# Homework Help: Max angular displacement from max angular velocity

1. Nov 4, 2009

### rdlcsh

1. The problem statement, all variables and given/known data

A thin rod of length 1.4m and mass .2kg is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 7.84 rad/s. Neglecting friction and air resistance, find how far above that position the center of mass rises.

m = .2 kg
l = 1.4 m
max angular velocity = 7.84 rad/s

I've figured out:
moment of inertia = .130666
max kinetic energy = 4.0157 J

2. Relevant equations

Work = integral of the torque from (theta-0 to theta-max)
tourqe = rF*sin(theta) = rmg*sin(theta)
tourqe = I(angular acceleration)

3. The attempt at a solution

I've tried integrating both tourqe equations above as well as integrating the second one subtracted from the first one. But i usually end up with something like:

sin(theta) = a number larger than 1

I think I am just integrating the wrong thing. I know this could probably be done with the conservation of engergy, but when I try that, I end up with a height that is greater than the lengh of the rod.

Thanks for any help!

2. Nov 4, 2009

### kuruman

Energy conservation is definitely the way to go with this. If you show your work, someone might be able to find out where you went wrong.

*** Edit ***
Actually, it seems that this problem is impossible. If the rod is allowed to swing from the "upside down" vertical position, the potential energy change of the center of mass with respect to the "right side up" vertical position is mgΔh = 0.2*9.8*1.4=2.744 J. This is less than the 4.016 J that you correctly calculated as the kinetic energy based on the angular speed given by the problem. It appears that the angular speed given by the problem is such that the rod will keep going round and round instead of reaching a maximum angle.

Last edited: Nov 4, 2009
3. Nov 4, 2009

### rdlcsh

That's the result I kept getting, but that's not the correct answer. I guessed that angular kinetic energy must not be the same as linear kinetic or potential energy. I was wrong in writing:

1/2 *I*w^2 = mgh

I still don't know how to do this problem. Any other ideas?

4. Nov 4, 2009

### rdlcsh

Sorry to bother everyone with this ridiculous problem. I've found out that the "correct" answer is 2.048 m, which is of course impossible. I guess that's what I get for going to a crappy college.

5. Nov 4, 2009

### kuruman

Good students learn and do well regardless of who's teaching them and in what place.

6. Nov 5, 2009

### rdlcsh

Sounds like you've either never had a really terrible class or never had a really fantastic teacher. Of course the most important thing is the student. But on the last two exams, the class averages were 39% and 30%. Is that really because the class is filled with bad students?

All I'm saying is that the last three days I spent trying to find an answer that didn't exist could have been better spent studying other things or learning new concepts.

Your comment is an insult to good schools everywhere. For anyone else who may be frustrated with their physics class, I highly reccomend MIT's Open CourseWare - ocw.mit.edu, especially Walter Lewin's classes.

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