Wednesday

July 30, 2014

July 30, 2014

Posted by **Ánfer** on Wednesday, October 30, 2013 at 11:36pm.

"

A potter's wheel, a thick stone disk of radius 0.50 m and mass 100 kg, is freely rotating at 50 rev/min. The potter can stop the wheel in 6.0 s by pressing a wet rag against the rim and exerting a radially inward force of 70 N. Find the effective coefficient of friction between wheel and wet rag.

α = Δω/Δt

Δω = ωf - ωi = 0 - 50 rev /min

Δω = - 50 rev /min [ 2 π rad / rev ] [ min / 60 s ] = - 5.24 rad / s

α = Δω/Δt = [ - 5.24 rad / s ] / 6.0 s

α = - 0.87 rad / s2

This angular acceleration is produced by a torque due to the friction force exerted by the potter with the wet rag.

τ = Iα

τ = r Ff

Ff = µ Fn = µ(70 N)

To evaluate this numerically, we need the numerical value of the moment of inertia for the potter's wheel.

I = (1/2) M R2 = 0.5 (100 kg) (0.50 m)2 = 12.5 kg m2

τ = Iα = (12.5 kg m2) ( 0.87 rad / s2 ) = 10.875 m N

τ = r Ff

or

Ff = τ / r = (10.875 m N) / (0.50 m) = 21.75 N

Ff = µ FN = µ(70 N)

or

µ = Ff / 70 N = 21.75 N / 70 N

µ = 0.32

"

I get most of it; just not this part. How does torque equal to <frictional force times radius>? They never explained.

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