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

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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.

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.

In this example, the torque (τ) is equal to the frictional force (Ff) times the radius (r) of the solid cylinder. The torque is the rotational equivalent of force in linear motion and is responsible for changing the angular velocity of an object.

In general, the torque on an object can be calculated by multiplying the force applied on the object by the perpendicular distance from the axis of rotation to the line of action of the force. In this case, the force applied is the frictional force between the wet rag and the wheel, and the radius is the distance from the axis of rotation (center of the wheel) to the point where the force is applied (the rim).

So, mathematically, the torque (τ) can be written as:

τ = r * Ff

Where:
τ = torque
r = radius
Ff = frictional force

The equation τ = r * Ff states that the torque is directly proportional to the force and the radius, meaning that if either the force or the radius increases, the torque will increase as well. In other words, the larger the force applied at a certain radius from the axis of rotation, the greater the torque produced.

In the given problem, the torque of the potter's wheel is produced by the friction force exerted by the potter using a wet rag. Torque can be defined as the rotational equivalent of force, and it is calculated as the product of the force applied perpendicular to the radius and the radius itself.

Consider a solid cylinder with radius 'r' and a frictional force 'Ff' acting on it at the rim. The torque 'τ' exerted by this force can be expressed as:

τ = r * Ff

Here, 'r' represents the distance from the axis of rotation (center of the wheel) to the point where the force is applied, and 'Ff' is the frictional force opposing the motion.

In the given problem, the moment of inertia of the potter's wheel is denoted by 'I', and the angular acceleration is denoted by 'α'. In this case, the torque 'τ' is also expressed as the product of the moment of inertia and the angular acceleration:

τ = I * α

Since torque remains constant in this case, we can equate these two expressions for torque:

I * α = r * Ff

By rearranging the equation, we can solve for the frictional force 'Ff':

Ff = (I * α) / r

Therefore, in this problem, the torque is equal to the frictional force multiplied by the radius of the solid cylinder.