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 ml2 / 3.a)3g/2lb)2l/3gc)3g/2l2d)mg1/2Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Explanation:

The initial potential energy of the rod when it is in the horizontal position is mgl/2.

When the rod is released, it starts to rotate about point A.

As it rotates, the potential energy is converted into kinetic energy.

At any instant, the angular velocity of the rod is ω and the linear velocity of the center of mass is v.

Using energy conservation, we can write:

mgl/2 = (1/2)(ml^2/3)(ω^2) + (1/2)mv^2

mgl/6 = (1/2)(ml^2/3)(ω^2) + (1/2)(ml/2)(ωl/2)^2

mgl/6 = (1/2)(ml^2/3)(ω^2) + (1/8)(ml^2/4)(ω^2)

mgl/6 = (5/24)(ml^2/3)(ω^2)

ω^2 = 3g/5l

The angular acceleration of the rod is the rate of change of angular velocity, which is given by:

α = ω/t

ω = αt

3g/5l = αt

α = 3g/5lt

The time taken for the rod to reach the vertical position is given by:

θ = (1/2)αt^2

π/2 = (1/2)(3g/5lt)t^2

t = (3/2)(2l/3g)^0.5

Substituting the value of t in the expression for α, we get:

α = 3g/2l

Hence, the initial angular acceleration of the rod is (Moment of inertia of rod about A is ml2 / 3.) ml^2/3 × 3g/2l = 3g/2l. Therefore, option A is correct.

Answer

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