Angular acceleration of a cylinder with a string

physmatics
Messages
16
Reaction score
0

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


-

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!
 
Physics news on Phys.org
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
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top