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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

Asked in CBSE AIPMT 2007 · Torque and angular acceleration

• (1) (mgl)/2
• (2) 3/2gl
• (3) (3g)/(2l)
• (4) (2g)/(3l)

Step-by-step solution

Why the other options are wrong

• (1) comes from this is the torque mg/l2 itself, not an angular acceleration; it still carries the mass, which must cancel
• (2) comes from multiplying by l instead of dividing by it; an angular acceleration must carry l⁻¹, since α=τ/I and I goes as l²
• (4) comes from flipping the 3 and the 2 in (mgl/2)/(ml²/3)=(3g)/(2l)

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