Question

A disc of radius $r$ is rotating about its centre with an angular speed ${\omega }_0$ it is gently placed on a rough horizontal surface. After what time it will be in pure rolling?

• A. $\frac{{\omega }_{0}r}{2\mu g}$
• B. $\frac{{\omega }_{0}r}{3\mu g}$
• C. $\frac{{\omega }_{0}r}{\mu g}$
• D. $\frac{3}{2}\frac{{\omega }_{0}r}{\mu g}$

Answer 1

The correct option is B $\frac{{\omega }_{0}r}{3\mu g}$

The acceleration due to friction is given as,

$a=\frac{f}{m}$

$=\frac{\mu mg}{m}$

$=\mu g$

Angular acceleration due to friction is given as,

$\alpha =\frac{\tau }{I}$

$=\frac{fR}{\frac{1}{2}m{R}^2}$

$=\frac{2\mu g}{R}$

The final angular velocity is given as,

$\omega ={\omega }_0-\alpha t$

$\omega ={\omega }_0-\frac{2\mu gt}{R}$

The final velocity is given as,

$v=at$

$v=\mu gt$

we know that,

$v=R\omega$

$\mu gt=R({\omega }_0-\frac{2\mu gt}{R})$

$t=\frac{{\omega }_{0}R}{3\mu g}$

Thus, the time is $\frac{{\omega }_{0}R}{3\mu g}$ after that the disc will be in pure rolling.