A thin ring of radius R is rolling without slipping on a horizontal surface with velocity v₀. The ring strikes a pin P at a height h above the surface and stops as well as stops rotating. Then h is equal to

Before the impact angular momentum of the ring about P, $\mathrm{Li}={\mathrm{mv}}_0\left(\mathrm{h}-\mathrm{R}\right)-\left({\mathrm{mR}}^2\right)\left(\frac{{\mathrm{v}}_0}{\mathrm{R}}\right)$

After the impact angular momentum of the ring about P, ${\mathrm{L}}_{\mathrm{f}}=0$

Since angular momentum about P is conserved,

${\mathrm{mv}}_0\left(\mathrm{h}-\mathrm{R}\right)-\left({\mathrm{mR}}^2\right)\left(\frac{{\mathrm{v}}_0}{\mathrm{R}}\right)=0$

$\Rightarrow \mathrm{h}=2\mathrm{R}$