A uniform rod of length L and mass M is hinged at one end and released from horizontal rest. What is its initial angular acceleration (I = (1/3)ML^2)?
Answer and explanation
Correct answer: B. alpha = 3g / (2L)
- Step 1: Torque about the pivot acts through the centre of mass at L/2: tau = M g (L/2). Step 2. Using tau = I alpha gives Mg(L/2) = (1/3)ML^2 alpha, so alpha = 3g / (2L). Answer: alpha = 3g / (2L). Why not the others: The moment of inertia of a rod about its end is (1/3)ML^2, not (1/12)ML^2.
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