Question

A circular hoop of mass $m$ and radius $R$ rest flat on a horizontal frictionless surface. A bullet, also of mass $m$, and moving with velocity $v$. Strikes the hoop and gests cmbedded in it. The thickness of the hoop is much smaller than $R$. The angular velocity with which the system rotates after the bullet strikes the hoop is

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

Answer 1

The correct option is C $\frac{2v}{3R}$

Given,

Total inertia = Inertia of disc + inertia of embedded bullet

Total inertia, $I=\frac{1}{2}m{R}^2+m{R}^2=\frac{3m{R}^2}{2}$

From conservation of angular momentum.

$mvR=I\omega$

$\omega =\frac{mv}{I}=\frac{mvR}{\frac{3m{R}^2}{2}}=\frac{2v}{3R}$

$\omega =\frac{2v}{3R}$

Hence, angular velocity is $\frac{2v}{3R}$