Solve Rotational Motion: Show Tension = 1/3 Disk Weight

[thenewbosco]

I have a diagram for this at http://snipurl.com/diagr

There is a uniform disk with radius R, mass M, it has a string wrapped around it and is attached to a fixed rod.
Part 1: Show the tension in the string is 1/3 the weight of the disk

what i have done for this is

$$Torque=Fd$$
and setting the force equal to Tension
$$Torque=TR$$
then $$\sum Torque=I\frac{a}{R}$$ and i used $$I=\frac{1}{2}MR^2$$for the moment of inertia

Then $$TR=\frac{1}{2}MRa$$
$$T=\frac{1}{2}Ma$$
and then acceleration is just g, so i get the tension equals 1/2 the weight. where have i gone wrong?

 

[arildno]

Access is forbidden to your link.

 

[thenewbosco]

help??