-
A uniform rod AB of length , 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
-
-
mgl 2 -
2 gl 2 -
3g 2l -
2g 3l
-
Correct Option: C
Weight of the rod will produce torque,
τ = mg × (i/2)
Also, τ = Iα
where, I is the moment of inertia = ml²/3
and α is the angular acceleration
| ∴ | α = mg × | ⇒ α = | |||
| 3 | 2 | 2l |
