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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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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