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

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(C) Let $V$ be the final velocity of the center of mass and $\omega$ be the angular velocity acquired by the rod.From the impulse-momentum theorem for

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$\frac{11}{7}u$

Vedclass · Answer

(C) Let $V$ be the final velocity of the center of mass and $\omega$ be the angular velocity acquired by the rod.From the impulse-momentum theorem for linear motion: $J = m(V - u) \quad ...(1)$From the angular impulse-momentum theorem about the center 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: $m(V - u) \cdot \frac{l}{4} = \frac{ml^2}{12} \omega \implies V - u = \frac{l}{3} \omega \implies V = u + \frac{l}{3} \omega \quad ...(3)$The velocity of point $P$ at distance $l/4$ from the center is given by $V_P = V + \omega \cdot \frac{l}{4}$.Given $V_P = 2u$,so $2u = V + \frac{l}{4} \omega \quad ...(4)$Substitute $V$ from $(3)$ into $(4)$: $2u = (u + \frac{l}{3} \omega) + \frac{l}{4} \omega$$u = \omega l \left(\frac{1}{3} + \frac{1}{4}\right) = \omega l \left(\frac{7}{12}\right) \implies \omega l = \frac{12}{7} u$Now,substitute $\omega l$ back into the expression for $V$: $V = u + \frac{1}{3} \left(\frac{12}{7} u\right) = u + \frac{4}{7} u = \frac{11}{7} u$.

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