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- Genie What actually produces the angular acceleration here?The weight Mg, acting at the centreThe normal force at the pivotStep 1Weight acts at L/2
- Genie Right. So the torque about the pivot is —τ = Mg · Lτ = Mg · (L/2)Step 2τ = MgL/2
- Genie And a rod about its END, not its centre — which moment of inertia?I = ⅓ ML²I = 1/12 ML²Step 3I = ML²/3
- Genie Put them together. α = τ / I gives you —α = 3g / 2Lα = 2g / 3LStep 4α = (MgL/2) ÷ (ML²/3)
- You solved itα = 3g / 2L
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