# Homework Help: Angular Energy/Velcoity

1. Dec 8, 2009

### LBenson

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

A uniform circular steel disc of mass 20kg and diameter 0.4m has a radius of gyration equal to 0.14m. It is lifted vertically 2.0m, and placed on an incline. The flywheel is then released and rolls down the incline without slipping. Find :

a) the potential energy of the disc at the top of the incline (Ans= 392.4J)
b) the maximum linear velocity of the disc as it reaches the bottom of the incline

2. Relevant equations

E=1/2*I*w^2 (I being Inertia and w being Angular Velocity)
I=m*k^2 (K being the radius of gyration)
V=w*r (V = Linear Velocity)

3. The attempt at a solution

Ok so i worked out a) with no problems. this next part if giving me a bit of trouble though.

Here are the values i have:

Mass = 20kg
Diameter=0.4m
P.E=392.4

Heres what i did:

First i worked out the inertia so so i can use the calculated value in the next bit

I=mk^2
I=20x0.14^2
I=0.392

Now because the disc doesn't slip i assumed the Potential Energy will = The energy at the bottom of the slope so i used

E=1/2*I*w^2

Re arrange for W

so
W^2=E/(1/2*I)
=392.4/0.196=2002.05
=(sqrt)2002.04

Now since i needed the Linear velocity i used the relationship equation of:

V=w*r
V=44.74x0.2
V=8.946m/s

So where am i going wrong?

Cheers for any help

2. Dec 8, 2009

### AEM

Conservation of energy says PE at top goes into PE at bottom plus KE of rotation at bottom plus KE of translation at the bottom (that's translational velocity of the center of mass).

Did you includethe last term?

3. Dec 8, 2009

### LBenson

I should probably include that the answer is 4.43 m/s i just dont know how to get it

4. Dec 8, 2009

### LBenson

Well i worked out the Angular Velocity at the bottom, is that what you're asking?

5. Dec 8, 2009

### pgardn

If I am reading this correctly...

The PE at the top of the incline would be converted into only translational (linear) KE if the object was just sliding, ie no friction. But the object is rolling. So the KE at the bottom of the incline is in two forms 1. Rotational KE 2. Translational KE

If you found omega you can find rotational KE. Subtract that from the PE at the top and you are only left with linear KE. So now just use 1/2mv^2 = linear KE (Joules) solve for v.