A uniform rod of mass m and length l is movin | vedclass

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(C) Let $V$ be the velocity of the centre of mass and $\omega$ be the angular velocity of the rod after the impulse $J$ is applied.From the impulse-mo

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$\frac{11}{7}u$

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(C) Let $V$ be the velocity of the centre of mass and $\omega$ be the angular velocity of the rod after the impulse $J$ is applied.From the impulse-momentum theorem for linear motion: $J = m(V - u) \quad ...(1)$From the angular impulse-momentum theorem about the centre of mass: $J \cdot \frac{l}{4} = I \omega = \left(\frac{ml^2}{12}\right) \omega \quad ...(2)$From equations $(1)$ and $(2)$,we get: $m(V - u) \cdot \frac{l}{4} = \frac{ml^2}{12} \omega \implies V - u = \frac{l \omega}{3} \implies V = u + \frac{l \omega}{3} \quad ...(3)$The velocity of point $P$ (at distance $\frac{l}{4}$ from the centre) is given by $V_P = V + \omega \cdot \frac{l}{4}$.Given $V_P = 2u$,we have: $2u = V + \frac{l \omega}{4} \implies V = 2u - \frac{l \omega}{4} \quad ...(4)$Equating the expressions for $V$ from $(3)$ and $(4)$: $u + \frac{l \omega}{3} = 2u - \frac{l \omega}{4}$$\frac{l \omega}{3} + \frac{l \omega}{4} = u \implies \frac{7l \omega}{12} = u \implies l \omega = \frac{12u}{7}$Substituting $l \omega$ into equation $(3)$: $V = u + \frac{1}{3} \left(\frac{12u}{7}\right) = u + \frac{4u}{7} = \frac{11u}{7}$.

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