Question 679: Centre of Mass and Rotational Motion

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 ml2/3, the initial angular acceleration of the rod will be

Accepted Answer

Torque about A,
$$	au  = mg 	imes {l \over 2} = {{mgl} \over 2}$$
$$	au  = I\alpha $$
$$ 	herefore $$ Angular acceleration,
$$\alpha  = {	au  \over I} = {{mgl/2} \over {m{l^2}/3}} = {{3g} \over {2l}}$$

Answer

$${{mgl} \over 2}$$

Answer

$${3 \over 2}gl$$

Answer

$${{3g} \over {2l}}$$

Answer

$${{2g} \over {3l}}$$

Centre of Mass and Rotational Motion

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 ml2/3, the initial angular acceleration of the rod will be

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