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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?