Posted by Ánfer on Wednesday, October 30, 2013 at 11:36pm.
How does torque equal to <frictional force times radius [of solid cylinder]>?
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
Ff = τ / r = (10.875 m N) / (0.50 m) = 21.75 N
Ff = µ FN = µ(70 N)
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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