Question

A circular disc of mass $m$ and radius $R$ is rotating on a rough surface having a coefficient of friction $\mu$ with an initial angular velocity $\omega$. Assuming a uniform normal reaction on the entire contact surface, the time after wisich the disc comes to rest is

• A. $\frac{\omega R}{\mu g}$
• B. $\frac{3\omega R}{4\mu g}$
• C. $\frac{1}{2}\frac{\omega R}{\mu g}$
• D. $\frac{\sqrt{3}}{2}\frac{\omega R}{\mu g}$

Answer 1

Answer: (B)

Consider a small element of disc of thickness $dx$ at a radius $x$.

Now $dN=\mu mg\cdot \frac{2\pi xdx}{\pi R^2}$
$=\frac{2\mu mgx}{R^2}dx$
$\Rightarrow d\tau =xdN=\frac{2\mu mg{x}^{2}dx}{R^2}$
$\Rightarrow \int d\tau =\frac{2\mu mg}{R^2}\underset{0}{\overset{R}{\int }}{x}^{2}dx=\frac{2}{3}\mu mgR=I\alpha =\frac{1}{2}m{R}^2\alpha$
$\Rightarrow \alpha =\frac{4}{3}\frac{\mu g}{R},$ If $t$ be the time for complete stop.
$\alpha t=\omega \Rightarrow t=\frac{\omega }{\alpha }=\frac{3\omega R}{4\mu g}$