View Single Post
Apr17-10, 09:08 PM
P: 5
1. The problem statement, all variables and given/known data
A potter's wheel having a radius 0.55 m and a moment of inertia 11.6 kgm2 is rotating freely at 55 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 66 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.

2. Relevant equations
for a solid wheel I = (1/2)m*R^2

3. The attempt at a solution
I started off converting 55rpm into 5.76 rad/s. I then divided 5.76rad/s by 8.00s to get the needed acceleration to stop the wheel and got -.72rad/s^2. Next i solved for the mass of the wheel by dividing the moment of inertia by .5*R^2 and got 7.018kg.

I figured by setting the force needed to stop the wheel equal to the force applied by the rag times the coefficient of friction i could divide the applied force on both sides and end up with the COF: m*a = F*u u=(m*a)/F u= (7.018kg*-.72rad/s^2)/66N = .07656
Phys.Org News Partner Science news on
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage