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

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(D) The principle of conservation of mechanical energy states that the total mechanical energy remains constant.At the lowest point,the rotational kin

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$\frac{1}{6}\frac{l^2\omega^2}{g}$

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(D) The principle of conservation of mechanical energy states that the total mechanical energy remains constant.At the lowest point,the rotational kinetic energy is maximum,and the potential energy of the centre of mass is minimum (taken as zero).At the highest point,the rotational kinetic energy is zero,and the potential energy of the centre of mass is maximum.Let $I$ be the moment of inertia of the rod about the axis passing through its end,which is $I = \frac{1}{3}ml^2$.Using the conservation of energy:$K_i + U_i = K_f + U_f$$\frac{1}{2}I\omega^2 + 0 = 0 + mgh$Substituting the value of $I$:$\frac{1}{2} 	imes (\frac{1}{3}ml^2) 	imes \omega^2 = mgh$$\frac{1}{6}ml^2\omega^2 = mgh$Solving for $h$:$h = \frac{l^2\omega^2}{6g}$

$A$ thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of

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