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{ml^{2}}{3}, $ the initial angular acceleration of the rod will be:
[AIPMT (S) 2007]
Options:
A) $ \frac{2g}{3l} $
B) $ mg\frac{l}{2} $
C) $ \frac{3}{2}gl $
D) $ \frac{3g}{2l} $
Show Answer
Answer:
Correct Answer: C
Solution:
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The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
$ I=\frac{ml^{2}}{3} $
where m is mass of rod and $ l $ is length.
Torque $ (\tau =I\alpha ) $ acting on centre of gravity of rod is given by
$ \tau =mg\frac{l}{2} $
or $ I\alpha =mg\frac{l}{2} $
or $ \frac{ml^{2}}{3}\alpha =mg\frac{l}{2} $
or $ \alpha =\frac{3g}{2l} $
