Question involving moment of inertia, rotation about a horizontal axis.

[xduckksx]

Homework Statement

A uniform cylinder, of radius 2a and moment of inertia $2Ma^2$ is free to rotate about its horizontal axis. A light, inextenzible string is wound round the cylinder and a particle of mass m is suspended on its free end. If the system is released from rest, find the acceleration of the particle.

Homework Equations

Torque(C)=Moment of inertia(I) x (angular accleration)$\alpha$.

The Attempt at a Solution

Resolving the tension in the string, we get T=mg-2ma(alpha)
Which gives upon calculating the angular acceleration (given that I=2Ma^2)
alpha=(mg)/(2am+Ma) which is somehow incorrect.

Can someone correct my error?

 

[kuruman]

You need to draw two separate free body diagrams, one for the hanging mass and one for the cylinder. Then you need to write two F = ma equations and one τ = Iα equation based on these two FBDs.