A potters wheela thick stone disk of radius 0500 m and mass

A potter\'s wheel-a thick stone disk of radius 0.500 m and mass 115 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 75.0 N. Find the effective coefficient of kinetic friction between the wheel and rag.

Solution

We have resistive force acting on the wheel, which would impart the wheel a negative angular acceleration.

Now, we know that the wheel takes 6 seconds to stop from a angular speed of 50 rev/min.

We will make use of the relation between the angular acceleration and the torque to form an equation involding the friction acting on the wheel as the friction is the one providing the angular acceleration.

So, angular speed = [50 x 2(3.14) / 60] rad/s = 5.233 rad/s

That is, angular acceleration = -5.233/6 = -0.8722 rad/s²

So the torque being provided by the friction acting between wheel and rag is Ineria/ angular acceleration

Or, Torque = [115 (0.25) /2] / 0.8722 = 16.4813 N-m

Now this torque is being provided by the friction

So, NR = 16.4813

or, = 16.4813 / 75*0.5 = 0.4395

Therefore the required coefficient of kinetic friction is 0.4395