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…

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 (ml^2)/3)

MARKS Content Team · Accepted Answer

Correct answer: (A) 3g/2l. The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is =(ml^2)/3 where, m is the mass and l its length. Torque τ=Iα acting on centre of gravity of rod is given by τ=mgl/2 also, (ml^2)/3·α=mgl/2 α=3g/2l

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 $\frac{m l^{2}}{3}$ )

Solution

More Rotational Motion Questions

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 ml23)

Solution

More Rotational Motion Questions

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 $\frac{m l^{2}}{3}$ )