Question

Problem 12.35 - Enhanced - with Feedback An 7.20-cm-diameter, 350 g solid sphere is released from rest at the top of a 2.00-m-long, 20.0 incline. It rolls, without slipping, to the bottom. Part A You may want to review (Pages 315 - 317). What is the sphere's angular velocity at the bottom of the incline? Express your answer with the appropriate units. Value Units Submit Request Answer Part B What fraction of its kinetic energy is rotational? V AEP O ?

Answer 1

Answers Given , Diameter d= 7.20 cm = 0.072019 radius r = 0.042 = 0.036m Mass m = 350 y = 0.350 kq Angle o = 20° la) What is the sphere's angular velocity at the bottom of the incline? let, Height of incline, H = 2 Sin 2o H = 0.684 m 1 loss of potential Enelgy a hain in kinetic Energy mgh = {m2 + hul where, I forsphere = // mre yágh = 1 ² + 14 12 2 mrtwe where = vw gh = 1 V² + / s v2

10*9.81 +0.684 7 67 look 7 v2 = 67 100k v2 = 9.58537 No 54.58597 jl= 3:096 wils There fore, 3:096 0.036 wwř - Bot 10 = 86 rad/s) past Bo what fraction of its kinetic Energy is rotational? As we seen above, Total tinetic Energy a 2 mua and W2 Rotational kinetic Energy z

Therefore, Rotatishal kit 1/5 not? Total ke 20.2857 = 28.57%. Retatichal ko Totul( 6 = 0.2857