Quote by mfb
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. :]