Calculating Kinetic Energy of a Rotational System with a Hanging Weight

[CaptainSFS]

Homework Statement

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?

Homework Equations

v=$$\omega$$R

I=$$\Sigma$$(m)(r)²

KE=(.5)(I)($$\omega$$)²

KE_total = KE_wheel + KE_weight

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! :)

 

[alphysicist]

Hi CaptainSFS,

Homework Statement

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?

Homework Equations

v=$$\omega$$R

I=$$\Sigma$$(m)(r)²

KE=(.5)(I)($$\omega$$)²

KE_total = KE_wheel + KE_weight

The Attempt at a Solution

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

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.

 

[CaptainSFS]

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

 

[alphysicist]

Glad to help!