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A massless rigid rod of length 3L has two masses attached at its ends: mass 5M at one end and mass M at the other. The rod is pivoted at point P located at a distance L from the 5M mass end (so 2L from the M mass end), with the rod initially held horizontal. When released, what is the instantaneous angular acceleration of the rod?

1. g / 2L
2. 7g / 3L
3. g / 13L
4. g / 3L

Correct answer: g / 13L

Solution

Let P be the pivot. 5M is at distance L from P (say left side, so falls downward creating a clockwise torque). M is at distance 2L from P (right side, creating counter-clockwise torque when 5M side goes down). Net torque = 5Mg*L - Mg*2L = 5MgL - 2MgL = 3MgL. Moment of inertia about P: I = 5M*L² + M*(2L)² = 5ML² + 4ML² = 9ML². Angular acceleration alpha = net torque / I = 3MgL / 9ML² = g/(3L).

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