Question

A 1.9 m radius cylinder with a mass of 531.1 kg rolls without slipping down a hill which is 56.5 meters high. At the bottom of the hill, what fraction of its total kinetic energy is invested in rotational kinetic energy?

Answer 1

si 9-MSAT sy 212238 = 91.3 mjö'x (186*(1-0) T=555*c final temparature. 8= 1.am m = 531.1 kg h 56.5m @VP? 1 = 4 mm 5,4x531-4x0.9% 1 = 958.64 kg-ugle maghput of k, = Uz fK2 a mgh to=o+(Amut 2 2 w²) We know w = 1 = 2mg2 nigh = 2 mut ļ (2 mm) 4 3 m g h = 1 mv ² + + mv = Bmw va 48h va 1435 – 449-8850.5 V=27.111 mls. (final velocity at the bottom) Total Kf = U, = mgh, = 531.1x9.8X 56.5 - 3 + k m ) KP Rotational = 3 mad = $ 10 mm= £7531/(87-17002 KErotational = 196,045-8J KErotational 196,045.8 KE total 894,070. 1 9 0.667 FKE rotational 2=0.667 KE total

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