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

Answer: 4.00

$f=\mu N$
$-R\int \mu Ndt=-\frac{m{R}^2{\omega }_0}{2}$
$R\mu \int Ndt=\frac{m{R}^2}{2}{\omega }_0=\frac{mR{V}_0}{2}$
$\int Ndt=mV-\left(-m{V}_0\right)$
$\int Ndt=m\left(V+V_0\right)$
$\int Ndt=2m{V}_0$
$2\mu Rm{V}_0=\frac{mR{V}_0}{2}$
$\mu =\frac{1}{4}$