Question

A circular hoop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also of mass m and moving with a velocity v, strikes the hoop and gets embedded 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. v/4R
• B. v/3R
• C. 2v/3R
• D. v/2R

Answer 1

The correct option is B

v/3R

Let system moves with $v_1$ after impact $\&$ rotates with $\omega$ anti-clockwise

Consider any point P on a straight horizontal line passing through the centre of the hoop 0 and parallel to $\overrightarrow{v}$

Angular momentum initial

${\overrightarrow{L}}_i=mvR\hat{j}$ (angular momentum of bullet about point p)

Angular momentum final

${\overrightarrow{L}}_f=\left\{m\right(v_1+\omega R)R+I_{CM}\omega \}\hat{j}$

${\overrightarrow{L}}_f=(m{v}_{1}R+2m\omega R^2)\hat{j}$

As there is no net external torque about p,

${\overrightarrow{L}}_i={\overrightarrow{L}}_f$

$\Rightarrow v_1+2\omega R=v$ - - - - - - (1)

Now by conservation of linear momentum,

$m{v}_1+m(v_1+\omega R)=mv$

$\Rightarrow 2v_1+\omega R=v$ ....(2)

Subtracting (2) From (1) we get

$v_1=\omega R$

substituting the above in equation (2)

$2\omega R+\omega R=v$

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

hence B is the correct option