Rotation and translation of a cylinder

[Hannisch]

Homework Statement

A homogenous cylinder with radius R and mass M is resting, with the axis vertical, on a horizontal surface, which can be asumed to be completely slippery. A string is partially wound around the cylinder. The string then runs over a pulley at the end of the table, in the same height as the winding, and then on to a weight with mass m hanging in its other end. Determine the angular acceleration and the tension in the string when the system is moving. The moment of intertia of the pulley can be neglected.

Homework Equations

F = ma

RF = Iα

The Attempt at a Solution

I'm going crazy with this, because it's not really that hard of a problem..

Okay, if I draw the force diagrams, I have the tension T on the cylinder, which is the force that will create its angular acceleration and I have T up and mg down on the mass m. The tension should also be the force that's accelerating the centre of mass of M. I'd say the translational acceleration of M needs to be the same as that for m.

m: mg - T = ma
M: T = Ma
rotation: RT = Iα

I for a cyldinder = MR²/2

from rotation: T = MRα/2

from M: a = T/M

a = Rα/2

Putting these in the m-equation: mg - MRα/2 = mRα/2

mg = Rα(M + m)/2

2mg/R(M + m) = α

And this is incorrect. There's something I'm missing and I can't see it - it is driving me insane. The correct answer is supposed to be α = 2mg/R(M + 3m). So close, yet sooo far away.

Could anyone tell me what I'm not seeing, please?

 

[ideasrule]

m: mg - T = ma
M: T = Ma
rotation: RT = Iα

This is your problem. The acceleration of the cylinder is not equal to the acceleration of the block. This is because the cylinder is letting out string as it spins, so the block has to fall faster than the cylinder's center of mass is moving.

 

[Hannisch]

Ah, okay. I thought I went wrong somewhere in the beginning. I'll try to do it during the day (there's no time right now, I need to go to my lecture), but thank you!