Block attached to string, wrapped around hollow cylinder and free to rotate

[Tina20]

Homework Statement

A 3.20 kg block is attached to a string that is wrapped around a 1.17 kg, 4.08 cm diameter hollow cylinder that is free to rotate on an axel through the center. The block is released 1.09 m above the ground. What is the speed of the block as it hits the ground?

Homework Equations

Ui = Kf
Mgh = Kcyl + Kblock
Mgh = 1/2Iw^2 + 1/2mv^2 , where w is angular speed and I is moment of inertia

Moment of Inertia of a hollow cylinder is MR^2
w = v/r

so,

Mgh = 1/2(MR^2)(v^2/r^2) + 1/2mv^2

The Attempt at a Solution

Mgh = 1/2(MR^2)(v^2/r^2) + 1/2mv^2
(3.2kg)(9.8)(1.09m) = 1/2(1.17kg)(v^2/(0.0204m^2) + 1/2(3.2)v^2
34.182 = 2.434x10-4 (v^2/4.1616x10-4) + 1.6v^2
140435.497 = v^2/4.16x10-4 + 1.6v^2
140435.497 = v^2( 1/4.16x10-4 + 1.6v)
140435.497 = v^2 (2402.92 + 1.6)
140435.497 = v^2 (2404.52)
58.4 = v^2
7.64 m/s = v


The answer is wrong unfortunately. I must be missing a vital piece of information in my equation?

 

[Quinzio]

1) On the second line of your solution there is 1 error.

2) Take a while to examine your equations.
Can you simplify this ?
Mgh = 1/2(MR^2)(v^2/R^2) + 1/2mv^2