Originally Posted by **ChipB** I guess there's a bit more to go here...
From $\displaystyle T = m\ddot x = \frac {I \ddot \theta}r $ you can relate $\displaystyle \dot \theta$ to $\displaystyle \dot x$ , and hence to $\displaystyle \dot y$. This should enable you to relate total vertical displacement of the weight (call it h) to the horizontal displacement of the spool, L. I think then using energy principles would bring it home:
$\displaystyle KE_{spool} + KE_{weight} = mgh$ |

I've worked the problem to have:

$\displaystyle mg - \frac {I \ ddot \theta}r = m( \ ddot x + r \ ddot \theta)$

I was given what I is equal to, so I can plug that in getting:

$\displaystyle mg - \frac {Md^2 \ddot \theta}8}r = m( \ ddot x + r \ ddot \theta)$

I'm still not sure how to go back to velocity of the weight though, How can I solve for \ddot x and have that be the acceleration of the falling weight?