Question

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 $I_0$)

• A. $\frac{mvr}{I_0+m{r}^2}$
• B. $\frac{2mvr}{I_0}$
• C. $\frac{v}{2r}$
• D. $\frac{mvr}{2I_0}$

Answer 1

The correct option is A $\frac{mvr}{I_0+m{r}^2}$

Let the final angular velocity of the "gun + table "system be $w$.

Given: velocity of bullet w.r.t gun is $v$.

Thus velocity of bullet w.r.t ground, $v^{\prime }=v-rw$

Applying conservation of angular momemtum: $L_i=L_f$

$(m{r}^2+I_o)w_o=I_{o}w-m{v}^{\prime }r$

$(m{r}^2+I_o)w_o=I_{o}w-m(v-rw)r$

$⟹w=\frac{(I_o+m{r}^2)w_o+mvr}{I_o+m{r}^2}$

Increase in angular velocity, $\Delta w=w-w_o=$$\frac{(I_o+m{r}^2)w_o+mvr}{I_o+m{r}^2}-w_o$

$⟹\Delta w=\frac{mvr}{I_o+m{r}^2}$