Pulley with rotational consideration

[endeavor]

An object descends from a pulley.
http://img501.imageshack.us/img501/9783/physics12qm.th.jpg
Use the conservation of mechanical energy to find the linear speed of the descending mass (m = 1.0kg) after it has descended a vertical distance of 2.0m from rest. (For the pulley, M = 0.30kg and R = 0.15m. Neglect friction and mass of the string)

I solved this question not using the conservation of mechanical energy. At first I tried using it though and I wound up getting:
v_cm = square root of (4/3 mgh/M)

However, I determined the acceleration using a pervious example:
a = 2mg/(2m + M)
and then used v² = 2ax
and got 5.8m/s which is the right answer...

but I don't know how to solve this question using the conservation of mechanical energy.

 

[vaishakh]

Let us take the body pulley Earth as our system. During the following event no external work is done on the system. So the mechanical energy of the system is conserved.
As the block slides an height h, then there is a loss of mgh amount of potential energy. Since there has been no loss of mechanical energy, the following potential energy is converted into Kinetic enegy. So you equate the change in potential energy to the change in Kinetic energy.

 

[endeavor]

yes, i understand that part. is kinetic energy in this case equal to:
1/2 I_cmw² + 1/2 Mv_cm²
and
I = 1/2MR²
so substituting v/R for w
mgh = 1/4 Mv_cm² + 1/2 Mv_cm²

but solving for v_cm here does not get me 5.8m/s