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Number47
Number47 is offline
#1
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
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