 # A potter’s wheel having a radius of 0.50 m and a moment of inertia of 12 kg m^2 is rotating freely 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 kinetic friction between the wheel and the wet rag.

This question aims to find the coefficient of kinetic friction between the wheel and the wet rag.

The opposition of any substantial body to its velocity change is defined as inertia. This involves changes in the direction of movement or the speed of the body. The moment of inertia is a quantifiable measure of a body’s rotational inertia, which means that the body possesses resistance to its rotation speed about an axis and which changes when the torque is applied. The axis can be internal or external, and may or may not be fixed.

The quantity of retarding force between the relative motion of two bodies is said to be sliding, moving friction, or kinetic friction. The movement of two surfaces incorporates kinetic friction as well. When a body on a surface is moved, it is subjected to a force whose direction is opposite to the direction of its motion. The force’s magnitude will be dependent on the coefficient of kinetic friction between two bodies. This is critical for comprehending the coefficient of kinetic friction. Rolling, sliding, static friction, etc are some examples of friction.  Also, kinetic friction incorporates a friction coefficient generally known as the coefficient of kinetic friction.

Let $\alpha$ be the angular acceleration, then:

$\alpha=\dfrac{w_f-w_i}{\Delta t}$

Since $w_f=0$, so that:

$\alpha=-\dfrac{w_i}{\Delta t}$

Let $\tau$ be the torque, then:

$\tau=I\alpha$

$\tau=-\dfrac{Iw_i}{\Delta t}$

Let $f$ be the friction force, then:

$f=-\dfrac{\tau}{r}$

Or $f=\dfrac{Iw_i}{r(\Delta t)}$

Here, $I=12\,kg\cdot m^2$, $w_i=50\,rev/min$, $r=0.50\,m$ and $\Delta t=60\,s$, and so the friction force will be:

$f=\dfrac{12\,kg\cdot m^2\times 50\,rev/min}{0.50\,m\times 60\,s}\times \dfrac{2\pi\, rad}{1 \,rev}\times \dfrac{1\,min}{60\,s}$

$f=21\,N$

Finally, let $\mu_k$ be the coefficient of friction, then:

$\mu_k=\dfrac{f}{f_n}$

$\mu_k=\dfrac{21\,N}{70\,N}$

$\mu_k=0.30$

## Example

A $3\,kg$ block is lying on a rough surface and a force of $9\, N$ is applied to it. The block is subjected to frictional forces as it moves across the surface. Suppose that the coefficient of friction is $\mu_k=0.12$, work out the magnitude of the friction force opposing the motion.

### Solution

Since $\mu_k=\dfrac{f}{f_n}$, so that:

$f=\mu_k f_n$

Here, $f_n$ is the normal force which can be calculated as:

$f_n=mg$

$f_n=(3\,kg)(9.81\,m/s^2)$

$f_n=29.43\,N$

And so, the kinetic friction force can be calculated as:

$f=(0.12)(29.43\,N)$

$f=3.53\,N$

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