Question

A circular wooden loop 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 loop at the bottommost point and gets embedded in it. The thickness of the loop is much smaller than $R$. The angular velocity with which the system rotates just after the bullet strikes the loop is

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

Answer 1

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

Assume that just after the collision, velocity of COM of the system is $V_{\text{COM}}$ and angular velocity about COM is $\omega$.


Just after collision:
Distance of COM of the system from $O$.
$y_{\text{COM}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$
$y_{\text{COM}}=\frac{m\left(0\right)+m\left(R\right)}{m+m}=\frac{R}{2}$
There is no external force on the system. Hence linear momentum will be conserved.
$mV+0=(m+m)V_{\text{COM}}$
$V_{\text{COM}}=\frac{V}{2}$ ... (1)
There is no torque on the system. Hence angular momentum will be conserved. Thus about COM,
$L_i=L_f$
where $L_i=mv\left(\frac{R}{2}\right)$
$L_f=L_{\text{Translational}}+L_{\text{Rotational}}$
Here, $L_{\text{Translational}}=0$ {because $V_{\text{COM}}$ passing through COM}
i.e $L_f=I_{COM}\omega$
MOI of system (loop + bullet) about COM
$I_{\text{COM}}={\left(I_{\text{loop}}\right)}_{\text{COM}}+{\left(I_{\text{bullet}}\right)}_{\text{COM}}$
$=(m{R}^2+m{\left(\frac{R}{2}\right)}^2)+m{\left(\frac{R}{2}\right)}^2$
$\Rightarrow I_{\text{COM}}=\frac{3m{R}^2}{2}$
Therefore, $L_i=L_f$
$\Rightarrow mV\left(\frac{R}{2}\right)=\left(\frac{3}{2}m{R}^2\right).\omega$
$\therefore \omega =\frac{V}{3R}$