How does torque equal to <frictional force times radius [of solid cylinder]>?

Question

"
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.

Answer 1

Torque is equal to frictional force times radius because torque is a measure of the rotational force that is applied to an object. The frictional force is the force that is applied to the object in order to cause it to rotate, and the radius is the distance from the center of rotation to the point where the force is applied. Therefore, the torque is equal to the frictional force multiplied by the radius.