ssm_ch10 - Chapter 10 Student Solutions Manual 13. We take...

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Chapter 10 – Student Solutions Manual 13. We take t = 0 at the start of the interval and take the sense of rotation as positive. Then at the end of the t = 4.0 s interval, the angular displacement is θω α =+ 0 1 2 2 tt . We solve for the angular velocity at the start of the interval: ( )( ) 2 2 1 2 1 2 2 0 120 rad 3.0 rad/s 4.0 s 24 rad/s. 4.0 s t t θα ω == = We now use = 0 + α t (Eq. 10-12) to find the time when the wheel is at rest: t =− 0 2 24 80 rad / s 3.0 rad / s s. . That is, the wheel started from rest 8.0 s before the start of the described 4.0 s interval. 21. (a) We obtain ( /m in . 200 60 20 9 rev / min)(2 rad / rev) s rad / s. π (b) With r = 1.20/2 = 0.60 m, Eq. 10-18 leads to (0.60)(20.9) 12.5 m/s. vr = (c) With t = 1 min, = 1000 rev/min and 0 = 200 rev/min, Eq. 10-12 gives = = o rev t 800 2 in . (d) With the same values used in part (c), Eq. 10-15 becomes = + = 1 2 1 2 200 100 1 600 o rev. b g t () ( ) 29. (a) Earth makes one rotation per day and 1 d is (24 h) (3600 s/h) = 8.64 × 10 4 s, so the angular speed of Earth is 5 4 2 rad 7.3 10 rad/s. 8.64 10 s π × × (b) We use v = ω r , where r is the radius of its orbit. A point on Earth at a latitude of 40° moves along a circular path of radius r = R cos 40°, where R is the radius of Earth (6.4 × 10 6 m). Therefore, its speed is 56 ( cos 40 ) (7.3 10 rad/s)(6.4 10 m)cos40 3.5 10 m/s. vR = × × ° = × 2
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(c) At the equator (and all other points on Earth) the value of ω is the same (7.3 × 10 –5 rad/s). (d) The latitude is 0° and the speed is 56 (7.3 10 rad/s)(6.4 10 m) 4.6 10 m/s. vR ==× × = × 2 33. The kinetic energy (in J) is given by KI = 1 2 2 , where I is the rotational inertia (in ) and is the angular velocity (in rad/s). We have kg m 2 == ( . 602 60 630 rev / min)(2 rad / rev) s/min rad / s. π Consequently, the rotational inertia is I K = 2 2 24400 12 3 2 ( (. .. J) rad / s) kg m 2 2 35. We use the parallel axis theorem: I = I com + Mh 2 , where I com is the rotational inertia about the center of mass (see Table 10-2(d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h = 0.50 m – 0.20 m = 0.30 m. We find IM L com 2 kg m kg m = × 1 12 1 12 056 10 4 67 10 2 2 2 .
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ssm_ch10 - Chapter 10 Student Solutions Manual 13. We take...

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