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

Answer 1

The correct option is A $\frac{V}{4R}$

Let after collision velocity of $COM$ of the loop along with bullet is $Ccm$

Now applying conservation of linear momentum

We get,

$mV=2m{V}_{c}m$

$\Rightarrow V_{cm}=\frac{V}{2}$

Considering the rotation motion C.M we get

$m.\frac{V}{2}.R=T_{cm}$

(Here only the angular momentum bullets is considered as other is zero)

$w=$ Angular velocity of composed system.

$I_c=$ moment of inertia of bullet and loop system.

=moment of inertia of bullet about the centre of mass + moment of inertia of bullet about COM

$=m{R}^2+m{R}^2=2m{R}^2$

so, we have,

$w=\frac{mVR}{2I_c}=\frac{mVR}{m{R}^2}$

$w=\frac{V}{4R}$