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 and gets embedded in it. The angular velocity with which the system rotates just after the bullet strikes the loop is?

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

Answer 1

Answer: (C)

$\frac{V}{3R}$

Center of mass of the loop- bullet system is $R/2$ from center of loop

By momentum consernation:

$mv=2m{V}_{cm}$

$V_{cm}=v/2$

Applying angular momentum conservation about the center of mass

$\frac{mVR}{2}=I_{c}w$

$I_c$ for ring$={mr}^2+m{\left(\frac{r}{2}\right)}^2$

$I_c$ for bullet$=m{\left(\frac{r}{2}\right)}^2$

Moment of inertia of the system, $I_c=\frac{6}{4}m{R}^2$

$=\frac{3}{2}m{R}^2$

$\frac{mvR}{2}=\frac{3}{2}m{R}^{2}w$

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