Moment of Inertia Problem: cylinder pivoting ona bearing

[delecticious]

Homework Statement

M, a solid cylinder (M=2.07 kg, R=0.135 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.570 kg mass. Starting from rest, the mass now moves a distance 0.350 m in a time of 0.490 s. Find Icm of the new cylinder.

Homework Equations

T -torque
I - moment of inertia
s = 1/2(v0 +vf)t
T/alpha = I
transational acceleration (at) = (alpha)r
translational force (Ft) = m(at)
T = Ft(r)

The Attempt at a Solution

I started out by solving for accleration using the time, distance and initial velocity given, but it's from there where I'm completely confused. To find I, I have to divide T by alpha, but if T is equal to m(at)r which can also be rewiten as mr^2(alpha) then won't the alpha's cancel out making fiding the acceleration useless? When I do that I get mr^2 = I but that can't be right. Anyone know what I'm doing wrong?

 

[Doc Al]

M, a solid cylinder (M=2.07 kg, R=0.135 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.570 kg mass. Starting from rest, the mass now moves a distance 0.350 m in a time of 0.490 s. Find Icm of the new cylinder.

I don't understand the problem. What new cylinder? (Did you present the problem exactly as given?)

Is that your figure or one that came with the problem?

 

[delecticious]

sorry, this problem has 4 different steps to it. The first step had a figure and the last 3 go with this figure. In the last step they changed the moment of inertia for the cylinder but everything else concerning the cylinder is the same, so disregard the "new" in front of cylinder if you like, it makes sense in the series of 4 steps but since you've only seen this part of the question it's just as well the only cylinder.