Question

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

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

Answer 1

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

$C_1{C}_2=\frac{R}{2}$
where, $C_2=$ combined centre of mass of ring and particle.
From conservation of angular momentum about $C_2$
$L_i=L_f$
$\therefore \frac{mv\cdot R}{2}=I\omega =[m{R}^2+\frac{m{R}^2}{4}+\frac{m{R}^2}{4}]\omega$
Solving this equation we get,
$\omega =\frac{v}{3R}$