- If we take moment at A then external torque will be zero.
Therefore, the initial angular momentum = the angular momentum after rotation stops (i.e. only leniar velocity exits)
$\text{Mv}\times\text{R}-\ell\omega=\text{Mv}_0\times\text{R}$
$\Rightarrow\text{MvR}-\frac{2}{5}\times\frac{\text{MR}^2\text{V}}{\text{R}}=\text{Mv}_0\text{R}$
$\Rightarrow\text{v}_0=\frac{\text{3V}}{5}$
- Again, after some time pure rolling starts
Therefore,
$\Rightarrow\text{M}\times\text{v}_0\times\text{R}=\Big(\frac{2}{5}\Big) \text{MR}^2\times\Big(\frac{\text{V}'}{\text{R}}\Big)+\text{Mv}'\text{R}$
$\Rightarrow\text{m}\times\Big(\frac{\text{3V}}{5}\Big)\times\text{R}=\Big(\frac{2}{5}\Big)\text{Mv}'\text{R}+\text{Mv}'\text{R}$
$\Rightarrow\text{V}'=\frac{3\text{V}}{7}$
