# ch10 - CHAPTER 10 SOLUTION FOR PROBLEM 21(a Use 1 rev = 2...

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C HAPTER 10 S OLUTION FOR P ROBLEM 21 (a) Use 1 rev = 2 π rad and 1 min = 60 s to obtain ω = 200 rev 1 min = (200 rev)(2 π rad / rev) (1 min)(60 s / min) = 20 . 9 rad / s . (b) The speed of a point on the rim is given by v = ω r , where r is the radius of the flywheel and ω must be in radians per second. Thus v = (20 . 9 rad / s)(0 . 60 m = 12 . 5 m / s. (c) If ω is the angular velocity at time t , ω 0 is the angular velcoty at t = 0, and α is the angular acceleration, then since the anagular acceleration is constant ω = ω 0 + α t and α = ω ω 0 t = (1000 rev / min) (200 rev / min) 1 . 0 min = 800 rev / min 2 . (d) The flywheel turns through the angle θ , which is θ = ω 0 t + 1 2 α t 2 = (200 rev / min)(1 . 0 min) + 1 2 (800 rev / min 2 )(1 . 0 min) 2 = 600 rev .

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C HAPTER 10 S OLUTION FOR P ROBLEM 41 Use the parallel-axis theorem. According to Table 10–2, the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by I com = M 12 ( a 2 + b 2 ) . A parallel axis through a corner is a distance h = 0 ( a/ 2) 2 + ( b/ 2) 2 from the center, so I = I com + Mh 2 = M 12 ( a 2 + b 2 ) + M 4 ( a 2 + b 2 ) = M 3 ( a 2 + b 2 ) = 0 . 172 kg 3 J (0 . 035 m) 2 + (0 . 084 m) 2 o = 4 . 7 × 10 4 kg · m 2 .
C HAPTER 10 S OLUTION FOR P ROBLEM 55 (a) Use constant acceleration kinematics. If down is taken to be positive and a is the acceleration of the heavier block, then its coordinate is given by y = 1 2 at 2 , so a = 2 y t 2 = 2(0 . 750 m) (5 . 00 s) 2 = 6 . 00 × 10 2 m / s 2 .

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