 cos 9◦ ft/sec and vy0 =293.3 sin 9◦ ft/s.
i ..0 600 ∆t 0.005 c 0.05 g 32.2vx0
.293.3 cos .9 deg
x0 0 vy0.293.3 sin .9 deg y0 0
vxi 1
xi 1
vyi 1
yi 1
vxi..c vxi ∆t
xi.vxi ∆t
vyi.g .c vyi ∆t
yi.vyi ∆t
0 200 400 600 800 1000
20
20
40
yi
xi
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 (soyf = 3 ft), xf = 20 ft, θ = 45◦ and hence
20 = v01√2t or t = 20
√2
v0
3 = −16.1t2 + v0 · 1√2t
Solving yields v0 =√
(16.1)80017
= v0 = 27.53 ft/s.
36
1.85 Using the projectile motion equations with x0 = y0, xf = 20 ft, yf = 3 ft andv0 = 30 ft/sec, equations (1.73) becomes
20 = (30) cos θt (1)
3 = −16.1t2 + (30) sin θt (2)
Substitution of t = 23 cos tθ
from (1) into (2) yields the transindental equation
(3) = −16.1 49 cos2 θ
+ 20 tan θ
Solving for θ yields θ = 33.7◦ and 64.8◦. Either angle will “work”, however thelower angle gives a trajectory up through the bottom of basket whereas the64.8◦ 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.
v 0 30 x 0 0 y 0 7 x f 20 y f 10
g 32.2
Initial guess for time and angle
θ .30 deg t f 3
Given
x f..v 0 t f cos θ x 0
y f.g
t f2
2..v 0 t f sin θ y 0
=Find ,θ t f1.132
1.568=
1.132
deg64.859
Angle in degrees
A second solution can be found with a lower angle.
v 0 30 x 0 0 y 0 7 x f 20 y f 10
g 32.2
Initial guess for time and angle These values are assumed less.
θ .10 deg t f 1
Given
x f..v 0 t f cos θ x 0
y f.g
t f2
2..v 0 t f sin θ y 0
=Find ,θ t f0.588
0.801=
0.588
deg33.69
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 = 120mph = 176 ft/s and yf = 0 (i.e., hits the ground) yields
xf = 176t (1)
0 = −16.1t2 + 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 (1)
yf = 11 = (v0 sin 80◦)t − 16.1t2 + 0 (2)
Substitute t = 6v0 cos 80
from (1) into to (2) to get
11 = 6 tan 80◦ − 16.162
v20 cos2 80◦
(3)
Solving yields v0 = 28.9 f/s. b) Repeating (a) with xf = 16 eq. (3) becomes
11 = 16 tan 80◦ − 16.1162
v20 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 = at
r= 12 ft/s2
140 ft=
0.08 rad/s2 a constant. dwdt
= 0.08 so that w − w0 = 0.08t or w = 0.08t + w0.From (1.84) an = rw2 = 25 = 150(w0 + 0.08ts)
2(1), where ts = time to slip.
Also at ts, a =√
a2t + a2
r =√
252 + 122 = 27.73 ft/s2 at slip. Solving (1), with
ω0 = 0 for ts yields ts = 10.08
(
25150
)1/2= 5.104 s.
1.89 This is a circular motion with r = 2 m and at(t) = 6 sin πt(m/s2). The particlestarts at rest so that θ(0) = ω(0) = 0. For circular motion v(t) =
∫ t0 atdt =
∫ t0 6 sin πt = 6
π(1 − cos πt). Also ar = v2
r= 1
2( 36
π2 )(1 − cosπt)2 = 18π2 (1 − cos π6)2.
From equation (1.81) taking the magnitude of a(t) yields
a(t) =√
a2t + a2
n =√
62 sin2 πt + [ 18π2 (1 − cos πt)2]2.
38
The plots follow:
t ..,0 0.001 2
v t .6
π1 cos .π t at t .6 sin .π t an t .18
π21 cos .π t 2
a t at t 2 an t 2
0 0.5 1 1.5 2
5
10
time s
vel
m/s
acc
el m
/s^2
a t
v t
t
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=0.5v 2;
at=6*sin(pi*t); a=sqrt(at 2+an 2)
ezplot(a,[0,2]), ezplot(v,[0,2])
1.90 From the solution to 1.89 a(t) = (36 sin3 πt + 183
π4 (cos πt− 1)4)1/2. Find ts whena(t) = 5. The answer can be seen from the plot given in figure S1.89 or fromsolving
52 = 36 sin2 πts + 182
π4 (cos πts − 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 = 301× 103 m
1hr
3600 sec·hr= 8.33 m/s. For circular
motion, equation (1.81) yields
an = v2
4= 8.33
200− 0.3469 m/s2 or an = 0.35 m/s2.
39
1.92 This is circular motion starting from rest so that θ(0) = ω(0) = 0, with r = 4m.
a) α(t) = 2t2 r/s2 so that dωdt
= 2t2 and ω − ω0 = 23t3
or w(t) = 23t3. Thus from
eq. 1.84 an = r(ω)2 = 4 · 49t6 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.
b) From eq. (1.82) d2sdt2
= ra = 8t2 so that dv = 8t2dt or v − 0 = 83t3. Thus
ds = 83t3dt and s(2) = 8
3
∫ 20 t3dt and the total distance traveled is s = 10.67 m.
1.93 Solution: α(t) = 3t2 − 2t rad/s2 with ω0 = θ0 = 0. Integrating yields dωdt
=
3t2 − 2t or∫ 20 dω =
∫ 60 3t2 − 2t or ω(t) = t3 − t2 rad/s. Likewise dθ = (t3 − t2)dt
so that θ(t) =∫ t0(t
3 − t2)dt = t4
4− t3
3or θ(t) = 1
4t4 − 1
3t3 rad.
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 =
∫ t0 x cos πxdx = 1
π2 [cos πt + πt sin πt − 1]so that ω(t) = (2 − 1
π2 ) + 1π2 (cos πt + πt sin πt) rad/s. Integrating again yields
θ − π6
= 1π3 [2 sin πt − πt − πt cos πt + 2π3t]
and
θ(t) = π6
+ 2t + 1π3 (2 sin πt − πt − πt cos πt) rad
1.95 Solution: ω0 = 3 rad/s and α(ω) = −2ω2 rad/s2. From eq. (1.87) α = ω dωdt
=
−2ω2. Solving yields∫ ω3
dωω
= −2∫ θ0 dθ, or ℓnω
3= −2θ. ω = 3e−2θ so that
dθdt
= 3e−2θ or∫ θ0 e2θdθ =
∫ t0 3dt = 3t. Evaluating the other integral yields
12e2θ|θ0 = 3t or e2θ − 1 = 6t
and 2θ = ℓn(6t + 1) or θ(t) = 12ℓn(6t + 1) and θ(10) = 1
2ℓn(61) = 2.055 rad.
1.96 Solution: ω0 = 2 rad/s and α(ω, t) = −0.01ω+4t rad/s2. Then dωdt
= −0.01ω+4twhich 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 solutionω(t) = e−0.01t [C + 40, 000e0.01t(0.0t − 1)]
Since ω(0) = 2, C = −3998 and
ω(t) = −3998e−0.01t + 4000(0.01t− 1)
ω(t) = −4000 − 3998e−0.01t + 40t rad/s = [0.01t − 1 + e−0.01t](4 × 104) rad/s
Since ω = dθdt
, integrating this ω(t) yields θ(t).
θ(t) = 400 + [0.005t2 − t − 0.01e−0.01t](4 × 104) rad
where 3998 has to be rounded to 4000.
40
1.97 The equation of motion can be written as dωdt
= 2 sin θ − 0.4ω. Since ω = dθdt
,this can be written as a second order nonlinear equation in θ:
θ + 0.4θ − 2 sin θ = 0 θ(0) = π6
and θ(0) = 0
which can be solved by numerical integration as suggested in equation (1.93).That is[
ωn+1
θn+1
]
=[
ωn + (2 sin θn − 0.4ωn)∆tθn + ωn∆t
]
,[
ω0
θ0
]
=[
0π6
]
which is plotted in the following Mathcad file:
i ..0 500 ω0 0 θ0
π6∆t 0.01
ωi 1
θi 1
ωi..2 sin θi
.0.4 ωi ∆t
θi.ωi ∆t
0 1 2 3 4 5
2
4
6
θi
.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
1.98 This is a second order, nonlinear differential equation in θ which can be solvednumerically by using the first order form suggested in equation (1.93). Fromthe given form of θ + 0.02θ|θ| − 3 cos θ = 0 θ(0) = π
6and θ(0) = 0. Equation
(1.93) becomes
[
ωn+1
θn+1
]
=[
ωn + (3 cos θn − 0.02ωn|ωn|]∆tθn + ωn∆t
]
which is plotted below in Mathcad:
i ..0 500 ω0 0 θ0
π6∆t 0.01
ωi 1
θi 1
ωi..3 cos θi
..0.02 ωi ωi ∆t
θi.ωi ∆t
0 1 2 3 4 5
1
2
3
θi
.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
1.99 Follow sample 1.18. Given r(t) = 3 cos ti + 3 sin tj + 4tk m, differentiationyields v(t) = −3 sin ti + 3 cos tj + 4k and a(t) = −3 cos ti − 3 sin tj so thatv(3) = i+ j+4k and a(3) = i− j. The unit tangent vector et is calculated fromeq. (1.97) to be
et(3) = v(3)|v(3)| = −0.085i − 0.594j + 0.8k
at(3) = (a · et)et = 0, an = a − at = a so that
en = an
|an| = a|a| = 0.99i − 0.141j
eb = et × en = 0.133i + 0.792j + 0.6k
1.100 From problem 1.99
v(t) = −3 sin ti + 3 cos tj + 4k and a(t) = −3 cos ti − 3 sin tj m/s2
Note magnitude of both v(t) and a(t) is constant.
1.101 From equation 1.100, ρ(t) = v2(t)|an(t)| where v2(t) is v(t) ·v(t) = 9 sin2 t+9 cos2 t+
16 = 9 + 16 = 25. Compute |an(t)|. From eq. (1.97) et = v
|v| = 15(−3 sin ti +
3 cos tj + 4k. Then from eq. (1.98)
at(t) = (a · et)et = 15[(−3 cos t)(−3 sin t) + (−3 sin t)(3 cos t) + (0)(4)]et = 0.
Now an = a − at = a − 0 = a. Thus |an(t)| = |a(t)| =√
32 cos2 t + 32 sin t = 3.Thus
ρ(t) = v2(t)|an(t)| = 1
3[25] = 8.33 m, a constant so that the motion is circular, moving
at a constant angular velocity.
1.102 Given r(t) = t2 i+3tj+10 sin tk m, successive differentiation yields the velocityand acceleration:
v(t) = r(t) = 2ti + 3j + 10 cos tk m/s
a(t) = r(t) = 2i − 10 sin tk m/s2
From eq. (1.100) the radius of curvature is ρ(t) = v2(t)|an|
Now v2 = v·v = 4t2+9+100 cos2 t. Following eq. (1.97) et(t) = v
|v| , at = (a·et)et,
an = a−at and ρ(t) = v2
|an| . Programming these formulations and evaluating ateach value of t yields
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
v
.2 t
3.10 cos t
a
2
0.10 sin t
etv
vat ..a et et
an a at
ρ.v v
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)
v = rer + rθeθ = ter + πt(1 + t2)eθ
a = (r − rθ2)er + (rθ + 2rθ)eθ = (1 − 2π2t2(1 + t2))er + (π(1 + t2) + 4t2π)eθ
To plot the motion define t : 0, 0.1...2, define r = 1+t2
2and θ = πt2. Then let
x = r cos θ and y = r sin θ which is plotted below.
44
The Mathcad code is:
t ..,0 0.01 2
r t1 t2
2θ t .π t2
x t r t 8 cos θ t y t .r t sin θ t
20 10 0 10 20
4
2
2
y t
x t
FIGURE S1.103
The Matlab code is:
EDU>t=linspace(0,2);
EDU>r=(1+t 2)/2;th=pi*t. 2;
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)| =√
t2 + π2t2(1 + t2)2 =
t√
1 + π2(1 + t2)2 = ds/dt so that
s − s0 =∫ 20
√
t2 + π2t2(1 + t2)2dt = 18.977 m.
1.105 Solution: r(t) = 2 so that r = r = 0 and there is circular motion. θ(t) = sin πtso that θ = π cos πt and θ = −π2 sin πt. From equations (1.103):
v(t) = rθeθ = 2π cos πteθ and
a(t) = −2π2 cos2 πter − 2π2 sin2 πteθ
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
r t 2
θ t sin .π t
0
30
6090
120
150
180
210
240270
300
330
2
1.5
1
0.5
0r t
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:dsdt
= |v(t)| =√
4π2 cos2 πt = |2π cos πt|s − s0 =
∫ 20 |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
r(t) − r(0) =∫ t0 r(x)dx =
∫ t0 2e−xdx = 2(1 − e−t)
r(t) − 1 = 2∫ t0(1 − e−x)dx so that r(t) = 1 + 2(t + e−t − 1) = −1 + 2t + 2e−t
Likewise
θ(t) − 0 = π∫ t0 dx = πt and θ(t) − 0 = π
∫ t0 xdx = πt2
2
46
Now from eqs. (1.103)
v(t) = 2(1 − e−t)er + [(2e−t + 2t − 1)πt]eθ
a(t) = [2e−t − (2e−t + 2t − 1)π2t2]er + [(2e−t + 2t − 1)π + 2(2 − 2e−t)πt]
1.108 Let r(t) = 1 + 2(t + e−t − 1) and θ(t) = πt2
2. Then define x(t) = r(t) cos θ(t),
y(t) = r(t) sin θ(t) and plot (Mathcad solution)
t ..,0 0.01 2r t 1 .2 t e t 1 θ t
.π t2
2
x t .r t cos θ ty t .r t sin θ t
3 2 1 0 1 2 3 4
4
2
2
y t
x t
FIGURE S1.108
The distance traveled is∫ 20 |v(t)|dt = 14.438 m
The Matlab code is:
EDU>linspace(0,2);
EDU>r=1+2*(t-exp(-t)-1);th=(pi*t. 2)/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 sin 2πt, to determine the accelerationrequires expressions for r, θ, r and θ. Since ω = 2π, dθ = 2πdt and θ = θ0 +2πt,θ = 2π, and θ = 0.
Likewise r = 2πra cos 2πt and r = −4π2ra sin 2πt. From eq. 1.103, ar =r − rθ2 = −4π2ra sin 2πt − (r0 + ra sin 2πt)4π2 = −4π2(2ra sin 2πt + r0) m/s2
aθ = rθ − 2rθ = (2)(2πra cos 2πt)(2π) = 8π2ra cos 2πt m/sec2
47
1.110 Solution: a) Let r0 = 2, ra = 1.5 < r0 and θ0 = 0, r(t) = 2 + 1.5 sin 2πt,θ(0) = 0 so θ(t) = 2πt. For one revolution let t = 0, 0.01...1 sec, then letx(t) = r(t) cos θ(t) and y(t) = r(t) sin θ(t) which is plotted below using Mathcad
t ..,0 0.01 2r t 2 .1.5 sin ..2 π t θ t ..2 π t
0
30
6090
120
150
180
210
240270
300
330
3210
r t
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)
b) The velocity is v(t) = rer + rθeθ so that:
|v(t)| = |√
r2 + r2θ2| = π√
1.5 cos2 2πt + (2 + 1.5 sin 2πt)2
so s =∫ 10 |v(t)|dt = 14.407 m
48
1.111 Solution: ar = 2t, aθ = cos(πz), r(0) = .5 m, θ(0) = 0. Since it starts fromrest θ(0) = r(0) = 0. From eq. (1.103) r − rθ2 = 2t, rθ + 2rθ = cos(πt)which is a system of 2 coupled 2nd order equations which are nonlinear andinhomogeneous. The initial conditions (4) are given above. To solve numericallyfollow sample 1.22 (use x = r, y = θ)
r = rθ2 + 2t x = r · y2 + 2tr = x
θ = −2 rθr
+ cos πtr
: y = −2xy/r + cos πtr
θ = y
The Euler formula becomes (tn+1 = tn + ∆t)
xn+1 = (rn · y2n + 2tn)∆t + xn
rn+1 = rn + xn∆tyn+1 = (−2xn · yn/rn + cos πtr
rn)∆t + yn
θn+1 = yn∆t + θn
x0
r0
y0
θ0
=
0.500
Once these are solved the polar coordinate r(t) and θ(t) are given by the digitalrecord for rn and θn.
49
1.112 The trajectory is obtained by plotting rn cos θn vs. rn sin θn from above. TheMathcad solution is:
Trajectory in meters
i . .0 2000 ∆t 0.001
a ,,,,v r ω θ t .2 t .r ω2
α ,,,,v r ω θ t .1
rcos .π t ..2 v ω
t .i ∆t
v
0r
0
ω0
θ
0
0.5
0
0
vi 1
ri 1
ωi 1
θi 1
.a ,,,,vi
ri
ωi
θi
ti
∆t vi
ri
.v
i
∆t
ωi
.α ,,,,vi
ri
ωi
θ
i
ti
∆t
θ .ωi
∆t
x .r cos θ yi
.ri
sin θ
0.5 1 1.5 2 2.5 3 3.50
0.2
0.4
0.6
0.8
yi
xi
0
i
i i i i
FIGURE S1.112
50
The equivalent Matlab code is:
function xdot=onept112(t,x);
xdot=[x(2);x(2)*x(3) 2+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 − rθ2 and aθ = rθ + 2rθ.
We are given r and θ which we need to integrate to get r, r, θ and θ. Firstconsider θ = 0 so that θ = constant = 1.5 rad/s. Integrating again yieldsθ(t) = 1.5t. Then the above becomes simply
ar = r − (1.5)2r, aθ = 2(1.5)r = 3r
Integrating r = 3 − 0.01r requires the solution of
r + 0.01r = 3
Which is a second order differential equation with particular solution rP = 3t0.01
.The homogeneous equation is r + 0.01r = 0 which has solution r1 = A andr2 = Beλt. Substitution yields
λ2 + 0.01λ = 0 or λ = −0.01 so the homogeneous solution is
rH(t) = A + Be−0.01t
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
r(0) = A + B = 0.4, r(0) = 0.01B + 300 = 0 or B = −30, 000
A = 30, 000.4
Thus
r(t) = −30, 000.4 − 30, 000e−0.01t + 300t m
Thus r = 300(1 − e−0.01t), r = 3e−0.01t so that
ar = 3e−0.01t − 2.25(30000.4− 30000e−0.01t + 300t)
aθ = (3)(30000.4) − 90, 000e−0.01t + 900t
51
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 .30000 e.0.01 t .300 t
θ t .1.5 t
0
30
6090
120
150
180
210
240270
300
330
200010000
r t
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 (assumingθ(0) = 0). Then r(t) = r(θ(t)) = 100 + 60 cos πt
4so that r(t) = −15π sin πt
4and
r = −15π2
4cos πt
4.
52
Thus from eq. (103)
v(t) = (−15π sin πt4)er + (π
4)(100 + 60 cos πt
4)eθ
and
a(t) =[
−154π2 cos πt
4−(
π4
)2 (
100 + 60 cos πt4
)
]
er +[
−152π2 sin πt
4
]
eθ
a(t) = −(100 + 120 cos πt4)π2
16er − [15π2
2sin πt
4]eθ
1.116 Here we need to find r(t) from the drawing and knowledge of θ(t). θ(t) =π4
sin πt, so that θ = π2
4cos πt and θ = −π
43 sin πt = −π2θ(t). From the drawing
300 = r(t) cos θ(t) so that
r(t) = 300 sec θ(t) = 300 sec (π4
sin πt)
r(t) = 75π2 tan(π4
sin πt) sec (π4
sin πt) cos πt
r(t) = 300[
π2
16sin2 πt tan(π
4sin πt) + π4
16cos2 πt(tan2(π
4sin πt) − 1)
]
sec(π4
sin πt)
Then
v = [75π2 tan(π4
sin πt)sec(π4
sin πt) cos πt]er + [75π2sec(π4
sin πt) cos πt]eθ
The aceleration becomes (in terms of θ, θ and θ)
a = 300 sec θ[(2 tan2 θ)θ2 + tan θθ]er + 300 sec θ[θ + 2 tan θθ2]eθ
53
1.117 Let θ be the angle between r(t) and the 60 mm line. Let β(t) be the anglebetween the 100 mm radius and the end of r(t). From this triangle
r(t) = 60 cos θ + 100 cosβ
and from the law of sines: (1) 60 sin θ = 100 sinβ. Differentiation of the lawof sines yields (2) 60 cos θθ = 100 cosββ. These last two expressions can beused to remove the β dependence. Differentiating r(t) and using (1) and (2) toremove the β term yields
r = −60 sin θθ − 100 sinββ = −60 cos θ tan βθ
where we note that eq. (1) can be used to remove the β dependence in favor ofθ. Now θ = constant = π is given, so r becomes
r = −60π sin θ − 60π cos θ tanβ, where β = sin−1(0.6 sin θ). From (1)
r = −60π2 cos θ + 60π2 sin θ tanβ − 602π2 cos2 θ100 cos3 β
which can be verified symbolically using one of the codes. With r, r and rgiven, θ = π, so that θ = πt and θ = 0, then a can be determined from equation(104):
ar = r − rθ2 and aθ = 2rθ = 2πr.
The Mathcad code for plotting this follows:
a r t ddr t .r t π2
a θ t ..2 dr t π
0 0.5 1 1.5 23000
2000
1000
0
1000
2000Accelerations vs time
time s
a r t
a θ t
t
Ac
ce
lera
tio
n i
n m
/s2
FIGURE S1.117
54
1.118 From the problem statement α(t) = 0.5 cos t. Integrating yields ω(t) = 0.5 sin t+ω0 = 0.5 sin t since it starts from rest. Integrating again (θ0 = 0) yieldsθ(t) = −0.5 cos t + C = −0.5 cos t + 0.5. Now recall the formulation of theprevious problem β(t) = a sin(0.6 sin(θ(t)). Then using the formulas developedin problem 1.117, ar and aθ can be determined. These are illustrated in thefigure where the derivatives are confirmed by symbolic computations.
t ..,0 0.01 7 θ t .0.5 cos t 0.5 ω t .0.5 sin t α t .0.5 cos t
β t asin .0.6 sin θ t r t .60 cos θ t 100 cos β t
First symbolicaly calculate the time derivative of r(t) and define it by rd(t)
d
d t.60 cos θ t 100 cos β t
..60 sin θ td
d tθ t ..100 sin β t
d
d tβ t
rd t ..60 sin θ td
d tθ t ..100 sin β t
d
d tβ t
Next symbolically calculate the second deriviative and denote it by rdd(t)
d
d t..60 sin θ td
d tθ t ..100 sin β t
d
d tβ t
..60 cos θ td
d tθ t
2
..60 sin θ td
d
2
2tθ t ..100 cos β t
d
d tβ t
2
..100 sin β td
d
2
2tβ t
rdd t ..60 cos θ td
d tθ t
2
..60 sin θ td
d
2
2tθ t ..100 cos β t
d
d tβ t
2
.100 sin β t
Now define the acceleration components in terms of r and its derivatives, theta and omega:
ar t rdd t .r t ω t2
aθ t ..2 rd t ω t .r t α t
0 2 4 6 8
100
50
50
100
ar t
aθ t
t
. d
d
2
2tβ t
FIGURE S1.118
55
1.119 The acceleration and velocity in cylindrical coordinates are given in eq. (1.109)and (1.111) and require r, r, r, z, z, θ and θ. Here r = 1.5, θ = π and z =0.5 cos 2πt so that r = r = 0, θ = 0, θ = 0, z = −π sin 2πt and z = −2π2 cos 2πt.Thus
v = 1.5πeθ − π sin(2πt)ez m/s
a = −1.5π2er − 2π2 cos(2πt)ez m/s2
The Mathcad code for generating this plot is:
i . .0 40 t .0.1 i
x .1.5 cos .π ti
yi
1.5 sin .π ti z
i.0.5 cos ..2 π t
i
10
11
0
1
00.2
0.20.4
,,x y z
i
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 dsdt
= |v| or s = s0 +∫ t0 |v|dt. Assuming
the particle starts out s0 = 0, s =∫ 20
√
(1.5π)2 + (π sin 2πt)dt = 10.398 m.
The total distance traveled can also be calculated using software. For example,
the Mathcad code for computing this distance follows:
v t
0
.1.5 π
.π sin ..2 π t
=d0
2
tv t 10.398
t . .,0 0.01 2
FIGURE S1.120
1.121 Solution: r(t) = R − zhR, = R − h−0.1ht
hR = R − R + 0.1Rt = 0.1Rt so that
r = R10
and r = 0. θ(t) = 2πt so that θ = 2π and θ = 0.
z(t) = h(1 − 0.1t) so that z = −0.1h and z = 0.
Thus
v = 0.1Rer + (0.1Rt)(2π)eθ − 0.1hez = 0.1Rer + 0.2πRteθ − 0.1hez, and
a = (−0.1Rt)4π2er + (R5· 2π)eθ = −0.4Rπ2ter + 0.4πReθ.
The particle reaches the bottom of the cone when z(t) = 0 or h(1 − 0.1t) = 0or at t = 10 sec. Since θ = 2πt, as 10 sec have pased, then θ has gone around10 times before it reaches the bottom (2π10 = 20π or 10 complete cycles).
1.122 Solution: for R = 3m and h = 5 m, v(t) = 0.3er + 0.6teθ − 0.5ez so that
|v| =√
(0.3)2 + (0.6t)2 + (0.5)2 as it takes 10 sec to travel to the bottom
s =∫ 100 |v|dt =
∫ 100
√
(0.3)2 + (0.6t)2 + (1.5)2dt = 94.67 m
1.123 From eq. (1.111) the given form of the differential equations are er : 3 =r − rθ2 (1), eθ : 2rθ = 0.1 (2) and from ez : z = −2. Since the particle startsfrom rest at zero, all the initial conditions are zero and since the z coordinate isdecoupled 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 equationswhich must be solved numerically. They are
r = rθ2 + 3
θ = −2 rθr
+ 0.1r
57
Putting these in 1st order form or Euler integration yields
rrθθ
=
0100
,
rrθθ
=
rθy2 + 3dr
0.1/r − 2drdθ/rdθ
The Euler equations then becomes (follow sample 1.22) as illustrated in
rn+1
drn+1
θn+1
dθn+1
=
rn + drn · ∆t(rn · dθ2
n + 3) · ∆t + drn
θn + dθn · ∆tdθn +
(
0.1rn
− 2drn·dθn
rn
)
· ∆t
,
r0
dr0
θ0
dθ0
=
0000
Figure S1.123 shows the integration along with a plot of the first 3 s usingMathcad.
i ..0 3000
∆t 0.001
ti.∆t i
ar ,,,r dr θ dθ 3 .r dθ2
aθ ,,,r dr θ dθ .1
r0.1 ..2 dr dθ
r0 1 dr0 0 θ0
0 dθ0
0
dri 1
ri 1
dθi 1
θi 1
dri.ar ,,,ri dri θ
idθ
i∆t
ri.dri ∆t
dθi
.aθ ,,,ri dri θi
dθi
∆t
θi
.dθi
∆t
0
30
6090
120
150
180
210
240270
300
330
1050
ri
θi
FIGURE S1.123
58
The required Matlab code is as follows:
function xdot=onept123(t,x);
xdot=[x(2);3+x(1)*x(4) 2;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),θ(t) = π2 cos πt and θ(t) = −π3 sin πt. Since φ(t) = π
2te−t, φ(t) = π
2(e−t − te−t)
and φ(t) = π2(−e−t + (−e−t + te−t)) = π
2e−t(t − 2). Now from equation (1.116)
v(t) = 0.2teR + π2(0.3 + 0.1t2)e−t(1 − t)eφ+[0.3 + 0.1t2][sin(πt
2e−t)](π2 cos πt)eθ
From Eq. (1.118)
a(t) = [0.2 − (0.3 + 0.1t2)(π2(1 − t)e−t)2 − (0.3 + 0.1t2)[sin(πt
2e−t)]2(π4 cos2 πt)]eR
+[(0.3 + 0.1t2)(π2(t − 2)e−t) + 2(0.2t)(π
2e−t(1 − t))
−(0.3 + 0.1t2)(π2 cos πt)2 sin(πt2e−t) cos(πt
2e−t)] eφ
+[(0.3 + 0.1t2)(−π3 sin πt) sin(πt2e−t) + 0.4t(π2 cos πt) sin(πt
2e−t)
+2(0.3 + 0.1t2)(π2e−t(1 − t))(π2 cos πt) cos(πt
2e−t]eθ
1.125 From Sample Problem 1.24 we have a = g sin φeφ, and we are given that R = 200mm, (Rθ)0 = 600 mm/s and that φ0 = π/2, θ0 = 0, g = 9810 mm/s2. Sinceh is constant (R = 200 ⇒ θ0 = 3), h = θ0 sin2 φ0 = 3 rad/s. Now fromthe 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 coupledequations
θ = hsin2 φ
and φ = h2 cos φsin2 φ
+ gR
sin φ
subject to the initial condition φ0 = π2, θ0 = 0, φ0 = 0.
59
The Euler formulation and a plot of the motion is given in the following Mathcadcode:
i ..0 800 ∆t 0.001 ti.∆t i h 3 R 200 g 9810
aφ φ.h2 cos φ
sin φ 2.g
Rsin φ
dθ φ3
sin φ 2
dφ0 0 φ0
π2
θ0 0
dφi 1
φi 1
θi 1
dφi.aφ φi ∆t
φi.dφi ∆t
θi.dθ φi ∆t
xi..R sin φi cos θi yi
..R sin φi sin θi zi.R cos φi
0 100
100
0
100
200
100
0
,,x y z
FIGURE S1.125
60
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)+(h 2*cos(p(n))/(sin(p(n)) 2 + (g/R)*sin(p(n))*dt;
p(n+1)=p(n)+x(n)*dt;
th(n+1)=th(n)+(3/sin(p(n)) 2)*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 θ0 = 0. Since θ0 and theconstant of motion θ sin2 φ = h must hold for all θ, h must be zero and thekinematic equations become φ = g
Rsin φ and θ = 0 which are numerically
integrated in the following figure (Mathcad):
i ..0 800 ∆t 0.001 ti.∆t i h 3 R 200 g 9810
aφ φ .g
Rsin φdθ φ 0
dφ0
0 φ0
π
2θ
00
dφi 1
φi 1
θi 1
dφi
.aφ φi
∆t
φi
.dφi
∆t
θi
.dθ φi
∆t
xi..R sin φ
icos θ
iyi
..R sin φi
sin θi
zi.R cos φ
i
00.20.40.60.81
100
0
100
200
100
0
,,x y z
FIGURE S1.126
61
The Matlab code is contained in the following file which when run will producean 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 = gR
sin φ for the case that the initialvelocity in the circumferential direction is zero (h = 0). We are given thatφ = 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 differentialequation becomes (φ = d2
dt2(π + β) = β)
β + gRβ = 0
which was solved analytically in Sample Problem 1.9. The solution is
β(t) = β0 cos(√
gRt) +
√
gRβ0 sin(
√
gRt)
where β0 and β0 are the initial angle and angular velocity given to the particle.
1.128 From the value of a in spherical coordinates, the following 3 equations result
12 = R − Rφ2 − Rθ2 sin2 φ
4 = Rφ + 2Rφ − Rθ2 sin φ cosφ
5 = Rθ sin φ + 2Rθ sin φ + 2Rφθ cos φ
which can be numerically integrated subject to the initial conditions
R(0) = 2θ(0) = 0
φ(0) = π/2
R(0) = 2θ(0) = −2φ(0) = 1/2
(since vθ = −4 = R(0) sinφ(0)θ(0))(since vφ = R(0)θ(0) + 2θ(0) + 1)
by direct comparison with v(0) = 2er + eφ − 4eθ.
62
Once one numerically integrated, the expressions x = R sin φ cos θ, y = R sin φ sin θand z = R cos φ can be used to construct a 3-D plot of the motion as describedin the figure that follows using Mathcad.
i ..0 1000 ∆t 0.001 ti.∆t i
aR ,,,R vφ vθ φ 12 .R vφ2 ..R vθ2 sin φ 2
aφ ,,,,R vR vφ vθ φ .1
R4 ..2 vR vφ ...R vθ2 sin φ cos φ
aθ ,,,,R vR vφ vθ φ .1.R sin φ
5 ...2 vR vφ sin φ ....2 R vφ vθ cos φ
vR0 2 R0 2 vφ0 0.5 φ0
π2
vθ0 2 θ0 0
vRi 1
Ri 1
vφi 1
φi 1
vθi 1
θi 1
vRi.aR ,,,Ri vφi vθi φi ∆t
Ri.vRi ∆t
vφi.aφ ,,,,Ri vRi vφi vθi φi ∆t
φi.vφi ∆t
vθi.aθ ,,,,Ri vRi vφi vθi φi ∆t
θi.vθi ∆t
xi..Ri sin φi cos θi yi
..Ri sin φi sin θi zi.Ri cos φi
1510
50
5
0
3
2
1
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) 2*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) 2+R(n)*vth(n) 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 xB = 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 thesame position when xB/A = 0. xB/A = 6t2 + 2 − (3t + 2) = 6t2 − 3t = 0 whent = 0 and when t = 1/2s. Since xB/A = 6t2 − 3t, vB/A = 12t − 3 so that theyhave 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 aposition 100 meters out. Starting the clock at 0 when car B crosses the 100m mark and taking the position x = 0 m as the reference point we can writexB = 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 20m/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 =∫ t10 20dt or xA(t) = 20t − 200 + xA(10)
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 timeat 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 = 700m.
1.132 Particle B reverses direction when the velocity is zero. Since xB = 12 + 18t −4.9t2, vB = 18 − 9.8t = 0 at tr = 18
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. Withrelative velocity vB/A(tc) = 16 − 9.8t = −19.83 m/s. since tc > tr, the particlescollide after B reverses direction as illustrated in the figure.
t ..,0 0.1 4
xB t 12 .18 t .4.9 t2 xA t 5 .2 t
0 1 2 3 4
10
20
30
xA t
xB t
t
FIGURE S1.132
1.133 Let car A be moving to the right with vA(0) = 60 mph = 88 ft/s, xA(0) = 0, andaA = −16 ft/s2. Let car B be moving to the left with xB(0) = 300, vB(0) = −88ft/s and aB = 15 ft/s2. Integrating each yields
vA = vA(0) − ∫ t0 16dt = 88 − 16t and
xA = xA(0) + 88t − 8t2 = 88t − 8t2
vB = vB(0) + 16∫ t0 dt = −88 + 16t
xB = xB(0) − 88t + 8t2 = 300 − 88t + 8t2
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 torest at vA = vB = 0 or 88 − 16t = 0 or t = 88
16= 5.5 sec. Thus the cars hit at
t = 2 sec. At that time
vB/A(t) = vB(2.109) − vA(2.109) = (−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) = 500ft, 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. Thecars meet when xB/A = 0 or when 500+85t−2t2 = 0. This yields two roots fort, one is negative and the other is t = 33.795s. Using the above formulas, thecars 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 + 1compute 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 = 3yields 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
2= xB − 3t2 − 1 or
xB(t) = 4.5t2 + 11 m.
1.136 Since the time is so large in seconds, leave this in mph, miles and hours. Forthe 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)(76)
= 2.27 hrs.
1.137 The pendulum falls as θB(t) = θ0 cos√
gℓt, with θB(0) = 0. Consider the vertical
(θ = 0) position as the reference point at which we would like to know the time
tc, i.e., solving θ0 cos√
gℓt = 0 for t yields
√
gℓtc = π
2or tc = π
2
√
ℓg
= 1.121 s,
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. Theparticle’s acceleration can be written as an = a m/s2 where a is an unknownconstant. Integrating yields vA = vA(0) + at = at. Integrating again yields:
xA(t) = xa(0) + a2t2 = −a
2t2 − 10. The value at collision for t will be tc = π
2
√
ℓg
so that a2t2 − 10 = 0 or a = 20
t2c= 80g
π2ℓ= 15.9 m/s2.
1.138 From eq. (1.77) vB = ℓθB = 5θB since the pendulum is in uniform circularmotion or radius ℓ = 5 m. Differentiation of the expression for θB(t) yields
θB = −θ0
√
gℓsin
(√
gℓt)
so that vB = ℓθ − θ0
√gℓ sin
(√
gℓt)
. Let tc be the
time of impact calculated in problem 1.37. Then from 1.37, vA = atc so thatvB/A = −θ0
√gℓ sin
(√
gℓt)
− 20t2c
tc. Substitute tc = π2
√
ℓg
and θ0 = π9
radians (20◦)
yields vB/A = π9
√gℓ + 40
π
√
gℓ
= 20.279 m/s.
1.139 The acceleration of each particle is given by
aA = −9.81 − 0.1vA (1) aB = −9.81 − 0.1vB (2)
with initial conditions vA(0) = 30 m/s, xA(0) = 100 m vB(0) = xB(0) = 0.Equation (1) becomes xA +0.1xA = −9.81 and (2) becomes xB +0.1xB = −9.81
These are second order, uncoupled linear differential equations with analyticalsolutions. The homogeneous solutions are xA = A, a constant and xA = Be0.1t
or xA(t) = A+Be−0.1t. The particular solution is Ct. Substitution of (xA)p = Ctinto (1) yields C = −98.1. Thus (xA)p = −98.1t and the total solution isxA = 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
xA(t) = 781 − 681e−0.1t − 98.1t and vA(t) = 68.1e−0.1t − 98.1.
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′ = −981 so that
xB(t) = 981(1 − e−0.1t) − 98.1t and vB(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 velocityat 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 separableand can be solved by hand. However, to gain a better conceptual understandingof the particle motion, the equations are solved by numerical integration usingthe following Mathcad code:
a v 9.81 .0.1 vi . .0 4060
∆t 0.001
t .i ∆t
vA
0yA
0vB
0yB
0
100
30
0
vAi 1
yAi 1
vBi 1
yBi 1
vAi
.a vAi
∆t
yAi
.vAi
∆t
vBi
.a vB i ∆t
yBi
.vB ∆t
0 1 2 3 4 50
50
100Position- time of A and B
time s
yA
yB
ti
=yA 29.241 =yB 29.252 The impact occurs 29.2 m above theoriginal position of B (taken as the origin).
=vB vA4055
19.999 The relative velocity at impact is 20 m/s
0
i
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
so that vs = vc + 1.5 = −0.1 − 0.01xs + 1.5 or dxs
dt+ 0.01xs = 1.4 which can
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) + C0] = 140 + C0e−0.01t.
Since xs(0) = 0, C0 = −140 and xs(t) = 140(1 − e−0.01t), xs(60) = 140(1 −e−0.6t) = 63.2 m.
. .0 14000
,x t 1.4 .0.01 x
∆t 0.01
x 0
x
i.v ,x
it ∆t
0 50 100 1500
50
100
150
x
i
ti
i
v
0
xi+1
FIGURE S1.140
69
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)0.1
= 125 s.
1.142 Given vA(t) = j m/s (constant), vB(0) = 0, aB(t) = 2[cos 30◦i + sin 30◦j] n/sa constant. Also rA(0) = 0, rB(0) = −4i + j m. From va(t) = 3j. rA(t) =rA(0) + 3tj = 3tj. Integration of aB(t) yields: vB(t) = vB(0) + 1.73ti + tj m/s= 1.73ti + tj. Integrating again yields
rB(t) = 1.732
t2 i + 12t2j + rB(0) = (0.866t2 − 4)i + (0.5t2 + 1)j
Thus rB/A(t) = (3t + 4 − 0.866t2)i − (0.5t2 + 1)j. For this to be zero, each
component must be zero, which cannot be because the j component is alwayspositive, 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
ddt
[(3 + 4 − 0.866t2)2 + (0.5t2 + 1)2] = 0
2(3t + 4 − 0.866t2)(3 − 1.73t) + 2(0.5t2 + 1)(t) = 0
which is cubic in t and has positive solutions. Then t = 0 or rB/A = 4i = j is
when the two particles are closest. As time evolves they move apart (rootsfound using Mathcad).
1.143 Using y up along north and x to the right along east, the airplane has relativespeed of vA/W = 200(cos θi + sin θj) mph where θ is the unknown direction.
The wind velocity is vw = 50(cos 45◦i + sin 45◦i) = 50√2i + 50√
2j = constant. We
would like vA = 200i. From the definition then
vA = vW + vA/W = (50)[(0.707)i + (0.707j)] + 200(cos θi + sin θ)j
which results in the two scalar equations
vA = 35.4 + 200 cos θ
0 = 35.4 + 200 sin θ
Solving yields: θ = −10.2◦ and vA = 232.2 mph.
70
1.144 Note that the flanker travels 40 ft (20 ft/s × 2 s = 40 ft) before the quarterback throws the ball at t = 0. Let xf(0) = 40i denote the initial position of thequarter back. For the ball
ab = −gk
so that
vb = v cos θ cos β i + (−gt + v sin θ)j + v cos θ sin βk
and
xb = [v cos θ cos βt + xb(0) i + [−gt2
2+ v sin θt + yb(0)]j + [v cos θ sin βt]k
where xb(0) = −30 ft (i.e. 30 yards behind the line of scrimmage) and yb(0) = 6ft. For the flanker xf = 20i, and integrating yields
xf(t) = (20t + xf(0))i + 8j + 60k
Setting xb = xf yields the three equations
from i: 20t + 40 = v cos θ cos βt − 30
from j: 8 = −gt2
2+ v sin θt + 6
from k: 60 = v cos θ sin β.
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)+403
= 27.375 yd
1.145 Following sample 1.28, let vs be the absolute velocity of the swimmer. With i
perpendicular to the shore and j up stream, vs = vi, where v is the unknownabsolute speed of the swimmer. Let θ be the angle that vs/w (swimmer relativeto the water) so that
vs/w = 6 cos θi + 6 sin θj (ft/s)
vw = −5j ft/s, so that
vs = vw + vs/w becomes
vi = −5j + 6 cos θi + 6 sin θj
from i: v = 6 cos θ
from j: 0 = 6 sin θ − 5
which is 2 equations in two unknowns: θ and v. Solving yields θ = 56.4◦,v = 3.3 ft/s. The time to travel 200 ft is t = d
v= 200
3.3= 60.3 sec.
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 path South 21.80 of East. Thewind direction for the 1200 miles was estimated to be South 450 East, during the next 600miles the wind direction is to the East and during the last 1200 miles, the wind is North450 East. The wind speed is estimated to be 50 mph.
The general relative motion equation for any segment of the flight is:v v vP w p w= + /
For the first leg of the flight, these vectors are:
v i j
v i j
v i j
p
w
p w
= −[ ]= −[ ]
= +[ ]
vp cos( . )ˆ sin( . )ˆ
cos( )ˆ sin( )ˆ
cos( )ˆ sin( )ˆ/
21 8 21 8
50 45 45
300 θ θ
The scalar equations of relative motion are solved using the Given-Find function inMathcad:
Make an initial guess for the velocity of the plane and the bearing of the planev p 300
θ .10 degGiven
.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
750 mi 250 mi 500 mi 1000 mi 500 mi Miami
Wind Vel E S 85 0 E E N 30 0 E S 300 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 bearing of the planev p 300
θ .10 degGiven
.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
=Find ,v p θ390.541
0.505=
0.505
deg28.934
vp = 391 mph θ = -30 0
Leg 2.
Make an initial guess for the velocity of the plane and the bearing of the planev p 300
θ .10 degGiven
.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 θ331.502
0.078=
0.078
deg4.469
73
1.148 Solution:
Seattle
San Francisco 45 0 450 New York
1200 mi
1200 mi 600 mi 1200 mi Miami
The flight is now to the West against the jet stream. The flight will be divided into threelegs.
Leg 1.
Make an initial guess for the velocity of the plane and the bearing of the planev p 200
θ .10 degGiven
.v p cos .10 deg .50 cos .45 deg .200 cos θ
.v p sin .10 deg .50 sin .45 deg .200 sin θ
=Find ,v p θ167.082
0.032=
0.032
deg1.833
The velocity of the plane and the course bearing are: vp = 167 mph θ = S 2 0 W
Leg 2.
Make an initial guess for the velocity of the plane and the bearing of the planev p 200
θ .10 degGiven
.v p cos .10 deg .50 cos .0 deg .200 cos θ
.v p sin .10 deg .50 sin .0 deg .200 sin θ
=Find ,v p θ150.571
0.131=
0.131
deg7.506
S 10 0 W
74
1.149 Solution:
21.80 Chicago
1200 mi
Memphis
750 mi 250 mi 500 mi 1000 mi 500 mi
Wind Vel E S 85 0 E E N 30 0 E S 300 E
The desired flight path from Chicago to Memphis is assumed to be S 800 E under a 100-mph wind N 300 E.
Make an initial guess for the velocity of the plane and the bearing of the planev p 300
θ .80 degGiven
.v p cos .80 deg .100 cos .30 deg .300 cos θ
.v p sin .80 deg .100 sin .30 deg .300 sin θ
=Find ,v p θ250.701
1.715=
1.715
deg98.262
The velocity of the plane and the course bearing are: vp = 251 mph θ = S 82 0 W
1.150 From sample 1.29 the time to reach the opposite shore is tf = 2dvB/w cos θ
= 40cos θ
sec. The expression for y(t) at y(tf) = 500 becomes
500 = 5[
(
10·cos θ200
)2403
3·cos3 θ− 210 cos θ
200· 402
2 cos2 θ
]
+ 10 sin θ 40cos θ
which is transendental in θ and has solution (see numerical root finder) θ = 56.43◦.
75
1.151 From sample 1.29 the angle is determined
θ = sin−1(
2vw
3vB/w
)
= sin(
(2)(2)(3)(10)
)
= 7.6◦
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 = 0.7become
θ + β = 90◦
−vs cos β = −0.3 cos θ
vs sin β = 0.7 − 0.3 sin θ
Solving using Mathcad yields: θ = 24.8◦, β = 64.3◦ and vs = 0.632 m/s.
1.153 We are given: xB = 2 sin πti + 5tj,
xG = 4t cos 30◦i + 4t sin 30◦j so that differentiation yields
vB = 2π cos πti + 5j, vG = 4 cos 30◦i + 4 sin 30j and
aB = −2π2 sin πti, aG = 0
From eq. 1.126, vG/B = vG − vB = (4 cos 30◦ − 2π cos πt)i + (4 sin 30◦ − 5)j
and aG/B = aG − aB = 2π2 sin πti
76
1.154 From the previous problem:
xG/A = (4t cos 30◦ − 2 sin πt)i + (4t sin 30◦ − 5t)j
so that the distance between the two as a function of time is
d(t) = |x(t)| = ((4t cos 30◦ − 2 sin πt)2 + (4t sin 30◦ − 5t)2)1/2
which is plotted below using Mathcad:
θ .30 deg t ..,0 0.1 3
xB t
.2 sin .π t.5 t
0
xG t
..4 t cos θ..4 t sin θ
0d t xB t xG t
0 1 2 3
5
10
15
Time s
Dis
tanc
e m
d t
t
FIGURE S1.154
1.155 Given xB = 2 sin πti + 5tj and
xG = 4t cos 30◦i + 4t sin 30 sin(
πt2
)
j,
xB/G = (2 sin πt − 4t cos 30◦)i + (5t − 4t sin 30◦ sin πt2)j
vB/G = (2π cos πt − 4 cos 30◦)i + (5 − 2πt sin 30◦ cos πt2− 4 sin 30◦ sin πt
2)j
aB/G = −2π2 sin πti + (−4π sin 30◦ cos πt2
+ π2t sin30◦ sin πt2)j
77
1.156 From xB/0 of the previous problem:
d(t) = |xB/G(t)| = [4π4 sin2 πt + (1 − 4π sin 30◦ cos πt2
+ πt sin 30◦ sin πt2)2]1/2
which is plotted in the figure using Mathcad.
θ .30 deg t ..,0 0.1 3
xB t
.2 sin .π t.5 t
0
xG t
..4 t cos θ
...4 t sin θ sin.π t
2
0 d t xB t xG t
0 1 2 3
10
20
30
Time s
Dis
tanc
e m
d t
t
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
1.157 From the given function
aA = ddt
(vA) = i + 1.224j− 1.224k − 0.2vA
which is a vector differential equation of the form
vA + 0.2vA = i + 1.224j− 1.224k = b
The solution of the homogeneous equation vAH= Ae−0.2t plus a particular
solution: vp = 10.2
(i + 1.224j− 1.224k) or
vA = Ae0.2t + 10.2
(i + 1.224j− 1.224k)
where A is evaluated from the listed condition vA(0) = 0 which yields
0 = A + 10.2
(i + 1.224j− 1.224k) or A = − 10.2
(1.224j− 1.224k)
Then
vA = 5(1 − e−0.2t)(i + 1.224j− 1.224k) m/s
Integration again yields
xA − xA(0) = 5(t + 5e−0.2t − 5)(i + 1.224j− 1.224k)
or
xA = 5(t + 5(e−0.2t − 1))(i + 1.224j− 1.224k)
Next consider aB = 3ti − 2 cos πtk m/s. Integrating yields
vB − vB(0) = 3t2
2i − 2 sin πt
πk where vB(0) = 0
Integrating again yields
xB − xB(0) = t3
2i − 2
π2 (1 − cos πt)k where xB(0) = 0
Now form
vB/A =[
3t2
2− 5(1 − e0.2t)
]
i − 6.12(1 − e0.2t)j
+[
6.12(1 − e−0.2t) − 2π
sin πt]
k
and
xB/A =[
t3
2− 5(t + 5(e−0.2t − 1))
]
i − [6.12(t + 5(e−0.2t − 1))] j
+[
1π2 (cos πt − 1) + 6.12(t + 5(e−0.2t − 1))
]
k
79
The figure represents a numerical solution and plots of the magnitude of xB/A
in Mathcad:
i ..0 3000 ∆t 0.001 ti.i ∆t
vx0 0 x0 0 vy0 0 y0 0 vz0 0 z0 0
ax ,vx t 1 .0.2 vx .3 t
ay ,vy t .2 0.0612 .0.2 vy
az ,vz t .0.621 2 .0.2 vz .2 cos .π t
vxi 1
xi 1
vyi 1
yi 1
vzi 1
zi 1
vxi.ax ,vxi ti ∆t
xi.vxi ∆t
vyi.ay ,vyi ti ∆t
yi.vyi ∆t
vzi.az ,vzi ti ∆t
zi.vzi ∆t
vi vxi2 vyi
2 vzi2 xi xi
2 yi2 zi
2
0 1 2 3
5
10
xi
ti
0 1 2 3
5
10
vi
ti
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=[t 3/2-5*(t-5*(exp(-0.2*b)-1));6.12*(t+5*(exp((-0.2*t)-1));(1/pi 2)*(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. UsingMathcad 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 byreplacing t with t + 1.5 in the expression for xB derived in Problem 1.157. Thisyields
xB = (t+1.5)3
3i − 1
π2 (1 − cos(π(t + 1.5))k
Then for t ≥ 4.5s,
xB/H =[
(t+1.5)3
2− 5(t + 5(e−0.2t − 1)
]
i − 6.12 [t + 5(e0.2t − 1)] j
+{
6.12 [t + 5(e−0.2t − 1)] − 2π2 (1 − cos(πt + 1.5))
}
k
and for 0 < t < 1.5
xB/A = (t+1.5)3
2i − 2
π2 (1 − cos(π(t + 1.5))k.
The Mathcad solution:
i ..0 3000 ∆t 0.001 ti.i ∆t
vx0 0 x0 0 vy0 0 y0 0 vz0 0 z0 0
ax ,vx t .1 .0.2 vx Φ t 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 .2 cos .π t
vxi 1
xi 1
vyi 1
yi 1
vzi 1
zi 1
vxi.ax ,vxi ti ∆t
xi.vx i ∆t
vyi.ay ,vyi ti ∆t
yi.vyi ∆t
vzi.az ,vzi ti ∆t
zi.vzi ∆t
vi vxi2
vyi2
vzi2
xi xi2
yi2
zi2
0 1 2 3
10
20
xi
ti
0 1 2 3
20
40
vi
ti
FIGURE S1.159
81
1.160 This is similar to sample problem 1.32. Using the coordinate system suggestedin Sample 1.32, yB denotes the distance to the bracket and xA to the man. Then
xA + 2yB = ℓ
2vB = −vA or vA = −2vB
Since vB = 1 ft/s, vA = −2 ft/s (down)
1.161 Use the pulley as a reference point, and let xA be the distance from the pulleyto A and xB the distance from the pulley to B. Then
xA + xB = constant and differentiating yields
xA = −vB, and aA = −aB.
1.162 Let vc be the cable velocity: vc = 0.5 m/s. Let xT denote the distance from thetree to the end of the truck. The length to the cable is xc + 2xT = ℓ so that
vc = −2vT or vT = −12vc = −0.25 m/s (right)
1.163 Let xA be the distance for the pulley to the man at A, xp be the distancebetween the top pulley and the bottom pulley. Let xB be the distance from thetop pulley to B and let the bracket for the bottom pulley be a height of c. Then
xA + 2xp + xB − c, but xp = (xB − c)
so that xA + 3xB =constant and vA = −3vB
with xB, xA pointed down, B traveling upwards yields
vB = −10 ft/s and vA = −3 · (−10) = 30 ft/s (down)
1.164 Let xA denote the distance from the top of the pulley to block A, yB denote thedistance from the top of the pulley down to the pulley holding B and y denotethe distance from the top of the pulley down to mass c (down position, to theright position). Then the length of the rope is
yc + 2yB + xA = constant
differentiation yields
vA + 2vB + vC = 0
or
vc = −2vB − 3 m/s and ac = −2aA − 1 m/s
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 forh = 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 =(
2025
) (
0.5 m/s2)
= 0.4 m/s, at y = 10 m.
1.166 Let xA extend from the fixed pulley above B to the left of pulley B to mass Aand let yB extend (positive) from the same pulley down to the mass at B. Letd be the distance between the two fixed pulleys. Then the length of the rope is
ℓ = 2yB + xA + (xA − d) + (xA − d)
or
const. = 2yB + 3xA.
Differentiate to get:
yB = −32
xA and yB = −32xA.
1.167 Use the top fixed pulleys as a reference and let xA denote the distance to massA, xB the distance to mass B, xc the distance to mass C and denote xp thedistance to tohe pulley that is free to move. All distances are vertical withpositive downward. There are two separate ropes. Let ℓ1 denote the from A tothe movable pulley so that
ℓ1 = xA + xB + (xB − xp) (1)
83
Let ℓ2 denote the length of the rope from mass C up to ground so that
ℓ2 = xc + 2xp (2)
Differentiation of (2) yields vp = −vc/2 which is substituted into the derivativeof (1) to yield
4vB + vc = −2vA m/s
1.168 Let the top pulley be the fixed frame of reference, and x denote the distancefrom the reference down (t) to the mass, and let xm be the distance from thereference down (t) to a point on the rope connecting to the motor. Them
ℓ = xm + 2x
and differentiation yields that
v = 12vm
From section 1.7 vm = rw = (0.2)(9.55 rad/s = 1.91 m/s and v = 12vm =
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 ofthe 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 = yB + 12
2xB xB√d2+x2
B
.
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 Crespectively. 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 yieldsvA = −vc − 2vB.
1.171 Solution:dsdt
= 20 m/s = constant so s = 20t and ds2
dt2= 0
θ(s) = 4 sin( s2000
) so that θ(t) = 4 sin(0.5t)
From equation 1.147 with x0 = y0 = 0
x(s) =∫ s0 cos(θ(η))dη y(s) =
∫ s0 sin θ(η)dη
84
The Mathcad solution:
s ..,0 10 6000θ s .4 sin
s
2000
x s d0
sucos θ u
y s d0
susin θ u
an s .202 dd s
θ s
4000 3000 2000 1000 0 1000
1000
1000
2000Road as viewed from above
km
km
y s
x s
0 2000 4000 6000
0.5
1
distance along road in km
norm
al a
ccel
erat
ion
m/s
^2
an s
s
FIGURE S1.171
From (1.146): an =(
dsdt
)2 dθ(s)ds
= (20)2 4200
cos(
s200
)
= 800 cos(
s200
)
y(s) and |an(s)| are plotted in the figure.
1.172 We are given that a(s) = 5 so that integrating vdv = 5ds yields v(s) =√
10sfor zero initial condition. Also since θ = 1 + s2
20002 ,dθds
= 2s10002 . But an(s) =
v(s)2 dθds
= (10s)( 2s20002 ). Thus |a| =
√
a2n + a2
t =√
(10s)2( 2s20002 )2 + 25. When
|a| = 8 the car spins out so that s must satisfy
82 = (10s)2( 2s20002 )2 + 25.
Solving for s yields: s2 =√
39(2000)2
20or s = 1, 118 m.
85
1.173 The given constant speed means that dsdt
= 60 mhour
· hour3600s
5280 ftm
= 88 ft/s. Todetermine the normal acceleration from equation (1.146) we need to calculate
(ds/dt)2 = 882 and dθ/ds. From the given value of θ: dds
(
s2000
+ e−5/1000)
=1
2000− 1
1000e−s/1000 Thus an = 882
1000
[
0.5 − e−s/1000]
= 7.744[
12− e−s/1000
]
ft/s2.
The Mathcad solution:v 88 s ..,0 10 5000
θ ss
2000e
s
1000an s .v
2 d
d sθ s
0 1000 2000 3000 4000 5000
2
4Magnitude of Normal Acceleration
distance traveled in f
ft/s
^2 an s
s
FIGURE S1.173
1.174 This requires the use of the 3-D formulation of equation 1.150. The componentof a normal to the above can be found by taking the dot product of the accel-eration with the unit normal vector. To define the unit normal vector can befound from an = a = at = a − a · t. First we need dθ
dsand dβ
ds. From the given
form: dθds
= 11000
(0.5 − e−s/1000), dβds
= π2
5000cos
(
πs5000
)
. The Mathcad solution:
v 88 s ..,0 10 5000
β s .π sin.π s
5000θ ss
2000e
s
1000
an s .v2 .d
d sθ s
2
cos β s2 d
d sβ s
2
0 1000 2000 3000 4000 5000
10
20
distance traveled in f
no
rma
l a
cc
ele
rati
on
in
ft/
s^2
an s
s
FIGURE S1.174
86
1.175 Solution:
v 88 s ..,0 10 5000 β s .π sin.π s
5000θ ss
2000e
s
1000
v s ..2 1.5 s
an s .v s 2 .dd s
θ s2
cos β s 2 dd s
β s2
0 1000 2000 3000 4000 5000
20
40
distance traveled in ft
norm
al a
ccel
erat
ion
in ft
/s^2
an s
s
FIGURE S1.175
1.176 Solution:
v 88 s ..,0 10 5000 β s .π sin.π s
5000θ ss
2000e
s
1000
ρ s1
.dd s
θ s2
cos β s 2 dd s
β s2
0 1000 2000 3000 4000 5000
1000
2000
3000
distance traveled in ft
radi
us o
f cu
rvat
ure
in f
t
ρ s
s
FIGURE S1.176
87
1.177 Solution:
s ..,0 1 50 θ ss
100 β s .π2
sin.π s
25
v 10
an s .v2 .dd s
θ s2
cos β s 2 dd s
β s2
0 10 20 30 40 50
10
20
distance traveled in m
norm
al a
ccel
erat
ion
in m
/s^2
an s
s
FIGURE S1.177
1.178 Solution:
s ..,0 1 100 θ ss
100 β s .π2
sin.π s
25
v s 10 .5 cos.π s
25
an s .v s 2 .dd s
θ s2
cos β s 2 dd s
β s2
at s .dd s
v s v s
a s at s 2 an s 2
0 20 40 60 80 100
20
40
60
distance traveled in m
tota
l acc
eler
atio
n in
m/s
^2
a s
s
FIGURE S1.178
88
CHAPTER 2
2.1 Given: µk = 0.1, W = 200 lb and P = 50 lb. The free body diagram yields:
N − mg − P sin 30 = 0 (sum of forces in y-direction) (1)
P cos 30 − f = max (sum of forces in x-direction) (2)
From (1) N = mg + P sin 30 = 200 + 50(12) = 225 lb
Friction force f = µkN = 0.1(225) = 22.5 lb
Therefore max = 50(√
32
) − 22.5 = 20.80 lb
⇒ ax = 20.80200/32.2
= 3.34 ft/s2
⇒ vx = 3.34t (assuming initially at rest)
⇒ x = 1.67t2 (assuming x(0) = 0)
Therefore t =√
1001.67
= 7.73 sec
F_
30o
f
N
x
y
FIGURE S2.1
2.2 Solution: µs = 0.4, m = 150 kg, µk = 0.3
T − f − mg sin θ = max
N − mg cos θ = may = 0
⇒ N = mg cos θ = 150(9.81) cos θ = 150(9.81) 100√1002+12
= 1471 N
T
mg100
f
N
x
o (1/100) = .573 0.01 rad~~θ = tan-1
θ
1
FIGURE S2.2
90
fmax = µsN = 0.4(1471) = 588.4 N
Therefore Tmax = fmax + mg sin θ = 588.4 + (150)(9.81) 1√1002+1
= 603 N
f = µkN = 0.3(1471) = 441 N
ax = 1m
(T − f − mg sin θ) = 1150
[
603 − 441 − 150(9.81) 1√1002+1
]
ax = 0.98 m/s2
Therefore vx = 0.98t m/s assuming zero initial velocity.
Therefore t = 50.98
= 5.1 sec
2.3 Examining the maximum friction µs = 0.2. This can produce an acceleration ofac = 0.2(32.2) = 6.44 ft/s2. Therefore the box slips because the force of staticfriction mac is less than that applied by the truck
ac = 0.15(32.2) = 4.83 ft/s2
ac/T = 4.83 − 10 = −5.17 ft/s2
(We could now determine where the crate falls off if we knew length of tuckbed.)
mg
f
N
FIGURE S2.3
2.4 Solution:
a)∑
F = ma
100 − 50 = 10032.2
a
a = 16.1 ft/s2
b) 100 − T = 10032.2
a
T − 50 = 5032.2
a
Therefore 50 = 15032.2
a ⇒ a = 10.73 ft/s2
c) 150 − T = 15032.2
a
91
T − 100 = 10032.2
a
⇒ 50 = 25032.2
a ⇒ a = 6.44 ft/s2
100 10050 50
150100
100
50 T
100
T
50
T
150
T
100
FIGURE S2.4
2.5 Solution:
N − mg = ma, or
N = mg + ma
for: a = +2 : N↑ = m(9.81 + 2) = 1181 N
for: a = −2.5 : N↓ = 100(9.81 − 2.5) = 731 N
N
+
mg
FIGURE S2.5
2.6 Solution:
−mg sin 30◦ + µmg cos 30◦ = ma along the conveyor. Or:
−0.5g + 0.2598g = a, thus:
a = −.24g.
92
v = −0.24gt + v0
x = −0.12gt2 + 0.2t
0.12(9.81)t2 − 0.2t − 10 = 0
t = 3s
Note that this slips immediately as µsg cos(30◦) < 0.5g
30o
10 9.8
30o
.
N
f
FIGURE S2.6
2.7 From S2.7a the constraint equation is
2(xA − xB) + (xc − xB) = constant
Thus 2xA − 3xB = 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)aA (3)
This is 3 equations in 3 unknowns which yields
aB = 1.765 m/s2, aA = 2.647 m/s2 and T = 23.533 N
C
X B
B A
x
B A3T
40g
N
100
20g
2T
N
(a) (b) (c)x
A
FIGURE S2.7
93
2.8 Constraints to motion are from figure S2.8a are :
xc + xp = ℓ1, therefore: ap = −ac.
And xA − xp + xB − xp = ℓ2, hence:
aA + aB − 2ap = 0, and thus aA + aB + 2ac = 0
C
P
B
A
x
FIGURE S2.8a
From the FBD’s of figure S2.8b:
25g − T2 = 25aA
T1 = 2T2
20g − T2 = 20aB
50g − 2T2 = 50ac.
But ac = −12(aA + aB)
Subtract 2x the first equation from third to get
50(ac − aA) = 0, or ac = aA.
From the constraint aB = −3aA. Subtract 2nd from 1st
5g = 85aA → aA = 0.577 m/s2
T2 = 231 N aB = −1.731 m/s2
T1 = 462 N ac = 0.577 m/s2
T2 T2 T1 T1
25g 20g2T2
50g
FIGURE S2.8b
94
2.9 From figure S2.9a, the constraints are
xA + 2xp = ℓ1, and
(xB − xp) + xB = ℓ2.
A 100 50
P
B
xA
xP
xP
FIGURE S2.9a
Differentiating yields aA = −2ap and 2aB = ap so that aA = −4aB
Next the FBD’s of Figure S2.9b yields:
100 · 9.81 − T1 = 100aA from A
T2 − 2T1 = 0 (assumed massless) from P
(50)(9.81) − 2T2 = 50aB from B
T1 T1 T12T
2
100(9.81) T2 50(9.81)
T2 =2T1
A P B
FIGURE S2.9b
These three equations, plus the constraints aA = −4aB form a system of fourequations and four unknowns which can be solved for aA. This yields
aA = 8.324 m/s2
Integrating vdv = aAdx from rest conditions yieldsv2
2= (9.324)(0.5) or v = 2.885 m/s
2.10 From S2.10, (100)a = 100x or
x = −50x
vdv = −50dxv2
2= −25(x2 − 0.32)
v =√
50(0.09 − x2)
dxdt
= 7.07√
0.09 − x2
95
∫ t0.3
dx√0.09−x2
=∫ t0 7.07dt
sin−1(
x0.3
)
|x0.3 = 7.07t
sin−1(
x0.3
)
− sin−1(1) = 7.07t
sin−1(
x0.3
)
= (7.07t + π2)
x = 0.3 sin(7.07t + π2)
x = 0.3 cos(7.07t)
Fs
x
FIGURE S2.10
Alternately from S2.10
−kx = mx where x(0) = 0, x(0) = 0.3m
m = 2 kg, k = 100 N/m, x(0) = 0.3
x + 50x = 0
Integrating Factor
x = Aeat, a2 + 50 = 0, a = ±7.07i
x = A sin(7.07t) + B cos(7.07t)
x(0) = 0.3 = B → B = 0.3
x(0) = 0 = 7.07A → A = 0
x = 0.3 cos(7.07t)
2.11 From the FBD of the system as a whole: P − 3mg sin α = 3ma, so that
a = 13m
[P − 3mg sin α]
From the FBD for car C:
P − TBC − mg sin α = ma = 13[P − 3mg sin α].
Solving yields
TBC = 2/3P .
Likewise a FBD of the system consisting of B and C taken together yields
TAB = 13P
96
A
B
C
mg
α
α
mg
mg
TBC
P
C
mg
P
FIGURE S2.11
2.12 From a FBD of the entire system as given in the top of Figure S2.12:
70 sin 20◦ − 70 cos 20◦(0.2) = 7032.2
a
a = 4.96 ft/s2
Now from the bottom of S2.12, the FBD yields
50 sin 20◦ − T − 0.2(50) cos 20◦ = 5032.2
(4.96)
So T = 0!
This result is expected, as the motion is independent of mass and both blocksslides independently.
20
20
o
20
50
o
20o
f
T50
FIGURE S2.12
2.13 Sled will start to move when
a) 500(t2 + 2t) = 0.5(200)(9.81) t2 + 2t − 1.962 = 0
t = −1 ±√
4+4(1.962)
2= −1 ±
√2.962
t = 0.721 s or t = −2.721
97
F(t)
f
FIGURE S2.13
b) From t = 0.721 to 6s
F (t) − 0.4(200)(9.81) = 200a
v = 1200
∫ 60.721[500(t2 + 2t) − 0.4(200)9.81]dt
= [0.833t3 + 2.5t2 − 3.924t]|60.721
= 248 m/s (555 mi/hr)
v(t) = 0.833t3 + 2.5t2 − 3.924t + 1.217
x(6) = [0.208t4 + 0.833t8 − 1.962t2 + 1.217t|60.721
x(6) = 386 m
c) From 6s on:
−0.4g = a
a = −3.924
v = −3.924t + 248
time where v = 0
t = 63.2s
x = −3.9242
t2 + 248t
= 7837 m
Total distance = 386 + 7837 = 8223 m or 8.2 km (5.1 mi)
98
2.14 Solution: fmax = 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
90(9.81)
f
1200N
A + B + C
220.7
1200
a=10.9 m/s2
60f=60a a=2.9 m/s
2A+B
f=0.3(60)(9.81) = 176.58
A+B slips
FIGURE S2.14a
b) Assume slip between each surface
20g
f 2
N 1
N1
f2
N 2
f1
30g
f2 N2
1200
f3
40g
N 3
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) = 40aB 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 sameacceleration)
99
c) Next C slips out from under A and B (the FBD is S2.14c). Then
0.25(60)(9.81) = 60aAB. Thus
aAB = 2.45 m/s2.
From the FBD on the bottom of Figure S2.14c:
1200 − 0.25(60)(9.81)
−0.25(90)(9.81) = 30ac
Thus the acceleration of the bottom block is
ac = 27.74 m/s2
60g
fAB
NAB
NAB
fAB
1200
f
N
30g
C
C
FIGURE S2.14c
2.15 Will the system slip?
20
40
30 400
FIGURE S2.15a
Is 400 ≥ 0.3(90)(9.81) = 265 N?
Yes, so the system slips.
A
B
f max=176.6
a=2.9 m/s2N
mg
FIGURE S2.15b
100
From the FBD of A and B:
400 − 0.25(90)9.81 = 90aT
aT = 1.99 m/s2
fmax = 176.6 which implies a = 2.9 m/s2
So A and B must move with C.
2.16 Solution:
N1
f1
30o
N2
f2
50g
N2
f2 f3
N3
10g
FIGURE S2.16a
Assume no slip and determine the required value of µs. From FBD of A:
N1 + µN2 cos 30◦ − N2 sin 30◦ = 0 (1)
µN1 + µN2 sin 30◦ + N2 cos 30◦ − 50g = 0 (2)
From the FBD of B:
−µN2 cos 30◦ + N2 sin 30◦ − µN3 = 0 (3)
−µN2 sin 30◦ − N2 cos 30◦ + N3 − 10g = 0 (4)
This is a system of four nonlinear equations in the four unknowns µs, N1, N2,and N3. Solved by Mathcad Eqs. 1-4 required µs = 0.242.
Since the µ required for equilibrium is larger than the 0.2 given the systemwill slip. Knowing it will slip, the equations of motion are (with zero accelera-tion into the walls)
N1 + µkN2 cos 30◦ − N2 sin 30◦ = 0
µkN1 + µkN2 sin 30◦ + N2 cos 30◦ − 50g = 50aA
−µkN2 cos 30◦ + N2 sin 30◦ − µkN3 = 10aB
−µkN2 sin 30◦ − N2 cos 30◦ + N3 − 10g = 0
µk = 0.15, this is a system of 5 unknowns N1, N2, N3, aA, aB and only 4equations. Hence we need a constraint of the motion. The acceleration of the
101
blocks in the normal direction at the contacting surface is the same (see S2.16b).Thus −aA cos 30◦ = aB sin 30◦ which provides the 5th equation. Solving these 5nonlinear equations in the 5 unknowns yields the desired acceleration (using acomputer code such as the MATLAB file at the end of this solution):
aA = −3.401 m/s2
aB = 5.891 m/s2
From d = a2t2, the time for box A to move 0.1 m is
t =√
(2)(0.1)3.401
= 0.242 s. During this time box B moves
xB(0.242) = 5.8912
(0.242)2 = 0.173 m
θθ
aA
aB
-aA cos 30o = aB
sin 30o
FIGURE S2.16b
This must be added to the distance the block B slides to the right due to itsseparation velocity. The velocity of separation is:
vB = 5.891 · 0.242 = 1.426 m/s
The deceleration of the block is −0.15 · 9.81 = −1.472 and the time when theblock stops is:
t = 1.4261.472
= 0.969 s
The total distance traveled is:
xB = −0.075 · 9.81(0.969)2 + 1.426 · 0.969 + 0.173 = 0.864 m
102
The MATLAB solution:
Main program to solve the 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(' N1= %6.4f',F(1)))
disp(sprintf(' N2= %6.4f',F(2)))
disp(sprintf(' N3= %6.4f',F(3)))
disp(sprintf(' 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 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 required for the 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 = ℓ2
0 + x2
where ∆ is the spring elongation.
m
x
l
l
o
Fs
N
mg
FIGURE S2.17
Thus
∆2 + 2ℓ0∆ − x2 = 0, or
∆ =−2ℓ0±
√4ℓ2
0+4x2
2= −ℓ0 ±
√
ℓ20 + x2. Using the positive root, the magnitude
of the spring force is
|Fs| = k[
√
ℓ20 + x2 − ℓ0
]
From the FBD:
mg − k[
√
ℓ20 + x2 − ℓ0
]
x√ℓ20+x2
= ma
Thus
d2xdt2
= g − km
[
√
ℓ20 + x2 − ℓ0
]
x√ℓ20+x2
a(x) = g − kxm
(
1 − ℓ0√ℓ20+x2
)
.
2.18 Solution:
From the figure
(ℓ + ∆)2 = x2 + ℓ2
ℓ2 + 2ℓ∆ + ∆2 = x2 + ℓ2
∆2 + 2ℓ∆ − x2 = 0
∆ = −2ℓ±√
4ℓ2+4x2
2=
√ℓ2 + x2 − ℓ (could write this directly)
mg − Fs sin θ − µkN = mx, sin θ = x√ℓ2+x2
N − Fs cos θ = 0, cos θ = ℓ√ℓ2+x2
104
θ
θ
Ν
F s
f
mg
x
l
FIGURE S2.18
N = k[√
ℓ2 + x2 − ℓ]
ℓ√ℓ2+x2
, Fs = k∆
N = kℓ[
1 − ℓ√ℓ2+x2
]
x = g − km
[
x(
1 − ℓ√ℓ2+x2
)
+ µkx|x|ℓ
(
1 − ℓ√ℓ2+x2
)]
x = g − km
(x + µkx|x|ℓ)(1 − 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
x0
0
0
a x g ..k
ml2 x2 l
x
l2 x2
vi 1
xi 1
vi
.a xi
∆t
xi
.vi
∆t
0 0.5 1 1.5 20.2
0
0.2
0.4
0.6Position vs time
Time s
xi
.i ∆t
2.20 First note that 30g ≥ 0.5(20)g, so it does in fact slip. The constraint is that
L = xA + xB + const
⇒ 0 = xA + xθ
⇒ 0 = aA + aB so that aA = −aB . (1)
From the FBD of (B):
30(9.8) − T = 30aB (2)
From the FBD of A:∑
Fx = maA becomes:
T − µkN = 20aA
T − µkmg = 20aA
T − 0.4(20)(9.81) = (20)aA (3)
Adding (2) and (3), using (1) yields
(30)(9.81) − (0.4)(20)(9.81) = 50aA
106
So that aA = 4.316 m/s2. Then d = aB
2t2 or
t =√
14.326
= 0.481 s.
20g
+
T
N
F
A
T
B
30g
+
(a) (b)
FIGURE S2.20
2.21 First determine the unit normal vector to the surface of the ice, n.
100 ft
z
x
y
45
30
t T
n
FIGURE S2.21
Then compute the component of the weight vector along n, denoted N . Theforce parallel to the ice will then be S = W −N , where W is the weight. Oncethe unit normal along S is determined, the motion can be treated as rectilinearalong the face of the ice. The following Mathcad code completes the solution.
107
T
cos .45 deg
0
sin .45 deg
t
0
cos .30 deg
sin .30 deg
W
0
0
130
nT t
T t
=n
0.655
0.378
0.655
Let us find the component of the weight that acts normal to the face of the ice.
N ..W n n
=N
55.714
32.167
55.714
The force parallel to the ice face causing her to slip is S
S W N
=S
55.714
32.167
74.286
We can first see if she slips by comparing |S| to µk|N|
=S 98.271 =.1.0 N 85.105
Therefore she slips.
The unit vector in the direction of the slip is s
sS
S
We can now treat the motion as rectilinear along the ice face in the direction s.
a .32.2
130S .0.8 N
=a 7.477 ft/s2
The distances she slides in the s direction is:
d = −100(t · s) = 66.144 ft
Now her velocity after sliding 66.144 ft can be determined.
v dvdx
= a, v =√
2ad, v = 31.45 ft/s. Yes, she will be injured.
108
2.22 Constraint: xA + xB = ℓ therefore aA = −aB
x B
x A
20
60
30o
Constraint
xA+xB
l=
... aA = -a B
FIGURE S2.22
From the FBD of A:
20g sin 30 + 0.2(20g) cos 30◦ − T = 20aA
From the FBD of B:
60g sin 30◦ − 0.2(20g) cos 30◦ − 0.1(80g) cos 30◦ − T = 60aB
Thus
132.1 − T = 20aA
192.4 − T = 60aB
Which along with the constraint equation is a system of 3 equations in 3 un-knowns. Solving yields
aB = 0.753 m/s2
aA = −0.753 m/s2
aA/B = aA − aB = −1.506 m/s2
A T
fA
NA
m Ag
(a)
B
m Ag
N B
f B
T
fA
N A
(b)
FIGURE S2.22a,b
109
2.23 The rope length yields: 2xA + yB = ℓ, so that 2aA + aB = 0
150
8 kgx
y
A 3 kg
2xA+ y
B= l
2aA+ aB = 0
FIGURE S2.23
A FBD of A yields:
150 − 2T = 8aA (1)
A FBD of B yields:
3 · (9.81) − T = 3aB (2)
Using the constraint, the 1st equation becomes
150 − 2T = −4aB
Multiply (2) by 2 yields
58.86 − 2T = 6aB
Subtracting these last two equations yields
−10aB = 91.14, so aB = −9.114 m/s, and aA = 4.557 m/s2
Also from (2): T = 3(9.81) − 3aB = 56.8 N
Solving
vB = aBt
for t yields:
t = 2.54.557
= 0.549 s
2.24 The constraint here is that xA + xB = constant, so that aA = −aB . Assumethat the system moves to the left.
From FBD of mass B
for y direction: NB − mBg cos β = 0, so NB = mBg cos β and fB = µkNB =µkmBg cos β
from x direction: −T + mBg sin β − fB = mBaB
Thus we have: aB = g(sinβ − µk cos β) − TmB
(1)
110
aB
+
T
f B = N B
N Bm
Bg
(a)
mA
g
aA+
T
fA= NA
NA
(b)
µk
µk
αβ
FIGURE S2.24a,b
From the FBD of mass A
from the y-direction: NA − mAg cos α = 0 or NA = mAg cos α
from the x-direction: −T + mAg sin α − µkmAg cos α = mAaA
Thus aA = xA = g(sin α − µk cos α) − TmA
. (2)
Now use the constraint aA+aB = 0 along with equations (1) and (2) to eliminatethe tension T . This yields
g(sinα − µk cos α) − TmA
+ g(sin β − µk cos β) − TmB
= 0
Solving yields T = g(
mAmB
mA+mB
)
(sin α + sin β − µk(cos α + cos β))
Thus the value of constant acceleration from (2) is:
xA = g(sinα − µk cos α) − g(
mAmB
mA+mB
)
[sin α + sin β − µk(cos α + cos β)]
2.25 Solution:
m
mga
y
x
N
µ s N
θ
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:
µsN = ma cos θ
Thus
µs(mg + ma sin θ) = ma cos θ
or
µsg = a(cos θ − µs sin θ)
or
a = µsgcos θ−µs sin θ
.
2.26 Solution:
m
mga
N
θ
f
x
y
FIGURE S2.26
From the FBD, the force sum in the x direction is:
µsN = ma cos θ
The force sum in the y direction is:
mg − N = ma sin θ
Thus:
N = m(g − a sin θ)
and
µsm(g − a sin θ) = ma cos θ
µsg = a(cos θ + µs sin θ)
a = µsgcos θ+µs sin θ
112
2.27 From the top drawing, the constraints are
B
A
x
y
2T1
20g
T1 2T2
3T 2
FIGURE S2.27
2yA + xp = ℓ1 and xB + 2(xB − xp) = ℓ2. Thus 3aB + 4aA = 0 (1)
From the FBD of the pulley, T1 = 2T2. Thus the FBD of A yields:
20(9.81) − 4T2 = 20aA (2)
The FBD of B yields:
−3T2 = 10aB → −3T2 = −1043aA so T2 = 40
9aA (3)
Substitution of (3) into (2) yields:
20(9.81) − 4(
409
)
aA = 20aA
or
(180 + 160)aA = 9(20)9.81
and
aA = 5.194 m/s2.
xA(t) = aA
2t2 thus
1.5 = 5.1942
t2, T2 = 23.1 N, t = 0.76 s, T1 = 46.2 N
2.28 From the drawing, the constraints are:
xA + xB + (xB − xp) = ℓ2 and 2xp + xc = ℓ1 combining yields
xA + 2xB + xc
2= const
Thus: 2aA + 4aB + ac = 0. Let T1 be the tension in cable 1 and T2 = T1/2 bethe tension in cable 2.
113
A
B
C
x
T1
10g
2T1
80g
2T 2T2
20gT1
A B CP
FIGURE S2.28
Now since a = 12at2, t =
√
2da
=√
(2)(4)4.5
= 1.33 s.
From the FBD of A: 10g − T1 = 10aA
From the FBD of B: 80g − 2T1 = 80aB
From the FBD of C: 20g − T1
2= 20ac
Rearranging these 4 equations in 4 unknowns yields:
10aA + T1 = 10g,
80aB + 2T1 = 80g
20ac + 12T1 = 20g
2aA + 4aB + ac = 0
In matrix form these become:
aA
aB
ac
T1
=
10 0 0 10 80 0 20 0 20 0.52 4 1 0
−1
10 · 9.8180 · 9.8120 · 9.81
0
which can be solved using a code to yield:
aA = −11.319, aB = 4.528, aC = 4.528,
T1 = 211.292 and T2 = T1
2= 105.646
Now since a = 12at2, t =
√
2da
=√
(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) = 30aT
Thus
114
N
f
20
10 100
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
NB
0.15N B NB
10g 100
0.15 NA
NA
FIGURE S2.30
From B
(0.15)(20)g = 20aB or aB = 1.47 m/s2
From A
100 − (0.15)(20)g − (0.15)(30)g = 10aA
So
aA = 2.643 m/s2
Block B slips.
2.31 Solution:
Note that aA = a cos θi + a sin θj and aB = xB i + yB j
so that
115
B
A
a
30o
NB
30g
y
x
NA30o
50g
NB
FIGURE S2.31
aB/A = aB − aA = (xB − a cos θ)i + (yB − a sin θ)y
Since aB/A cannot have a component in the j direction we have the constraintthat
yB = a sin θ (1)
From the FBD of B we have
(x direction) 0 = mBxB so that xB = 0
(y direction) NB − mBg = mB yB (2)
From the FBD of A we have
(x direction) −NA sin θ = mAxA = mAa cos θ (3)
(y direction) −NB − mAg + NA cos θ = mAa sin θ (4)
This yields a system of 4 equations in the 4 unknowns yB, a, NA and NB whichcan be written in matrix form as
1 − sin θ 0 030 0 0 −10 50 cos θ sin θ 00 50 sin θ − cos θ 1
yB
aNA
NB
=
0−(30)(9.81)
0−(50)(9.81)
This has solution
a = −6.824 m/s2, yB = −3.411 m/s2, NA = 591.0 N, NB = 191.9 N.
The relative acceleration is
aB/A = (xB − a cos θ)i
which has magnitude
0 + (6.824)(.866) = 5.91 m/s2.
Thus
t =√
2da
=√
2(.1)5.91
= 0.184 s
116
2.32 Solution:
y
x
mg
N
Fs
α
l
α
FIGURE S2.32
The spring force is
Fs = k∆.
From the geometry
∆ =√
ℓ2 + x2 − ℓ.
Summing forces in the x direction (see S2.32) yields
mg cos α − k(√
ℓ2 + x2 − ℓ) x√ℓ2+x2
− µkNx|x| = mx
Summing forces in the y direction yields:
N − mg sin α − k(√
ℓ2 + x2 − ℓ) ℓ√ℓ2+x2
= 0
Therefore
x = 1m
{
mg cos α − kx(
1 − ℓ√ℓ2∗x2
)
− µkx|x|
[
mg sin α + kℓ(
1 − ℓ√ℓ2+x2
)]}
or
x = 1m
{
mg cos α −(
kx + kℓµkx|x|
) (
1 − ℓ√ℓ2+x2
)
− µkx|x|mg sin α
}
117
2.33 Solution:
k 40:= µk 0.2:= L 0.2:= m 5:= g 9.81:= α 30 deg⋅:=
i 0 4000..:=
∆t 0.001:=
v0
x0
0.00001
0
:=
a x v,( )1
mm g⋅ cos α( )⋅ k x µk L⋅
v
v⋅+
⋅ 1L
L2
x2
+
−
⋅− µk m⋅ g⋅ sin α( )⋅v
v⋅−
⋅:=
vi 1+
xi 1+
vi
a xi
vi
,( )∆t⋅+
xi
vi∆t⋅+
:=
0 1 2 3 40
1
2
3
xi
i ∆t⋅
118
2.34 Solution:
x
50 ft
(a)
50x
x
110
(b)
FIGURE S2.34a,b
First determine her velocity when the bungee cord becomes taut
x = +g, but vdv = +gdx, so that v0 =√
2 · g · ℓ = 62.16 ft/s
Now the FBD is illustrated in S2.34b and yields −kx + mg = mx, with v = v0.
Thus a = v dvdx
= g − km
x
Integrating both sides yields:∫ 0v0
vdv =∫ d0
[
g − km
x]
dx
−gℓ = gd − k2m
d2
d =−g±
√g2+ 2kgℓ
m−km
= 77.893 ft (positive root)
Therefore 150 − 60 − 77.9 = 12.1 ft above the ground.
Also, note from the quadratic formula above (using k = c/ℓ) that d is propor-tional to ℓ. Thus, Fx = kd = cd
ℓis independent of ℓ. For the numbers given,
Fx = 5 · 77.9 = 389.5 lb.
2.35 Solution:
cv
mg
y
x
FIGURE S2.35
From the FBD, the sum of the forces in the y-direction when the skydiver hasreached her terminal velocity (i.e. zero acceleration) yields:
cv − mg = 0
119
Solving for c = 4.0A = mgv
.
Thus A = 14.0
· (60)(9.81)9
= 16.4 m2
2.36 From the FBD of the man there are 3 forces acting on him: gravity, the dragforce, which shuts “off” after 10 meters, and the buoyancy force which shuts offwhen he stops. The force summation yields
−Fb − Fc + mg = mx
Now Fc = cv from 10 m or Fc = cv < x − 10 >0
Likewise FB = −mg from 13m or < x − 13 >0
Thus
x = − cvm
< x − 10 >0 −g < x − 13 >0 +g
Fc Fb
mg+
FIGURE S2.36
The force at impact is mx| = mg − cv, where v =√
2ad and x = 10, thus
Fimpact = 70 · 9.81 − 500√
(2)(10)(9.81) = −6313 N. The depth beneath the
surface that the diver travels is d = 13.195-10 =3.195 m (see Mathcad soln).
m 70:= c 500:= ∆t 0.001:= g 9.81:= i 0 5000..:=
a v x,( ) gc v⋅
mΦ x 10−( )⋅
v
v⋅− g Φ x 13−( )⋅−:=
v0
x0
0
0
:=v
i 1+
xi 1+
vi
a vi
xi
,( )∆t⋅+
xi
vi∆t⋅+
:=
0 5 10 150
10
20
vi
xi
x3057
13.194=
x3058
13.195=
x5000
13.195=
120
2.37 This can be solved by “hand” or by using the singularity function and numeri-cally integrating.
By hand: for the first 5 seconds:
a(t) = 0.7t,
v(t) = 0.72
t2
s(t) = 0.76
t3
At t = 5, we have v5 = 8.75 m/s, s5 = 14.58 m.
After the first 5 seconds a = −2 m/s and integrating yields (starting the clockover at t = 0)
v(t) = −2t + v5 = −2t + 8.75
s(t) = −2t2
2+ 8.75t
Solving v(t) = 0 for t yields t = 8.752
= 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.
The total distance traveled is 33.8 meters.
i . .0 940
∆t 0.01
g 9.81 m 1200
t .i ∆t
a t ..Φ 5 t 0.7 t .Φ t 5 2
v
0x
0
0
vi 1
xi 1
vi
.a ti
∆t
xi
.v ∆t
0 5 10
20
20
40
vi
xi
ti
0
i
i
121
2.38 The free body diagram given in S2.38:
+
Fk
Fd
mg
FIGURE S2.38
Here Fd = cx < x − 35 >0< v > is the damping force applied by the matand Fk = k(x − 35)2 < x − 35 >0 is the spring force applied by the mat. Theequation of motion obtained by summing forces in the vertical direction is justmx = mg − Fd − Fk. Two design criteria must be met: total force must beless than 600 lb and the total mat displacement must be less than 6 ft. Thefollowing Mathcad code computes the displacement by numerical integration,and can be evaluated for various values of c and k simply by changing thedefinition statements for these constants. The solution is shown for one possiblecombination of c and k that satisfy the design criteria.
g 32.2:= m120
g:= k 32:= c 20:=
a v x,( ) g Φ x 20−( )k
m⋅ x 20−( )
2⋅− Φ x 20−( )
c
m⋅ v⋅ Φ v( )⋅−:=
∆t 0.001:= i 0 8000..:= ti
i ∆t⋅:=
v0
x0
0
0
:=v
i 1+
xi 1+
vi
a vi
xi
,( )∆t⋅+
xi
vi∆t⋅+
:=
Fi
m a vi
xi
,( )⋅:=
0 5
500
0
Fi
i ∆t⋅
0 5 100
20
xi
ti
122
2.39 The sum of forces in the vertical direction ignoring gravity is
mx = 50000[< 3 − t >0 +∑4
n=1 < t − 6n >< 6n + 3 − t >]
The following Mathcad code integrates the equation of motion and display thedisplacement and velocity:
m 500:= F 50000:=
a t( )F
mΦ 3 t−( )
1
4
n
Φ t 6 n⋅−( ) Φ 6 n⋅ 3+ t−( )⋅( )∑=
+
⋅:=
∆t 0.01:= i 0 2700..:= ti
i ∆t⋅:=
v0
x0
100
0
:=v
i 1+
xi 1+
vi
a ti( )∆t⋅+
xi
vi∆t⋅+
:= v2700
1.604 103
×=
x2700
2.3 104
×=
0 10 20 30
0
2000
vi
ti
0 10 20 300
2 .104
xi
ti
2.40 A FBD of the ball is given in S2.40a.
20o
y
x
mg
F
FIGURE S2.40
The problem is solved by setting up the computational solution and then per-forming a trial and error procedure to find the required force. The Mathcadsolution is shown for one possible value of F (roughly the minimum possiblevalue). The initial conditions are vx(0) = −120 ft/s, x(0) = 0, y(0) = 4 ft (thisis assumed to be the level of the ball when it is struck, based on an averagebatter size and general location of the strike zone) and vy(0) = 0. The forcerequired is found to be about 128 lb.
123
W5
16:= g 32.2:= m
W
g:=
F 13200 m⋅:= F 128.106=
ax t( )F
mΦ 0.015 t−( )⋅ cos 20 deg⋅( )⋅:= ay t( ) g−
F
mΦ 0.015 t−( )⋅ sin 20 deg⋅( )⋅+:=
∆t 0.001:= i 0 4550..:=
ti
i ∆t⋅:=
vx0
x0
vy0
y0
120−
0
0
4
:=
vxi 1+
xi 1+
vyi 1+
yi 1+
vxi
ax ti( )∆t⋅+
xi
vxi∆t⋅+
vyi
ay ti( )∆t⋅+
yi
vyi∆t⋅+
:=x4534
354.065=
y4534
7.137 104−
×=
0 100 200 300 4000
50
100
y i
xi
2.41 A free body diagram of the box is given in the figure.
The forces are Fk = kx < x − 4 >0, f = µk cos θ v|v|
The FBD yields the following two equations∑
Fy : N − mg cos 45◦ = 0∑
Fx : mx = −f − Fk + mg sin 45◦
124
f
N
mg
y
Fk
FIGURE S2.41
The following Mathcad code integrates to find the solution.
µ:= 0.6
i . .0 5000 ∆t 0.001
m 2 g 9.81 k 500θ .45 deg
t .i ∆t
a ,v x .1
m..m g sin θ ..µ cos θ
v
v.Φ x 4 .k x 4
0 1 2 3 4 5
5
5
vi
xi
t
v
0x
0
0
0v
i 1
xi 1
v
i
.a ,v
i
x i ∆t
x .v ∆t
ii
i
i
125
The equivalent MATLAB code is:
function yprime = ds2pt41(t,y) yprime(1,:) = y(2,:);
% friction forceif y(2) = = 0,
Ef = 0;else
Ef = 4.162*y(2,:)/abs(y(2,:)):end
% spring forceif 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 forcesFk with Fk + Fc where Fc = cv(x − 4). Thus the equation of motion can bewritten as: ma = mg(sin θ − µk cos θ v
|v|)− < x − 4 >◦ (k(x − 4) + cv)
The following MATLAB code can be used to solve for the velocity and displace-ment using ODE for plotting the response.
function yprime = ds2pt41(t,y)
yprime(1,:) = y(2,:);
%friction forceif y(2) = = 0,Ff = 0;elseEf = 4.162*y(2,:)/abs(y(2,:));end
% spring and damper forceif y(1) < =0,Fc = 0;Fk = 0;elseFc = 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:= m 2:= g 9.81:= θ 45 deg⋅:=
i 0 5000..:= ∆t 0.001:= ti
i ∆t⋅:=
a v x, c,( )1
mm g sin θ( ) µk cos θ( )⋅
v
v⋅−
⋅
⋅ Φ x 4−( ) k x 4−( )⋅ c v⋅+[ ]⋅−
⋅:=
v0
x0
0
0
:=v
i 1+
xi 1+
vi
a vi
xi
, 20,( )∆t⋅+
xi
vi∆t⋅+
:=
0 2 4 60
2
4
6
xi
ti
v0
x0
0
0
:=v
i 1+
xi 1+
vi
a vi
xi
, 200,( )∆t⋅+
xi
vi∆t⋅+
:=
0 2 4 60
2
4
6
xi
ti
These codes are run several times for different values of c until one is foundthat keeps the box from losing contact with the spring. A value of c = 20 verynearly keeps it in contact, but a value as high as c = 200 is needed to truly keepthe box in contact with the spring.
127
2.43 Assuming positive in the direction of the expanding bag:
30 m/s
vB
FIGURE S2.43
Fd = 5 Ns/m ·vH/B m/s
vH/B = 30 + vB
5(30 + vB) = 500
vB = 70 m/s
2.44 Solution:
Fa Fs
mg
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 thatthe spring forces does not act until x = 5.
The MATLAB code for solving this consists of the following m-file, then usingODE and PLOT to compute and plot the solution x(t). The m-file is
function yprime = ds2pt41(t,y)
yprime(1,:) = y(2,:);
% viscous forceFd = 0.1*y(2,:)*abs(y(2,:));
% spring forceif y(1) < =0,Fk = 0;elseFk = 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 g 9.81
i . .0 10000
∆t 0.001
t .i ∆t
a ,v x .m g ..k Φ x 5 x 5 ..c v v
v
0x
0
0
vi 1
xi 1
v
i
.a ,v
i
xi ∆t
x .v ∆t
0 2 4 6 8 100
5
10
15Position vs time
time s
xi
.i ∆t
ii
i
0
2.45 Solution:
ρ 130
N
FIGURE S2.45
ρ = 100 ft v = 60 mph = 88 ft/s∑
Fn = man = mv2
ρ
N − 130 = 13032.2
v2
100
129
Therefore N − 130 =(
13032.2
)
882
100
N = 443 lb up
2.46 Solution:
mg
500 ft
FIGURE S2.46
To fly the hill N = 0
mg = m v2
500, v =
√32.2 × 500 = 126.9 ft/s = 86.5 mph
2.47 Solution:
mg
ρf
Nβ β
FIGURE S2.47
N cos β − mg − f sin β = 0, N sin β + f cos β = mv2
p
f = µsN N = mgcos β−µ sinβ
mv2
ρ= mg (sinβ+µ cos β)
(cos β−µ sin β)
v =√
ρg sinβ+µ cos βcos β−µ sinβ
2.48 Solution: 25 mph = 36.7 ft/s. Therefore 36.72
100= 32.2
[
sinβ+0.3 cos βcos β−0.3 sinβ
]
0.418(cosβ − 0.3 sinβ) = sin β + 0.3 cos β
β = 6.02◦
130
2.49 Solution:
v
Pmg
L
FIGURE S2.49
amax = 5g = v2
pv = 700 mph = 1027 ft/s
ρ = (1027)2
5(32.2)= 6551 ft
2.50 Solution:
β β
mg
FIGURE S2.50
mg sin β = mRβ wdw = gR
sin βdβ
w2
a|ww0
= gR(− cos θ)β
0
w2 = 2gR
(1 − cos β) + w20
At separation:
mg cos β = mv2
Rv2
R2 = 2gR
(1 − cos β) +v2
0
R2
g cos β = 2g(1 − cos β) +v2
0
R3g cos β = 2g +
v2
0
R
β = cos−1[
23
+v2
0
3gR
]
131
2.51 Solution:
N
mg
θ
θ
FIGURE S2.51
The following equations can be written:
mg cos θ = mRθ (1)
N − mg sin θ = mRθ2 (2)
Eq (1) can be rewritten as ω dωdθ
= gR
cos θ (where ω = θ).
Integrating this gives:ω2
2= g
R(sin θ − sin θ0)
Thus:
ω =√
2gR
(sin θ − sin θ0)
and
v =√
2gR (sin θ − sin θ0)
2.52 Solution:
mRθ = mg cos θ − f θ|θ|
−mg sin θ + N = mRθ2
Therefore mRθ = mg cos θ − µkθ|θ| [mg sin θ + mRθ2]
Canceling the mass
θ = gR
cos θ − µk
Rθ|θ| [g sin θ + Rθ2]
132
2.53 Solution:
The mass cancels out of the equation of motion and values have been assumed for theradius and the coefficient of kinetic friction. The Mathcad code is:
i . .0 2000
∆t 0.001 g 9.81 µ k 0.15 R 0.3
t .i ∆t
α ,ω θ .g
Rcos θ ..
µ k
R
ω
ω.g sin θ .R ω
2
ω0
θ
0
.10 deg
ωi 1
θi 1
ωi
.α ,ωi
θ ∆t
θ .ωi
∆t
0 0.5 1 1.5 20
1
2
3Angular Position vs time
Time s
θi
ti
i
0
i
i
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)==0yprime(2) = -mu*(y(2) 2+gg*sin(y(1)))+gg*cos(y(1));elseyprime(2) = -mu*(y(2) 2+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:
Fs = kR(2 cos θ2− 1)
Ff = µk|N | θ|θ|
From Sample 2.12, the FBD yields upon summing forces in the normal andtangential directions∑
Fn = −N + mg sin θ + Fs cos θ2
= mRθ2
N = mg sin θ + kR(2 cos θ2− 1) cos θ
2− mRθ2
∑
Ft = −Ff + Fs sin θ2− mg cos θ = mRθ
The equation of motion is thus:
mRθ = −µk
∣
∣
∣mg sin θ + kR(2 cos θ2− 1) cos θ
2− mRθ2
∣
∣
∣
θ|θ|
+kR(2 cos θ2− 1) sin θ
2− mg cos θ
θ = −µk
∣
∣
∣
gR
sin θ + km
(2 cos θ2− 1) cos θ
2− θ2
∣
∣
∣
θ|θ|
+ km
(2 cos θ2− 1) sin θ
2− g
Rcos θ
When the slider is in equilibrium, θ = 0 and θ = 0. The Mathcad solutionfollows.
The solution shown here is for a coefficient of kinetic friction of 0.18, a valuethat was chosen for a typical response. The spring would be unloaded whenθ = 120◦ but there would still be the gravitational acceleration acting on theslider. It may be observed that the slider is oscillating about a slighter higherangle (roughly 123◦), which is the equilibrium angle.
Note the slow start of the slider as it overcomes the friction force. The springforce is high when θ = 30◦ but this contributes to a high normal force andtherefore 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 animpossibility. In this formulation the friction term in the equation of motion ismultiplied by θ/|θ| to make sure the friction always opposes motion; this term iszero when the angular velocity is zero so friction will never initiate motion. Todetermine whether motion would occur in a particular case, determine whetherthe spring force minus the gravitational force in the initial configuration isgreater than the coefficient of static friction times the normal force. That wouldbe a separate calculation as the friction force in this equation of motion is zerountil motion starts.
134
m 1:= k 600:= R 0.2:= g 9.81:= µk 0.18:=
N ω θ,( ) m g⋅ sin θ( )⋅ k R⋅ 2 cosθ
2
⋅ 1−
⋅ cosθ
2
⋅+ m R⋅ ω2
⋅−:=
α ω θ,( ) g−
Rcos θ( )⋅
k
m2 cos
θ
2
⋅ 1−
⋅ sinθ
2
⋅+µk
m R⋅N ω θ,( )⋅
ω
ω⋅−:=
i 0 15000..:= ∆t 0.0001:= ti
i ∆t⋅:=
ω0
θ0
0
π
6
:=ωi 1+
θ i 1+
ωi α ωi θ i,( )∆t⋅+
θ i ωi ∆t⋅+
:=
0 0.5 1 1.50
100
200
θi
deg
ti
f θ( ) g−
Rcos θ( )⋅
k
m2 cos
θ
2
⋅ 1−
⋅ sinθ
2
⋅+:=
x 120 deg⋅:=root f x( ) x,( )
deg123.35=
135
2.55 We are given α = constant and so ω = αt.
r 2ω
rα
FIGURE S2.55
f = m(rαet + rω2en) is the total force. The total acceleration is
µsmg = m√
(rα)2 + (rω2)2
µs = 1g
√
(rα)2 + (r(αt)2)2 = rαg
√1 + α2t4
2.56 Solution:
A
ω_
B
RA
RB
FIGURE S2.56
RA + RB = ℓ
T = mBRBω2 , T = mARAω2
Therefore mBRB = mARA.
RA = ℓ − RB
RB = mA
mBℓ − mA
mBRB
RB = mA
mA+mBℓ
RA = mB
mA+mBℓ
136
2.57 Solution:
R
e
θ
<
t
e
<
n
e<
n
mg
N
FIGURE S2.57
Sum forces along t: N cos θ−mg = 0 sum forces along n : N sin θ = mRω2 sin θ
Therefore N = mRω2, mRω2 cos θ = mg, θ = cos−1(
gRω2
)
.
2.58 First-determine the required velocity at top.
mg
N
n
<
FIGURE S2.58a
From the figure S2.58a:
N + mg = mv2
R, N = 1
2mg
v =√
32gR or ωT =
√
32
gR.
θ
θ
R
ω0mg
N
FIGURE S2.58b
137
From S2.58b:
mRθ = −mg sin θ (sum of tangential forces)∫ ωTω0
ωdω =∫ π0
gR
sin θdθ
ω2
T−ω2
0
2= + g
Rcos θ|π0 = −2g
R
ω20 = 4g
R+ 3
2gR
= 112
gR
v0 =√
112gR
2.59 From Problem 2.58
n
<
N
mg
FIGURE S2.59
N − mg =mv2
0
R= m · 11
2g or N = mg + 11
2mg
N = 132
mg
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:
yA = ℓθ2 sin θ+g cos θ sin θγ+sin2 θ
,
where
γ = mA/mB
For ℓ = 0.2 m, θ = 30◦, θ(0) = 0, mA = 3 kg and mB = 2 kg:
yA = 2.427 m/s (acceleration of block A).
The acceleration of block B is:
aB = aAj + ℓθ cos θj− ℓθ sin θi
From the solution to problem 2.61:
(mA + mB sin2 θ)θ + mB θ2 cos θ sin θ + (mA + mB)gℓsin θ = 0
For the numbers given:
θ = −35.036 rad/s2.
And:
aB = 2.427j− 6.068j + 3.504i = 3.504i− 3.641j m/s.
139
2.61 The FBD’s are given in the figure:
m Ag
y
mB
g
(t)θ
θ
TN
m A gT
mB
g
θx
FIGURE S2.61
In the y-direction for block A:
(1) T sin θ = mAyA
and in the radial and tangential directions for block B:
(2) −T + mBg cos θ = mB(−ℓθ2 + yA sin θ)
(3) −mBg sin θ = mBℓθ + mB yA cos θ
Substitution of (2) into (1) yields
sin θ{mBℓθ2 − mByA sin θ + mBg cos θ} = mAyA
or
ℓθ2 sin θ + g cos θ sin θ = (mA
mB+ sin2 θ)yA
Thus we have
(4) yA = ℓθ2 sin θ+g cos θ sin θγ+sin2 θ
.
where
γ = mA/mB
Substitute 4 into 3:
ℓθ + ℓθ2 sin θ+g cos θ sin θγ+sin2 θ
· cos θ = −g sin θ.
(γ + sin2 θ)θ + θ2 cos θ sin θ + gℓcos2 θ sin θ + γ g
ℓsin θ + g
ℓsin3 θ = 0
which can also be written as:
(1 + sin2 θγ
)θ + 1γθ2 cos θ sin θ + g
γℓcos2 θ sin θ + g
ℓsin θ + g
γℓsin3 θ = 0
When mA >> mB (γ very large) this becomes the pendulum equation:
θ + gℓsin θ = 0
Resubstituting for γ into the original equation and simplifying:
(mA + mB sin2 θ)θ + mB θ2 cos θ sin θ + (mA + mB)gℓsin θ = 0
140
2.62 The free body diagrams are given in S2.62:
mA
g
N
T
β
T
mBg
y
x
A: B:
FIGURE S2.62
From the solution to problem 2.63 for θ, with θ = 0 and θ = 0:
θ = −g sinβ cos β(mA+mB)ℓ[mA+mB sin2(β]
Writing the equation of motion in the x-direction for block A in the coordinatesystem shown in S2.62:
mAaA = mAg sin β + T sin β
Writing the equation of motion in the y-direction for block B in the coordinatesystem shown in S2.62:
T cos β − mBg cos β = mBℓθ sin β
Solving the previous equation for T :
T = mBg + mBℓθ tanβ
Substituting this gives the acceleration of block A (which is in the x-direction):
aA = mA+mB
mAg sin β + mB
mAℓθ sin β tan β
And the acceleration of B is:
aB = aAi + ℓθ cos β i + ℓθ sin β j
141
2.63 Solution:
mA
g
NT
β
θ
θ
T
mBg
y
x
A: B:
FIGURE S2.63
T sin(β + θ) + mAg sin β = mAaA
−T sin θ = mBaBx
T cos θ − mBg = MBaBy
aB = aA + aB/A
aA = aA(cos β i − sin β j) + ℓθ2(− sin θi + cos θj)
(1) T sin(β + θ) + mAg sin β = mAaA
(2) −T sin θ = mB[aA cos β + ℓθ cos θ − ℓθ2 sin θ]
(3) T cos θ − mBgg = MB[−aA sin β + ℓθ sin θ + ℓθ2 cos θ]
Therefore aA = TmA
sin(β + θ) + g sin β
Substituting into (2) and (3)
2) −T sin θ = mB
[
TmA
sin(β + θ) cos β + g sin β cos β + ℓθ cos θ − ℓθ2 sin θ]
T[
sin θ + mB
mAsin(β + θ) cos β
]
= −mBgsinβ cos β − mBℓθ cos θ + mBℓθ2 sin θ
T = −mB [g sin β cos β+ℓθ cos θ−ℓθ2 sin θ]
sin θ+mBmA
sin(β+θ) cos β
3) T cos θ = mBg − mB[aA sin β − ℓθ sin θ − ℓθ cos θ]
T [cos θ + mB
mAsin(β + θ) sin β] = mB[q(1 − sin2 β) + ℓθ sin θ + ℓθ2 cos θ]
T = mB [g(1−sin2 β)+ℓθ sin θ+ℓθ2 cos θ]
cos θ+mBmA
sin(β+θ) sinβ
Equating two expressions for T
−[cos θ + mB
mAsin(β + θ) sin β] · [g sin β cos β + ℓθ cos θ − ℓθ2 sin θ]
= [sin θ + mb
mAsin(β + θ) cos β] · [g cos2 β + ℓθ sin θ + ℓθ2 cos θ]
Therefore g sin β cos β cos θ + ℓθ cos2 θ − ℓθ2 sin θ cos θ
+g cos2 β sin θ + ℓθ sin2 θ + ℓθ sin θ cos θ
+mB
mAsin(β + θ)[g sin2 β cos β + ℓθ cos θ sin β − ℓθ2 sin θ sin β]
+mB
mAsin(β + θ)[g cos2 β cos β + ℓθ sin θ cos β + ℓθ2 cos θ cos β]
142
g sin(β + θ) cos β + ℓθ + mB
mAsin(β + θ)[g cos β + ℓθ sin(β + θ)+ ℓθ2 cos(β + θ)] = 0
g sin(β + θ) cos β(mA+mB
mA) + ℓθ[1 + mB
mAsin2(β + θ)]
+ℓθ2 mB
mAsin(β + θ) cos(β + θ) = 0
θ = −g sin(β+θ) cos β(mA+mB)+ℓθ2mB sin(β+θ) cos(β+θ)ℓ[mA+mB sin2(β+θ)]
2.64 Solution: (in Mathcad)
i . .0 2000
∆t 0.001
t .i ∆t
m A 3
m B 2
g 9.81
β 0L 0.5
ddθ ,θ dθ
...m A m B g sin β θ cos β ....m B L dθ sin β θ cos β θ
.L m A.m B sin β θ
2
dθ
0θ
0
.20 deg
dθi 1
θi 1
dθi.ddθ ,θ
idθ ∆t
θ .dθ ∆t
0 0.5 1 1.5 240
20
0
20
40Angle- time relationship
time s
θi
deg
ti
i
i
i
i
0
2
143
2.65 Solution: (in Mathcad)
i . .0 2000
∆t 0.001
t .i ∆t
m A 3
m B 2
g 9.81
β .30 degL 0.5
ddθ ,θ dθ
...m A m B g sin β θ cos β ....m B L dθ sin β θ cos β θ
.L m A.m B sin β θ
2
dθ
0θ
0
.20 deg
dθi 1
θi 1
dθ
i
.ddθ ,θi
dθ
i
∆t
θ .dθ ∆t
0 0.5 1 1.5 2100
50
0
50Angle- time relationship
time s
θi
deg
ti
i
0
ii
2
2.66 Solution:
an = 3 · g = v2
pv = 200 mph = 293 ft/s
p = (293)2
96.6= 889 ft
From the solution of 2.47:
an = g(sin β+µ cos β)cos β−µ sinβ
.
3(cos β − µ sin β) − (sin β + µ cos β) = 0
−(3µ + 1) sin β + (3 − µ) cosβ = 0
Therefore tanβ = 3−µ3µ+1
µ = 0.85
β = 31.2◦.
144
2.67 The Mathcad code to solve this problem is:
N 917:= n 0 N..:= ∆t 0.001:= tn
n ∆t⋅:= L 3:= g 9.81:=
v0
s0
θ0
dθ0
0
0
70− deg⋅
0
:=
vn 1+
sn 1+
θn 1+
dθn 1+
vn
g sin θn( )⋅ 0.3 g cos θn( )⋅ dθn vn( )
2⋅+
⋅+
∆t⋅−
sn
vn∆t⋅+
70− deg⋅ 1
sn
L
3
−
⋅
3 70⋅ deg⋅s
n( )2
L3
⋅
:=
sN
3.001= vN
4.227= tN
0.917=
N 817:= n 0 N..:= ∆t 0.001:= tn
n ∆t⋅:= L 3:= g 9.81:=
v0
s0
θ0
dθ0
0
0
70− deg⋅
0
:=
vn 1+
sn 1+
θn 1+
dθn 1+
vn
g sin θn( )⋅ 0.0 g cos θn( )⋅ dθn vn( )
2⋅+
⋅+
∆t⋅−
sn
vn∆t⋅+
70− deg⋅ 1
sn
L
3
−
⋅
3 70⋅ deg⋅s
n( )2
L3
⋅
:=
sN
2.999= vN
6.668= tN
0.817=
The child’s velocity is 4.227 m/s at the bottom of the slide when the coefficientof kinetic friction is 0.3, and 6.668 m/s when there is no friction. It takes thechild 0.917 seconds to reach the bottom of the slide when the coefficient ofkinetic 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); %positions = y(2); %speed
L = 3; %length of slidemu = 0.3; %kinetic coefficient of friction
g = 9.81; %acceleration due to gravity
theta = -1.222*(1-(s/L) 3); %angletheta = 1.222*3 *(s/L) 2/L; %dtheta/ds
% differential equationsyprime(1,:) = s ;yprime(2,:) = - mu*(s 2*theta + g*cos(theta))*sign(s )-g*sin(theta);
% v x and v yyprime(3,:) = s *cos(theta);yprime(4,:) = s *sin(theta);
return
146
2.68 The Mathcad code to solve the problem:
The height of the slide must first be determined numerically as shown below.
x 0 0
The height was originally choosen to be 3 m, and was reduced to 2.854 m so that thebottom would be 0.6 m from the ground.
y 0 2.854
s . .,0 0.1 3
θ s ..70 deg 1s
3
3
x s x 0 d0
s
ζcos θ ζ
y s y 0 d0
s
ζsin θ ζ
=y 3 0.6
0 0.5 1 1.5 20.5
1
1.5
2
2.5
3Slide Contour
horizontal distance m
y s
x s
The solution for the child falling from the top is straight forward as the only force actingon the child is gravitational acceleration.
y(t) = -g m/s
: 2
y(t) = - gt m/s.
y(t) = + 2.854 m
The time when the child hits the ground is t = 0.763s and the velocity is 7.485 m/s.
vert
ica
l dis
tan
ce m
- gt2
2
147
2.69 Solution:
N 6263
g 9.81
µ 0.5
α .47 degL 46i . .0 N
∆t 0.001
t .i ∆t
a ,,θ dθ v .g sin θ .µ .g cos θ .v dθ
x
0y
0
20
=s6263
46.005
0 10 20 30 4010
0
10
20Slide Profile
Horizontal Position m
yi
xi
The slide provides a very small velocity at the bottom and the time on the slide is 6.263 s.A longer time on the slide may be obtained by changing the initial angle and thecoefficient of friction as well as the length of the slide.
v0
s0
θ
0dθ
0
0
α
0
0
vi 1
si 1
θi 1
dθi 1
vi.a ,,θ dθ
i
v ∆t
si
.v t
.α 1s
i
L
2
..2 αs
i
L2
i i i
∆
=v 3.4536263
=t 6.2636263
0 xi 1
yi 1
x
i
.cos θi si 1
si
y .sin θi
si 1
si
i
i
2
148
2.70 Solution:
θ(s) = (s/25000) cos(s/500), for 0 < s < 3000dsdt
= 100 km/hr = 1003.6
= 27.8 m/s
dθds
= 125000
[cos(s/500) − s500
· sin( s500
)], N(s) = cos(θ(s)) + v2
gdθds
∆s 1:= i 0 3000..:= si
i ∆s⋅:=
θ s( )s
25000cos
s
500
⋅:=
x0
y0
0
0
:=xi 1+
yi 1+
xi
cos θ si( )( )∆s⋅+
yi
sin θ si( )( )∆s⋅+
:=
0 500 1000 1500 2000 2500 3000100
50
0
50
y i
xi
dθ s( )1
25000cos
s
500
⋅s
25000sin
s
500
⋅1
500⋅−:=
v 1001000
3600⋅:= v 27.778=
Ni
cos θ si( )( ) v
2
9.81dθ s
i( )⋅+:=
0 500 1000 1500 2000 2500 3000
1
1.02
Ni
xi
149
2.71 Solution:
s . .,0 10 7000
θ s .s
4000cos
s
4001
dθ s .1
4000cos
s
400.
s
400sin
s
400v 66
g 32.2a max
.0.5 g
a n s .v dθ s
0 1000 2000 3000 4000 5000 6000 700020
0
2018.941213
15.547668
a n s
a max
.7 1030 s
Yes, the occupant slips near the end of the ride.
1
4000
2.72 Solution:
0 < s < 20
θ = s sin−1[0.2 − 0.5 cos( s10
)]
s = 1− m/s∑
Fn = F = ms2θ′
Fmax = µsN = µsmg > ms2θ′
µsg > s2θ′ no slip condition
µs > s2θ′
g= 1020.2571
9.81= 2.62
θ = 5 sin−1[0.2 − 0.5 cos( s10
)]
θ′ = 5 1√1−u2
u′
where u = 0.2 − 0.5 cos( s10
)
u′ = 0.5 sin( s10
) · 110
max θ′(s) = 0.2571 r/m
@ s = 17.17 m
150
The following is a Mathcad analysis:
v 10:= s 0 20..:=
θ s( ) 5 asin 0.2 0.5 coss
10
⋅−
⋅:=
dθ s( )sθ s( )
d
d:=
an s( ) v2
dθ s( )⋅:=
0 10 200
20
40
an s( )
s
µ s( )an 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) = −0.100π2 cos(πt)
θ(t) = π6
sin(πt)
θ(t) = π2
6cos(πt)
θ(t) = −π3
6sin(πt)
v(t) = −0.100π sin(πt)er + (0.300 + 0.100 cosπt)π2
6cos2(πt)eθ
a(t) = [−0.100π2 cos(πt) − (0.3 + 0.1 cos(πt))π4
36cos2 πt]er
+[(0.3 + 0.1 cosπt)(−π3
6sin πt) + 2(−0.1π sin(πt))(π2
6cos πt)]eθ
F = marer + maθ eθ
= m{−0.100π4
36cos3(πt) − 0.300π4
36cos2(πt) − 0.100π2 cos(πt)}er
−0.3mπ3
6sin(πt){1 + cos(πt)}eθ
151
2.74 It is given that:
F(t) = (3t2 − 1)er + cos(πt6)eθ, m = 1
Therefore upon comparison with eq. (2.39) in the book:
m(r − rθ2) = 3t2 − 1
m(rθ + 2rθ) = cos(πt6).
This gives the following differential equation to solve:
r = rθ2 + 1m
(3t2 − 1)
θ = −2 rrθ + 1
mrcos(πt
6)
With initial conditions
r(0) = 2 θ(0) = 0 r(0) = 0 θ(0) = 0
The solution in MATLAB is given in the following file:
function yprime = ds2pt74(t,y)
m = 1;Fr = 3*t 2-1;Fth = cos(t*pi.6);
r = y(1); % radial positionth = y(2); % angular positionv = y(3); % radial velocityw = y(4); % angular velocity
yprime(1) = v;yprime(2) = w;yprime(3) = Fr/m + r*w 2;yprime(4) = (1/r)*(Fth/m-2*v*w);
return
152
The Mathcad solution follows:
m 1:=
ddr r dθ, t,( ) r dθ2
⋅1
m3 t
2⋅ 1−( )⋅+:=
ddθ dr r, dθ, t,( ) 2− dr⋅ dθ⋅
r
1
m r⋅cos
π t⋅
6
⋅−:=
i 0 2000..:= ∆t 0.001:= ti
i ∆t⋅:=
dr0
r0
dθ0
θ0
0
2
0
0
:=
dri 1+
ri 1+
dθi 1+
θ i 1+
dri
ddr ri
dθi, ti
,( )∆t⋅+
ri
dri∆t⋅+
dθi ddθ dri
ri
, dθi, ti
,( )∆t⋅+
θ i dθi ∆t⋅+
:=
xi
ri
cos θ i( )⋅:= yi
ri
sin θ i( )⋅:=
1.5 2 2.5 3 3.5
2
0
y i
xi
2.75 The force acting on m along r is:
Differentiating the given value of θ yields:
Fr = −k(r − 0.2)er.
θ = 0.1t2
θ = 0.2t
θ = 0.2
Using equation 2.39 yields:
m(r − rθ2) = −k(r − 0.2)
m(rθ + 2rθ) = 0
The governing differential equations and initial conditions are:
θ = −24rθ
θ = −2rrθ
k = 400N/m m = 4kg
θ(0) = 0 r(0) = 0 r(0) = 0.2 θ(0) = 0
153
The MATLAB code is:
function yprime = ds2pt75(t,y)
m = 4;k = 40;L = 0.2;
th = 0.1*t 2;thdot = 0.2*t;
yprime(1) = y(2);yprime(2) = y(1)*thdot 2-(k/m)*(y(1)-L);
return
The Mathcad solution is:
i . .0 10000
∆t 0.001k 40
m 4
t .i ∆t
θ .0.1 t 2
dθ .0.2 t
ddr ,r dθ .r dθ2 .
k
mr 0.2
dr0
r0
0
0.2
dri 1
ri 1
dri
.ddr ,ri
dθi
∆t
ri
.dri
∆t
0
30
6090
120
150
180
210
240270
300
330
0.3
0.2
0.1
0
Trajectory Plot0.33073
0.2ri
θi
Starting point
i i
ii
i
154
2.76 Solution:
θ
µFs
f = - k mg(k v___
v ||)
FIGURE S2.76
v = (rex + rθeθ). The unit vector along v for computing f is:
v
|v| = rex+rθeθ√r2+r2θ2
Equation (2.39) from the book yields:
m(r − rθ2) = −k(r − 0.2) − µrmg r√(r2+r2θ2)
or
r = rθ2 − km
(r − 0.2) − µk · g · r√r2+r2θ2
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*t 2; thdot = 0.2*t; v = sqrt(y(2) 2+(y(1)*thdot) 2);
if v==0,yprime(1) = y(2);yprime(2) = y(1)*thdot 2-(k/m)*(y(1)-1);elseyprime(1) = y(2);yprime(2) = y(1)*thdot 2-(k/m)*(y(1)-1)-mu*g*y(2)/v;end
return
155
The Mathcad solution is:
k 20
m 4
µ 0.2g 9.81i . .0 10000
∆t 0.001
t .i ∆t
θ t .0.1 t2
dθ t .0.2 t
ddr ,,,r dr dθ t .r dθ .k
mr 0.2 ..µ g
dr
dr 2 .r dθ2
dr0
r0
0
0.2
dri 1
ri 1
dri
.ddr ,,,ri
dri
dθ ti ti
∆t
ri
.dri
∆t
0
30
60
90
120
150
180
210
240
270
300
330
0.4
0.2
0
Particle movement 0-10 s0.515584
0.199999ri
θ ti
Starting pointof the particlet=0 s
i
2
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)
r(θ) = 0.200(sin θ)θ
r(θ) = 0.200[(cos θ)θ2 + (sin θ)θ]
156
The equation of motion in the radial direction is:
Nr − k(r − 0.100) = m(r − rθ2), where θ = 1.5. Thus:
Nr = k[0.200(2 − cos θ) − 0.100] + m[0.200 cos θ(1.5)2 − 0.200(2 − cos θ)(1.5)2]
= 650(0.300 − 0.200 cos θ) + 4[0.400 cos θ(1.5)2 − 0.200(2)(1.5)2]
Thus
Nr = 191.4 − 126.4 cos θ.
2.78 We are given θ = 2 − cos θ, so that
θ = sin θθ = sin θ(2 − cos θ)
Thus
Nr = k(r − 0.100) + m(r − rθ2)
r(t) = 0.200[cos θ(2 − cos θ)2 + sin2 θ(2 − cos θ)]
rθ2 = 0.200(2 − cos θ)(2 − cos θ)2
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. 2;
k = 650; m = 4; l = 0.1;
Nr = m*(rddot-r.*thdot. 2)+k*(r-l);
return
The following computes the solution in Mathcad:
157
θ 0π
48, 4 π⋅..:=
Fs θ( ) 650 0.2 2 cos θ( )−( )⋅ 0.1− ⋅:=
ddr θ( ) 0.2 cos θ( ) 2 cos θ( )−( )2⋅ sin θ( )( )2 2 cos θ( )−( )⋅+ ⋅:=
rωω θ( ) 0.2 2 cos θ( )−( )3⋅:=
Nr θ( ) Fs θ( ) 4 ddr θ( ) rωω θ( )−( )⋅+:=
0 200 400 600 8000
100
200
300
Nr θ( )
θ
deg
2.79 Following the solution of sample Problem 2.17 the following codes are used tonumerically 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);
% velocitiesyprime(1,:) = rdot;yprime(2,:) = thdot;
% accelerationsyprime(3,:) = r*thdot 2-(k/m)*(r-r0)+g*cos(theta);yprime(4,:) = (-2*rdot*thdot-g*sin(theta))/r;
return
In Mathcad the code is:
158
k 1600
g 9.81N 2400
∆t 0.0005i . .0 N
v0
r0
ω0
θ
0
0.3
0
.30 deg
m:= 3.6 kg
The mass was adjusted by trial and error to produce the required path of motion.
a ,,,v r ω θ .r ω2 .g cos θ .
k
mr 0.3
α ,,,v r ω θ .1
r.g sin θ ..2 v ω
vi 1
ri 1
ωi 1
θi 1
vi
.a ,,,vi
ri
ωi
θi
∆t
ri
.vi
∆t
ωi
.α ,,,vi
r
i
ωi
θi
∆t
θ .ωi
∆t
xi
.ri
sin θ
yi
0.3 .ri
cos θ
-0.05 0.1 0.15
0
0.05Pendulum Trajectory
horizontal position (m)
yi
xi
0
i
i
i
0.05-0.1-0.15
-0.05
2.80 Solution:
159
mg
θk(r-L)
θ
FIGURE S2.80
From the free body diagram, the equations of motion can be written:
−k(r − L) + mg cos θ = mr − mrθ2
r = rθ2 + g cos θ − km
(r − L)
and:
−mg sin θ = 2mrθ + mrθ
θ = −grsin θ − 2 r
rθ
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-t 2);yprime(1) = rdot;yprime(2) = thdot;yprime(3) = r*thdot 2+Fr/m;yprime(4) = (-2*rdot*thdot+Fth/m)/r;return
The numerical solution in Mathcad is given on the following page. Note that thesolution (pendulum angle versus time) to Sample Problem 2.13 is also computedand plotted (in blue). The two solutions differ when k = 60 but are in goodagreement when k = 600. The length of the pendulum is also plotted as afunction of time showing the degree to which the mass is oscillating on thespring (note that the oscillations are very small when k = 600, i.e. in that casethe pendulum is behaving approximately as if the spring were a rigid bar orstring).
160
m 2:= L 2:= g 9.81:=
ddr r dθ, θ, k,( ) r dθ2
⋅ g cos θ( )⋅+k
mr L−( )⋅−:=
ddθ dr r, dθ, θ,( ) g−
rsin θ( )⋅ 2
dr dθ⋅
r⋅−:=
i 0 60000..:= ∆t 0.0001:= ti
i ∆t⋅:=
dr0
r0
dθ0
θ0
0
L
0
30 deg⋅
:=
dri 1+
ri 1+
dθi 1+
θ i 1+
dri
ddr ri
dθi, θ i, 60,( )∆t⋅+
ri
dri∆t⋅+
dθi ddθ dri
ri
, dθi, θ i,( )∆t⋅+
θ i dθi ∆t⋅+
:=
dβ0
β0
0
30 deg⋅
:=dβi 1+
βi 1+
dβig
Lsin βi( )⋅ ∆t⋅−
βi dβi ∆t⋅+
:=
0 2 4 650
0
50
θi
deg
β i
deg
ti
0 2 4 62
2.5
3
ri
ti
dr0
r0
dθ0
θ0
0
L
0
30 deg⋅
:=
dri 1+
ri 1+
dθi 1+
θ i 1+
dri
ddr ri
dθi, θ i, 600,( )∆t⋅+
ri
dri∆t⋅+
dθi ddθ dri
ri
, dθi, θ i,( )∆t⋅+
θ i dθi ∆t⋅+
:=
0 2 4 650
0
50
θi
deg
β i
deg
ti
0 2 4 61.5
2
2.5
3
ri
ti
161
2.81 The free-body diagram is:
R
mgN
θ
θ
FIGURE S2.81
Using equation (2.39) in the book the equations of motion are:
−N + mg cos θ = m(r − rθ2)
−mg sin θ = m(rθ + 2rθ)
Subject to: θ(0) = v0
R
we have that: θ = ωdωdθ
= − rR
sin θ, or:∫ ωv0/R ωdω =
∫ θ0 − g
Rsin θdθ
12(ω2 − v2
0
R2 ) = rR(cos θ − 1), thus: ω2 = 2g
R(cos θ − 1) +
v2
0
R2
The normal force will be a minimum at θ = 180◦
N = −mg + mRω2
Therefore ω2 ≥ rR
= 2gR
(−2) +v2
0
R2
v0 =√
5gR
2.82 The free-body diagram is:
mg
N
θ
f
FIGURE S2.82
From the diagram and equation (2.39) in the book:
−N + mg cos θ = −mRθ2
162
−mg sin θ − µk |N | θ|θ| = mRθ
Solving for the normal force:
N = mg cos θ + mRθ2
And substituting this into the equation for angular acceleration:
θ = − gR
sin θ − µk
R
∣
∣
∣g cos θ + Rθ2∣
∣
∣
θ|θ|
The computer solution in Mathcad is (independent of mass):
R 2:= µk 0.2:= v0 5:= g 9.81:=
α ω θ,( ) g−
Rsin θ( )⋅
µk
Rg cos θ( )⋅ R ω
2⋅+⋅
ω
ω⋅−:=
i 0 5000..:= ∆t 0.001:= ti
i ∆t⋅:=
ω0
θ0
v0
R
0
:=ωi 1+
θ i 1+
ωi α ωi θ i,( )∆t⋅+
θ i ωi ∆t⋅+
:=
0 2 4 650
0
50
100
θi
deg
ti
2.83 From the geometry in Figure S2.83a:
R/2
L
R
θ
β
FIGURE S2.83a
From the law of cosines:
L2 = R2 + R2
4− R2 cos(θ)
L = R2
√5 − 4 cos θ
163
And:R2
4= R2 + L2 − 2RL cos β
R2
4= R2 + R2
4(5 − 4 cos θ) − R2
√5 − 4 cos θ cos β
cos β = 2−cos θ√5−4 cos θ
From the law of sines:sin θL
= sin βR/2
sin β = sin θ√5−4 cos θ
The spring force is:
Fs = k(L − L0) = kR2
[√5 − 4 cos θ − 1
4
]
Components of the spring force are:
Fsn = −Fs cos β = −kR2
[
1 − 14√
5−4 cos θ
]
(2 − cos θ)
Fsθ = −Fs sin β = −kR2
[
1 − 14√
5−4 cos θ
]
sin θ
N
Fs
θe
> >
en
β
FIGURE S2.83b
From the free-body diagram in S2.83b, the equation of motion is:
mRd2θdt2
= −kR2
[
1 − 14√
5−4 cos θ
]
sin θ
d2θdt2
= − k2m
[
1 − 14√
5−4 cos θ
]
sin θ
164
2.84 Solution:
N
Fs
θe
> >
en
β
mg
θ
FIGURE S2.84
With the addition of the weight as shown in S2.84, the equation of motionbecomes:
mRd2θdt2
= −kR2
[
1 − 14√
5−4 cos θ
]
sin θ − mg cos θ
d2θdt2
= − k2m
[
1 − 14√
5−4 cos θ
]
sin θ − gR
cos θ
2.85 Solution:
N
Fs
θe
> >
en
βf
FIGURE S2.85
The friction force is:
f = −µkNθ|θ|
With the addition of the friction force as shown in S2.85, the equation of motionbecomes:
mRd2θdt2
= −kR2
[
1 − 14√
5−4 cos θ
]
sin θ − µkNθ|θ|
d2θdt2
= − k2m
[
1 − 14√
5−4 cos θ
]
sin θ − µkNmR
θ|θ|
where:
N = kR2
[
1 − 14√
5−4 cos θ
]
(2 − cos θ) − mRθ2
165
2.86 Solution:
N
Fs
θe
> >
en
β
mg
θf
FIGURE S2.86
With the addition of both the weight and the friction force as shown in S2.86,the equation of motion becomes:
mRd2θdt2
= −mg cos θ − kR2
[
1 − 14√
5−4 cos θ
]
sin θ − µkNθ|θ|
d2θdt2
= − gR
cos θ − k2m
[
1 − 14√
5−4 cos θ
]
sin θ − µkNmR
θ|θ|
where:
N = mg sin θ + kR2
[
1 − 14√
5−4 cos θ
]
(2 − cos θ) − mRθ2
A Mathcad code for integrating the equation of motion is as follows with somechoice 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:= µ 0.3:= g 9.81:= R 1:=
f θ( ) 11
4 5 4 cos θ( )⋅−⋅−:=
N dθ θ,( ) kR
2⋅ f θ( )⋅ 2 cos θ( )−( )⋅ m R⋅ dθ
2⋅− m g⋅ sin θ( )⋅+:=
ddθ dθ θ,( ) k−
2 m⋅f θ( )⋅ sin θ( )⋅
g
Rcos θ( )⋅−
µ
m R⋅N dθ θ,( )⋅
dθ
dθ⋅−:=
i 0 40000..:= ∆t 0.0001:= ti
i ∆t⋅:=
dθ0
θ0
0
30 deg⋅
:=dθi 1+
θ i 1+
dθi ddθ dθi θ i,( )∆t⋅+
θ i dθi ∆t⋅+
:=
0 2 4100
50
0
50
θi
deg
ti
2.87 We are given:
M = 5.98 × 1024kg
r = 6380 × 103 m
G = 66.7 × 10−12 m3/kg · s2
F
m
FIGURE S2.87
GmMr2 = mv2
r
v =√
GMr
=√
5.98×66.7500+6380
× 103 = 7614 m/s
167
2.88 Solution:
Rω2m = GMmR2
ω =√
GMR3
The period τ is
τ = 2πω
= 2π√GM
R3
2
2.89 Solution:
τ = 23 hrs 56 min. = 86.16 ×103 s.
τ = 2π√GM
r3/20
r3/20 = τ
√GM
2π= 273.87 × 109 m
r0 = 42210 km.
r = r0 − R = 42210 − 6380
r = 35830 km
v =√
GMr0
= 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
v4 =√
MGRA
= 7.506 × 103 m/s
R1 = 6380 + 400 = 6780 km
v1 =√
MGR1
= 7.670 × 103 m/s
From Eq. (2.67) for a conic section:1r
= A(1 + e cos θ)
For elliptic orbit r0 ⇒ θ = 0, rapogee ⇒ θ = 180
Therefore rapogee
r0= (1+e)
(1−e)
Solving for e: e = (ra/r0)−1(ra/r0)+1
ra
r0= 7080
6780= 1.044
e = 0.022
h is the angular momentum/unit mass = r0v0. From eq. 2.69
e = h2
GMr0− 1 =
r0v2
0
GM− 1
The velocity to go into elliptical orbit is
v2 =√
GMR1
(1 + e) = 7.754 × 103 m/s
168
Since angular momentum is concerned in elliptical orbit
v3ra = v2r0, or v3 = v267807080
= 7.411 × 103 m/s
Therefore the first boost v1 → v2 is 7.670 × 103 m/s or 7.754 × 103 m/s
The second boost v3 → v4 is
7.411 × 103 → 7.506 × 103 m/s.
2.91 From Problem 2.89
R4 = 42, 170 km
v4 = 3.075 × 103 m/s
From Problem 2.90
R1 = 6780 km,
v1 = 7.670 × 103 m/s
e = R4/R1−1R4/R1+1
= 0.723
v2 =√
GMR1
(1 + e) = 10.070 × 103 m/s
v3 = R1
R4v2 = 1.619 × 103 m/s
Therefore v1 → v2: 7.670 × 103 → 10.070 × 103 m/s
v3 → v4 : 1.619 × 103 → 3.075 × 103 m/s
2.92 For an elliptical orbit
e = 0.12 vp = 4000 m/s1r0
= GMh2 [1 + e], h = r0v0
r0 = GMv2
0
(1 + e), M = 5.98 × 1024 kg, G = 66.7 × 10−r m2/kg · s2
r0 = 5.98×66.7×1012
16×106 (1 + 0.12) = 27.92 × 106 m
Therefore k = 111.68 × 109 m2/s
At the apogee1ra
= GMh2 (1 − e)
1ra
= 5.98×66.7×10!2
111.683×1018 (0.88) = 2.8 × 10−8
ra = 35.7 × 106 m va = kra
= 3128 m/s
Altitude = ra − re = (35.7 − 6.378) × 106 = 29, 332 km
169
2.93 Solution:
For a parabolic arc: e = 11r
= GMk2 (1 + e cos θ)
r = 200, 000, 000, θ = 140◦
cos 140◦ = −0.766
Therefore h2 = 2 × 108 × 5.98 × 1024 × 66.7 × 10−12(1 − 0.766)
h = 136.6 × 109
d = hv
= 136.6×109
2.222×103
v
d
FIGURE S2.93
d = 23, 395 km
Re = 6380 km
Altitude = d − Re = 17, 015 km
2.94 Solution:
ra = 6380 × 103 + 392.62 × 103 = 6772.62 × 103 m
rp = 6380 × 103 + 385.42 × 103 = 6765.42 × 103 m
1rp
= GMh2 [1 + e], e = h2
GMr0
− 1
τ = 92.34 · 60 = 5.54 × 103 seconds
h = πτ(ra + rp)
√rarp = 5.196 × 1010
e = h2
GMrp− 1 = 5.108 × 10−4
2.95 Final velocity (Mir’s velocity at the perigee), using values from 2.94:
v4 =√
GMrp
(1 + e) = 7.687 × 103 m/s
R1 = 6380 + 100 = 6480 km
v1 =√
MGR1
=√
5.98×66.76.480
× 103 = 7.846 × 103 m/s
170
The Hohmann transfer-elliptic intercept orbit
R1 = 6480 km, rp = 6765 km
e2 =rpR1
−1rpR1
+1= 0.022
The velocity to go into this orbit is
v3 =√
GMR1
(1 + e2) = 7932 × 103 m/s
Conservation of momentum:
v3 = 7932 × 103 × R1
rp= 7598 × 103 m/s
v1 → v2 7.846 → 7.932 km/s
v3 → v4 7.598 → 7.687 km/s
2.96 Solution: v4 =√
MGR4
, R4 = 6380 + 150 = 6530 km
v4 = 7816 m/s
R1 = 6765 km v1 = 7687 m/s
rpra
FIGURE S2.96
e =4arp
−1rarp
+1, ra = 6765, rp = 6530 ra
rp= 1.036,
e = 0.018,
v2 to go into elliptic transfer orbit
v2 =√
GMR
(1 − e) = 7618 m/s
v3 = 7618 × ra
rp= 7892 m/s
Retrofiring
v1 → v2 7687 → 7618 m/s
v3 → v4 7892 → 7816 m/s
171
2.97 Solution:
mg
fw
Nw
N b
fb
β
FIGURE S2.97
w = −mgk
Nω = −NωeR
fω = +µNω(−eeθ cos β + k sin β)
Nb = Nb(eθ sin β + cos βk)
fb = µNb(−eθ cos β + k sin β)
The equations of motion are
−Nω = −mrθ2
−µNω cos β + Nb sin β − µNb cos β = mrθ
−mg + µNω sin β + Nb cos β + µNb sin β = mz
Constraint: a = a(+ cos βeθ − sin βk) − rθ2er
−µmrθ2 cos β + Nb(sin β − mu cos β) = ma cos β (1)
−mg + µmkθ2 sin β + Nb(cos β + µ sin β) = −ma sin β (2)
Mulp (1) by sin β and (2) by cosβ and add.
−mg cos β + Nb(cos2 β − µ sin β cos β + sin2 β − µ sin β cos β) = 0
Therefore Nb = mg cos β1−2µ sinβ cos β
a = 1m
(−µmrθ2 + mg(sin β−µ cos β)1−2µ sinβ cos β
)
Therefore
θ = α = 1mr
[
−µmrθ2 cos β + mg cos β(sin β−µ cos β)1−2µ sinβ cos β
]
z = 1m
[
−mg + µmrθ2 sin β + mg cos β(cos β+µ sin β)1−2µ sinβ cos β
]
z = +µrθ2 sin β − g sinβ(sin β−µ cos β)1−2µ sinβ cos β
The Mathcad solution:
172
i . .0 5000
∆t 0.001
t .i ∆t
g 9.81
β .30 deg
µ 0.1r 0.5
α ω .1
r....µ r ω ω cos β ..g cos β
sin β .µ cos β
1 ...2 µ sin β cos β
a ω ....µ r ω ω sin β ..g sin βsin β .µ cos β
1 ...2 µ sin β cos β
0 1 2 3 4 55
0
5Velocity in z direction
time s
vi
.r ωi
ti
0 2 4 60
20
40
θi
ti
vi 1
zi 1
ωi 1
θi 1
vi
.a ωi
∆t
zi
.vi
∆t
ωi
.α ωi
∆t
θ .ωi
∆ti
v0
z0
ω0
θ
0
0
0
00
i
173
2.98 We are given:
vθ = rθ, r = 1.5, h = 50, vθ0= 20
1.5
a = (r − rθ2)er + (2rθ + rθ)eθ + zk
mg
N
FIGURE S2.98
F = −N ex − mgk
r = r = 0
N = mrθ2
0 = rθ
−mg = mz
θ = 0, θ = θ0, θ = θ0t
z = −g, z = −gt, z = −12ft2 + z0
The following MATLAB code plots this:
clear allformat short eformat compact
t = linspace(0,2);
g = 9.81;
v0 = 20;R = 1.5;w0 = v0/R;
theta = w0*t;z = -(g/2)*t. 2;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
t
i
i
100
ω20
1.5
x .1.5 cos .ω ti
y .1.5 sin .ω ti
zi
..1
29.81 t
i
2 50
3-D trajectory
1 0 110
1
35
40
45
50
,,x y z
175
2.99 Solution:
v =
0r · ωvz
f = −µNv/|v|f = −µmrθ2√
r2ω2+v2g
(rωeθ + vzk)
mrθ = − µmrθ2√r2ω2+v2
3
rθ, z = −g − µrθ2vz√r2ω2+v2
z
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) 2+zdot 2);vprime(1) = thdot;yprime(2) = zdot;yprime(3) = -mu*R*thdot 3/v;yprime(4) = -mu*R*thdot 2*zdot/v-g;
return
176
The Mathcad code follows:
i . .0 390
∆t 0.01
t .i ∆t g 9.81 µ 0.2 r 1.5
α ,vz ω..µ r ω
3
.r ω2
vz2
a ,vz ω g...µ r ω
2vz
.r ω2
vz2
vz0
z0
ω0
θ
0
50
20
1.5
0
vzi 1
zi 1
ωi 1
θi 1
vzi
.a ,vzi
ωi
∆t
zi
.vzi
∆t
ωi
.α ,vz
i
ωi
∆t
θ .ωi
∆t
n . .0 391
x .r cos θn
y .r sin θn
zn
zn
3-D Trajectory
1 0 1101
0
20
40
,,x y z
177
2.100 This follows 2.99 and the codes are the same with different initial condition.The Mathcad version follows:
i . .0 390
∆t 0.01
t .i ∆t g 9.81 µ 0.2 r 1.5
α ,vz ω..µ r ω
3
.r ω2
vz2
a ,vz ω g...µ r ω
2vz
.r ω2
vz2
vz0
z0
ω0
θ
20
30
20
1.5
0
vzi 1
zi 1
ωi 1
θi 1
vzi
.a ,vzi
ωi
∆t
zi
.vzi
∆t
ωi
.α ,vz i ωi
∆t
θ .ωi
∆t
n . .0 391
x .r cos θn
y .r sin θn
zn
zn
0
i
n
n
178
2.101 We have vθ = v0 cos θ and vz = v0 sin β
mg
N
yz
x
voβ
θ
FIGURE S2.101
N = −N eR
W = mg(cos θer − sin θeθ)
mz = 0 (1)
−mrθ2 = −N + mg cos θ (2)
mrθ = −mg sin θ (3)
Eq. (3) may be written
ω dωdθ
= −g/r sin θ
Integrating yields:ω2
2|ωω0
= g/r cos θ|θ0ω2 − ω2
0 = 2gr(cos θ − 1) (4)
From Eq. 2, minimum velocity can be determined when θ = 180◦ and N = 0.The angular velocity ω2
180 = g/r. From (4)
g/r − ω20 = −7g
rTherefore ω0 =
√
5gr
vθ0=
√4gr = v0 cos β, v0 =
√5gR
cos β.
2.102 From the previous problem the forces are
N = −N er
W − mg(cos θer − sin θeθ)
f = −µN√(rω)2+vz2
(rωeθ + vzez)
rω0 = v0 cos β
vz0 = v0 sin β
−N + mg cos θ = −mrω2
179
mz = −µNvz/√
(rω)2 + vz2
mrθ = −mg sin θ − µNrω/√
(rω)2 + vz2
From 2.101 ω =√
g/r when θ = 180◦
N = mg cos θ + mrω2
Therefore z = −µ(g cos θ+rω2)vz√(rω)2+vz2
θ = −rrsin θ − µ(g cos θ+rω2)ω√
(rω)2+(vz)2
The minimum initial velocity of 10.6 m/s was determined by trial and errorusing the following codes. The motion damps out after 4 seconds and theparticle 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) 2+zdot 2);N = m*g*cos(th)+m*R*thdot 2;
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
t .i ∆t
µ 0.2 β .50 deg g 9.81 v0 10.6 r 0.6
V ,vz ω .r ω2
vz 2
a ,,vz ω θ ..µ .g cos θ .r ω2 vz
V ,vz ω
α ,,vz ω θ .g
rsin θ ..µ .g cos θ .r ω
2 ω
V ,vz ω
vz0
z0
ω0
θ
.v0 sin β
0
.v0cos β
r
0
vzi 1
zi 1
ωi 1
θi 1
vzi
.a ,,vzi
ωi
θi
∆t
zi
.vz
i
∆t
ωi
.α ,,vz ωi
θ ∆t
θi
.ωi
∆t
ω ming
r
=ω min 4.044
0
i
i
i
181
0 100 200 300 400 50010
0
10
20
angular position degrees
ωi
θ i
deg
0 1 2 3 4 50
200
400
600Angle vs time
time s
θi
deg
ti
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:
3
β
x
y
N
mg
θ
FIGURE S2.103
N = −N er
W = mg(cos β sin θer + cos β cos θeθ + sin βk)
The equations of motion are:
−N + mg cos β sin θ = −mrθ2
mg cos β cos θ = mrθ
mg sin β = mz
The data is:m = 2 kg θ(0) = 0r = 1 mβ = 20◦
The z equation can be quickly integrated:
z = g sin βz = 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
t .i ∆t
β .20 degr 1g 9.81
α θ ..g
rcos β cos θ
ω0
θ
0
0
ωi 1
θi 1
ωi
.α θi
∆t
θ .ωi
∆t
n . .0 2500
zn
..g sin βtn
2
2
0 0.5 1 1.5 2 2.5100
0
100
200Angular position vs time
time s
θi
deg
ti
0 0.5 1 1.5 2 2.50
5
10
15Position down sluiceway
time s
z i
ti
0
i
i
184
2.104 We are given: µk = 0.3
Friction acts in a direction opposite to the velocity
v = rωeθ + vzk
N = −N er
W = mg(cos β sin θer + cos β cos θeθ + sin βk)
f = −Nµ v|v|
The equations of motion are:
−N + mg cos β sin θ = −mrω2
mg cos β cos θ − µN rω√(rω)2+vz2
= mrθ
mg sin β − µN vz√(rω)2+vz2
= mz
Therefore N = m(g cos β sin θ + rω2) and
α(vz, ω, θ) = 1r
[
g cos β cos θ − µNm
rω√(rω)2+vz2
]
a(vz, ω, θ) = g sin β − µNm
vz√(rω)2+vz2
.
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) 2 + zdot 2);N = m*g*cos(beta)*sin(th)+m*R*thdot 2;
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 . .0 6000
∆t 0.001
t .i ∆t
β .20 deg r 1 g 9.81
N ,ω θ ..g cos β sin θ .r ω2
α ,,vz ω θ .1
r..g cos β cos θ ..µ N ,ω θ
.r ω
.r ω2
vz2
a ,,vz ω θ .g sin β ..µ N ,ω θvz
.r ω2
vz2
vz0
z0
ω0
θ
0
0
0
0
vzi 1
zi 1
ωi 1
θi 1
vzi
.a ,,vzi
ωi
θi
∆t
zi
.vz ∆t
ωi
.α ,,vz ωi
θi
∆t
θi
.ωi
∆t0
i
i
i
0 1 2 3 4 5 60
100
200
time s
θi
deg
ti
0 1 2 3 4 5 60
10
20Position along Sluiceway
time s
z i
ti
186
2.105 We have m = 2 kg, R = 1.5 m, v(0) = 10eθ, φ(0) = 90
θ
φ
N
mg
FIGURE S2.105
The forces are:
N = −N eR
W = mg(sin φeφ − cos φeR)
The coordinate accelerations are
aR = R − Rφ2 − R sin2 φθ2
aφ = Rφ + 2Rφ − R sin φ cosφθ2
aθ = R sin φθ + 2Rθ sin φ2Rφθ cos φ
Therefore −N − mg cos φ = −mR(φ2 + sin2φθ2)
mg sin φ = mR(φ − sin φ cosφθ2)
0 = R sin θθ + 2Rφθ cos φ
From sample problem 2.22, the last equation
θ = sin φ0vθ0
R sin2 φ= 10
R sin2 φ
φ = gR
sin φ +cos φv2
θ0sin2 φ0
R2 sin3 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) 2);yprime(3) = -(g/R)*sin(ph) + 100*cos(ph)/(R 2*sin(ph) 3);
return
187
The Mathcad code is (first define R and g in the code):
∆t 0.001
t .i ∆t
ddφ φ .gsin φ
R
.100 cos φ
.R 2 sin φ3
dφ0
φ0
0
.90 deg
dφi 1
φi 1
dφi
.ddφ φi
∆t
φi
.dφi
∆t
dθ φ10
.R sin φ2
θ 0
θi 1
θ .dθ φi
∆t
x ..R sin φi
cos θ
y ..R sin φi
sin θ
zi
.R cos φi
=φ0
1.571
3-D Trajectory
10
11
0
1
0.40.3
0.20.1
0
,,x y z
i
i
i
i
0
i
i
188
2.106 We are given: R = 1.5, θ0 = 101.5
, θ0 = 0, φ0 = 90◦
v = Rφeφ + R sin φθeθ
N = −N eR
W = mg(sin φeφ − cos φer)
f = −µN v
|v|
−N − mg cos φ = −mR(φ2 + sin2 φθ2)
mg sin φ − µN Rφ|v| = mR(φ − sin φ cos φθ2)
−µN R sinφθ|v| = mR(sin φθ + 2φθ cos φ)
The mass drops out.
N/m = R(φ2 + sin2 φθ2) − g cos φ
Therefore
φ = rR
sin φ − µR(φ2 + sin2 φθ2) φ|v| + sin φ cos φθ2
θ = 1sin φ
[
−µNm
sinφθv
− 2φθ 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(phdot 2+sin(ph) 2*thdot 2);N = g*cos(ph)+v 2/R;
yprime(1) = y(3);yprime(2) = y(4);yprime(3) = - (g/R)*sin(ph) -mu*Nphdot/v+sin(ph)*cos(ph)*thdot 2;yprime(4) = -mu*N*thdot/v-2*phdot*thdot/tan(ph);
return
189
The Mathcad code is:
R 1.5 g 9.81 µ 0.3i . .0 4000
∆t 0.001
t .i ∆t
N ,,dφ φ dθ .R dφ2 .sin φ
2dθ .g cos φ
v ,,dφ φ dθ .R dφ2 ..R sin φ dθ
2
ddφ ,,dφ φ dθ .g
Rsin φ ..µ N ,,dφ φ dθ
dφ
v ,,dφ φ dθ..sin φ cos φ dθ
2
ddθ ,,dφ φ dθ .1
sin φ...µ N ,,dφ φ dθ sin φ
dθ
v ,,dφ φ dθ...2 dφ dθ cos φ
dφ0
φ0
dθ0
θ
0
.90 deg
10
1.5
0
dφi 1
φi 1
dθi 1
θi 1
dφi
.ddφ ,,dφi
φi
dθi
∆t
φi
.dφi
∆t
dθ .ddθ ,,dφi
φi
dθi
∆t
θ .dθi
∆t
x ..R sin φi
cos θ
y ..R sin φi
sin θ
zi
.R cos φi
i
i
i
0
i
i
i
i
2.107 Solution:
z
yx
xz
y
_ R
N
mg
ψ
ψ
e
w_
ϕ
ϕ
FIGURE S2.107
190
N = −N eφ
W = mg(− cos φee + sin φeφ)
f = −µN v
|v|
aR = R − Rφ2 − R sin2 φθ2
aφ = Rφ + 2Rφ − R sin φ cos φθ2
aθ = R sin φθ + 2Rθ sin φ + 2Rφθ cos φ
φ = const
v = ReR + R sin φθeθ
−mg cos φ − µN Rv
= m(R − R sin2 φθ2)
−N + mg sin φ = −mR sin φ cos φθ2
−µN R sinφθv
= m(R sin φθ + 2Rθ sin φ)
N/m(dθ, R) = g sin φ + R sin φ cos φθ2
v(dR, R, dθ) =√
dR2 + (R sin φdθ)2
R = −g cos φ + R sin2 φθ2 − µNm
R/v
θ = −µ vm
θv− 2Rθ
R
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(rdot 2+(r*sin(phi)*thdot) 2);N = g*sin(phi)+r*sin(phi)*cos(phi)*thdot 2;
yprime(1) = rdot;yprime(2) = thdot;yprime(3) = -mu*N*rdot /v - g*cos(phi)+r*sin(phi) 2*thdot 2;yprime(4) = -mu*N*thdot/v - 2*rdot*thdot/r;
return
191
The Mathcad code is:
i . .0 6000
∆t 0.001
t .i ∆t
φ .30 deg
g 9.81 µ 0.0
R02
cos φN ,dθ R .g sin φ ...R sin φ cos φ dθ
2
v ,,dθ dR R dR2 ..R sin φ dθ2
ddR ,,dθ dR R .g cos φ ..R sin φ2
dθ ..µ N ,dθ RdR
v ,,dθ dR R
ddθ ,,dθ dR R ..µ N ,dθ Rdθ
v ,,dθ dR R.2
.dR dθ
R
dR 0
R0
dθ
0θ
0
R0
10
.R0 sin φ
0
dRi 1
Ri 1
dθi 1
θi 1
dR .ddR ,,dθ
i
dR Ri
∆t
R
i
.dRi
∆t
dθ
i
.ddθ ,,dθ
i
dR
i
Ri
∆t
θ .dθi ∆t
x ..R sin φ cos θ
y ..R sin φ sin θ
zi
.Ri
cos θ
i
0
i
i
ii
i
i
i
i
i
ii
iii
Path of Motion
202
0
5
0
,,x y z
192
2.108 Use the analysis and codes of 2.107 with a new friction coefficient. The codesare essentially the same, so only the Mathcad plot is shown.
Path of Motion
05050
0
50
100
0
100
,,x y z
2.109 Solution:
mg
N
FIGURE S2.109
N = N eR
W = mg(− cos φeR + sin φeφ)
V = Rφeφ + R sin φθeθ
f = −µN v
|v|
N − mg cos φ = −m(Rφ2 + R sin2 φθ2)
mg sin φ − µN Rφv
= m(Rφ − R sin φ cosφθ2)
−µN R sinφθv
= m(R sin φθ + 2Rφθ cos φ)
193
v(dφ, φ, dθ) = R ·√
dφ2 + (sin(φ)dθ)2
N/m = g cos φ − R(φ2 + sin2 φθ2)
φ = gR
sin φ − µNm
φv
+ sin φ cosφθ2
θ = −µNm
θv− 2φθ cos φ
sinφ
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(phdot 2+(sin(ph)*thdot) 2);N = g*cos(ph) -v 2/r;
yprime(1) = phdot;yprime(2) = thdot;yprime(3) = -mu*N*phdot/v + (g/4)*sin(ph) +sin(ph)*cos(ph)*thdot 2;yprime(4) = -mu*N*thdot/v -2*phdot*thdot/tan(ph);
return
194
The Mathcad code is:
i . .0 900
∆t 0.001
t .i ∆t
R 6
g 9.81
µ 0.0
v ,,dφ φ dθ .R dφ2 .sin φ dθ
2
N ,,dφ φ dθ .g cos φ .R dφ2 .sin φ dθ
2
ddφ ,,dφ φ dθ .g
Rsin φ ..µ N ,,dφ φ dθ
dθ
v ,,dφ φ dθ..sin φ cos φ dθ
2
ddθ ,,dφ φ dθ ..µ N ,,dφ φ dθdθ
v ,,dφ φ dθ...2 dφ dθ
cos φ
sin φ
dφ0
φ0
dθ
0θ
0
0.5
6
.10 deg
4
.6 sin .10 deg
0
dφi 1
φi 1
dθi 1
θi 1
dφi
.ddφ ,,dφi
φi
dθi
∆t
φi
.dφi
∆t
dθ i.ddθ ,,dφ
iφ
idθ
i∆t
θ .dθi
∆t
i
i
0 200 400 600 800 10005
0
5
10
N ,,dφi
φi
dθi
in 866
=N ,,dφn
φn
dθ 3.95 103
=φ
866
deg41.541
=θ
deg75.735
The particle will fall off the surface when the normal force goes to zero and the positioncoordinates are shown.
n
866
195
2.110 Use the same codes and analysis as in 2.109. The MATLAB code becomes:
clear allformat short 3format 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. 2+(sin(ph).*thdot). 2);N = g*cos(ph( - v. 2/r;
figure(1),plot(t,N)
196
The Mathcad code is:
i . .0 2000
∆t 0.001
t .i ∆t
R 6
g 9.81
µ 0.2
v ,,dφ φ dθ .R dφ2 .sin φ dθ
2
N ,,dφ φ dθ .g cos φ .R dφ2 .sin φ dθ
2
ddφ ,,dφ φ dθ .g
Rsin φ ..µ N ,,dφ φ dθ
dθ
v ,,dφ φ dθ..sin φ cos φ dθ
2
ddθ ,,dφ φ dθ ..µ N ,,dφ φ dθdθ
v ,,dφ φ dθ...2 dφ dθ
cos φ
sin φ
dφ0
φ0
dθ
0θ
0
0.5
6
.10 deg
4
.6 sin .10 deg
0
dφi 1
φi 1
dθi 1
θi 1
dφi
.ddφ ,,dφi
φi
dθi
∆t
φi
.dφi
∆t
dθ i.ddθ ,,dφ
iφ
idθ
i∆t
θ .dθi
∆ti
0 500 1000 1500 200010
0
10
N ,,dφi
φi
dθi
in 1816
=N ,,dφn
φn
dθ 0.011
=φ
n
deg51.168
=θ
n
deg113.121
The mass stays on the surface for a greater length of time when friction is present. If thefriction is too great, the particle will not leave the surface before it stops.
n
-
197
2.111 The data is:
y x
zmg
ϕ
θ
FIGURE S2.111
φ(0) = 30◦ θ(0) = 0 m = 3 kg R(0) = 2 vθ = 0.5, therefore θ(0) = 0.5.
Fs = −k(R − R0)eF
W = mg(cos φeR − sin φeφ)
The equation of motion become:
m[R − Rφ2 − R sin2 φθ2] = mg cos φ − k(R − 2)
m[Rφ + 2Rφ − sin φ cos φθ2] = −mg sin φ
m[R sin φθ + 2Rθ sin φ + 2Rφθ cos φ] = 0, or:
R = g cos− km
(R − 2) + Rφ2 + R sin2 φθ2
φ = − gR
sin φ − 2RφR
+ sin φ cos φθ2
θ = −2RθR
− 2φθ cos φsinφ
The last equation can be written1
R sin φddt
[R2 sin2 φθ] = 0
Therefore
R2 = sin2 φθ = h
h = R20 sin2 φ0θ0 = R0 sin φ0vθ0
For this case h = 2 sin 30◦(0.5) = 0.5
θ = hR2 sin2 φ
The differential equations may be written as
R = g cos φ − km
(R − 2) + Rφ2 + R sin2 φh2
R4 sin4 φ
R = g cos φ − km
(R − 2) + Rφ2 + h2
R3 sin2 φ
φ = − gR
sin φ − 2RφR
+ sin φ cos φh2
R4 sin4 φ
198
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)) 2;
yprime(1) = rdot;yprime(2) = phdot;yprime(3) = thdot;yprime(4) = -(k/m)*(r-R)+g*cos(ph) + r*phdot + 4*sin(ph) 2*thdot 2;yprime(5) = -(g/r)*sin(ph) -2*(rdot/r)*phdot + sin(ph)*cos(ph)*thdot 2;
return
199
The Mathcad code is:
h 0.5 g 9.81 k 200 m 3
i . .0 5000
∆t 0.001
t .i ∆t
ddR ,,,dR R dφ φ .g cos φ .k
mR 2 .R dφ 2 h2
.R 3 sin φ2
ddφ ,,,dR R dφ φ .g
Rsin φ .2
.dR dφ
R
.cos φ h2
.R 4 sin φ3
dR0
R0
dφ0
φ0
0
2
0
.30 deg
dRi 1
Ri 1
dφi 1
φi 1
dR i.ddR ,,,dR
iR
i
dφi
φi
∆t
Ri
.dR
i
∆t
dφi
.ddφ ,,,dR
i
Ri dφi
φi
∆t
φi
.dφi
∆t
θ 0
θi 1
.h
.Ri
2 sin φi
2∆t θi
x ..R sin φi
cos θ yi
..R sin φi
sin θ zi
.Ri
cos φiiii i i
0
i . .0 5000
i
200
Path of motion
1
0
1
1
0.5
0
0.5
1
2
1
0
1
,,y x z
2 1 0 1 20.5
0
0.5
yi
xi
2 1 0 1 22.2
2
1.8
1.6
zi
xi
201
2.112 The MATLAB code is:
clear allformat short eformat 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 = (pi 2/36000)*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)*v 2*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 . .,0 10 6000
W 6000
g 32.2
θ ss
2000e
s
1000
β s .π
12sin
.π s
3000
dθ s1
2000
1
1000e
s
1000
dβ s .π
36000cos
.π s
3000
Γ s
..sin θ s cos β s dθ s ..cos θ s sin β s dβ s
..cos θ s cos β s dθ s ..sin θ s sin β s dβ s
.cos β s dβ s
v s 88
F n s ..W
gv s 2 Γ s
0 1000 2000 3000 4000 5000 60000
500
1000
Position ft
F n s
s
203
2.113 Solution:
a = 2, 0 ≤ s ≤ 3000
v =√
2as
v(3000) = 109.5 ft/s (75 mph)
a = −1 3001 ≤ s ≤ 6000∫ v109.5 vdv =
∫ s3000 ads
v2
2− (109.5)2
2= −s + 3000
v =√
109.52 − 2(s − 3000)
The Mathcad code follows:
s . .,0 10 6000
W 6000
g 32.2
θ ss
2000e
s
1000
β s .π
12sin
.π s
3000
dθ s1
2000
1
1000e
s
1000
dβ s .π
36000cos
.π s
3000
Γ s
..sin θ s cos β s dθ s ..cos θ s sin β s dβ s
..cos θ s cos β s dθ s ..sin θ s sin β s dβ s
.cos β s dβ s
v s .Φ 3000 s .4 s .Φ s 3001 109.52 .2 s 3000
F n s ..W
gv s 2 Γ s
0 1000 2000 3000 4000 5000 60000
500
1000
1500
Position ft
F n s
s
204
2.114 Using the analysis of 2.113 the MATLAB code is:
clear allformat short eformat 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 = (pi 2/36000)*cos(pi*s/3000);
avec = [...a.*cos(t).*cos(b)+v. 2.*(-sin(t).*cos(b).*t -cos(t).*sin(b).*b ),...a.*sin(t).*cos(b)+v. 2.*( cos(t).*cos(b).*t -sin(t).*sin(b).*b ),...a.*sin(b)+v. 2.*cos(b).*b ];
mag a = sqrt(sum((avec.*avec)’)’);
figure(1),plot(s,mag a)xlabel(’distance along road (ft)’)ylabel(’acceleration (ft/sˆ2)’)grid
205
The Mathcad code is:
s . .,0 10 6000
W 6000
g 32.2
θ ss
2000e
s
1000
β s .π
12sin
.π s
3000
dθ s1
2000
1
1000e
s
1000
dβ s .π
36000cos
.π s
3000
Γ s
..sin θ s cos β s dθ s ..cos θ s sin β s dβ s
..cos θ s cos β s dθ s ..sin θ s sin β s dβ s
.cos β s dβ s
v s .Φ 3000 s .4 s .Φ s 3001 109.52 .2 s 3000
a n s ..1 v s 2Γ s
a t s .Φ 3000 s 2 Φ s 3001
a s a n s 2 a t s 2
0 1000 2000 3000 4000 5000 60002
4
6
8Acceleration Magnitude vs position
Position ft
a s
s
206
2.115 Solution:
fN
mg
FIGURE S2.115
F = −mgk + N n − µt = ma
n = Γ|Γ| f = −µN v
|v|
t = cos θ(s) cos β(s)i + sin θ(s) cos β(s)j + sin β(s)k]
N = mgk · n + m · v2
p
a(s) = −mgk·t−µNm
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). 2);
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) + v 2*norm(Gvec);
yprime(1) = y(2);yprime(2) = -g*(kvec’*tvec) - mu*N/m;
return
The Mathcad code is:
m 30
µ 0g 9.81
dθ sπ
20
θ s.π s
20
dβ s ..π
302
s
10
β s .π
31
s
10
2
Γ s
..sin θ s cos β s dθ s ..cos θ s sin β s dβ s
..cos θ s cos β s dθ s ..sin θ s sin β s dβ s
.cos β s dβ s
t s
.cos θ s cos β s
.sin θ s cos β s
sin β s
n sΓ s
Γ s
k
0
0
1
N ,v s ...m g k n s ..m v Γ s
a ,v s ..g k t s ..µ
mN ,v s
v
v
208
i . .0 1580
∆t 0.001
t .i ∆t
v0
s0
0
0
vi 1
si 1
v .a ,v
i
si
∆t
si
.v ∆t
=s1580
10.008
0 0.5 1 1.5 20
5
10
15Position on slide vs time
time s
s
i
ti
=v 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.
i
i i
1580
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). 2);
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) + v 2*norm(Gvec);
yprime(1) = y(2);yprime(2) = -g*(kvec’*tvec) - mu*N/m;
return
210
The Mathcad code is:
m 30
µ 0.2
g 9.81
dθ sπ
20
θ s.π s
20
dβ s ..π
302
s
10
β s .π
31
s
10
2
Γ s
..sin θ s cos β s dθ s ..cos θ s sin β s dβ s
..cos θ s cos β s dθ s ..sin θ s sin β s dβ s
.cos β s dβ s
t s
.cos θ s cos β s
.sin θ s cos β s
sin β s
n sΓ s
Γ s
k
0
0
1
N ,v s ...m g k n s ..m v Γ s
a ,v s ..g k t s ..µ
mN ,v s
v
v
211
i . .0 1712
∆t 0.001
t .i ∆t
v0
s0
0
0
vi 1
si 1
v i.a ,v
i
si
∆t
si
.v ∆t
=s1712
10.001
0 0.5 1 1.5 20
5
10
15Position on slide vs time
time s
si
ti
=v 7.9661712
i
i
212
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