Question

A particle of mass $m$ moving with speed $v$ collide inelastically with the stationary ring and gets embedded in it. Find the angular velocity with which the system rotates after the collision. Assume that the thickness of the ring is much smaller than its radius.

• A. $\frac{v}{2R}$
• B. $\frac{v}{3R}$
• C. $\frac{2v}{3R}$
• D. $\frac{3v}{4R}$

Answer 1

The correct option is B $\frac{v}{3R}$

We know body under free rotation, rotates about its centre of mass (C.O.M)
Therefore, system of ring and mass after collision will rotate about its C.O.M.
Since, the mass of both ring and particle is same, therefore new COM will lie in between them i.e. at a distance $R/2$ from the centre of ring.
According to question,



Since, there is no external torque acting, therefore applying law of conservation of angular momentum.
$L_i=L_f$
$mv{r}_{\perp }=I\omega =(I_{ring}+I_{particle})\omega$
$\frac{mvR}{2}=(m{R}^2+\frac{m{R}^2}{4}+\frac{m{R}^2}{4})\omega$
$[\because I_{ring}=I_o+m{d}^2]$
$\Rightarrow \omega =\frac{v}{3R}$