# Slipping and rotation

A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?

Someone had solved the problem following way

\begin{align*}
\mathrm{df} &=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx} \\

\implies \mathrm{d} \tau &=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}} \\
\implies\int \mathrm{d} \tau &=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x}
\end{align*}
Again, \begin{align*}
\tau &=\frac{2}{3} \mu \mathrm{mgR} \\
\tau &=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha \\
\implies \alpha &=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}} \\
\alpha \mathrm{t}=\omega \implies \mathrm{t}&=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}}
\end{align*}

Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other way to solve the problem?

$$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$

## 1 answer

I know this is a bit late, but if you still haven't figured it out, here goes:

Firstly, let the mass per unit area of the disk be $\sigma = \frac{m}{\pi R^2}$.

Consider a small element of area $dA$ at a distance $x$ from the center of the disk. The mass of this element is $dm = \sigma dA$. The frictional force on this element is $dF = \mu (dm) g = \mu \sigma g \times dA$

The torque on this small element is $d\tau = x \times dF = \mu \sigma g x \times dA$. (note that by $\times$ I just mean normal multiplication, not the cross product or anything).

Now let us consider a thin circular ring with thickness $dx$ at a distance $x$ from the center (I mean to say a ring of radius $x$ whose center is the center of the disk). The net torque $\tau$ on this ring would be the integral of all the small torques on the small elements of area $dA$ lying on this thin ring.

$$\tau = \int d\tau = \mu \sigma g x \int dA = \mu \sigma g \times (2 \pi x^2 dx)$$

Note that in this integral $x$ is a constant, since $x$ is the distance from the center for all the small elements on the circular ring we have taken under consideration.

As $dx \to 0$, we have infinite such rings on the disk. So now we need to integrate over all these rings. The equation above matches the second equation of your post.

I think you should be able to take it from here. The important thing is to keep in mind what we are doing: first we find the net torque on a thin ring situated at a distance $x$ from the center which I've done above, and then we integrate over all such rings.

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