A potter's wheel having a radius 0.45 m and a moment of inertia of 14.9 kg · m2 is rotating freely at 49 rev/min. The potter can stop the wheel in 8.0 s by pressing a wet rag against the rim and exerting a radially inward force of 72 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.

1 answer

49 rev/min * 1 min/60 s * 2 pi radians/rev
= 5.13 radians/s

angular acceleration alpha = 5.13 rad/s / 8 s
= .641 radians/s^2

M = torgue or moment = I alpha
= 14.9 * .641 = 9.56 Newton meters

F = M/radius = 9.56/.45 = 21.2 Newtons

so
21.2 = mu * 72
mu = .295