homework 10 solutions

# homework 10 solutions - Chapter 10 1 The problem asks us to...

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Unformatted text preview: Chapter 10 1. The problem asks us to assume vcom and a) are constant. For consistency of units, we write 5280 ﬁ/mi _ = 7480 ﬁ/min . 60mm/h vcom = (85h) [ Thus, with Ax = 60 ﬂ , the time of ﬂight is t = Ax/v = (60 ft) /(7480 ft/min) = 0.00802 min. com During that time, the angular displacement of a point on the ball’s surface is 6 = wt = (1800 rev/min) (0.00802 min) z 14 rev . 5. Applying Eq. 2—15 to the vertical axis (with +y downward) we obtain the free-fall time: Ay=v0yt+lgt2 :>t= 2(10 “0 =1.4 s. 2 9.8 m/s2 Thus, by Eq. 10-5, the magnitude of the average angular velocity is a) _ (2.5 rev) (272 rad/rev) avg 1.4 s =11 rad/s. 12. (a) We assume the sense of rotation is positive. Applying Eq. 10-12, we obtain a)=a)0+at :> a=W=9.0x103rev/min2. (12/ 60) min (b) And Eq. 10-15 gives 6? = gm)0 + co)t = %(1200 rev/min + 3000 rev/min) min] = 4.2 ><102 rev. 35. Since the rotational inertia of a cylinder is I = % MR2 (Table lO-2(c)), its rotational kinetic energy is K :1le =1MR2w2. 2 4 (a) For the smaller cylinder, we have 1 K1 :20 .25 kg)(0.25 m)2(235 rad/s)2 =1 .08><103 J z 1. 1x103 J. (b) For the larger cylinder, we obtain 1 K2 =Z(1.25 kg)(0.75 m)2(235rad/s)2 =9.7l><103 1:9.7x103 J. 45. [We take a torque that tends to cause a counterclockwise rotation from rest to be posmve and a torque tending to cause a clockwise rotation to be negative. Thus, a positive torque of magnitude r1 F1 sin 61 is associated with 131 and a negative torque of magnitude r2F2 sin 62 is associated with 132. The net torque is consequently r = 13171 sin 6?, —r2F2 sin 62. Substituting the given values, we obtain r = (1.30 m)(4.20 N) sin75° — (2.15 m)(4.90 N) sin 60° = —3.85 N - m. 63. We use E to denote the length of the stick. Since its center of mass is E/ 2 from either end, its initial potential energy is émgﬂ, where m is its mass. Its initial kinetic energy is zero. Its ﬁnal potential energy is zero, and its ﬁnal kinetic energy is §Ia)2, where I is its rotational inertia about an axis passing through one end of the stick and a) is the angular velocity just before it hits the ﬂoor. Conservation of energy yields lmg€=11a12 :>a)= 2 2 I The free end of the stick is a distance 2 from the rotation axis, so its speed as it hits the ﬂoor is (from Eq. 10-18) ~ ‘ . 3 v = aw = 1’ mg! . I Using Table 10-2 and the parallel—axis theorem, the rotational inertial is I = §m€2 , so v: 3ge= 3(9.8m/s2)(l.00m)= 5.42m/s. ...
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• Spring '08
• Richardr

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