Question

A circular hoop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also a mass m and moving with 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) \[3V/4R\]

Answer 1

Hint: This problem involves collision and there is no loss of energy during the collision. Also, there is no friction present on the ground. It makes a simple problem of energy conservation.

Complete step by step answer:
In this problem, the bullet strikes the hoop and after the collision, it comes to rest. Since there is no loss of energy in the form of heat or sound, the entire energy of a bullet is transferred to the road. Hence, we can use the law of conservation of linear and angular momentum to solve this problem because the second body starts to rotate around its axis; the hoop is uniform and hence the mass is uniformly distributed.
Mass of bullet, = m
Mass of hoop= m
The radius of the hoop = R
The initial velocity of bullet = v
Initial velocity of hoop = 0
We need to find the final angular velocity of the bullet.
Using law of conservation of linear momentum, after the collision the bodies start rotating, so,
Momentum before collision= momentum after collision
\[mv+0=(m+m){{v}_{cm}}\]
\[{{v}_{cm}}=\dfrac{v}{2}\]
Now, By conservation of angular momentum,
\[\begin{align}
 & L=mvr \\
 & =mv\times \dfrac{R}{2} \\
\end{align}\]
Since the axis of rotation is at half the radius.
Inertia for the hoop, \[{{I}_{1}}=m{{r}^{2}}+m{{[\dfrac{r}{2}]}^{2}}\]
Inertia for the bullet, \[{{I}_{2}}=m{{[\dfrac{r}{2}]}^{2}}\]
Moment of inertia of the system= \[{{I}_{1}}+{{I}_{2}}\]
= \[\dfrac{3}{2}m{{R}^{2}}\],
now applying the conservation law,
$ mV\dfrac{R}{2}=\dfrac{3}{2}m{{R}^{2}}\omega \\
\therefore \omega =\dfrac{V}{3R} \\
$

So, the value of the angular velocity of the system comes out to be \[\omega =\dfrac{V}{3R}\]. Hence option B is correct.

Note:
We have to take the moment of inertia of the body given, we should keep in mind different bodies have different moments of inertia. Also, the moment of inertia depends upon the axis of rotation around which it is being calculated.