(B) By the law of conservation of energy, the potential energy lost by the rod is equal to the rotational kinetic energy gained by it.Initially, the c

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(B) By the law of conservation of energy, the potential energy lost by the rod is equal to the rotational kinetic energy gained by it.Initially, the center of mass of the rod is at a height $h = l/2$ from the ground.Potential Energy $(PE)$ lost = $mgh = mg(l/2)$.Rotational Kinetic Energy $(KE)$ gained = $(1/2)I\omega^2$, where $I$ is the moment of inertia of the rod about the hinge $A$.The moment of inertia of a rod about one end is $I = ml^2/3$.Equating $PE$ and $KE$: $mgl/2 = (1/2)(ml^2/3)\omega^2$.$mgl/2 = (ml^2/6)\omega^2$.$\omega^2 = (mgl/2) \times (6/ml^2) = 3g/l$.Therefore, $\omega = \sqrt{3g/l}$.

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 uniform rod $AB$ of mass $m$ and length $l$ is hinged at one end $A$ to the ground level. Initially, the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its $B$ end strikes the ground is $(g = \text{acceleration due to gravity})$

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A
$\sqrt{2g/l}$
B
$\sqrt{3g/l}$
C
$\sqrt{mg/l}$
D
$\sqrt{mg/3l}$

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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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