A potter’s wheel—a thick stone disk of radius 0.500 m
and mass 100 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 70.0 N. Find the effective coefficient of kinetic
friction between wheel and rag
The Attempt at a Solution
Actually, I could get the answer using angular acceleration and torque, but I could do the same thing with Newtonian Laws and I want to know why.
Tangential velocity of the wheel:
50 rev/ min = 2.618 m/s
u = 2.618 m/s, v = 0, a = ?, t = 6
v = u + at
0 = 2.618 + 6a
a = 0.436 m/s^2
F = ma = (100)(0.436) = 43.6N
The friction coefficient
43.6 = (coefficient)(70)
coefficient = 0.623, which is exactly the double of the correct answer.