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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:

1. mgℓ / 2
2. 3 / 2 gℓ
3. 3g / 2ℓ
4. 2g / 3ℓ

Correct answer: 3g / 2ℓ

Solution

The torque about point A due to the weight of the rod is τ = (mg)(ℓ/2), as the center of mass is at the midpoint of the rod. Using τ = Iα, where I = mℓ²/3, we get α = τ/I = (mgℓ/2) / (mℓ²/3) = 3g / 2ℓ.

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