A rod of mass m and length l is placed on a s | vedclass

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(B) Let $v_1$ be the velocity of the shot with respect to the ground and $\omega$ be the angular velocity of the rod.The velocity of the gun (at the e

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$\frac{6}{43} \frac{v}{l}$

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(B) Let $v_1$ be the velocity of the shot with respect to the ground and $\omega$ be the angular velocity of the rod.The velocity of the gun (at the end of the rod) is $v_g = \omega \frac{l}{2}$ in the opposite direction to the shot.The velocity of the shot relative to the gun is $v_{rel} = v_1 - (-v_g) = v_1 + \omega \frac{l}{2} = v$.Thus,$v_1 = v - \omega \frac{l}{2}$.By the principle of conservation of angular momentum about the pivot (centre of the rod):Initial angular momentum $L_i = 0$.Final angular momentum $L_f = \frac{m}{10} v_1 \frac{l}{2} - I_{total} \omega = 0$.Here,$I_{total} = I_{rod} + I_{gun} = \frac{ml^2}{12} + m(\frac{l}{2})^2 = \frac{ml^2}{12} + \frac{ml^2}{4} = \frac{ml^2}{3}$.So,$\frac{m}{10} (v - \omega \frac{l}{2}) \frac{l}{2} = \frac{ml^2}{3} \omega$.$\frac{1}{20} (v - \omega \frac{l}{2}) = \frac{l}{3} \omega$.$v - \omega \frac{l}{2} = \frac{20l}{3} \omega$.$v = \omega l (\frac{20}{3} + \frac{1}{2}) = \omega l (\frac{40+3}{6}) = \omega l \frac{43}{6}$.$\omega = \frac{6v}{43l}$.

$A$ rod of mass $m$ and length $l$ is placed on a smooth floor and pivoted at its centre. $A$ gun of mass $m$ (without a shot) is mounted at one end. It fires a shot of mass $\frac{m}{10}$ with velocity $v$ relative to the gun. The angular velocity of the rod will be:

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