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11.1

a. μstanθ1+(mr2/ICM)μstanθ1+(mr2/ICM); inserting the angle and noting that for a hollow cylinder ICM=mr2,ICM=mr2, we have μstan60°1+(mr2/mr2)=12tan60°=0.87;μstan60°1+(mr2/mr2)=12tan60°=0.87; we are given a value of 0.6 for the coefficient of static friction, which is less than 0.87, so the condition isn’t satisfied and the hollow cylinder will slip; b. The solid cylinder obeys the condition μs13tanθ=13tan60°=0.58.μs13tanθ=13tan60°=0.58. The value of 0.6 for μSμS satisfies this condition, so the solid cylinder will not slip.

11.2

From the figure, we see that the cross product of the radius vector with the momentum vector gives a vector directed out of the page. Inserting the radius and momentum into the expression for the angular momentum, we have
l=r×p=(0.4mi^)×(1.67×10−27kg(4.0×106m/s)j^)=2.7×10−21kg·m2/sk^l=r×p=(0.4mi^)×(1.67×10−27kg(4.0×106m/s)j^)=2.7×10−21kg·m2/sk^

11.3

Isphere=25mr2,Icylinder=12mr2Isphere=25mr2,Icylinder=12mr2; Taking the ratio of the angular momenta, we have:
LcylinderLsphere=Icylinderω0Isphereω0=12mr225mr2=54LcylinderLsphere=Icylinderω0Isphereω0=12mr225mr2=54. Thus, the cylinder has 25%25% more angular momentum. This is because the cylinder has more mass distributed farther from the axis of rotation.

11.4

Using conservation of angular momentum, we have
I(4.0rev/min)=1.25Iωf,ωf=1.01.25(4.0rev/min)=3.2rev/minI(4.0rev/min)=1.25Iωf,ωf=1.01.25(4.0rev/min)=3.2rev/min

11.5

The Moon’s gravity is 1/6 that of Earth’s. By examining Equation 11.12, we see that the top’s precession frequency is linearly proportional to the acceleration of gravity. All other quantities, mass, moment of inertia, and spin rate are the same on the Moon. Thus, the precession frequency on the Moon is
ωP(Moon)=16ωP(Earth)=16(5.0rad/s)=0.83rad/s.ωP(Moon)=16ωP(Earth)=16(5.0rad/s)=0.83rad/s.

Conceptual Questions

1.

No, the static friction force is zero.

3.

The wheel is more likely to slip on a steep incline since the coefficient of static friction must increase with the angle to keep rolling motion without slipping.

5.

The cylinder reaches a greater height. By Equation 11.4, its acceleration in the direction down the incline would be less.

7.

All points on the straight line will give zero angular momentum, because a vector crossed into a parallel vector is zero.

9.

The particle must be moving on a straight line that passes through the chosen origin.

11.

Without the small propeller, the body of the helicopter would rotate in the opposite sense to the large propeller in order to conserve angular momentum. The small propeller exerts a thrust at a distance R from the center of mass of the aircraft to prevent this from happening.

13.

The angular velocity increases because the moment of inertia is decreasing.

15.

More mass is concentrated near the rotational axis, which decreases the moment of inertia causing the star to increase its angular velocity.

17.

A torque is needed in the direction perpendicular to the angular momentum vector in order to change its direction. These forces on the space vehicle are external to the container in which the gyroscope is mounted and do not impart torques to the gyroscope’s rotating disk.

Problems

19.

v CM = R ω ω = 66.7 rad/s v CM = R ω ω = 66.7 rad/s

21.

α = 3.3 rad / s 2 α = 3.3 rad / s 2

23.

I CM = 2 5 m r 2 , a CM = 3.5 m / s 2 ; x = 15.75 m I CM = 2 5 m r 2 , a CM = 3.5 m / s 2 ; x = 15.75 m

25.

positive is down the incline plane;
aCM=mgsinθm+(ICM/r2)ICM=r2[mgsin30aCMm]aCM=mgsinθm+(ICM/r2)ICM=r2[mgsin30aCMm],
xx0=v0t12aCMt2aCM=2.96m/s2,xx0=v0t12aCMt2aCM=2.96m/s2,
ICM=0.66mr2ICM=0.66mr2

27.

α=67.9rad/s2α=67.9rad/s2,
(aCM)x=1.5m/s2(aCM)x=1.5m/s2

29.

W = −1080.0 J W = −1080.0 J

31.

Mechanical energy at the bottom equals mechanical energy at the top;
12mv02+12(12mr2)(v0r)2=mghh=1g(12+14)v0212mv02+12(12mr2)(v0r)2=mghh=1g(12+14)v02,
h=7.7m,h=7.7m, so the distance up the incline is 22.5m22.5m.

33.

Use energy conservation
12mv02+12ICylω02=mghCyl12mv02+12ICylω02=mghCyl,
12mv02+12ISphω02=mghSph12mv02+12ISphω02=mghSph.
Subtracting the two equations, eliminating the initial translational energy, we have
hCyl=v02ghCyl=v02g,
hSph=5v026ghSph=5v026g,
Therefore the difference in height is
hCylhSph=v026g=5.0m/s2(6)9.8m/s2=0.43m.hCylhSph=v026g=5.0m/s2(6)9.8m/s2=0.43m.
Thus, the hollow sphere, with the smaller moment of inertia, rolls up to a lower height of 1.00.43=0.57m.1.00.43=0.57m.

35.

The magnitude of the cross product of the radius to the bird and its momentum vector yields rpsinθrpsinθ, which gives rsinθrsinθ as the altitude of the bird h. The direction of the angular momentum is perpendicular to the radius and momentum vectors, which we choose arbitrarily as k^k^, which is in the plane of the ground:
L=r×p=hmvk^=(300.0m)(2.0kg)(20.0m/s)k^=12,000.0kg·m2/sk^L=r×p=hmvk^=(300.0m)(2.0kg)(20.0m/s)k^=12,000.0kg·m2/sk^

37.

a. l=45.0kg·m2/sk^l=45.0kg·m2/sk^;
b. τ=10.0N·mk^τ=10.0N·mk^

39.

a. l1=−0.4kg·m2/sk^l1=−0.4kg·m2/sk^, l2=l4=0l2=l4=0,
l3=1.35kg·m2/sk^l3=1.35kg·m2/sk^; b. L=0.95kg·m2/sk^L=0.95kg·m2/sk^

41.

a. L=1.0×1011kg·m2/sL=1.0×1011kg·m2/s; b. No, the angular momentum stays the same since the cross-product involves only the perpendicular distance from the plane to the ground no matter where it is along its path.

43.

a. v=gtj^,r=di^,l=mdgtk^v=gtj^,r=di^,l=mdgtk^;
b. F=mgj^,τ=dmgk^F=mgj^,τ=dmgk^; c. yes

45.

a. mgh=12m(rω)2+1225mr2ω2mgh=12m(rω)2+1225mr2ω2;
ω=51.2rad/sω=51.2rad/s;
L=16.4kg·m2/sL=16.4kg·m2/s;
b. ω=72.5rad/sω=72.5rad/s;
L=23.2kg·m2/sL=23.2kg·m2/s

47.

a. I=720.0kg·m2I=720.0kg·m2; α=4.20rad/s2α=4.20rad/s2;
ω(10s)=42.0rad/sω(10s)=42.0rad/s; L=3.02×104kg·m2/sL=3.02×104kg·m2/s;
ω(20s)=84.0rad/sω(20s)=84.0rad/s;
b. τ=3.03×103N·mτ=3.03×103N·m

49.

a. L=1.131×107kg·m2/sL=1.131×107kg·m2/s;
b. τ=3.77×104N·mτ=3.77×104N·m

51.

ω = 28.6 rad / s L = 2.6 kg · m 2 / s ω = 28.6 rad / s L = 2.6 kg · m 2 / s

53.

Lf=25MS(3.5×103km)22πTfLf=25MS(3.5×103km)22πTf,
(7.0×105km)22π28days=(3.5×103km)22πTfTf=28days(3.5× 10 3km)2(7.0× 10 5km)2=7.0×104day=60.5s(7.0×105km)22π28days=(3.5×103km)22πTfTf=28days(3.5× 10 3km)2(7.0× 10 5km)2=7.0×104day=60.5s

55.

f f = 2.1 rev / s f 0 = 0.5 rev / s f f = 2.1 rev / s f 0 = 0.5 rev / s

57.

r P m v P = r A m v A v P = 10.2 km / s r P m v P = r A m v A v P = 10.2 km / s

59.

a. Idisk=5.0×10−4kg·m2Idisk=5.0×10−4kg·m2,
Ibug=2.0×10−4kg·m2Ibug=2.0×10−4kg·m2,
(Idisk+Ibug)ω1=Idiskω2,ω2=14.0rad/s(Idisk+Ibug)ω1=Idiskω2,ω2=14.0rad/s
b. ΔK=0.014JΔK=0.014J;
c. ω3=10.0rad/sω3=10.0rad/s back to the original value;
d. 12(Idisk+Ibug)ω32=0.035J12(Idisk+Ibug)ω32=0.035J back to the original value;
e. work of the bug crawling on the disk

61.

Li=400.0kg·m2/sLi=400.0kg·m2/s,
Lf=500.0kg·m2ωLf=500.0kg·m2ω,
ω=0.80rad/sω=0.80rad/s

63.

I0=340.48kg·m2I0=340.48kg·m2,
If=268.8kg·m2If=268.8kg·m2,
ωf=25.33rpmωf=25.33rpm

65.

a. L=280kg·m2/sL=280kg·m2/s,
If=89.6kg·m2If=89.6kg·m2,
ωf=3.125rad/sωf=3.125rad/s; b. Ki=437.5JKi=437.5J,
Kf=437.5JKf=437.5J

67.

Moment of inertia in the record spin: I0=0.5kg·m2I0=0.5kg·m2,
If=1.1kg·m2If=1.1kg·m2,
ωf=I0Ifω0ff=155.5rev/minωf=I0Ifω0ff=155.5rev/min

69.

Her spin rate in the air is: ff=2.0rev/sff=2.0rev/s;
She can do four flips in the air.

71.

Moment of inertia with all children aboard:
I0=2.4×105kg·m2I0=2.4×105kg·m2;
If=1.5×105kg·m2If=1.5×105kg·m2;
ff=0.3rev/sff=0.3rev/s

73.

I0=1.00×1010kg·m2I0=1.00×1010kg·m2,
If=9.94×109kg·m2If=9.94×109kg·m2,
ff=3.32rev/minff=3.32rev/min

75.

I=2.5×10−3kg·m2I=2.5×10−3kg·m2,
ωP=1.17rad/sωP=1.17rad/s

77.

a. LEarth=7.06×1033kg·m2/sLEarth=7.06×1033kg·m2/s,
ΔL=5.63×1033kg·m2/sΔL=5.63×1033kg·m2/s;
b. τ=1.4×1022N·mτ=1.4×1022N·m;
c. The two forces at the equator would have the same magnitude but different directions, one in the north direction and the other in the south direction on the opposite side of Earth. The angle between the forces and the lever arms to the center of Earth is 90°90°, so a given torque would have magnitude τ=FREsin90°=FREτ=FREsin90°=FRE. Both would provide a torque in the same direction:
τ=2FREF=1.1×1015Nτ=2FREF=1.1×1015N

Additional Problems

79.

aCM=310gaCM=310g,
v2=v02+2aCMxv2=(7.0m/s)22(310g)x,v2=v02+2aCMxv2=(7.0m/s)22(310g)x, v2=0x=8.34m;v2=0x=8.34m;
b. t=vv0aCM,t=vv0aCM, v=v0+aCMtt=2.38sv=v0+aCMtt=2.38s;
The hollow sphere has a larger moment of inertia, and therefore is harder to bring to a rest than the marble, or solid sphere. The distance travelled is larger and the time elapsed is longer.

81.

a. W=−500.0JW=−500.0J;
b. K+Ugrav=constantK+Ugrav=constant,
500J+0=0+(6.0kg)(9.8m/s2)h500J+0=0+(6.0kg)(9.8m/s2)h,
h=8.5m,d=17.0mh=8.5m,d=17.0m;
The moment of inertia is less for the hollow sphere, therefore less work is required to stop it. Likewise it rolls up the incline a shorter distance than the hoop.

83.

a. τ=34.0N·mτ=34.0N·m;
b. l=mr2ωω=3.6rad/sl=mr2ωω=3.6rad/s

85.

a. dM=3.85×108mdM=3.85×108m average distance to the Moon; orbital period 27.32d=2.36×106s27.32d=2.36×106s; speed of the Moon 2π3.85×108m2.36×106s=1.0×103m/s2π3.85×108m2.36×106s=1.0×103m/s; mass of the Moon 7.35×1022kg7.35×1022kg,
L=2.90×1034kgm2/sL=2.90×1034kgm2/s;
b. radius of the Moon 1.74×106m1.74×106m; the orbital period is the same as (a): ω=2.66×10−6rad/sω=2.66×10−6rad/s,
L=2.37×1029kg·m2/sL=2.37×1029kg·m2/s;
The orbital angular momentum is 1.22×1051.22×105 times larger than the rotational angular momentum for the Moon.

87.

I=0.135kg·m2I=0.135kg·m2,
α=4.19rad/s2,α=4.19rad/s2, ω=ω0+αtω=ω0+αt,
ω(5s)=21.0rad/s,ω(5s)=21.0rad/s, L=2.84kg·m2/sL=2.84kg·m2/s,
ω(10s)=41.9rad/s,ω(10s)=41.9rad/s, L=5.66kg·m/s2L=5.66kg·m/s2

89.

In the conservation of angular momentum equation, the rotation rate appears on both sides so we keep the (rev/min) notation as the angular velocity can be multiplied by a constant to get (rev/min):
Li=−0.04kg·m2(300.0rev/min), Lf=0.08kg·m2ffff=−150.0rev/min clockwiseLi=−0.04kg·m2(300.0rev/min), Lf=0.08kg·m2ffff=−150.0rev/min clockwise

91.

I0ω0=IfωfI0ω0=Ifωf,
I0=6120.0kg·m2I0=6120.0kg·m2,
If=1180.0kg·m2If=1180.0kg·m2,
ωf=31.1rev/minωf=31.1rev/min

93.

Li=1.00×107kg·m2/sLi=1.00×107kg·m2/s,
If=2.025×105kg·m2If=2.025×105kg·m2,
ωf=7.86rev/sωf=7.86rev/s

Challenge Problems

95.

Assume the roll accelerates forward with respect to the ground with an acceleration aa. Then it accelerates backwards relative to the truck with an acceleration (aa)(aa).

The forces on a cylinder on a horizontal surface are shown. The cylinder has radius R and moment of inertia one half m R squared and is centered on an x y coordinate system that has positive x to the right and positive y up. Force m g acts on the center of the cylinder and points down. Force N points up and acts at the contact point where the cylinder touches the surface. Force f sub s points to the right and acts at the contact point where the cylinder touches the surface.


Also, Rα=aaI=12mR2 Fx=fs=maRα=aaI=12mR2 Fx=fs=ma,
τ=fsR=Iα=IaaRfs=IR2(aa)=12m(aa) τ=fsR=Iα=IaaRfs=IR2(aa)=12m(aa),
Solving for aa: fs=12m(aa)fs=12m(aa); a=a3a=a3,
xx0=v0t+12at2;d=13at2;t=3da;xx0=v0t+12at2;d=13at2;t=3da;
therefore, s=1.5ds=1.5d

97.

a. The tension in the string provides the centripetal force such that Tsinθ=mrω2Tsinθ=mrω2. The component of the tension that is vertical opposes the gravitational force such that Tcosθ=mgTcosθ=mg. This gives T=5.7NT=5.7N. We solve for r=0.16mr=0.16m. This gives the length of the string as r=0.32mr=0.32m.
At ω=10.0rad/sω=10.0rad/s, there is a new angle, tension, and perpendicular radius to the rod. Dividing the two equations involving the tension to eliminate it, we have sinθcosθ=(0.32msinθ)ω2g1cosθ=0.32mω2gsinθcosθ=(0.32msinθ)ω2g1cosθ=0.32mω2g;
cosθ=0.31θ=72.2°cosθ=0.31θ=72.2°; b. linitial=0.08kg·m2/slinitial=0.08kg·m2/s,
lfinal=0.46kg·m2/slfinal=0.46kg·m2/s; c. No, the cosine of the angle is inversely proportional to the square of the angular velocity, therefore in order for θ90°,ωθ90°,ω. The rod would have to spin infinitely fast.

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