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 m l 2 3 )

Question

Aicharya · Accepted Answer

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

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

where, $m$ is the mass and $l$ its length.

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

$τ = m g \frac{l}{2}$

also, $\frac{m l^{2}}{3} \cdot α = m g \frac{l}{2}$

$α = \frac{3 g}{2 l}$

A uniform rod of length l and mass m is free to rotate in a vertical plane about A . The rod initially in…

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