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## Homework Statement

A rigid rod of mass 3.50 kg and length of 2.70 m rotates in a vertical (x,y) plane about a frictionless pivot through its center. Particles m1 (mass=4.70 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 46.1o with the horizontal.

## Homework Equations

I = (1/12)ML^2 + MR^2 + MR^2

τ = Iα

## The Attempt at a Solution

So I tried using the moment of inertia to find the I of the rod and each mass, then tried plugging that into the torque equation, but I can't seem to figure out what the torque equation is supposed to be. I tried to have τ=m1gL1sinθ-m2gL2sinθ, but that didn't work out. Can anyone help me?