View Single Post
Nov28-12, 11:26 AM
P: 441
Quote Quote by mfb View Post
That does not work.
[tex]dK=\frac{1}{2}Id(\omega^2) = I\omega d\omega[/tex]
This can be integrated, but where is the point in taking the derivative if you integrate afterwards?
Thank you for your reply.

So that would mean that I can simply take the final and initial ω values. I will already have K from the work done by the block, and the values of ω, so the only unknown in the equation would be the moment of inertia.

Thanks! Hope this is a correct analysis. :]