Download Solutions Chapter 12

Chapter 12, Conceptual Questions
12.1
As suggested by the figure, we will assume that the larger sphere is more massive. Then the center of gravity
would be at point a because if we suspend the dumbbell from point a then the counterclockwise torque due to the large
sphere (large weight times small lever arm) will be equal to the clockwise torque due to the small sphere (small weight
times large lever arm).
Look at the figure and mentally balance the dumbbell on your finger; your finger would have to be at point a.
The sun-earth system is similar to this except that the sun’s mass is so much greater than the earth’s that the center of
mass (called the barycenter for astronomical objects orbiting each other) is only 450 km from the center of the sun.
12.2.
To double the rotational energy without changing ω requires doubling the moment of inertia. The moment of
inertia is proportional to R 2 so R must increase by
2.
1 2
1
Iω . For a disk, I = MR 2 . Since the mass is the same for all three
2
2
disks, the quantity R 2ω 2 determines the ranking. Thus K a = K b > K c .
12.3.
The rotational kinetic energy is K rot =
12.4. No. The moment of inertia does not have any dependence on a quantity that indicates an object is rotating,
such as ω or α , so an object does not have to be rotating to have a moment of inertia.
12.5.
Mass that is further away from the axis of rotation contributes more to the moment of inertia I = ∫ r 2 dm. Here,
r is the distance from the axis of rotation to the mass element dm. Note r 2 is always positive. For a rod, there is more
mass further away from an axis through the rod’s end than one through its middle.
4
3
12.6. Because sphere 2 has twice the radius, its mass is greater by a factor of 23 = 8, since m = π r 3 ρsteel . The
added mass is also distributed further from the center and, so, I ∝ mr 2 leads to I 2 ∝ ( 8m1 )( 2r1 ) = 32 I1.
2
12.7. It will be easier to rotate the solid sphere because the hollow sphere’s mass is generally distributed further
from its center. If you roll both simultaneously down an incline, the solid sphere will win.
12.8.
τ e > τ a = τ b > τ c = τ d > τ f The torque τ = rF sin θ . We must calculate each torque:
L
L
2
F sin 45° =
LF
2
4
τ e = L ( 2F )
τa =   F
 2
τd =
 L
 
τb =  F
4
L
2
τ c =   F sin 45° =
LF
4
2
τ f = LF sin 0° = 0
12.12. The block attached to the solid cylinder hits first. The solid cylinder has a smaller moment of inertia since
more of its mass is closer to the rotation axis, so has less resistance to a change in its rotational motion. The torque
applied by the string attached to the block makes the solid cylinder change its rotation and unwind the string faster.
12.13. The moment of inertia for the tuck position is smaller than that of the pike position. Since the angular
momentum of the diver is conserved, any initial angular velocity is increased more when the diver moves to the tuck
position relative to the pike position.
12.14. The angular momentum L of disk b is larger than the angular momentum of disk a. Calculate L for each:
1
L1 = I1ω1 = mr12ω1
2
21
1

1

L2 = I 2ω 2 = m ( 2r12 )  ω1  = 2  mr12ω1  = 2 L1
2
2
2




103
Chapter 12, Exercises and Problems
12.1. Model: A spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:
Solve: The speed v = rω , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2π /60 rad/s = 6π rad/s. Thus, v =
(0.70 m)(6π rad/s) = 13.2 m/s.
Assess: A speed of 13.2 m/s ≈ 26 mph for the hands is a little high, but reasonable.
12.2. Model: Assume constant angular acceleration.
 2π rad  min 
Solve: (a) The final angular velocity is ω f = ( 2000 rpm ) 

 = 209.4 rad/s. The definition of angular
 rev  60 s 
acceleration gives us
α=
∆ω ω f − ω i 209.4 rad/s − 0 rad/s
=
=
= 419 rad/s
∆t
∆t
0.50 s
The angular acceleration of the drill is 4.2 × 10 2 rad/s.
(b)
1
2
θ f = θ i + ωi ∆t + α ( ∆t ) = 0 rad + 0 rad +
2
1
2
( 419 rad/s )( 0.50 s ) = 52.4 rad
2
 rev 
The drill makes (52.4 rad) 
 = 8.3 revolutions.
 2π rad 
12.4. Model: Assume constant angular acceleration.
Visualize:
 2π rad  min 
The initial angular velocity is ω i = ( 60 rpm ) 

 = 2π rad/s.
 rev  60 s 
The angular acceleration is
Solve:
104
α=
ωf − ωi
∆t
=
0 rad/s − 2π rad/s
= −0.251 rad/s 2
25 s
The angular velocity of the fan blade after 10 s is
ω f = ω i + α ( t − t0 ) = 2π rad/s+ ( −0.251 rad/s 2 ) (10 s − 0 s ) = 3.77 rad/s
The tangential speed of the tip of the fan blade is
vt = rω = ( 0.40 m )( 3.77 rad/s ) = 1.51 m/s
(b)
1
2
2
The fan turns 78.6 radians = 12.5 revolutions while coming to a stop.
θ f = θ i + ωi ∆t + α ( ∆t ) = 0 rad + ( 2π rad/s )( 25 s ) +
1
( −0.251 rad/s2 ) ( 25 s )2 = 78.6 rad
2
12.5. Model: The earth and moon are particles.
Visualize:
Choosing xE = 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the
distance from the center of the earth.
Solve:
24
22
8
m x + mM xM ( 5.98 × 10 kg ) ( 0 m ) + ( 7.36 × 10 kg )( 3.84 × 10 m )
xcm = E E
=
mE + mM
5.98 × 1024 kg + 7.36 × 10 22 kg
= 4.67 × 106 m
Assess: The center of mass of the earth-moon system is called the barycenter, and is located beneath the surface of
the earth. Even though xE = 0 m the earth influences the center of mass location because mE is in the denominator of
the expression for xcm .
12.6. Visualize: Please refer to Figure EX12.6. The coordinates of the three masses mA , mB , and mC are (0 cm,
0 cm), (0 cm, 10 cm), and (10 cm, 0 cm), respectively.
Solve: The coordinates of the center of mass are
xcm =
mA xA + mB xB + mC xC (100 g)(0 cm) + (200 g)(0 cm) + (300 g)(10 cm)
=
= 5.0 cm
(100 g + 200 g + 300 g)
mA + mB + mC
ycm =
mA yA + mB yB + mC yC (100 g)(0 cm) + (200 g)(10 cm) + (300 g)(0 cm)
=
= 3.3 cm
mA + mB + mC
(100 g + 200 g + 300 g)
12.7. Visualize: Please refer to Figure EX12.7. The coordinates of the three masses mA , mB , and mC are (0 cm,
10 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively.
Solve: The coordinates of the center of mass are
xcm =
mA xA + mB xB + mC xC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(10 cm)
=
= 6.7 cm
mA + mB + mC
(200 g + 300 g + 100 g)
ycm =
mA yA + mB yB + mC yC (200 g)(10 cm) + (300 g)(10 cm) + (100 g)(0 cm)
=
= 8.3 cm
mA + mB + mC
(200 g + 300 g + 100 g)
12.8. Model: The balls are particles located at the ball’s respective centers.
Visualize:
105
Solve:
The center of mass of the two balls measured from the left hand ball is
xcm =
(100 g )( 0 cm ) + ( 200 g )( 30 cm ) = 20 cm
100 g + 200 g
The linear speed of the 100 g ball is
 2π rad  min 
v1 = rω = xcmω = ( 0.20 m )(120 rev/min ) 

 = 2.5 m/s
 rev  60 s 
12.9. Model: The earth is a rigid, spherical rotating body.
Solve:
The rotational kinetic energy of the earth is K rot = 12 Iω 2 . The moment of inertia of a sphere about its diameter
(see Table 12.2) is I = 52 M earth R 2 and the angular velocity of the earth is
ω=
2π rad
= 7.27 × 10−5 rad/s
24 × 3600 s
Thus, the rotational kinetic energy is
1 2

K rot =  M earth R 2 ω 2
2 5

1
= (5.98 × 1024 kg)(6.37 × 106 m) 2 (7.27 × 10−5 rad/s) 2 = 2.57 × 10 29 J
5
12.10. Model: The triangle is a rigid body rotating about an axis through the center.
Visualize: Please refer to Figure EX12.10. Each 200 g mass is a distance r away from the axis of rotation, where r is
given by
0.20 m
0.20 m
= cos30° ⇒ r =
= 0.2309 m
r
cos30°
Solve: The moment of inertia of the triangle is I = 3 × mr 2 = 3(0.200 kg)(0.2309 m) 2 = 0.0320 kg m 2 . The frequency
of rotation is given as 5.0 revolutions per s or 10π rad/s. The rotational kinetic energy is
K rot. =
1 2 1
Iω = (0.0320 kg m 2 )(10.0π rad/s) 2 = 15.8 J
2
2
12.11. Model: The disk is a rigid body rotating about an axis through its center.
Visualize:
Solve:
The speed of the point on the rim is given by vrim = Rω . The angular velocity ω of the disk can be
determined from its rotational kinetic energy which is K = 12 Iω 2 = 0.15 J. The moment of inertia I of the disk about its
center and perpendicular to the plane of the disk is given by
106
1
1
I = MR 2 = (0.10 kg)(0.040 m) 2 = 8.0 × 10−5 kg m 2
2
2
2(0.15 J)
0.30 J
2
⇒ω =
=
⇒ ω = 61.237 rad/s
I
8.0 × 10−5 kg m 2
Now, we can go back to the first equation to find vrim . We get vrim = Rω = (0.040 m)(61.237 rad/s) = 2.4 m/s.
12.12. Model: The baton is a thin rod rotating about a perpendicular axis through its center of mass.
Solve:
The moment of inertia of a thin rod rotating about its center is I =
1
ML2 . For the baton,
12
1
2
( 0.400 kg )( 0.96 m ) = 0.031 kg m 2
12
The rotational kinetic energy of the baton is
I=
2
K rot =
1 2 1

 2π rad  min  
Iω = ( 0.031 kg m 2 )  (100 rev/min ) 

  = 1.68 J
2
2
 rev  60 s  

12.13. Model: The structure is a rigid body rotating about its center of mass.
Visualize:
We placed the origin of the coordinate system on the 300 g ball.
Solve: First, we calculate the center of mass:
xcm =
(300 g)(0 cm) + (600 g)(40 cm)
= 26.67 cm
300 g + 600 g
Next, we will calculate the moment of inertia about the structure’s center of mass:
I = (300 g)( xcm ) 2 + (600 g)(40 cm − xcm ) 2
= (0.300 kg)(0.2667 m) 2 + (0.600 kg)(0.1333 m) 2 = 0.032 kg m 2
Finally, we calculate the rotational kinetic energy:
1 2 1
 100 × 2π

Iω = (0.032 kg m 2 ) 
rad/s  = 1.75 J
2
2
 60

2
K rot =
12.14. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the
rotation axis passes through mass A and is perpendicular to the page.
Visualize: Please refer to Figure EX12.14.
∑ mi xi = mA xA + mB xB + mC xC + mD xD
Solve: (a) xcm =
mA + mB + mC + mD
∑ mi
=
ycm =
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.057 m
100 g + 200 g + 200 g + 200 g
mA yA + mB yB + mC yC + mD yD
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m)
= 0.057 m
700 g
(b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is
=
I A = ∑ mi ri 2 = mA rA2 + mB rB2 + mC rC2 + mD rD2
i
= (0.100 kg)(0 m) 2 + (0.200 kg)(0.10 m) 2 + (0.200 kg)(0.1414 m) 2 + (0.200 kg)(0.10 m) 2 = 0.0080 kg m 2
12.15. Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:
107
Solve: (a) xcm =
∑m x
∑m
i i
=
i
=
ycm =
mA xA + mB xB + mC xC + mD xD
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.057 m
100 g + 200 g + 200 g + 200 g
mA yA + mB yB + mC yC + mD yD
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m)
= 0.057 m
700 g
(b) The moment of inertia about a diagonal that passes through B and D is
=
I BD = mA rA2 + mC rC2
where rA = rC = (0.10 m)cos 45° = 7.07 cm and are the distances from the diagonal. Thus,
I BD = (0.100 kg)rA2 + (0.200 kg) rC2 = 0.0015 kg m 2
Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.
12.16. Model: The three masses connected by massless rigid rods is a rigid body.
Visualize: Please refer to Figure EX12.16.
xcm =
Solve: (a)
∑m x
∑m
i i
i
ycm =
∑m y
∑m
i
i
=
=
(0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m)
= 0.060 m
0.100 kg + 0.200 kg + 0.100 kg
(0.100 kg)(0 m) + (0.200 kg)
(
0.100 kg + 0.200 kg + 0.100 kg
i
)
(0.10 m) 2 − (0.06 m) 2 + (0.100 kg)(0 m)
= 0.040 m
(b) The moment of inertia about an axis through A and perpendicular to the page is
I A = ∑ mi ri 2 = mB (0.10 m) + mC (0.10 m) 2 = (0.100 kg)[(0.10 m) 2 + (0.10 m) 2 ] = 0.0020 kg m 2
2
(c) The moment of inertia about an axis that passes through B and C is
I BC = mA
(
(0.10 m) 2 − (0.06 m) 2
)
2
= 0.00128 kg m 2
Assess: Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .
12.17. Model: The door is a slab of uniform density.
Solve: (a) The hinges are at the edge of the door, so from Table 12.2,
1
2
I = ( 25 kg )( 0.91 m ) = 6.9 kg m 2
3
(b) The distance from the axis through the center of mass along the height of the door is
 0.91 m

d =
− 0.15 m  = 0.305 m. Using the parallel–axis theorem,
2


1
2
2
I = I cm + Md 2 = ( 25 kg )( 0.91 m ) + ( 25 kg )( 0.305 cm ) = 4.1 kg m 2
12
Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass.
12.18. Model: The CD is a disk of uniform density.
Solve: (a) The center of the CD is its center of mass. Using Table 12.2,
108
1
1
2
MR 2 = ( 0.021 kg )( 0.060 m ) = 3.8 × 10−5 kg m 2
2
2
(b) Using the parallel–axis theorem with d = 0.060 m,
I cm =
I = I cm + Md 2 = 3.8 × 10−5 kg m 2 + ( 0.021 kg )( 0.060 m ) = 1.14 × 10−4 kg m 2
2
12.19. Visualize:
Solve: Torque by a force is defined as τ = Fr sin φ where φ is measured counterclockwise from the r vector to the
F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N
force:
(30 N)r1 sinφ1 + (20 N)r2 sinφ2 = (30 N)(0.02 m) sin ( − 90°) + (20 N)(0.02 m) sin (90°)
= ( − 0.60 N m) + (0.40 N m) = −0.20 N m
Assess: A negative torque causes a clockwise acceleration of the pulley.
12.21. Visualize:
Solve:
The net torque on the spark plug is
τ = Fr sin φ = −38 N m = F (0.25 m)sin(−120°) ⇒ F = 176 N
That is, you must pull with a force of 176 N to tighten the spark plug.
Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for
the net torque.
12.22. Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center.
Solve: The net torque on the axle is
109
τ = FA rA sin φA + FB rB sin φB + FC rC sin φC + FD rD sin φD
= (30 N)(0.10 m)sin(−90°) + (20 N)(0.050 m)sin 90° + (30 N)(0.050 m)sin135° + (20 N)(0.10 m)sin 0°
= −3 N m + 1 N m + 1.0607 N m = −0.94 N m
Assess: A negative torque means a clockwise rotation of the disk.
12.23. Model: The beam is a solid rigid body.
Visualize:
( )
The steel beam experiences a torque due to the gravitational force on the construction worker FG
and the
C
gravitational force on the beam FG . The normal force exerts no torque since the net torque is calculated about the point
( )
B
where the beam is bolted into place.
( )
Solve: The net torque on the steel beam about point O is the sum of the torque due to FG
C
( )
and the torque due to FG
B
.
The gravitational force on the beam acts at the center of mass.
τ = (( FG )C )(4.0 m)sin(−90°) + (( FG ) B )(2.0 m)sin(−90°)
= −(70 kg)(9.80 m/s 2 )(4.0 m) − (500 kg)(9.80 m/s 2 )(2.0 m) = −12.5 kN m
The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is 12.5 kN
m.
12.24. Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass.
Visualize:
Solve: (a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm:
τ = τ ball + τ arm = ( mb g ) rb sin 90° + ( ma g )ra sin 90°
= (3.0 kg)(9.8 m/s 2)(0.70 m)+(4.0 kg)(9.8 m/s 2)(0.35 m) = 34 N m
(b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line, so
τ = τ ball + τ arm = ( mb g ) rb sin 45° + ( ma g )ra sin 45°
= (3.0 kg)(9.8 m/s 2)(0.70 m)(0.707) + (4.0 kg)(9.8 m/s 2)(0.35 m)(0.707) = 24 N m
12.25. Solve: τ = Iα is the rotational analog of Newton’s second law F = ma. We have τ = (2.0 kg m 2 )(4.0 rad/s 2 ) =
8.0 kg m 2 /s 2 = 8.0 N m.
110
α = τ /I , a graph of the angular acceleration looks just like the torque graph with the
12.26. Visualize: Since
numerical values divided by I = 4.0 kg m 2 .
Solve: From the discussion about Figure 4.47
ω f = ω i + area under the angular acceleration α curve between ti and tf
The area under the curve between t = 0 s and t = 3 s is 0.75 rad/s. With ω1 = 0 rad/s, we have
ω f = 0 rad/s + 0.75 rad/s = 0.75 rad/s
12.27. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the
center of mass. Assume that the size of the balls is small compared to 1 m.
Visualize:
We placed the origin of the coordinate system on the 1.0 kg ball.
Solve: The center of mass and the moment of inertia are
xcm =
(1.0 kg)(0 m) + (2.0 kg)(1.0 m)
= 0.667 m and ycm = 0 m
(1.0 kg + 2.0 kg)
I about cm = ∑ mi ri 2 = (1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 = 0.667 kg m 2
We have ω f = 0 rad/s,
becomes
tf − ti = 5.0 s, and ω i = −20 rpm = −20(2π rad/60 s) = − 23 π rad/s, so ω f = ωi + α (tf − ti )
2π
 2π

0 rad/s =  −
rad/s  + α (5.0 s) ⇒ α =
rad/s 2
3
15


Having found I and α , we can now find the torque τ that will bring the balls to a halt in 5.0 s:
2
 2π
 4π
τ = I about cmα =  kg m 2 
rad/s 2  =
N m = 0.28 N m
3
 15
 45
The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction.
12.28. Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is
perpendicular to the disk.
Solve: To determine
the
torque
(τ)
needed
to
take
the
plastic
disk
from
ω i = 0 rad/s
to
ω f = 1800 rpm = (1800)(2π ) / 60 rad/s = 60π rad/s in tf − ti = 4.0 s, we need to determine the angular acceleration (α ) and
the disk’s moment of inertia (I ) about the axle in its center. The radius of the disk is R = 10.0 cm. We have
1
1
I = MR 2 = (0.200 kg)(0.10 m) 2 = 1.0 × 10−3 kg m 2
2
2
ω − ω i 60π rad/s − 0 rad/s
=
= 15π rad/s 2
ω f = ω i + α (tf − ti ) ⇒ α = f
4.0 s
tf − ti
111
Thus, τ = Iα = (1.0 × 10−3 kg m 2 )(15π rad/s 2 ) = 0.047 N m.
12.29. Model: The compact disk is a rigid body rotating about its center.
Visualize:
Solve: (a) The rotational kinematic equation ω1 = ω 0 + α (t1 − t0 ) gives
200π
 2π 
(2000 rpm) 
rad/s 2
 rad/s = 0 rad + α (3.0 s − 0 s) ⇒ α =
60
9


The torque needed to obtain this operating angular velocity is
 200π

rad/s 2  = 1.75 × 10−3 N m
 9

τ = Iα = (2.5 × 10−5 kg m 2 ) 
(b) From the rotational kinematic equation,
1  200π
1
θ1 = θ 0 + ω 0 (t1 − t0 ) + α (t1 − t0 ) 2 = 0 rad + 0 rad + 
2
2 9
100π
= 100π rad =
revolutions = 50 rev
2π
Assess:
2

rad/s 2  ( 3.0 s − 0 s )

Fifty revolutions in 3 seconds is a reasonable value.
12.30. Model: The rocket attached to the end of a rigid rod is a rotating rigid body. Assume the rocket is small
compared to 60 cm.
Visualize: Please refer to Figure EX12.30.
Solve: We can determine the rocket’s angular acceleration from the relationship τ = Iα . The torque τ can be found
from the thrust (F) using τ = Fr sin φ . The moment of inertia (I) can be calculated from equations given in Table 12.2.
Specifically, I = I rod about one end + I rocket becomes
1
1
M rod L2 + ML2 = (0.100 kg)(0.60 m) 2 + (0.200 kg)(0.60 m) 2
3
3
= 0.012 kg m 2 + 0.072 kg m 2 = 0.0840 kg m 2
⇒α =
τ
I
=
Fr sin φ (4.0 N)(0.60 m)sin(45°)
=
= 20 rads/s 2
I
0.0840 kg m 2
Assess: The rocket will accelerate counterclockwise since α is positive.
12.31. Model: The rod is in rotational equilibrium, which means that τ net = 0.
Visualize:
As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the
figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end
of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your
thumb on top (pushing down) and your forefinger underneath (pushing up).
112
Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this
point, so it exerts no torque. To prevent rotation, the pin’s normal force npin exerts a positive torque (ccw about the left
end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force on the
rod acts at the center of mass, so
τ net = 0 N m = τ pin − (0.40 m)(2.0 kg)(9.8 m/s 2) − (0.80 m)(0.50 kg)(9.8 m/s 2)
⇒ τ pin = 11.8 N m
12.32. Model: The massless rod is a rigid body.
Visualize:
Solve:
To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (τ net = 0 Nm). We have
( Fnet ) y = (40 N) − (100 N) + (60 N) = 0 N,
so the object is in translational
equilibrium. Measuring τ net about the left end,
τ net = (60 N)(3.0 m)sin( +90°) + (100 N)(2.0 m)sin(−90°) = −20 N m
The object is not in equilibrium.
12.33. Model: The object balanced on the pivot is a rigid body.
Visualize:
Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium.
Solve: There are three forces acting on the object: the gravitational force FG acting through the center of mass of
1
the long rod, the gravitational force FG acting through the center of mass of the short rod, and the normal force P
( )
( )
2
on the object applied by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is
− ( FG )1 − ( FG )2 + P = 0 N ⇒ P = ( FG )1 + ( FG )2 = (1.0 kg)(9.8 m/s 2 ) + (4.0 kg)(9.8 m/s 2 ) = 49 N
Measuring torques about the left end, the equation for rotational equilibrium τ net = 0 Nm is
Pd − w1 (1.0 m) − w2 (1.5 m) = 0 Nm
⇒ (49 N)d − (1.0 kg)(9.8 m/s 2 )(1.0 m) − (4.0 kg)(9.8 m/s 2 )(1.5 m) = 0 N ⇒ d = 1.40 m
Thus, the pivot is 1.40 m from the left end.
12.34. Model: The see-saw is a rigid body. The cats and bowl are particles.
Visualize:
113
Solve:
The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point.
τ net = 0 = ( FG )1 ( 2.0 m ) − ( FG )2 ( d ) − ( FG ) B ( 2.0 m )
m2 gd = m1 g ( 2.0 m ) − mB g ( 2.0 m )
d=
( m1 − mB )( 2.0 m ) = ( 5.0 kg − 2.0 kg )( 2.0 m ) = 1.5 m
m2
4.0 kg
Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined
mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat.
12.35. Solve: (a) According to Equation 12.35, the speed of the center of mass of the tire is
vcm = Rω = 20 m/s ⇒ ω =
vcm 20 m/s
 60 
2
=
= 66.67 rad/s = ( 66.7 ) 
 rpm = 6.4 × 10 rpm
R 0.30 m
 2π 
(b) The speed at the top edge of the tire relative to the ground is vtop = 2vcm = 2(20 m/s) = 40 m/s.
(c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.
12.36. Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass distribution.
Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass. The
moment of inertia of the can about the center of mass is 12 MR 2, where R is the radius of the can. Also vcm = Rω ,
where ω is the angular velocity of the can. The total kinetic energy of the can is
1
1
1
11
 v 
2
2
K = K cm + K rot = Mvcm
+ I cmω 2 = Mvcm
+  MR 2  cm 
2
2
2
2 2
 R 
3
3
2
= Mvcm
= (0.50 kg)(1.0 m/s) 2 = 0.38 J
4
4
2
12.37. Model: The sphere is a rigid body rolling down the incline without slipping.
Visualize:
The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down.
Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation
will be K f = U i . The kinetic energy consists of both translational and rotational energy. This means
Kf =
1
1
1 2
1

2
I cmω 2 + Mvcm
= Mgh ⇒  MR 2 ω 2 + M ( Rω ) 2 = Mgh
2
2
2 5
2

7
⇒ MR 2ω 2 = Mg (2.1 m)sin 25°
10
⇒ω =
10
7
g (2.1 m)(sin 25°)
=
R2
10
7
(b) From part (a)
114
g (2.1 m)(sin 25°)
= 88 rad/s
(0.04 m) 2
K total =
1
1
7
1
12
1

2
I cmω 2 + Mvcm
= MR 2ω 2 and K rot = I cmω 2 =  MR 2 ω 2 = MR 2ω 2
2
2
10
2
2 5
5

2 2
1
MR ω
K
1 10 2
⇒ rot = 57
= × =
K total 10 MR 2ω 2 5 7 7
12.47. Visualize: Please refer to Figure EX12.47.
Solve:
L = r × mv = (3.0iˆ + 2.0 ˆj ) m × (0.1 kg)(4.0ˆj ) m/s
= 1.20(iˆ × ˆj ) kg m 2 /s + 0.8( ˆj × ˆj ) kg m 2 /s = 1.20kˆ kg m 2 /s + 0 kg m 2 /s
= 1.20kˆ kg m 2 /s or (1.20 kg m 2 /s, out of page)
12.49. Model: The disk is a rotating rigid body.
Visualize: Please refer to Figure EX12.49.
Solve: From Table 12.2, the moment of inertial of the disk about its center is
1
1
I = MR 2 = (2.0 kg)(0.020 m) 2 = 4.0 × 10 −4 kg m 2
2
2
The angular velocity ω is 600 rpm = 600 × 2π /60 rad/s = 20π rad/s. Thus, L = Iω = (4.0 × 10−4 kg m 2 )(20π rad/s) =
0.025 kg m 2 /s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the
− x direction. Consequently,
L = −0.025 iˆ kg m 2 /s = (0.025 kg m 2 /s, into page)
12.56. Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin.
Visualize: Please refer to P12.56.
Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1 , and mass m2 .
Using Table 12.2 for the moment of inertia of the rod, we get
2
I rod = I rod about center + I m1 + I m2 =
=
1
 L
L
ML2 + m1   + m2  
12
 2
4
2
1
1
1
L2  M
m 
ML2 + m1L2 + m2 L2 =  + m1 + 2 
12
4
16
4 3
4 
Assess: With m1 = m2 = 0 kg, I rod 121 ML2 , as expected.
12.61. Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium
( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m). We also apply the model of static friction.
Visualize:
Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other hand, the
floor exerts both the normal force n1 and the static frictional force fs . The gravitational force FG on the ladder acts
through the center of mass of the ladder.
Solve: The x- and y-components of Fnet = 0 N are
∑F
x
= n2 − f s = 0 N ⇒ f s = n2
∑F
y
115
= n1 − FG = 0 N ⇒ n1 = FG
The minimum angle occurs when the static friction is at its maximum value fs max = µ s n1. Thus we have
n2 = f s = µs n1 = µ s mg . We choose the bottom corner of the ladder as a pivot point to obtain τ net , because two forces
pass through this point and have no torque about it. The net torque about the bottom corner is
τ net = d1mg − d 2n2 = (0.5L cosθ min ) mg − ( L sinθ min ) µs mg = 0 N m
⇒ 0.5cosθ min = µ s sin θ min ⇒ tan θ min =
0.5
µs
=
0.5
= 1.25 ⇒ θ min = 51°
0.4
12.63. Model: The structure is a rigid body.
Visualize:
Solve:
We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv because the
pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque. For
the beam to stay in equilibrium, the net torque about this point is zero. We can write
τ about left end = −( FG ) B (3.0 m) − ( FG ) W (4.0 m) + (T sin150°)(6.0 m) = 0 N m
Using ( FG ) B = (1450 kg)(9.8 m/s 2 ) and ( FG ) W = (80 kg)(9.8 m/s 2 ), the torque equation can be solved to yield
T = 15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.
12.71. Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force
on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block m1
to the right, but a leftward force on the outer edge of the pulley.
Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T − m2 g = m2a2 . Using the constraint
−a2 = + a1 = a, we have T = m1a and −T + m2 g = m2 a. Adding these equations, we get m2 g = ( m1 + m2 ) a, or
a=
m2 g
mm g
⇒ T = m1a = 1 2
m1 + m2
m1 + m2
(b) When the pulley has mass m, the tensions (T1 and T2 ) in the upper and lower portions of the string are different.
Newton’s second law for m1 and the pulley are:
T1 = m1a
and
T1R − T2 R = − Iα
116
We are using the minus sign with α because the pulley accelerates clockwise. Also, a = Rα . Thus, T1 = m1a and
T2 − T1 =
I a aI
=
R R R2
Adding these two equations gives
I 

T2 = a  m1 + 2 
R


Newton’s second law for m2 is T2 − m2 g = m2 a2 = − m2 a. Using the above expression for T2 ,
I 
m2 g

a  m1 + 2  + m2 a = m2 g ⇒ a =
2
R
m
m
+


1
2 + I /R
Since I = 12 mp R 2 for a disk about its center,
a=
m2 g
m1 + m2 + 12 mp
With this value for a we can now find T1 and T2 :
T1 = m1a =
Assess:
m1m2 g
m1 + m2 + 12 mp
T2 = a ( m1 + I / R 2 ) =
1
m2 g
1  m2 ( m1 + 2 mp ) g

m
+
m
=
1
p
( m1 + m2 + 12 mp )  2  m1 + m2 + 12 mp
For m = 0 kg, the equations for a, T1 , and T2 of part (b) simplify to
a=
m2 g
m1 + m2
and T1 =
m1m2 g
m1 + m2
and T2 =
m1m2 g
m1 + m2
These agree with the results of part (a).
12.74. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline
without slipping. We also assume that the coefficient of rolling friction is zero.
Visualize:
The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational
potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy.
Solve: The conservation of energy equation K f + U gf = K i + U gi is
1
1
1
1
2
M (v1 ) 2cm + I cm (ω1 ) 2 + Mgy1 = M (v0 ) cm
+ I cm (ω 0 ) 2 + Mgy0
2
2
2
2
1
1 2
1
1
(v ) 2

2
2
0 J + 0 J + Mgy1 = M (v0 )cm
+  MR 2  (ω 0 )cm
+ 0 J ⇒ Mgy1 = M (v0 ) 2cm + MR 2 0 2cm
2
2 3
2
3
R

5
(v ) 2
5
5 (5.0 m/s) 2
2
⇒ gy1 = (v0 )cm
⇒ y1 = 6 0 cm =
= 2.126 m
6
g
6 9.8 m/s 2
The distance traveled along the incline is
y1
2.126 m
=
= 4.3 m
sin 30°
0.5
Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the
incline is approximately 10 mph.
s=
12.81. Model: The angular momentum of the satellite in the elliptical orbit is a constant.
Visualize:
117
Solve: (a) Because the gravitational force is always along the same direction as the direction of the moment arm
vector, the torque τ = r × Fg is zero at all points on the orbit.
(b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is
perpendicular to r at points a and b, so β = 90° and L = mvr. Thus
r 
Lb = La ⇒ mvb rb = mva ra ⇒ vb =  a  va
 rb 
30,000 km
30,000 km
ra =
− 9000 km = 6000 km and rb =
+ 9000 km = 24,000 km
2
2
 6000 km 
⇒ vb = 
 (8000 m/s) = 2000 m/s
 24,000 km 
(c) Using the conservation of angular momentum Lc = La , we get
r 
mvc rc sin β c = mva ra ⇒ vc =  a  va /sin β c
 rc 
rc = (9000 km) 2 + (12,000 km) 2 = 1.5 × 107 m
From the figure, we see that sin β c = 12,000 15,000 = 0.80. Thus
 6000 km  (8000 m/s)
vc = 
= 4000 m/s

0.80
 15,000 km 
12.89. Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating
about its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum of
the car and track is conserved.
Visualize:
Solve:
The toy car’s steady speed of 0.75 m/s relative to the track means that
vc − vt = 0.75 m/s ⇒ v c = vt + 0.75 m/s,
where vt is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum implies
that
Li = Lf
0 = I cω c + I tω t = ( mr 2 )ω c + ( Mr 2 )ω t = mω c + M ω t
The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track. Since
v
v
ω c = c , ω t = t , we have
r
r
118
0 = mvc + Mvt
⇒ m ( vt + 0.75 m/s ) + Mvt = 0
⇒ vt = −
M
( 0.200 kg )
( 0.75 m/s ) = −
( 0.75 m/s ) = −0.125 m/s
m+M
( 0.200 kg + 1.0 kg )
The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the track
is
v ( 0.125 m/s )
ωt = t =
= 0.417 rad/s clockwise.
r
0.30 m
In rpm,
 rev  60 s 
ω t = ( 0.417 rad/s ) 


 2π rad  min 
= 4.0 rpm
Assess: The speed of the track is less than that of the car because it is more massive.
12.91. Model:
Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The
mechanical energy of the marble is conserved.
Visualize:
Solve: The ball’s center of mass moves in a circle of radius R − r. The free-body diagram on the marble at its highest
position shows that Newton’s second law for the marble is
mv12
R−r
The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the
normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration needed
for the circular motion. For n → 0 N,
mg + n =
mg =
mv 2
⇒ v12 = g ( R − r )
(R − r)
Since rolling motion requires v12 = r 2ω12 , we have
ω12 r 2 = g ( R − r ) ⇒ ω12 =
g (R − r)
r2
The conservation of energy equation is
1
1
( K f + U gf ) top of loop = ( K i + U gi )initial ⇒ mv12 + Iω12 + mgy1 = mgy0 = mgh
2
2
Using the above expressions and I = 52 mr 2 the energy equation simplifies to
1
1 2
 g (R − r ) 
mg ( R − r ) +   mr 2 
 + mg 2( R − r ) = mgh ⇒ h = 2.7 ( R − r )
2
2
2 5
 r

119