SATHEE: Rotational Motion Question 63

Rotational Motion Question 63

Question: A uniform rod AB of length $l$ and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is $\frac{m l^{2}}{3},$ the initial angular acceleration of the rod will be:

[AIPMT (S) 2007]

Options:

A) $\frac{2 g}{3 l}$

B) $m g \frac{l}{2}$

C) $\frac{3}{2} g l$

D) $\frac{3 g}{2 l}$

Show Answer

Answer:

Correct Answer: C

Solution:

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is

$I = \frac{m l^{2}}{3}$

where m is mass of rod and $l$ is length.

Torque $(τ = I α)$ acting on centre of gravity of rod is given by

$τ = m g \frac{l}{2}$

or $I α = m g \frac{l}{2}$

or $\frac{m l^{2}}{3} α = m g \frac{l}{2}$

or $α = \frac{3 g}{2 l}$

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Rotational Motion Question 63

Question: A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ml23, the initial angular acceleration of the rod will be:

Options:

Answer:

Solution:

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Question: A uniform rod AB of length $l$ and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is $\frac{m l^{2}}{3},$ the initial angular acceleration of the rod will be: