A horizontal turn table in the form of a disc of radius `r` carries a gun at `G` and rotates with angular velocity `omega_(0)` about a vertical axis passing through the centre `O`. The increase in angular velocity of the system if the gun fires a bullet of mass `m` with a tangential velocity `v` with respect to the gun is (moment of inertia of gun `+` table about `O` is `I_(0)`

Options

A. `(mvr)/(I_(0)+mr^(2))`
B. `(2mvr)/(I_(0))`
C. `v/(2r)`
D. `(mvr)/(2I_(0))`

Answer

A

Explanation

Given that `I_(0)` is the moment of inertia of table and gun and m the mass of bullet.
Initial angular momentum of system about centre:
`L-i=(I_(0)+mr^(2))omega_(0)`……..i
Let `omega` be the angular velocity of the table affter the bullet is fired.
Final angular momentum
`L_(f)=I_(0)omega-m(v-romega)r`.........ii
Where `(v-romega)` is absolute velocity of bullet to the right.
Equating i and ii we get `(omega-omega_(0))=(mvr)/(I_(0)+mr^(2))`
This is also the increase in angular velocity.