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:
-
2g/3l
-
mgl/2
-
3gl/2
-
3g/2l
Solution
D.
3g/2l
The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
l = ml3/3
where m is a mass of rod and l is the length.
Torque acting on centre of gravity of rod is given by