Angular acceleration of a cylinder with a string

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Homework Statement


A cylinder with radius r = 0.6 m hangs in a horisontal frictionless axis. A string is winded around it and a constant force, F = 50.0 N, is acting on the string from t1 = 0 s to t2 = 2.00 s. During this time, L = 5 m of the string unwinds. The system starts from rest.
a) Angular acceleration, alpha = ?
b) Angular velocity, omega2 at t2 = ?
c) Final kinetic energy = ?
d) Moment of inertia, I = ?

Homework Equations


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The Attempt at a Solution


The only thing that actually troubles me is finding alpha, because I think that once I have it, everything else is easily solved. (am I right thinking that omega2 = alpha*t2 ? Since alpha is constant, I cannot imagine it being in any other way)
Now, I've tried using r x F = I*alpha, but as I don't know the mass of the cylinder, that equation doesn't help me. (I think that once I know alpha, r x F = I*alpha is the equation to use to find I) I've used the fact that the work done by the force is W = F * L = 2500 Nm.
By using
W = \int F*v dt
with boundaries t1 = 0 to t2 = 2, a = alpha*r and F = m*a I get 2*F*alpha*r = W. From this I get alpha = 41.7 rad/s2. I'm not sure this is correct though, and I would have expected alpha to be a bit smaller...
With alpha = 41.7 rad/s2 and t2 = 2 s I get omega2 = 41.7*2 = 83.4 rad/s. I also used W = T2 - 1, where T is kinetic energy. Since omega1 = v1 = 0, T1 = 0. T2 = (I*(omega2)2)/2. With the moment of inertia for a cylinder, I = (m*r2)/2 I get m(r2*omega22)/4 = m*alpha*r*L, and m cancels out. Solving for omega2 I get 37.3 rad/s, which is not what I got using omega2 = alpha*t2. Can anyone explain this to me? I'm sure I got a lot wrong, and the fact that I'm in my summer house without a single physics book doesn't help.

Thanks a lot for even reading this!
 
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Check the work: F=50 N, s=5 m.

ehild
 
Woah... That's quite embarrasing!
Anyway, I'm happy that that was the only mistake I made... Thank you, ehild!
 
It was a nice work. You are welcome.

ehild
 
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