Question

A disc of radius $R$ and mass $m$ is projected on to a horizontal floor with a backward spin such that its centre of mass speed is $v_o$ and angular velocity is ${\omega }_o$. If the minimum value of ${\omega }_o$ so that the disc eventually returns back is $\frac{x{v}_o}{R}$, find the value of $x$.

Answer 1

Initially, there is forward slipping. Therefore, friction is backwards and maximum
Let velocity becomes zero in time $t_1$ and angular velocity becomes zero in time $t_2$
Then, $0=v_0-a{t}_1$ or

$t_1=\frac{v_0}{a}=\frac{v_0}{\mu g}$ ....(i)
and $0={\omega }_0-\alpha t_2$ or

$t_2=\frac{{\omega }_0}{\alpha }$ Here,

$\alpha =\frac{\mu mgR}{\frac{1}{2}m{R}^2}=\frac{2\mu g}{R}$

$\therefore t_2=\frac{{\omega }_{0}R}{2\mu g}....\left(ii\right)$
Disk will return back when

$t_2>t_1$

or $\frac{{\omega }_{0}R}{2\mu g}>\frac{v_0}{\mu g}$

or ${\omega }_0>\frac{2v_0}{R}$