Rotational Motion Question 216
Question: A horizontal turn table in the form of a disc of radius $r$ carries a gun at $\mathrm{G}$ 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 0 is $I_0$ )
Options:
A) $\frac{mvr}{I _{0}+mr^{2}}$
B) $\frac{2mvr}{I _{0}}$
C) $\frac{v}{2r}$
D) $\frac{mvr}{2I _{0}}$
Show Answer
Answer:
Correct Answer: A
Solution:
[a] 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
$L_i(I_0+m r^2) \omega_0$ - (i) Let $w$ be the nagular velocity of table after the bullet is fired.
Final angular momentum
$L f=(I_0 \omega-m(v-r \omega)) r.$ - (ii) Where $(v-r \omega)$ is absolute velocity of bullet to the right.
Equations (i) and (ii);
we get $(\omega-\omega _0)=\frac{m v r}{I _0+m r^2} $
BETA
