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P: 441
 Quote by mfb That does not work. $$dK=\frac{1}{2}Id(\omega^2) = I\omega d\omega$$ This can be integrated, but where is the point in taking the derivative if you integrate afterwards?

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. :]