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CHAPTER EIGHT SOLUTIONS
CHAPTER EIGHT SOLUTIONS
8.12
Taking torques about the left end of the scaffold, we have
T1(0) - (700 N)(1 m) - (200 N)(1.5 m) + T2(3 m) = 0.
From which,
1
Then, from ΣFy = 0, we have
T1 + T2 - 700 N - 200 N = 0.
Since T2 = 333 N, we find
(a)
x
(a)
(b)
200 N
3m
V
8.16
2
2m
Tx = T⋅ cos60° = 0.5 T
3m
(c)
T
___________________________
See the diagram below:
Ty = T⋅sin60° = 0.866T
700 N
H
(b)
1m
1.5 m
T1 = 567 N.
8.14
700 N
T
T2 = 333 N.
200 N
80 N
If x = 1 m, then
Στleft end = (-700 N)(1 m) - (200 N)(3 m) - (80 N)(6 m) + (0.866T)(6 m). Equating this to
zero gives: T = 342 N.
From ΣFx = 0,
H = 0.5 T = 171 N.
From ΣFy = 0,
V = 980 N - 0.866T = 683 n.
If T = 900 N
Στleft end = (-700 N)(x) - (200 N)(3 m) - (80 N)(6 m) + (779.4 N)(6 m). Equating this to zero
and solving for x gives:
x = 5.13 m.
From Στ = 0 about the pivot shown, we have (Tsin30°)d - (200 N)d = 0
where d is the length of the beam.
The forces H, V, and Tx produce
Pivot
no torque about the pivot point.
30°
Solving for T, We find:
T=
400 N.
H
Tx
From ΣFx = 0, we have
d
H - Tcos30° = 0.
V
T = 400 N, so
H = 346 N
(to right).
From ΣFy = 0,
V + Tsin30° - 200 N = 0 , which yields
V = 0.
104
T
y
200 N
CHAPTER EIGHT SOLUTIONS
8.12 We choose the pivot point at the left end, and use Στ = 0.
-(700 N)(0.5 m) - (294 N)(1 m) + (T1sin40°)(2 m) = 0, which yields
T1 = 501 N.
T
700 N
2
Now use ΣFx = 0.
T1cos40° - T3 = 0.
0.5 m
With T1 = 501 N, this gives
T3 = 384 N.
T3
1m
1m
Finally, ΣFy = 0 gives
W = 294 N
T2 + T1sin40°- 994 N = 0.
From which, T2 = 672 N.
T1
T
1y
40°
200 N
T1x
8.13
F = Tension force in Deltoid Muscle,
t
F = Force exerted on arm by shoulder joint
s
F = F sin12°
ty
t
Arm
0.08 m
F
sx
θ
Ftx = F cos12°
t
Point O
(Shoulder Joint)
0.29 m
W
F
sy
Fsy
ΣτO = Ftsin 12°(0.08 m) -(41.5 N)(0.29 m) = 0.
From which Ft = 724 N (Tension in deltoid muscle)
ΣFy = 0 gives
- Fsy + Ftsin 12° - 41.5 N = 0, yielding,
ΣFx = 0 gives
Fsx = Ftcos 12° and
Fsx = 709 N.
Therefore,
and
F
sx
Fs
Fsy = 109 N.
(Fsx)2 + (Fsy)2 = 717 N,
Fsy
tanθ =
= 0.1539, or
θ = 8.75°.
Fsx
Fs =
8.14
Using the fact that the scaffold is in translational equilibrium, we have
ΣFy = TL - 750 N - 345 N - 500 N - 1000 N + TR = 2595 N, or
TL + TR = 2595 N
(1)
where TL and TR are the tensions in the left and right cables, respectively. Since the scaffold is
also in rotational equilibrium, we may write,
Στleft end = -(750 N)(1 m) - (345 N)(1.5 m) - (500 N)(2 m)
- (1000 N)(2.5 m) + (TR)(3 m) = 0,
or TR = 1589 N. Then equation (1) gives TL = 1006 N.
8.15
Choosing the pivot point at the point O shown, Στ = 0
becomes
(50 N)(7.5 cm) + T(0) - R(3.5 cm) = 0.
Thus, R = 107 N.
R
C = 50 N
3.5 cm
T
7.5 cm
Now, apply ΣFy = 0 to obtain
-50 N + T - 107 N = 0
and T = 157 N.
O
105
Pivot
CHAPTER EIGHT SOLUTIONS
Pivot
d
5
Ty
Tx
d
2
Cx d2
V
Cy
C
Fy
V
Fx
F
8.16 Let us first resolve all forces into components parallel to and perpendicular to the leg, as shown. Use
Στ = 0 about the pivot indicated.
Ty(d/5) - Cy(d/2) - Fy(d) = 0, where
d is the length of the lower leg.
Cy = C sinV = (30 N)sin40° = 19.3 N, and
Fy = FsinV = (12.5 N)sin40° = 8.03 N.
Thus, Ty = 88.5 N,
but Ty = T sin25°.
So, T = 209 N.
8.17
Στ)point o = 0 gives
(-1200 N)
L
3L
3L
cos 65°+(Tcos25°)(
sin65°)+(Tsin25°)(
cos65°)-(2000 N)(Lcos65°).
2
4
4
From which,
T = 1465 N.
From ΣFx = 0,
H = Tcos 25° = 1328 N (toward right).
From ΣFy = 0,
V = 3200 N - Tsin25°= 2581 N (upward).
Tx = Tcos25°
Ty = Tsin25°
L
3L/4
Ty
Tx
2000 N
L/2
8.18
1200 N
We call the tension in the cord at the left end of the sign, T1 and the
65°
H
tension in the cord near the middle of the sign, T2, and we choose our
Point O
pivot point at
V
the point where T1 is attached.
T1
T2
2
Στpivot = 0 = (-W)(0.5 m) + T2(0.75 m) = 0, so,
T2 = W
3
0.75 m
From ΣFy = 0, T1 + T2 - W = 0.
Substituting the expression for T2 and solving, we find
1
T1 = W.
3
0.5 m W
106
CHAPTER EIGHT SOLUTIONS
8.19
Use ΣFy = 0:
or
F1 - 500 N - 800 N = 0,
F1 = 1300 N.
Now, apply ΣFx = 0: f - F2 = 0,or f = F2.
The lever arm for the 500 N force is
(7.5 m)cos60° = 3.75 m
The lever arm for the 800 N force is
dcos60° = d/2 .
For F2, the lever arm is (15 m)sin60° = 13 m.
F2
(1)
(15 m)sin60°
(2)
800 N
F1
pivot
500 N
60° f
(7.5m)cos60°
d cos60°
Using Στ = 0 with pivot pt. at base of the ladder,
-(500
N)(3.75 m) - (800 N)(d/2) + F2(13 m) = 0
(3)
(a) When d = 4 m, Equation (3) gives:
F2 = 267 N
Equation (2) then gives f = 267 N, and
F1 = 1300 N from above.
(b) If d = 9 m: Equation (3) gives F2 = 421 N, Equation (2) yields f = 421 N,
and Equation (1) gives
F1 = 1300 N.
If the ladder is ready to slip, µs = f/F1 = (421 N)/1300 N = 0.324.
8.20
Use ΣFy = 0:
F1 - 200 N - 800 N = 0.
F1
Thus, F1 = 1000 N.
The friction force, f, at the base of the ladder is
f = µ sF 1
(8 m)sin50°
when the ladder is on the verge of slipping. Thus,
f=
F
1
0.6(1000 N) = 600 N.
800 N
Now use ΣFx = 0: f - F2 = 0, or F2 = f = 600 N.
pivot
200 N
50° f
Finally, use Στ = 0 with the pivot pt. at the base of the ladder.
The lever arm for the 200 N force is
(4mcos50°) =
(4 m)cos50°
2.57 m.
d cos50°
The lever arm for the 800 N force is
dcos50° = 0.643d, where d is the distance from the base of the ladder up to the position of the
person. Finally, the lever arm for the force F2 is
(8 m) sin50° = 6.13 m.
We have:
-(200 N)(2.57 m) - (800 N)(.643d) +(600 N)(6.13 m) = 0
giving, d = 6.2 m.
8.21
Total torque about the elbow gives
-(0.33 m)(2 kg)(9.8 m/s2) + FBcos75°(0.08 m) = 0, or
8.22
First realize that the normal force N supports the
entire weight of the body. Thus, N = w = 700 N.
Also, observe the diagram at right and see that α = 0
since T is perpendicular to the end of the rod in the
model. From the conditions for equilibrium, we
have ΣFx = 0 = -Tsinθ + Rsin15° = 0,
or
Tsinθ = Rsin15°
(1)
Also, ΣFy = 0 = N - Rcos15° + Tcosθ = 0,
or (with N = 700 N),
Tcosθ = Rcos15° - 700 N. (2)
FB = 312 N.
T
5 cm
1
N = W = 700 N
Στpivot = 0 ⇒
- N(18 cm)cosθ + (7
cm)T = 0, or with N = 700 N,
T = (1800 N)cosθ.
(3)
(1800 N)sinθcosθ
Substitute (3) into (1) to obtain
R=
.
(4)
sin15
107
7 cm
2
8 cm
α=θ
θ
R
15°
Choose
Pivot here
CHAPTER EIGHT SOLUTIONS
Now substitute (3) and (4) into equation (2) to find
(1800 N)sinθcosθcos15°
(1800 N)cos2θ =
- 700 N, which reduces to
sin15°
cos2θ + 0.3889 = 3.732sinθcosθ. Square both sides of the equation to obtain
cos4θ =
2
2
2
0.7778cos θ + 0.15123 = 13.9282sin θcos θ.
Using the identity
sin2θ = 1 - cos2θ, this reduces to
cos4θ - 0.8809cos2θ + 0.01013 = 0
(a quadratic equation in cos2θ).
2
The quadratic formula gives
cos θ = 0.86926, or cos2θ = 0.0117
From these, we find that cosθ = ± 0.93234 or cosθ = ± 0.108
which gives
θ = 21.2°, θ = 158.8°, θ = 83.8°, and θ = 96.2°. Of these solutions, θ = 21.2° is
the only one that is physically possible.
Then (with θ = 21.2°), equation (3) yields T = 1678 N, and equation (4) gives
R=
2345 N.
8.23
8.24
8.25
Considering the torques about the point at the bottom of the bracket yields:
F(0.06 m) = 0, so
F = 66.7 N.
(0.05 m)(80 N) -
⎛1.1x10-10 m⎞⎟ = 0.55 X 10-10 m from the center of mass, we have I = Σmr2,
Since each atom is ⎜
2
⎝
⎠
or
I = 2 x (2.32 X 10-26 kg)(0.55 X 10-10 m)2 = 1.4 X 10-46 kg m2.
Set L = 4 m, r = 0.5 m, M = 0.5 kg, m = 5 kg. Using the
+α
free-body diagram of the mass, we obtain (taking down as
I
positive),
r
ΣFy = mg - T = ma.
(1)
+a
The torque on the spool gives
1
Support
T
Στcenter = Tr = Iα = Mr2α, or
Force
2
1
T=
Mrα.
(2)
2
Since the string does not slip on the spool, rα = a, and equation (2) becomes,
T=
T
M
W = Mg
1
Ma.
2
Putting this into equation (1), gives,
1
mg
(5 kg)(9.8 m/s2)
mg - Ma = ma, or a =
=
= 9.33 m/s2.
2
M
(5 kg + 0.25 kg)
m+
2
1
Since the mass starts from rest,
L = voyt + at2, gives
2
2(4 m)
= 0.926 s as the time for the mass to fall 4 m.
9.33 m/s2
a 9.33 m/s2
Also, α = =
= 18.66 rad/s2, so at t - 0.926 s,
r
0.5 m
ω = ω0 + αt = 0 + (18.66 rad/s)(0.926 s) = 17.3 rad/s.
t=
8.26
(a)
(b)
(c)
2L
=
a
I = Σmiri2
First, apply the above equation about the x axis. We have,
Ix = (3 kg)(9 m2)+(2 kg)(9 m2)+(2 kg)(9 m2)+(4kg)(9 m2) = 99.0 kg m2.
About the y axis, we have
Iy = (3 kg)(4 m2) + (2 kg)(4 m2) + (2 kg)(4 m2) + (4kg)(4 m2) = 44.0 kg m2.
The distance, r, (from an axis through O and perpendicular to the page) out to each of the
masses is found from the pythagorean theorem: r =
108
(2 m)2 + (3 m)2 =
13 m2 ,
CHAPTER EIGHT SOLUTIONS
and the moment of inertia is
I0 = (3 kg)(r2) + (2 kg)(r2) + (2 kg)(r2) + (4kg)(r2), or
= (11 kg)(13 m2) = 143 kg m2.
8.27
We use τ = Iα .
τx = Ixα = (99 kg m2)(1.5 rad/s2) = 149 N m
τy = Iyα = (44 kg m2)(1.5 rad/s2) = 66.0 N m
τ0 = I0α = (143 kg m2)(1.5 rad/s2) = 215 N m
109
CHAPTER EIGHT SOLUTIONS
8.28
We have already found (problem 8.26) Iy = 44.0 kg m2, and we know
τ
(20 N m)
τ = Iα. Thus,
α= =
= 0.455 rad/s2, and
I
(44 kg m2)
ω = ω0 + αt = 0 + (0.455 rad/s2)(3 s) = 1.36 rad/s.
8.29
Let θ = 30°, ω0 = 60.0 rad/s and R = 0.03/2 = 0.015 m.
The forces acting on the coin parallel to the incline give
ΣFparallel = - f - mgsinθ = ma = mRα,
(1)
where f is the friction force between the coin and incline. The net torque about the coin's center
of mass is
1
1
Στcenter = fR = Iα = MR2α, or
f = MRα.
2
2
1
Equation (1) then becomes
MRα- mgsinθ = mRα, yielding
2
2gsinθ
2(9.8 m/s2)sin30°
α=== -217.8 rad/s2.
3R
(3)(0.015 m)
Therefore
a = rα = (0.015 m)(-217 rad/s2) = - 3.27 m/s2.
2
2
Then v = v 0 + 2as, with v0 = rω0 = (0.015 m)(60 rad/s) = 0.9 m/s
and v = 0, gives: 0 = (0.9 m/s)2 + 2(-3.27 )s. This gives
s = 0.124 m as the distance the coin goes up the incline.
0.25
= 0.125 m, and L = 4 m. A study of the forces parallel to the incline yields,
2
ΣF|| = mgsinθ - f = ma (1)
where f is the friction force between sphere and incline. The net torque about the center of the
sphere gives
2mR2 ⎛ a ⎞
a
Στcenter = fR = Iα =
, (α = , sphere rolls without slipping)
3 ⎝R⎠
R
2ma
so that f =
. Therefore equation (1) becomes,
3
2ma
3gsinθ
mgsinθ = ma, yielding a =
= 2.94 m/s2.
3
5
If the sphere starts from rest, then its center of mass moves a distance
1
2L
8
L = at2, which gives t =
=
= 1.7 s as the required time.
2
a
2.94
8.30
Let θ = 30°, R =
8.31
(a)
(b)
1
1
mr2 = (0.85 kg)(4 X 10-2 m) 2 = 6.8 X 10-4 kg m2.
2
2
τnet = Iα = (6.8 X 10-4 kg m2) (66 rad/s2) = 4.49 X 10-2 kg m2.
The torque exerted by the fish = Fr, so the net torque is
τnet = Fr - 1.3 Nm = 4.49 X 10-2 kg m2.
1.345 Nm
From which,
F=
= 34 N.
4 X 10-2 m
1
1
θ = ω0t + αt2 = 0 + (66 rad/s2)(0.5 s) 2 = 8.25 rad, and
2
2
-2
s = rθ = (4 X 10 m)(8.25 rad) = 0.33 m = 33 cm.
I=
110
CHAPTER EIGHT SOLUTIONS
8.32
8.33
1
1
mr2 = (100 kg)(0.5 m) 2 = 12.5 kg m2,
2
2
ω0 = 50 rev/min = 5.24 rad/s
ω - ω0 0 - 5.24 rad/s
α=
=
= - 0.873 rad/s2
t
6s
τ = Iα = (12.5 kg m2)(-0.873 rad/s2) = -10.9 Nm
Also, the magnitude of the torque is given by fr = 10.9 N m, where f is the force of friction.
10.9 Nm
Therefore,
f=
= 21.8 N, and
0.5 m
f
21.8 N
f = µkN yields
µk = =
= 0.31.
N
70 N
I=
We first calculate the moment of inertia as
1
1
I = MR2 = (150 kg)(1.5 m) 2 = 168.8 kg m2.
2
2
Then we note that ωf = 0.5 rev/s = π rad/s, and calculate α as,
ω = ω0 + αt
or π rad/s = 0 + α (2 s). Thus,
Then τ = Iα becomes Fr = Iα, or F =
8.34
8.35
α = π/2 rad/s2.
Iα (169 kg m2)(π/2 rad/s2)
=
= 177 N.
r
1.5 m
The moment of inertia of the pulley is
1
1
I=
MR2 = (5 kg)(0.6 m) 2 = 0.90 kg m2.
2
2
First we apply Newton's second law to the falling bucket of mass 3 kg
(weight 29.4 N). We have
29.4 N - T = (3 kg)a.
(1)
Now apply τ = Iα to the pulley. We have
a
I a 0.90 kg m2
TR = Iα = I ⎛ ⎞ . Thus, T = 2 =
a. (2)
⎝R⎠
R
(.6 m)2
We can solve (1) and (2) simultaneously to find a. We find,
(a) a = 5.35 m/s2 downward.
1
(b) To find how far it drops, we use y = v0t + at2.
2
1
y = 0 + (5.35 m/s2)(4 s) 2 = 42.8 m
2
a 5.35 m/s2
(c) α = =
= 8.91 rad/s2.
R
0.6 m
The resultant torque is given by
(120 N)(0.811 m) - (100 N)(0.811 m) = 16.2 N m
1
1
The moment of inertia is: I = mr2 = (2.1 kg)(0.811 m) 2 = 0.691 kg m2.
2
2
τ
16.2 Nm
Then, τ = Iα gives
α= =
= 24 rad/s2.
I 0.691 kg m2
111
R
T
T
a
3 kg
29.4 N
CHAPTER EIGHT SOLUTIONS
8.36
The free body diagram is shown at the right.
The resultant torque about the center of the pulley is
(1.14 X 104 N)r - T1r = (79.8 kg m4)α
T1 - 9800 N = (1000 kg)a
(2)
Since there is no slipping, we also have
a = rα = (0.762 m)α.
Solving (1), (2), and (3) simultaneously,
we find a = 1.41 m/s2.
8.37
8.38
(3)
(a)
Newton's second law for the cylinder is
mg - T = ma .
(1)
1 2
(b) τ = Iα becomes:
Tr = mr α.
(2)
2
(c) Solve (2) for T and substitute α = a/r. We find:
mr2a ma
T= 2 =
.
(3)
2
2r
Finally, substitute (3) into (1) to find,
ma
2
mg = ma which gives
a = g.
2
3
(a)
(b)
r
(1)
From Newton's second law, applied in the vertical
direction, we have
r = 0.762 m
α
T
R
1
4
T2 =1.14 x10 N
T1
1000 kg
a
9800 N
T
r
a
mg
We first determine the angular acceleration due to the applied torque and the frictional torque
by writing
τapplied + τfriction = Iα1 with τapplied + τfriction = 36 N m.
Now in six seconds, ω changes from 0 to 10 rad/s.
Using ω = ω0 + α1t, we have 10 rad/s = 0 + α1(6 s), giving
α1 = 1.67 rad/s2. Then, 36 N m= (1.67 rad/s2)I, or I = 21.6 kg m2.
When the applied torque is removed, we have τfriction = Iα2.
Using ω = ω0 + α2t we have 0 = 10 rad/s + α2(60 s), so
α2 = -0.167 rad/s2. Therefore,
(c)
τfriction = (21.6 kg m2)(-0.167 rad/s2) = - 3.6 N m, and
⎥τfriction⎥ = 3.6 N.
The angular displacement during the first 6 s is
0 + 10 rad/s
θ1 = ω1ave(Δτ) =
(6 s) = 30 rad, and that during the last 60 s is
θ2 =
2
10 rad/s
ω2ave(Δτ)2 =
(60 s) = 300 rad.
2+0
θ = θ1 + θ2 = 30 rad + 300 rad = 330 rad = 52.5 rev.
112
CHAPTER EIGHT SOLUTIONS
8.39
The initial angular velocity of the wheels is zero, and the final angular velocity is
v 50.0 m/s
ωf = =
= 40.0 rad/s.
r
1.25 m
ωf - ω0
40.0 rad/s - 0
Thus, α =
=
= 83.3 rad/s2.
t
0.48 s
τcenter of a wheel = fr, so τ = Iα gives
Iα (110 kg m2)(83.3 rad/s2)
=
= 7.33 X 103 N, and
r
1.25 m
f
7.33 X 103 N
µk =
=
= 0.524.
N
1.4 X 104 N
f=
8.40
We begin by using v = Rω, where R is the radius of the sphere.
1
2
ω2 2MR2ω2
1
Then,
KErot = Iω2 =( MR2)
=
= Mv2.
2
5
2
10
5
The total kinetic energy is given by
1
1
7Mv2
KEtot = KErot + KEcm =
Mv2 +
Mv2 =
.
5
2
10
KErot
Then,
= (1/5)/(7/10) = 2/7.
KEtot
8.41
The moment of inertia of a long thin rod about an axis through one end is I =
1
ML2. The total
3
rotational kinetic energy is given as
1
1
KErot =
I ω2 +
I ω2 , with
2 h h 2 m m
m hL 2h
(60 kg)(2.7 m)2
Ih =
=
= 145.8 kg m2, and
3
3
m m L 2m
(100 kg)(4.5 m)2
Im =
=
= 675 kg m2.
3
3
(2π rad) 1 h
In addition,
ωh =
= 1.454 X 10-4 rad/s, while
(12 h) 3600 s
(2π rad) 1 h
ωm =
= 1.745 X 10-3 rad/s. Therefore,
(1 h) 3600 s
1
1
KErot = (145.8)(1.454 X 10-4) 2 + (675)(1.745 X 10-3 ) 2 = 1.03 X 10-3 J.
2
2
8.42
(a)
(b)
8.43
I = Σmr2 = (4 kg)(3 m)2 + (2 kg)(2 m)2 + (3 kg)(4 m)2 = 92.0 kg m2
1
1
KE = Iω2 = (92.0 kg m2)(2 rad/s) 2 = 184 J.
2
2
v4 = r4ω = (3 m)(2 rad/s) = 6 m/s, v2 = r2ω = (2 m)(2 rad/s) = 4 m/s,
v3 = r3ω = (4 m)(2 rad/s) = 8 m/s.
1
1
1
1
KE = Σ mv2 = (4 kg)(6 m/s) 2 + (2 kg)(4 m/s) 2 + (3 kg)(8 m/s) 2
2
2
2
2
KE = 184 J.
Using W = ΔKE = KEf - KEi =
2W
2(3000 J)
we get I = 2 =
ω
(200 rad/s)2
1
Iω2 - 0,
2
= 0.15 kg m2 .
113
CHAPTER EIGHT SOLUTIONS
8.44
The moment of inertia of the cylinder is
1
1
I = mr2 = (81.6 kg)(1.5 m) 2 = 91.8 kg m2, and
2
2
the angular acceleration of the merry-go-round is found as
α = τ/I = (F r)/I = (50 N)(1.5 m)/(91.8 kg m2) = 0.817 rad/s2.
At t = 3 s, we find the angular velocity , ω = ω0 + αt, to be
ω = 0 + (0.817 rad/s2)(3 s) = 2.45 rad/s, and
1
1
KE = Iω2 = (91.8 kg m2)(2.45 rad/s) 2 = 280 J.
2
2
1
1
mv2cm = Iω2, gives
2
2
2
1
1
2mR
1
1
mv2cm =
ω2 = m(Rω)2 = mv2t , where vt is the tangential speed of a point
2
2 5
5
5
on the equator of the ball. Cancelling the mass gives
v2cm 2
vcm
=
, or
= 0.63.
2
5
vt
v t
8.45
Setting KEtrans = KErot , or
8.46
(a)
(b)
(c)
8.47
1
1
mv2 = (10 kg)(10 m/s) 2 = 500 J.
2
2
1
1 1
v2
1
KErot = Iω2 = ( mr2)( 2 ) = (10 kg)(10 m/s) 2 = 250 J.
2
2 2
4
r
KEtotal = KEtrans + KErot = 750 J.
KEtrans =
Let I be the moment of inertia of the object about the point of rotation. Then τ = Iα so that τ
constant implies, α is constant.
Then
ω22 = ω12 + 2α(θ2 - θ1), or
ω22 - ω12 = 2αΔθ
1
1
1
Also, W = ΔKE = Iω22 Iω21 = I(ω22 - ω21) = Iα(Δθ) = τ(Δθ)
2
2
2
8.48 The forces on a rolling object on an incline of slope angle θ are the force of friction, f, the normal
force, N, and its weight, mg. We apply Newton's second law with the x axis along the plane, to
find,
mgsinθ - f = ma.
(1)
+Y
N
Now use, τ = Iα , or fr = Iα = I(a/r).
Therefore,
f = (I/r2)a.
(2)
Substitute into (1) for the frictional force f from (2) and solve the
resulting equation for a.
f
gsinθ
a=
.
I
+X
(1 + 2)
mr
θ
This is the linear acceleration of the center of mass of the rolling
W = mg
2
2
object. For a solid sphere, I = mr2 or (I/mr2) = .
5
5
From this, we see that a = gsinθ/1.4.
I
1
Using the same approach for a solid cylinder,
= , and a = gsinθ/1.5
2
2
mr
Finally, for a hollow ring, (I/mr2) =, and a = gsinθ/2.
Thus, we find asphere > acylinder > aring, so the sphere wins and the ring comes in last.
8.49
(a)
The angular velocity of the flywheel is 5000 rev/min = 524 rad/s, and its moment of inertia
1
1
is, I = mr2 = (500 kg)(2 m) 2 = 103 kg m2.
2
2
114
CHAPTER EIGHT SOLUTIONS
(b)
Therefore, the stored energy is
1
1
KE = Iω2 = (103 kg m2)(524 rad/s) 2 = 1.37 X 108 J.
2
2
A 10 hp motor is equivalent to 7460 W. Therefore, if energy is used at the rate of 7460 J/s,
the stored energy will last for
t = E/P = 1.37 X 108 J/7.46 X 103 J/s = 1.84 X 104 s = 5.10 h.
8.50
Work done = Fs = (5.57 N)(0.8 m) = 4.46 J, and
1
1
1
Work = ΔKE = Iωf2 Iω02 = Iωf2. (ω0 = 0 since top starts from rest.)
2
2
2
1
Thus, 4.46 J = (4 X 10-4 kg m2) ωf2 and from this,
ωf = 149 rad/s.
2
8.51
I=
2
2
mr2 = (5.98 X 1024 kg)(6.38 X 106 m) 6 = 9.74 X 1037 kg m2
5
5
2π rad
The intial period = 24 h = 86,400 s, and
ω0 =
= 7.27 X 10-5 rad/s.
86400 s
1
Therefore, the initial kinetic energy is KEi = Iω02 = 1.57 X 1029 J.
2
2π rad
The final period = 24 h + 1 min = 86,460 s, ωf =
= 7.267 X 10-5 rad/s.
86460 s
1
1
The energy available = ΔKE =
Iω02 Iωf2.
2
2
ΔKE = 3.57 X 1026 J, and time =
Substituting the appropriate values gives,
3.57 X 1026 J
=
= 1.8 X 106 y
2 X 1020 J/y
8.52
energy available
rate of energy use
Let us first consider conservation of energy, with the zero level for potential energy at the base of
the ramp. We have
1
1
1
1
mvi2 + Iωi2 + mgyi = mvf2 + Iωf2 + mgyf, or
2
2
2
2
1
1
mvi2 + Iωi2 + 0 = 0 + 0 + (mg)(s)sin20°, where s is the distance the hoop rolls up the
2
2
ramp. For a hoop, I = mr2. Thus,
Iωi2 = mr2ωi2 = m(rωi)2 = mvi2.
Therefore, our equation for conservation of energy becomes
1
1
mvi2 + mvi2 = (mg)(s)sin20°, or, solving for s, we have
2
2
s = vi2/gsin20°. Since vi = rωi = (3 m)(3 rad/s) = 9 m/s, we find
s = (9 m/s)2/(9.8 m/s2)sin20° = 24 m.
8.53
(a)
(b)
The angular velocity of the earth on its axis is
ω = 2π rad/day = 7.27 X 10-5 rad/s. We assume the earth is a uniform sphere of radius R
and calculate its moment of inertia as
2
2
I = mR2 = (5.98 X 1024 kg)(6.38 X 106 m) 2 = 9.74 X 1037 kg m2.
5
5
The angular momentum is found as
L = Iω = (9.71 X 1037 kg m2)(7.27 X 10-5 rad/s) = 7.08 X 1033 J s.
For the orbiting earth, ω = 2π rad/yr = 1.99 X 10-7 rad/s, and the angular momentum is given
by
L = Iω = mr2ω = (5.98 X 1024 kg)(1.49 X 1011 m)2 (1.99 X 10-7 rad/s),
L = 2.66 X 1040 J s.
115
CHAPTER EIGHT SOLUTIONS
8.54
The total angular momentum about the center point is given by
m hL 2h
(60 kg)(2.7 m)2
L = Ihω h + Im ω m
with
Ih =
=
= 145. 8 kg m2, and Im =
3
3
m m L 2m
(100 kg)(4.5 m)2
=
= 675 kg m2.
3
3
(2π rad) 1 h
In addition,
ωh =
= 1.454 X 10-4 rad/s, while
(12 h) 3600 s
(2π rad) 1 h
ωm =
= 1.745 X 10-3 rad/s. Thus,
(1 h) 3600 s
L = (145.8 kg m2)(1.454 X 10-4
rad
rad
) + (675 kg m2)(1.745 X 10-3
), or
s
s
L = 1.2 kg m2/s.
8.55
Let M and m be the mass of the cylinder and putty, respectively, and let R be the radius of the
cylinder. Then the total angular momentum before the putty hits the cylinder is
MR2
(10 kg)(1 m)2(7 rad/s)
Lbefore = I0ω0 =
ω0 =
= 35 kg m2/s.
2
2
After the putty hits the cylinder, we have
⎛MR2 + mr2⎞⎟ ω
Lafter = Iω + mr2(ω) = ⎜
⎝ 2
⎠
2
⎛(10 kg)(1 m) + (0.25 kg)(.9 m)2⎞⎟ ω = (5.20 kg m2)(ω).
= ⎜
2
⎝
⎠
Using Lbefore = Lafter , we have (5.20 kg m2)ω = 35 kg m2/s, so
ω = 6.73 rad/s.
8.56
We use conservation of angular momentum, with Ii = I, and If = 0.9I
I iω i
I
Thus, Ifωf = Iiωi gives ωf =
=
ω = 1.111ωi.
0.9I i
ωf
KEf - KEi⎞
The % change in kinetic energy = ⎛⎜
100%. Substituting appropriate values, gives
⎝ KEi ⎟⎠
%change in KE = 11.1%.
The kinetic energy has increased because she has had to do work to pull her arms inward.
8.57
First compute some needed constants:
mass of people = mp = (5.5 x 109)(70 kg) = 3.85 X 1011 kg
I(due to people) = Ip = mpR2E = (3.85 x 1011 kg)(6.37 x 106 m)2 = 1.56 x 1025 kg m2 .
I for Earth alone:
2
2
IE = MER2E = (5.98 X 1024 kg)(6.37 X 106 m) 2 = 9.71 X 1037 kg m2.
5
5
2π rad 1 day
Original angular speed of Earth: ω0 =
= 7.27 X 10-5 rad/s.
1 day 86400 s
The original angular momentum of the system (Earth plus people) is
L0 = IEω0 + Ipω0 = (IE + Ip)ω0
The tangential speed of a point on the equator of the Earth is v = REω so when the people start
running, their tangential speed is
v' = v + 2.5 m/s = REω + 2.5 m/s, and their angular speed is
v'
2.5 m/s
ω' =
=ω+
, where ω is the new angular speed of the Earth.
RE
RE
The angular momentum of the system is now given by
116
CHAPTER EIGHT SOLUTIONS
Lf = Learth + Lpeople = IEω + Ipω' = IEω + Ip(ω +
2.5 m/s
),or
RE
Ip
(2.5 m/s) .
RE
Conserving angular momentum(Lf = L0) gives:
Ip
(2.5 m/s)Ip
ω
ω = ω0 (2.5 m/s) , or
=1.
RE(IE + Ip)
ω0
ω0RE(IE + Ip)
Lf = (IE + Ip)ω +
ω
(2.5)(1.56 X 1025)
=1 ,
ω0
(7.27 X 10-5)(6.37 X 106)(9.71 X 1037 + 1.56 X 1025)
ω
= 1 - 8.67 X 10-16. Thus, the ratio of the length of the new day to the old day is
ω0
T
2π ω0
1
1
=
=
=
≈ 1 + 8.67 X 10-16, and
T0
ω 2π
ω
1 - 8.67 X 10-16
ω0
the length of the new day is T ≈ T0(1 + 8.67 X 10-16)
T = T0 + T0(8.67 X 10-16) = 24 h + (86,400 s)(8.67 X 10-16), or
T ≈ 24 h + 7.5 X 10-11 s. Therefore, the day is lengthened by approximately 7.5 X 10-11 s.
8.58
The total angular momentum is given by
Itotal = Iweights + Istudent = 2(mr2) + (3 kg m2)
Before: r = 1 m. Thus, Ii = 2(3 kg)(1 m)2 + (3 kg m2) = 9 kg m2.
After: r = 0.3 m. Thus, If = 2(3 kg)(0.3 m)2 + (3 kg m2) = 3.54 kg m2.
We now use conservation of angular momentum
Ifωf = Iiωi, or
Ii
9
(a) ωf = ωi =
(0.75 rad/s) = 1.9 rad/s.
If
3.54
1
1
(b) KEi = Iiωi2 = (9 kg m2)(0.75 rad/s) 2 = 2.5 J,
2
2
1
1
KEf = Ifωf2 = (3.54 kg m2)(1.91 rad/s) 2 = 6.4 J.
2
2
8.59
Using conservation of angular momentum, we have
Lapogee = Lperihelion, or (mr2a)ωa = (mr2p)ωp. Thus,
va
vp
(mr2a) = (mr2p)
, giving
ra
rp
rp
0.59 AU
rava = rpvp , or va =
v =
(54 km/s) = 0.91 km/s.
ra p 35 AU
117
CHAPTER EIGHT SOLUTIONS
8.60
(a)
The moment of inertia of the system is given by
1
1
I = Iman + Iwheel = mr2 + MR2 = (80 kg)r2 + (25 kg)(2 m) 2.
2
2
When the man is at a distance of r = 2 m from the axis, the equation above gives
Ii = 370
kg m2 , and when the man moves to a
point 1 m from the center (r = 1 m), the moment of inertia becomes,
If = 130 kg m2 . The initial angular velocity of the system is
ωi = 0.2 rev/s = 1.26 rad/s, and we find the final angular velocity via conservation of angular
momentum, as
Ii
370
ωf = ωi =
(1.26 rad/s) = 3.58 rad/s.
If
130
1
1
(b) The change in kinetic energy is KEf - KEi = Ifωf2 - Iiωi2 , or
2
2
1
1
ΔΚΕ = (130 kg m2)(3.58 rad/s) 2 - (370 kg m2)(1.26 rad/s) 2 = 540 J.
2
2
This difference results from work done by the man on the system as he walks inward.
8.61
I0 = mr02 = (0.12 kg)(0.4 m)2 = 1.92 X 10-2 kg m2
If = mrf2 = (0.12 kg)(0.25 m)2 = 7.5 X 10-3 kg m2, and
v0
I0
0.8 m/s
ω0 =
=
= 2 rad/s. Now, use conservation of angular momentum: ωf =
ω =
r0
0.4 m
If 0
1.92 X 10-2 kg m2
(2 rad/s) = 5.12 rad/s.
7.5 X 10-3 kg m2
1
1
The work done = ΔKE =
I ω 2 - I ω 2. Substituting the appropriate values found earlier, we
2 f f 2 0 0
have: work done = 5.99 X 10-2 J.
8.62
(a)
The table turns opposite to the way the woman walks, so its angular momentum cancels that
of the woman. From conservation of angular momentum, we have
Lf = Li = 0, so
Lf = Iwωw + Itableωtable = 0, and
ωtable = -
Iw
Itable
ωwoman = -
m w r2
vwoman
x
, or
Itable
r
(60 kg)(2 m)2
1.5 m/s
X
= - 0.36 rad/s. (0.36 rad/s CW)
2m
500 kg m2
1
1
work done = ΔKE = KEf - 0 =
mwomanv2woman + Iω2table
2
2
1
1
= (60 kg) (1.5 m/s) 2 + (500 kg m2)(0.36 rad/s) 2 = 99.9 J.
2
2
ωtable = -
(b)
8.63 (a) From conservation of angular momentum: (I1 + I2)ω = I1ω0, or
I1
ω=
ω
I1 + I2 0.
1
1
(b) Kf = (I1 + I2) ω2 , and
K i = I1ω 02 ,
2
2
Kf
I1
so
= (after some algebra) =
which is less than 1.
Ki
I1 + I2
8.64
Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand.
Then the total torque (taking CCW as positive) of these hands about the center of the clock is
given by
118
CHAPTER EIGHT SOLUTIONS
Lh
Lm
-g
τ = -mhg
sinθh - mmg
sinθm = (mhLhsinθh + mmLmsinθm) .
2
2
2
If we take t = 0 at 12 o'clock, then the angular positions of the hands at time t are θh = ωht, where
π
ωh = rad/h and θm = ωmt, where ωm = 2π rad/h.
6
9.8 m/s2
π
Therefore, τ = [(60 kg)(2.7 m)sin( t) + (100 kg)(4.5 m)sin 2πt]
2
6
π
or τ = -2.205 X 103 N m[0.36sin( t) + sin2πt], where time, t is expressed in hours.
6
(a) At 3:00, t = 3 h, so that
π
τ = -2.205 X 103 N m[[ 0.36sin ]+ sin6π] = -793.8 N m.
2
15
(b) At 5:15, t = 5 h +
h= 5.25 h, and substitution as above gives:
60
τ = - 2508.8 N m.
(c) At 6:00
τ = 0 N m.
(d) At 8:20
τ = -1163.7 N m.
(e) At 9:45
τ = -2938.4 N m.
8.65
We will use conservation of energy in the form
1
1
1
1
mvi2 + Iωi2 + mgyi = mvf2 + Iωf2 + mgyf
2
2
2
2
1
1 2
0 + 0 + mg(6 m)sin37° = m(0.2 m)2ω2 + ⎛ ⎞ m(0.2)2ω2 + 0
2
⎝2⎠ 5
2
where we have used vf = rω and I = mR2.
5
The mass cancels from the equation, and we find the angular velocity to be
ω = 36 rad/s.
8.66
The free body diagram is shown at the right.
Ry
Ty = Tsin12°
Στ = 0 , yields
Spine
R
2L/3
x
L
2L
- (350 N) + (Tsin 12°)( ) - (200 N)L = 0.
2
3
Tx = Tcos12°
L/2
From which,
T = 2705 N.
200 N
Compression force along spine = Rx, and we find this
350 N
Pivot
L
from ΣFx = 0, which gives Rx = Tx = T cos12° =
2646 N.
8.67
Use the lower left hand corner as the pivot point, and apply Στ = 0.
-(10 N)(.15 m) - (T1cos50°)(.15 m) + (T1sin50°)(0.3 m) = 0
Which gives,
T1 = 11.2 N.
Now use ΣFx = 0:
-F + (11.2)cos50° = 0,
and
F = 7.23 N.
Finally, ΣFy = 0 yields
T2 - 10 N + (11.2 N)sin50° = 0
so, T2 = 1.39 N.
8.68
The free body diagram is shown.
For a maximum value of x, the
rod is on the verge of slipping, so
f = µkN = 0.5 N.
Now, use ΣFx = 0
N = Tx = Tcos37°, so
N = 0.799 T.
2m
N
f
119
T
2
0.3 m
0.15 m
0.15 m
Ty = Tsin37°
= 0.602 T
Tx = Tcos37°
= 0.799 T
W
W
T1x
10 N
2m
x
T1y
F
CHAPTER EIGHT SOLUTIONS
Therefore,
f = 0.5 N. = 0.399 T .
0.399 T + 0.602 T = 2W which reduces to
Taking torques about the left end, we have
-Wx = W(2 m) + Ty (4 m) = 0, which gives
8.69
From ΣFy = 0, f + Ty - 2w = 0, or
T = 2W.
x = 2.8 m.
The free body diagram is shown at the right: ΣFy = 0
gives
Point
A
Ay - 29,400 N - 98,000 N = 0, so
Ax
Ay = 127,400 N.
Summing torques about point A, gives
1m
Bx(1 m) - 29,400 N(2 m) - 98,000 N(6 m) = 0,
and Bx = 646,800 N.
B
x
Finally, ΣFx = 0 gives Ax = Bx = 646,800 N.
The resultant force for A can be found from the
Pythagorean theorem, and an appropriate trig function, to be
FA = 6.59 X 105 N at an angle of 78.9° to the left of vertical.
Ay
98,000 N
W = 29,400 N
2m
6m
8.70
The initial moment of inertia of the system is: I0 = Σmiri2 = (4M)(1 m)2.
The moment of inertia of the system after the spokes are shortened is
If = Σmfrf2 = (4M)(0.5 m)2 .
Ii
We conserve angular momentum as:
ωf = ωi = (4)(2 rev/s) = 8.0 rev/s.
If
8.71
From Newton's second law, we have
mgsin37° - T = ma, which gives
2
(12 kg)(9.8 m/s )sin37° - T = (12 kg)a, or
T = 70.56 N - (12 kg)a.
a
2 m/s2
Also
α= =
= 20 rad/s2
(2)
r
0.1 m
(a) Thus, from (1),
T = 70.56 N - (12 kg)(2 m/s2) = 46.6 N.
τ
Tr
(46.6 N)(0.1 m)
(b) From τ = Iα, we have: I =
=
=
= 0.233 kg m2.
20 rad/s
α
α
8.72
(1)
ω = 0 + (20 rad/s2)(2 s) = 40 rad/s.
(c)
ω = ω0 + αt becomes:
(a)
The center of mass of the wheel and the mass m move with the same velocity, v. Using
conservation of energy, we have
1
1
v
1
mgH = Iω2 + Mv2 + mv2 , where
ω=
, so
2
2
R
2
I
M
m 2
2 kg m2
8 kg 4 kg 2
+
+
]v = [
+
+
]v = (10 kg)v2, or
2
2
2
2
2
2R
2(0.5 m)2
mgH (4 kg)(9.8 m/s2)(2 m)
=
= 7.84 m2/s2 and v = 2.8 m/s.
10 kg
10 kg
mgH = [
(b)
Consider the mass and write Δy = vyave(t) =
t=
(c)
v2 =
v+0
t, so
2
2Δy
2(2 m)
=
= 1.43 s.
v
2.8 m/s
Since the center of mass moves a distance equal to H, the angular displacement is
2m
1 rev
= 4 rad
= 0.637 rev.
0.5 m
2π rad
120
θ=
H
=
r
CHAPTER EIGHT SOLUTIONS
8.73 Since the ladder is about to slip,
f = (fs)max = µsN at each contact point.
Because the ladder is still (barely) in equilibrium: ΣFx = 0,
which gives
ff - Nw = 0, or
Nw
µsN f = Nw giving
Nf=
.
µs
N f, N w =
Nf - mg + fw = 0, or the above to eliminate
L/
2
µsmg
.
(1)
1 + µ 2s
Στlower end = 0 gives
L
- mg cosθ + NwLsinθ + fwLcosθ = 0,
2
mg
which can be written as + Nwtanθ + µsNw = 0, or
2
mg = 2Nw(tanθ + µs).
(2)
Substituting equation (1) into equation (2) gives
2 µsmg(tanθ + µs)
mg =
, which reduces to
1 + µ 2s
1 + µ2s = 2µstanθ + 2µ2s, or
With
θ = 60°, this becomes
which has one positive solution:
8.74
NW
L/
2
ΣFy = 0 ⇒
fW = µsNW
θ
NF
µ2s + (2tanθ)µ2s - 1 = 0.
µ2s + 3.646µs - 1 = 0,
µs = 0.268.
(a) and (b)
First, we write down the second law for the 2 kg mass.
T1 - 19.6 N = (2 kg)a.
(1)
Now, we write the second law for the 5 kg mass.
49 N - T2 = (5 kg)a.
(2)
Finally, τ = Iα for the pulley becomes
(T2 - T1)(0.5 m) = (5 kg m2)(a/0.5 m)
or T2 - T1 = (20 kg)a.
(3)
Equations (1), (2), and (3) can be solved simultaneously to give
a = 1.1 m/s2,
T1 = 22 N, and T2 = 44 N.
121
W = mg
fF = µs NF
CHAPTER EIGHT SOLUTIONS
8.75
(a) and (b)
First, we write down Newton's second law for the 3 kg mass.
T2 = (3 kg)a.
(1)
The second law for the 4 kg mass is
39.2 N - T1 = (4 kg)a.
(2)
Now apply τ = Iα to the pulley (axis of rotation at its center)
(T1 - T2)r = (0.5 kg m2)(a/r), or
T1 - T2 = (5.56 kg)a.
(3)
Equations (1), (2) and (3) can be solved simultaneously to give:
a = 3.12 m/s2,
T1 = 26.7 N, and T2 = 9.37 N.
8.76
(a) and (b)
1
a
MR2 ,
2
R
where a is the acceleration of the falling masses.
For each of the falling masses:
ΣF = mg - T = ma.
Combining Equations (1) and (2), we find:
Mmg
4mg
T=
, and a =
.
M + 4m
(M + 4m)
For the cylinder; Στ = (2T)R =
122
(1)
(2)