KE of a rotational system

1. Oct 25, 2008

CaptainSFS

1. The problem statement, all variables and given/known data

A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 5 kg weight is hanging on the string. The system of the weight and disk is released from rest.

a) When the 5 kg weight is moving with a speed of 1.7 m/s, what is the kinetic energy of the entire system?

2. Relevant equations

v=$$\omega$$R

I=$$\Sigma$$(m)(r)2

KE=(.5)(I)($$\omega$$)2

KEtotal = KEwheel + KEweight

3. The attempt at a solution

KE(wheel) = (.5)((15)(.25)2)((1.7)/(.25))2 = 21.675

+

KE(weight) = (.5)(5)(1.7)2 = 7.225

= 28.9 [This answer is incorrect though, I'm unsure as to what I'm doing incorrectly.]

Any help is much appreciated, thanks! :)

2. Oct 25, 2008

alphysicist

Hi CaptainSFS,

This line is incorrect. It is saying that the rotational inertia I for this uniform disk is mr2, which is not true. There should be a table in your book that gives the formula for I for different shapes.

3. Oct 27, 2008

CaptainSFS

Thanks, :P. It needed that constant for a disk (1/2). Thanks for your help. :)

4. Oct 28, 2008