Deriving work-kinetic energy theorem for rotational motion

[gimak]

Homework Statement

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

Homework 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

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?

 

[Born]

Are you allowed to use calculus?

 

[ehild]

Homework Statement

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

Homework 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

[/QUOTE]

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

ehild

 

[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?

 

[Born]

Or τ = I*α = I*a/r maybe...

and you should think of your s as Δs

 

[ehild]

How is it wrong? Force is torque times radius, is it not.

It is not. Check your textbooks.

ehild

 

[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).