- #1

Tina20

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## 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?