Linear acceleration of blocks on unbalanced suspended rod

[moose726]

Homework Statement

Two blocks each of mass m are suspended from the ends of a rigid weightless rod of length L1+L2 with L1 = 25 cm (left of fulcrum) and L2 = 83 cm (right of fulcrum). The rod is held in the horizontal position and then released.

(a) What is the linear acceleration of the block closer to the fulcrum as it starts to move?
m/s2
(b) What is the linear acceleration of the other block as it starts to move?

Homework Equations

$$\Sigma$$Torque = I$$\alpha$$

The Attempt at a Solution

$$\Sigma$$Torque = (.25m)(m)-(.83m)(m) = I$$\alpha$$
I have no idea where to go from here. Is the rotational inertia the mass of the entire system multiplied the distance from the fulcrum to the center of mass squared I=(MR(cm)^2)? Do I use a=$$\alpha$$R and Force equations?

 

[rl.bhat]

Hi moose726, welcome to PF.
Torque is produced due to the weight of the mass.(T = mg*R)
Find net torque and moment of inertia of the rod about the fulcrum. From that find the angular acceleration. And using the relation a = αR, find the linear acceleration.

 

[moose726]

Thanks a lot for the information and welcome, I finally got the answer.