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 $d x$ at a radius $x$.

Now $d N=\mu m g \cdot \frac{2 \pi x d x}{\pi R ^2} $
$=\frac{2 \mu m g x}{ R ^2} d x$
$\Rightarrow d \tau=x d N=\frac{2 \mu m g x^2 d x}{R^2} $
$\Rightarrow \int d \tau=\frac{2 \mu m g }{R^2} \int\limits_0^R x^2 d x=\frac{2}{3} \mu m g R=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.
$\alpha t=\omega \Rightarrow t=\frac{\omega}{\alpha}=\frac{3 \omega R }{4 \mu g }$