Question

A circular ring of mass $m$ and radius $R$ rests flat on a horizontal smooth surface as shown in the figure. A particle of mass $m$, and moving with a velocity $v$, collides inelastically $(e=0)$ with the ring. Find the angular velocity with which the system rotates after the particle strikes the ring.

Answer 1

Given: A circular ring of mass m and radius R rests flat on a horizontal smooth surface as shown in the figure. A particle of mass m, and moving with a velocity v, collides in elastically (e=0) with the ring.

To find the angular velocity with which the system rotates after the particle strikes the ring.

Solution:

By momentum conservation :

$mv=2m{V}_{cm}$

$V_{c}m=\frac{v}{2}$

Applying angular momentum conservation about the centre of mass.

$m\frac{v}{2}r=I\omega$

$I_{ring}=m{r}^2+m{\left(\frac{r}{2}\right)}^2$

moment of inertia of the system $I=\frac{5}{4}m{r}^2$

Hence $m\frac{v}{2}r=\frac{5}{4}m{r}^2\omega ⟹\omega =\frac{2v}{5r}$

is the angular velocity with which the system rotates after the particle strikes the ring