## Physics: Principles with Applications (7th Edition)

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