I don't think the direction matters, it rarely does.
So how would I find the torque? They're both going downwards, but I know that the larger mass is going to exert more of a torque. So would that make the torque:
T = r*m1g*cos(θ) - r*m2g*cos(θ)
Alright so for moment of inertia, I had:
I = (1/12)(3.5)(2.7)^2 + 4.7(.675)^2 + 2.6(.675)^2
And then my torque was:
T = (.675)(9.81)(4.7)sin(46.1) + (.675)(9.81)(4.7)sin(46.1)
I'm not quite sure about the torque calculation being correct.
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...