A uniform thin rod of length L and mass m is | vedclass

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(D) When an impulse $P$ is applied at one end of the rod,the linear momentum of the center of mass is $P = mv$,so the velocity of the center of mass i

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$\frac{2 P^{2}}{m}$

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(D) When an impulse $P$ is applied at one end of the rod,the linear momentum of the center of mass is $P = mv$,so the velocity of the center of mass is $v = \frac{P}{m}$.The angular impulse about the center of mass is $J = P 	imes \frac{L}{2}$. Since $J = I\omega$,where $I = \frac{mL^2}{12}$ is the moment of inertia about the center of mass,we have $\omega = \frac{P(L/2)}{mL^2/12} = \frac{6P}{mL}$.The total kinetic energy $K$ is the sum of translational and rotational kinetic energy: $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Substituting the values: $K = \frac{1}{2}m\left(\frac{P}{m}ight)^2 + \frac{1}{2}\left(\frac{mL^2}{12}ight)\left(\frac{6P}{mL}ight)^2$.$K = \frac{P^2}{2m} + \frac{1}{2}\left(\frac{mL^2}{12}ight)\left(\frac{36P^2}{m^2L^2}ight) = \frac{P^2}{2m} + \frac{3P^2}{2m} = \frac{4P^2}{2m} = \frac{2P^2}{m}$.

$A$ uniform thin rod of length $L$ and mass $m$ is lying on a smooth horizontal table. $A$ horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

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