Angular Acceleration of A Rigid Rod

1. The problem statement, all variables and given/known data
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.


2. Relevant equations
I = (1/12)ML^2 + MR^2 + MR^2
τ = Iα

3. 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?

 

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.

 

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)