1. The problem statement, all variables and given/known data Two identical disks with mass m and radius r are connected via a massless wire of length L which is winded up around both disks. Disk B is connected to the ceiling and is free to rotate around its axis. Disk A is besides disk B and will fall due to the gravitational force with acceleration of g. Visualization http://i.imgur.com/LZAk170.png While disk A falls it will unwind. Question 1 is to find the acceleration of disk A - assuming it moves straight down. Question 2 is to find the speed of disk A when the massless wire has reached its maximum extension. Question 3 is to explain what happens when the wire has reached its maximum extension. 2. Relevant equations Fg=mg, ΣF=ma Torque=Fr, Torque=Iα (I is moment of inertia and α is angular acceleration) I=½mr2 α=atan/r (rolling without slipping) Conservation of energy: K1+U1=K2+U2 Kinetic energy(translational + rotational): K=½mvcm2 + ½Iω2 vcm=rω (rolling withouth slipping) Potential energy: U=mgh 3. The attempt at a solution I'm stuck at question 2 but to know if i'm correct I have also put my answer for question 1 here: For question one I divided the situation in two parts. First part is to determine the tangular acceleration of disk B. The force applied on disk B is Fg from disk A which is mg. By equalising both torque terms you get mgr = ½mr2*atan/r. Which gives atan=2g. The second part is to find the downwards acceleration of disk A. There are two forces applying on disk A and that is the tension of the rope T and the Fg. Force T is the only force applying torque on disk A. So we can write ΣF=mg-T=macm. To solve this first we equalise the torque equations again: ½mr2*acm/r = Tr. This gives T=½macm. Subbing this in the force equation you get: g-½acm=acm --> acm=2/3g. The total downwards acceleration of disk A is atan+acm=8/3g. For question 2 I started with the conservation of energy. The starting position is when both disks are at rest at the same height. There is no kinetic energy and only potential energy with height L. When disk A falls the potential energy will turn into kinetic energy of both translation and rotation. Disk B stays at the same height but will also have kinetic energy of only rotation. Because disk B stays at the same height the potential energy of disk B will cancel out. So we get: UA1 = KA2 + KB2. mgL = ½mvcm2 + ½IωA2 + ½IωB2. The speed at which disk A moves down due to the rotation of disk B is vtan=rωB. The speed at which disk A moves down due to its own rotation is vcm=rωA. After subbing everything in you get: gL=¾vcm2+¼vtan2. I know the total speed of disk A going down is vtan+vcm but because of those fractions it's not that easy to sub it in. I have a feeling the solution is really easy but I cannot see it. Any help would be much appreciated. For question 3 i have a feeling disk B will start the oscillate from right to left and disk A will move along with it as a pendulum.