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A rigid rod of mass 8.70 kg and length 2.30 m rotates in a vertical (x,y) plane about a frictionless pivot through its center. Particles m1 (mass=4.30 kg) and m2 (mass=2.60 kg) are attached at the ends of the rod. Determine the size of the angular acceleration of the system when the rod makes an angle of 50.1° with the horizontal.

I tried getting the moment of inertia using Irod = 1/12ML^2 + the Is of both the particles. Then I found the weight difference and used that to compute torque in torque = r x F. Finally I used those two answers to get angular acceleration. Of course this is wrong... can someone help me please? Or at least tell me what I did wrong... I think I may be confused about the angle between r and F. Anyway, any help would be apperciated! Thanks.

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# Rod + 2 Particle masses = angular acceleration

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