Question

The figure shown is just before collision, the velocity of centre of uniform disc is $v_0$ vertically downward & $\omega_{0}$ is the angular velocity as shown. If is found that collision is elastic and after collision disc stops rotating then if coefficient of friction is $\left(\frac{1}{P}\right)$ then the value of $P$ is ? If $v_{0}=R\omega_0$. ($R$ is radius of disc)

Answer 1

$ f =\mu N $
$-R \int \mu N d t =-\frac{m R^{2} \omega_{0}}{2}$
$R \mu \int N d t=\frac{mR^{2}}{2} \omega_{0}=\frac{m R V_{0}}{2}$
$\int N d t=m V-\left(-m V_{0}\right)$
$\int N d t=m\left(V+V_{0}\right)$
$\int N d t=2 m V_{0}$
$2 \mu RmV _{0}=\frac{ mRV _{0}}{2}$
$\mu=\frac{1}{4}$