# Deriving work-kinetic energy theorem for rotational motion

1. Aug 14, 2014

### gimak

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

How to derive W=(.5*I)*((ωf^2)-(ωi^2)) from W=τ*(θ)

2. Relevant equations

The two equations seen above (big W is work; little w is rotational velocity, τ is torque, alpha=rotational acceleration).

s=r*θ a=r*(alpha) (ωf^2)-(ωo^2)=2*(alpha)*(θ) F=r*τ F=ma

3. The attempt at a solution

W=T*(θ)=(F/r)*(s/r)=(r^-2)*F*s=(r^-2)*ma*s=(r^-2)*m*(r*alpha)((ωf^2)/(2*alpha))*r
=0.5*m*ωf^2

Obviously this isn't right. What went wrong?

2. Aug 14, 2014

### Born

Are you allowed to use calculus?

3. Aug 15, 2014

### ehild

[/QUOTE]

The formula in red is wrong. How is the torque defined?

ehild

4. Aug 15, 2014

### gimak

response to ehlid

How is it wrong? Force is torque times radius, is it not. Oh, I can't use scalar multiplication on this. Do I have to use the cross product: F=t x r?

5. Aug 15, 2014

### Born

Or τ = I*α = I*a/r maybe...

and you should think of your s as Δs

6. Aug 15, 2014

### ehild

It is not. Check your textbooks.

ehild

7. Aug 15, 2014

### gimak

correction

Shoot! I'm sorry!!! T=r*F. I got it backwards!!! If I plug in the right equation, the first equation will eventually turn into 0.5*m*(r^2)*(ωf^2)=I*(ωf^2).

Last edited: Aug 15, 2014