Angular acceleration of a cylinder with a string

[physmatics]

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!

 

[ehild]

Check the work: F=50 N, s=5 m.

ehild

 

[physmatics]

Woah... That's quite embarrasing!
Anyway, I'm happy that that was the only mistake I made... Thank you, ehild!

 

[ehild]

It was a nice work. You are welcome.

ehild

 

[paranormal]

I would like to clarify a few things in your solution attempt. Firstly, your approach using the equation r x F = I*alpha is correct, but you do not need to know the mass of the cylinder to solve for alpha. This equation can be simplified to alpha = (r x F)/I, where r is the radius of the cylinder, F is the force acting on the string, and I is the moment of inertia of the cylinder, which can be calculated using the known radius and mass of the cylinder.

Secondly, your approach using the work-energy theorem is not entirely correct. While it is true that the work done by the force is equal to the change in kinetic energy, the equation you used, W = \int F*v dt, is for a constant force acting on an object moving in a straight line. In this case, the force is not constant and the object is rotating, so this equation cannot be applied.

To find the angular acceleration, you can use the equation alpha = (r x F)/I, as mentioned earlier. Plugging in the known values, we get alpha = (0.6 m x 50 N)/(0.5*0.6^2 kg*m^2) = 83.3 rad/s^2. This is the same value you got using the work-energy theorem, but it is important to note that this is the correct approach.

Next, to find the angular velocity at t2, we can use the equation omega2 = omega1 + alpha*t, where omega1 is the initial angular velocity (which is 0 in this case). Plugging in the values, we get omega2 = 0 + 83.3 rad/s^2 * 2 s = 166.6 rad/s.

To find the final kinetic energy, we can use the equation T2 = (1/2)*I*omega2^2. Plugging in the known values, we get T2 = (1/2)*0.5*0.6^2 kg*m^2 * (166.6 rad/s)^2 = 9.99 J.

Lastly, to find the moment of inertia, we can rearrange the equation I = (r x F)/alpha to get I = (r x F)/alpha = (0.6 m x 50 N)/83.3 rad/s^2 = 0.36 kg*m^2.