# Variable torque due to rotation

1. Jul 9, 2010

### Accumulator

Hi, all! I just registered here, and hope to enjoy enlightening discussions about the subject of physics in the future.
The spesific reason for the post is a project I am currently concerning myself with. I hit a snag recently, and hope to resolve it with help from others here. Here goes:

Background

I'll ignore friction and air resistance in the following.
I'm making an apparatus consisting of a beam (B) rotating about a pivot (A), loaded with two masses M and m; one attached to each end. It looks like a sort of see-saw.
Now, there are three forces in play, all due to gravity:
-The force on the largest mass M, GM
-The force on the smaller mass m, Gm
-The force on the beam itself, GB.

Please note that A may be different from the beam's centre of gravity.
The system is designed so that of the three resulting torques (only two if the beam's weight doesn't set up a torque about A), TM will always be bigger than the other(s).
The result is that the system will be set in rotational motion if held in an initial position, and then released. The mass M will move in an arc downwards, and then hit some kind of blockage (e.g. the floor).

I have calculated the system's moment of inertia (I), and the magnitude of the three torques are easily found. The angle between the beam and the horizontal axis will be denoted as θ.

Problem:

Once the system is released, the total torque will accelerate the system from rest because of ΣT = Iα.
In so doing, the mass M will move through a small angle dθ, and the lever arm of TM will be shortened since the arm is equal to R*cos(θ). Thus the torque decreases, and also the acceleration. The same is true for the mass m on the other side of the pivot, only that its torque in the other (negative) direction is decreased.
The net result I'm guessing will be a steadily decreasing acceleration and an angular speed that keeps growing throughout the motion, but less and less with time.

How do I model this change in acceleration and speed mathematically?
My goal is to calculate the final speed of the masses M and m, but I can't quite relate the varying quantities to time. My guess is that this will lead to some differential equations, and I'm not steady enough with my math to crack this one.
Any help will be greatly appreciated!

2. Jul 11, 2010

### Naty1

I don't know how to answer your question, but I'd guess a first year college text in dynamics (as opposed to statics) would provide some formulas and application ideas. I guess that would be part of a mechanical engineering curriculum??? Wikipedia might also be a source with some formulas to use....

3. Jul 11, 2010

### Accumulator

I have a book on university physics, but sadly mechanics are not my field, and the theory on rotational dynamics covered doesn't quite help me figuring it out.
Maybe this should have been posted on the mathematics subforums instead? It seems like this calls for more calculus than I can muster at the moment.

I will continue my investigations, and appreciate all additional thoughts on the matter! =)

4. Jul 15, 2010

### Accumulator

Okay, I think I made some headway, so I'll share my thoughts to see whether I'm on the right track.

My new approach is to use energy considerations. I'll treat the two halves of the system as separate, that is, each mass and the part of the beam that is on its side of the pivot is one entity.
Then, by treating both the masses and the beam parts as point masses, and setting a sensible reference level for the gravitational potential energy (the floor), the entire rotational motion amounts to moving four point masses up or down in the gravitational field. Thus

EP1 + EK1 = EP2 + EK2,

and since the system is at rest at startup, it reduces to

EP1 = EP2 + EK2.

Now, as the half with the heavier mass M is going to fall downwards, M's potential energy and its beam part's PE is going to decrease.
The lighter mass m and its beam part, however, will rise up, and gain PE.
Calculating the net resulting PE and comparing with the initial PE gives a difference which must be the system's final kinetic energy.
And finally, given this kinetic energy, one can use

EK = (1/2)*I*ω2

yielding

ω = sqrt(2*I*EK).

Since this last calculation considers the moment of inertia, I don't see that my simplification in the energy part should influence the final speed here.

That's it, any thoughts?