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

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(D) According to the principle of conservation of mechanical energy,the initial rotational kinetic energy is converted into the gravitational potentia

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

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(D) According to the principle of conservation of mechanical energy,the initial rotational kinetic energy is converted into the gravitational potential energy of the centre of mass at the maximum height.$TE_{i} = TE_{f}$$\frac{1}{2} I \omega^{2} = mgh$Here,$I$ is the moment of inertia of the rod about the axis passing through its end,which is $I = \frac{1}{3} m l^{2}$.Substituting this into the equation:$\frac{1}{2} 	imes (\frac{1}{3} m l^{2}) \omega^{2} = mgh$$\frac{1}{6} m l^{2} \omega^{2} = mgh$Canceling $m$ from both sides:$h = \frac{1}{6} \frac{l^{2} \omega^{2}}{g}$

$A$ 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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