A potter\'s wheel -- a thick stone disk of radius 0.500 m and mass 125 kg -- is

Question

A potter's wheel -- a thick stone disk of radius 0.500 m and mass 125 kg -- is freely rotating at 50.0 rev/min. The potter can stop the wheel in 6.00 s by pressing a wet rag against the rim and exerting a radially inward force of 69.0 N. Find the effective coefficient of kinetic friction between the wheel and the rag.

Explanation / Answer

R- 0.500m M-125kg wi-50 rev/min t- 6s N-69N Moment of inertia of a wheel: I=MR^2 / 2 &its angular acceleration form wf = wi + at is, a= wf-wi / t => a= -wi/t since wf=0 the fricition is f = mu * N and torque = f * R SO, torque = | I * a | f * R = | (1/2MR^2) (-wi/t) | mu * N * R = (wi*M*R^2)/ 2t solve for mu then convert it by using 2pi = 1 rev & 1 min = 60s

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