(C) The rod oscillates about one end. The moment of inertia of the rod about an axis passing through one end is $I = \frac{ml^2}{3}$.By the principle

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

(C) The rod oscillates about one end. The moment of inertia of the rod about an axis passing through one end is $I = \frac{ml^2}{3}$.By the principle of conservation of mechanical energy,the rotational kinetic energy at the lowest point is equal to the gravitational potential energy gained by the centre of mass at the maximum height $h$:$\frac{1}{2} I \omega^2 = mgh$Substituting the value of $I$:$\frac{1}{2} \left( \frac{ml^2}{3} \right) \omega^2 = mgh$$\frac{1}{6} ml^2 \omega^2 = mgh$Solving for $h$:$h = \frac{l^2 \omega^2}{6g}$

Class 11 Physics System of Particles and Rotational Motion Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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$A$ 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:

A
$\frac{1}{6} \frac{l \omega}{g}$
B
$\frac{1}{2} \frac{l^2 \omega^2}{g}$
C
$\frac{1}{6} \frac{l^2 \omega^2}{g}$
D
$\frac{1}{3} \frac{l^2 \omega^2}{g}$

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System of Particles and Rotational Motion

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Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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$A$ rod of length $1 \ m$ is held vertically and allowed to fall such that its lower end remains fixed on the ground. The velocity of the other end of the rod when it hits the ground is ......... $m/s$ $(g = 9.8 \ m/s^2)$.

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$A$ 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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$A$ thin uniform rod of mass $1 \ kg$ and length $1 \ m$ is hinged at one end to the ground. It originally stands vertically and is allowed to fall to the ground. If the rod hits the ground with angular speed $\omega$, then the correct statement is (Assume $g = 10 \ m \ s^{-2}$):

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