NEET2006PhysicsRotational MotionActual
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 the rod about A is m l^2 3 ).
Options
- Am g l 2
- B3 g 2 l
- C2 l 3 g
- D3 g 2 l^2
Correct answer
B. 3 g 2 l
Step-by-step solution
Here torque =m g ( 1 2 ) M. I. of rod about A is: I= m l^2 3 Angular acceleration of the rod is = I = m g ( 1 2 ) m l^2 3 = 3 g 2 l .