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{ml^{2}}{3} $ )
[AIPMT (S) 2006]
Options:
A) $ \frac{3g}{2l} $
B) $ \frac{2l}{3g} $
C) $ \frac{3g}{2l^{2}} $
D) $ mg\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{ml^{2}}{3} $
where m is mass of rod and $ l $ its 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} $
$ \therefore $ $ \alpha =\frac{3g}{2l} $
BETA
