Question

Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests on a support that serves as an axis of rotation. The block on the right has a mass of 5.4 kg. Determine the magnitude of the angular acceleration when the system is allowed to rotate.



So far I found that the torque on the right side is 74.088 clockwise (5.4 x 9.8) (1.4)
and the torque on the left is 70.56 counterclockwise
(12 x 9.8) (0.6)

But how do I dind the angular acceleration. I don't think I can solve for the moment of inertia because the axis of rotation is not in the center so it does not fit the formula (1/12)ML^2

PLEASE HELP, what do I do next?

Answer 1

Is it not simply T/I=angular acceleration :/

Answer 2

T = (12.8)(9.8)(0.6) - (5.4)(9.8)(1.4) = 1.176 NM

T = Ia
I = mr^2
1.175 = (5.4)(1.4)^2(a) + (12)(0.6)^2(a)

a = 0.079 rad/s^2