Question

A circular disc of mass m and radius R is placed on horizontal surface of coefficient of friction $\mu$. The disc is given an angular velocity $\omega$. Plane of disc is horizontal and there is no initial velocity of the center of mass.

• A. Net torque acting on disc is $\frac{\mu mgR}{3}$
• B. Net friction force acting on disc is $\mu mg$
• C. Time after which disc stops rotating is $\frac{3wR}{4\mu g}$
• D. Center of mass of disc will not displace

Answer 1

Answer: (C), (D)

The correct options are
C Time after which disc stops rotating is $\frac{3wR}{4\mu g}$
D Center of mass of disc will not displace


$df=\mu N=\mu \left(dm\right)g$
Say, $\sigma$ is the areal density, then
$df=\mu \sigma 2\pi rdrg$

The small torque,
$d\tau =rdf=\mu \sigma 2\pi g{r}^{2}dr$

$\tau =\int d\tau =\mu g\sigma 2\pi {\int }_0^R{r}^{2}dr$

$\tau =\mu g\frac{m}{\pi R^2}2\pi \frac{R^3}{3}=\frac{2}{3}\mu gmR$

$\tau =I\alpha \Rightarrow \frac{2}{3}\mu gmR=\frac{m{R}^2}{2}\alpha \Rightarrow \alpha =\frac{4\mu g}{3R}$

$w_{\text{final}}=w_{\text{initial}}+\alpha t\Rightarrow w=\frac{4\mu g}{3R}t$

$t=\frac{3wR}{4\mu g}$

Net friction is zero, so COM continues to stay at rest.