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10.1

a. 40.0rev/s=2π(40.0)rad/s40.0rev/s=2π(40.0)rad/s, α=ΔωΔt=2π(40.0)0rad/s20.0s=2π(2.0)=4.0πrad/s2α=ΔωΔt=2π(40.0)0rad/s20.0s=2π(2.0)=4.0πrad/s2; b. Since the angular velocity increases linearly, there has to be a constant acceleration throughout the indicated time. Therefore, the instantaneous angular acceleration at any time is the solution to 4.0πrad/s24.0πrad/s2.

10.2

a. Using Equation 10.11, we have 7000rpm=7000.0(2πrad)60.0s=733.0rad/s,7000rpm=7000.0(2πrad)60.0s=733.0rad/s,
α=ωω0t=733.0rad/s10.0s=73.3rad/s2α=ωω0t=733.0rad/s10.0s=73.3rad/s2;
b. Using Equation 10.13, we have
ω2=ω02+2αΔθΔθ=ω2ω022α=0(733.0rad/s)22(73.3rad/s2)=3665.2radω2=ω02+2αΔθΔθ=ω2ω022α=0(733.0rad/s)22(73.3rad/s2)=3665.2rad

10.3

The angular acceleration is α=(5.00)rad/s20.0s=0.25rad/s2α=(5.00)rad/s20.0s=0.25rad/s2. Therefore, the total angle that the boy passes through is
Δθ=ω2ω022α=(5.0)202(0.25)=50radΔθ=ω2ω022α=(5.0)202(0.25)=50rad.
Thus, we calculate
s=rθ=5.0m(50.0rad)=250.0ms=rθ=5.0m(50.0rad)=250.0m.

10.4

The initial rotational kinetic energy of the propeller is
K0=12Iω2=12(800.0kg-m2)(4.0×2πrad/s)2=2.53×105JK0=12Iω2=12(800.0kg-m2)(4.0×2πrad/s)2=2.53×105J.
At 5.0 s the new rotational kinetic energy of the propeller is
Kf=2.03×105JKf=2.03×105J.
and the new angular velocity is
ω=2(2.03×105J)800.0kg-m2=22.53rad/sω=2(2.03×105J)800.0kg-m2=22.53rad/s
which is 3.58 rev/s.

10.5

I parallel-axis = I center of mass + m d 2 = m R 2 + m R 2 = ( 1 3 ) m R 2 I parallel-axis = I center of mass + m d 2 = m R 2 + m R 2 = ( 1 3 ) m R 2

10.6

The angle between the lever arm and the force vector is 80°;80°; therefore, r=100m(sin80°)=98.5mr=100m(sin80°)=98.5m.

The cross product τ=r×Fτ=r×F gives a negative or clockwise torque.

The torque is then τ=rF=−98.5m(5.0×105N)=−4.9×107N·mτ=rF=−98.5m(5.0×105N)=−4.9×107N·m.

10.7

a. The angular acceleration is α=20.0(2π)rad/s010.0s=12.56rad/s2α=20.0(2π)rad/s010.0s=12.56rad/s2. Solving for the torque, we have iτi=Iα=(30.0kg·m2)(12.56rad/s2)=376.80N·miτi=Iα=(30.0kg·m2)(12.56rad/s2)=376.80N·m; b. The angular acceleration is α=020.0(2π)rad/s20.0s=−6.28rad/s2α=020.0(2π)rad/s20.0s=−6.28rad/s2. Solving for the torque, we have iτi=Iα=(30.0kg-m2)(−6.28rad/s2)=−188.50N·miτi=Iα=(30.0kg-m2)(−6.28rad/s2)=−188.50N·m

10.8

3 MW

Conceptual Questions

1.

The second hand rotates clockwise, so by the right-hand rule, the angular velocity vector is into the wall.

3.

They have the same angular velocity. Points further out on the bat have greater tangential speeds.

5.

straight line, linear in time variable

7.

constant

9.

The centripetal acceleration vector is perpendicular to the velocity vector.

11.

a. both; b. nonzero centripetal acceleration; c. both

13.

The hollow sphere, since the mass is distributed further away from the rotation axis.

15.

a. It decreases. b. The arms could be approximated with rods and the discus with a disk. The torso is near the axis of rotation so it doesn’t contribute much to the moment of inertia.

17.

Because the moment of inertia varies as the square of the distance to the axis of rotation. The mass of the rod located at distances greater than L/2 would provide the larger contribution to make its moment of inertia greater than the point mass at L/2.

19.

magnitude of the force, length of the lever arm, and angle of the lever arm and force vector

21.

The moment of inertia of the wheels is reduced, so a smaller torque is needed to accelerate them.

23.

yes

25.

|r||r| can be equal to the lever arm but never less than the lever arm

27.

If the forces are along the axis of rotation, or if they have the same lever arm and are applied at a point on the rod.

Problems

29.

ω = 2 π rad 45.0 s = 0.14 rad/s ω = 2 π rad 45.0 s = 0.14 rad/s

31.

a. θ=sr=3.0m1.5m=2.0radθ=sr=3.0m1.5m=2.0rad; b. ω=2.0rad1.0s=2.0rad/sω=2.0rad1.0s=2.0rad/s; c. v2r=(3.0m/s)21.5m=6.0m/s2.v2r=(3.0m/s)21.5m=6.0m/s2.

33.

The propeller takes only Δt=Δωα=0rad/s10.0(2π)rad/s−2.0rad/s2=31.4sΔt=Δωα=0rad/s10.0(2π)rad/s−2.0rad/s2=31.4s to come to rest, when the propeller is at 0 rad/s, it would start rotating in the opposite direction. This would be impossible due to the magnitude of forces involved in getting the propeller to stop and start rotating in the opposite direction.

35.

a. ω=25.0(2.0s)=50.0rad/sω=25.0(2.0s)=50.0rad/s; b. α=dωdt=25.0rad/s2α=dωdt=25.0rad/s2

37.

a. ω=54.8rad/sω=54.8rad/s;
b. t=11.0st=11.0s

39.

a. 0.87rad/s20.87rad/s2;
b. θ=12,600radθ=12,600rad

41.

a. ω=42.0rad/sω=42.0rad/s;
b. θ=220radθ=220rad; c. vt=42m/sat=4.0m/s2vt=42m/sat=4.0m/s2

43.

a. ω=7.0rad/sω=7.0rad/s;
b. θ=22.5radθ=22.5rad; c. at=0.10m/s2at=0.10m/s2

45.

α=28.6rad/s2α=28.6rad/s2.

47.

r = 0.78 m r = 0.78 m

49.

a. α=−0.314rad/s2α=−0.314rad/s2,
b. ac=197.4m/s2ac=197.4m/s2; c. a=ac2+at2=197.42+(−6.28)2=197.5m/s2a=ac2+at2=197.42+(−6.28)2=197.5m/s2
θ=tan−1−6.28197.4=−1.8°θ=tan−1−6.28197.4=−1.8° in the counterclockwise direction from the centripetal acceleration vector

51.

ma=40.0kg(4.75m/s2)=190Nma=40.0kg(4.75m/s2)=190N
The maximum friction force is μSN=0.6(40.0kg)(9.8m/s2)=235.2NμSN=0.6(40.0kg)(9.8m/s2)=235.2N so the child does not fall off yet.

53.

vt=rω=1.0(2.0t)m/sac=vt2r=(2.0t)21.0m=4.0t2m/s2at(t)=rα(t)=rdωdt=1.0m(2.0)=2.0m/s2.vt=rω=1.0(2.0t)m/sac=vt2r=(2.0t)21.0m=4.0t2m/s2at(t)=rα(t)=rdωdt=1.0m(2.0)=2.0m/s2.
Plotting both accelerations gives

Figure shows a linear acceleration in meters per second squared plotted as a function of time in seconds. Centripetal starts at the origin of the coordinate system and grows exponentially with time. Tangential is positive and remains constant with time


The tangential acceleration is constant, while the centripetal acceleration is time dependent, and increases with time to values much greater than the tangential acceleration after t = 1s. For times less than 0.7 s and approaching zero the centripetal acceleration is much less than the tangential acceleration.

55.

a. K=2.56×1029J;K=2.56×1029J;
b. K=2.68×1033JK=2.68×1033J

57.

K = 434.0 J K = 434.0 J

59.

a. vf=86.5m/svf=86.5m/s;
b. The rotational rate of the propeller stays the same at 20 rev/s.

61.

K = 3.95 × 10 42 J K = 3.95 × 10 42 J

63.

a. I=0.315kg·m2I=0.315kg·m2;
b. K=621.8JK=621.8J

65.

I = 7 36 m L 2 I = 7 36 m L 2

67.

v = 7.14 m / s . v = 7.14 m / s .

69.

θ = 10.2 ° θ = 10.2 °

71.

F = 30 N F = 30 N

73.

a. 0.85m(55.0N)=46.75N·m0.85m(55.0N)=46.75N·m; b. It does not matter at what height you push.

75.

m 2 = 4.9 N · m 9.8 ( 0.3 m ) = 1.67 kg m 2 = 4.9 N · m 9.8 ( 0.3 m ) = 1.67 kg

77.

τ n e t = −9.0 N · m + 3.46 N · m + 0 3.38 N · m = −8.92 N · m τ n e t = −9.0 N · m + 3.46 N · m + 0 3.38 N · m = −8.92 N · m

79.

τ = 5.66 N · m τ = 5.66 N · m

81.

τ = 57.82 N · m τ = 57.82 N · m

83.

r × F = 4.0 i ^ + 2.0 j ^ 16.0 k ^ N · m r × F = 4.0 i ^ + 2.0 j ^ 16.0 k ^ N · m

85.

a. τ=(0.280m)(180.0N)=50.4N·mτ=(0.280m)(180.0N)=50.4N·m; b. α=17.14rad/s2α=17.14rad/s2;
c. α=17.04rad/s2α=17.04rad/s2

87.

τ = 8.0 N · m τ = 8.0 N · m

89.

τ = −43.6 N · m τ = −43.6 N · m

91.

a. α=1.4×10−10rad/s2α=1.4×10−10rad/s2;
b. τ=1.36×1028N-mτ=1.36×1028N-m; c. F=2.1×1021NF=2.1×1021N

93.

a = 3.6 m / s 2 a = 3.6 m / s 2

95.

a. a=rα=14.7m/s2a=rα=14.7m/s2; b. a=L2α=34ga=L2α=34g

97.

τ = P ω = 2.0 × 10 6 W 2.1 rad / s = 9.5 × 10 5 N · m τ = P ω = 2.0 × 10 6 W 2.1 rad / s = 9.5 × 10 5 N · m

99.

a. K=888.50JK=888.50J;
b. Δθ=294.6revΔθ=294.6rev

101.

a. I=114.6kg·m2I=114.6kg·m2;
b. P=104,700WP=104,700W

103.

v = L ω = 3 L g v = L ω = 3 L g

105.

a. a=5.0m/s2a=5.0m/s2; b. W=1.25N·mW=1.25N·m

Additional Problems

107.

Δ t = 10.0 s Δ t = 10.0 s

109.

a. 0.06rad/s20.06rad/s2; b. θ=16.7revolutionsθ=16.7revolutions

111.

s = 405.26 m s = 405.26 m

113.

a. I=0.363kg·m2I=0.363kg·m2;
b. I=2.34kg·m2I=2.34kg·m2

115.

ω = 6.68 J 4.4 kgm 2 = 1.23 rad / s ω = 6.68 J 4.4 kgm 2 = 1.23 rad / s

117.

F = 23.3 N F = 23.3 N

119.

α = 190.0 N-m 2.94 kg-m 2 = 64.4 rad / s 2 α = 190.0 N-m 2.94 kg-m 2 = 64.4 rad / s 2

Challenge Problems

121.

a. ω=2.0t1.5t2ω=2.0t1.5t2; b. θ=t20.5t3θ=t20.5t3; c. θ=−400.0radθ=−400.0rad; d. the vector is at −0.66(360°)=−237.6°−0.66(360°)=−237.6°

123.

I = 2 5 m R 2 I = 2 5 m R 2

125.

a. ω=8.2rad/sω=8.2rad/s; b. ω=8.0rad/sω=8.0rad/s

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