A tape,pulley,disk and undisclosed mass.

[MathematicalPhysicist]

the question is:
A disk of mass M and radius R unwinds from a tape wrapped around it. the tape passes over a frictionless pulley, and a mass m is suspended from the other end. assume that the disk drops vertically.
1. relate the acclerations of m and the disk, a and A, respectively to the angualr accelration of the disk.
(the answer clue reveals that the naswer is: if A=2a, then alpha=3A/R.
my answer is that the accleration of the disk equals -a+\alphaR=A and then i get that if A=2a then alpha equals 3a/R, but it's the opposite it should be 3A/R, which i don't see how to arrive at this.

anyway, in the attached file there's a pic of this, the above left pic.
thanks in advance.

[chaoseverlasting]

I dunno if I'm doing this wrong, but this is what I get:

Let the tension in the string be T, acceleration of the disk be A and of the mass be a and R be the radius of the disk.

For the small mass m:

T-mg=ma ---1

For the disk:

Mg-T=MA

TR=I\alpha

I=\frac{MR^2}{2}

A=R\alpha

Solving these, A=2g/3, T=g/3, a=\frac{g(M-3m)}{3m}. What did I do wrong?

[MathematicalPhysicist]

the problem is that it's not given to you that the disk rolls without slippering, if it were so, then obviously we would have A=R*(alpha).

[chaoseverlasting]

But if the rope is wound tightly across the disk, then it must roll without slipping as there is no other option. It cant slip (across what?). Therefore a=r(alpha) must hold.

[chaoseverlasting]

In any case, the velocity of the string along the tension must be the same at all the points on the string.

[MathematicalPhysicist]

the problem is that's not a string but a tape!