Rotational Motion - 2 Disc System

[HoodedFreak]

Homework Statement

Two metal disks, one with radius R1 = 2.50 cm and
mass M1 = 0.80 kg and the other with radius R2 = 5.00 cm and
mass M2 = 1.60 kg, are welded together and mounted on a frictionless
axis through their common center, as in Problem 9.77.
(a) A light string is wrapped around the edge of the smaller disk,
and a 1.50-kg block is suspended from the free end of the string.
What is the magnitude of the downward acceleration of the block
after it is released? (b) Repeat the calculation of part (a), this time
with the string wrapped around the edge of the larger disk. In which
case is the acceleration of the block greater? Does your answer
make sense?

Homework Equations

T = I * α
a = α * R
I of a cylinder = MR^2/2
T = F * l

The Attempt at a Solution

a) The block is enacting a torque, which is equal to the force of gravity times the lever arm.
T = 1.5g * R
where R is the radius of the smaller cylinder
The moment of inertia of the system of cylinders is MR^2/2 for the smaller cylinder + (2M)*(2R)^2/2 for the bigger cylinder = 9MR^2/2

T = I*α ⇒ 1.5gR = 9MR^2/2 * α
3g = 9MR * α
3g = 9Ma
a = g/3M = 4.09 m/s^2

The answer however is a = 2.8

What did I do wrong?

 

[Doc Al]

The block is enacting a torque, which is equal to the force of gravity times the lever arm.
T = 1.5g * R

Do not assume that the tension in the string (which is what exerts the torque) is equal to the weight of the block.

 

[HoodedFreak]

Do not assume that the tension in the string (which is what exerts the torque) is equal to the weight of the block.

I see, I did that and I got the right answer, thanks alot!

I did:
T - 1.5g = m*a
T*R = 9MR^2α/2

Then I solved them to get a = 2.88 which is the answer.
and repeated the same process for b) to get a = 6.13