Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 38

Answer

a. $0.50 kg \cdot m^2$ b. $ 2.4 \times 10^{-2} N \cdot m$

Work Step by Step

a. Treat the ball like a point mass. $$I=MR^2=(0.350 kg)(1.2m)^2\approx 0.50 kg \cdot m^2$$ b. The net torque is zero because the angular acceleration is zero, so the torque needed has the same magnitude as the torque caused by air friction. (0.020 N)(1.2m) = $ 2.4 \times 10^{-2} N \cdot m$.
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