# 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*}

In the last line that's not $u$ that's $\mu$. Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other to solve the problem?

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

## 1 answer

If the object were sliding the friction force would be $\mu N$ where $N$ is the normal contact force. $N=Mg$ so the first part $\mu Mg$ is to do with the total friction force.

The problem is that it's rotating, and torque that slows the disc (from a given part of the disc) depends on the the distance from the middle that each part of the disc is.

For example a square cm area at the edge has more slowing effect than a square cm near the middle.

To get things ready for the later integration the person has found a formula for the force from a circular segment a distance $x$ from the middle, of width $dx$ and circumference $2\pi x$. The area of this is $2\pi x dx$

So the second part of the formula is needed to find the right fraction of the total friction force that should be allocated to the circular segment at radius $x$, that's why it's also been divided by the total area $\pi R^2$.

Then the later integration moves onto finding the torque from that segment and that depends on the distance from the middle, hence the extra $x$...

## 1 comment

This is a great example of a question best suited to the 'Problems' category, so I've decided to move it here. If people disagree with this decision, I can move it back — Mithrandir24601 8 days ago