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

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(C) The rod is oscillating about one of its ends. The moment of inertia of the rod about an axis passing through one end is $I = \frac{1}{3} m l^2$.Us

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

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(C) The rod is oscillating about one of its ends. The moment of inertia of the rod about an axis passing through one end is $I = \frac{1}{3} m l^2$.Using the principle of conservation of mechanical energy,the maximum rotational kinetic energy at the lowest point is equal to the gain in gravitational potential energy at the maximum height $h$ of the centre of mass.$\frac{1}{2} I \omega^2 = mgh$Substituting the value of $I$:$\frac{1}{2} (\frac{1}{3} m l^2) \omega^2 = mgh$$\frac{1}{6} m l^2 \omega^2 = mgh$$h = \frac{1}{6} \frac{l^2 \omega^2}{g}$

$A$ thin rod of mass $m$ and length $l$ is oscillating about a horizontal axis through its one end. Its maximum angular speed is $\omega$. Its centre of mass will rise up to a maximum height of:

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