SATHEE: Rotational Motion Question 62

Rotational Motion Question 62

Question: A uniform rod of length $l$ and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is $\frac{m l^{2}}{3}$ )

[AIPMT (S) 2006]

Options:

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

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

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

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

Show Answer

Answer:

Correct Answer: A

Solution:

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

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

where m is mass of rod and $l$ its 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 $\frac{m l^{2}}{3} α = m g \frac{l}{2}$

$∴$ $α = \frac{3 g}{2 l}$

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Rotational Motion Question 62

Question: A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is ml23 )

Options:

Answer:

Solution:

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Question: A uniform rod of length $l$ and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is $\frac{m l^{2}}{3}$ )