Itemizing P.M.

[Michael]

No Jim this explanation dosn't cover all aspects.

> And if a rod or string were to re-connect then the weight would immediately swing around the new axle point at the same velocity as the straight line and the previous rotation.

I meant radial kinetic energy, I should have used the term radial velocity I suppose.
Radial velocity, the speed at which a weight is traveling radially. Okay, when a weight is released it, as you said, is no longer traveling in a radial path. If a weight that was traveling in a straight line were to connect to a wheel at a point that was 90 degrees to the wheel then there would be no loss if the wheel it connected to was also traveling at the same speed. Here's a visual example.

Wheel is 10 feet. Weight is traveling horizontally to the wheel and connects to the wheel at the 12 oclock position.

What we are working with though is the reconnection of a weight thats been let go and then reconnects a distance a bit further down from the axel on the same wheel.
There are different means when a weight is released from the same wheel and this discussion is not in trying to cover those at the moment but, for instance, with a horizontal wheel the wheel will have to do some work against the thrown weight to move it outwards, or it can be said that the weight interacts with the wheel in a way that prevents it from traveling in a straight line and makes it move outwards. When the weight reconnects on this system there should be a loss for two reasons.

(1) because of this last sentence and,
(2) because the weight doesn't reconnect at a 90 degree angle, there will be absorbtion and shock losses. This is the math Jonathan is working on.

There is still something very interesting about all of this that I'll post on later.

Michael

[Jonathan]

The math I'm trying to do covers all of it, not just the last part there. I think I found what was messing me up before, but I took a break today when I got struck trying to solve for r(@):
(d^2)r(@)=((d@)^2)-k)r(@)+kr_o
....dt^2......((.dt)......)...)
where @ is theta, k and r_o are a constants, _ means subscript, ... are for spacing, and one parenthesis above another is an attempt at a parenthesis that is twice the normal height.

[Jonathan]

Never mind solving for r(@). I did the math and found that the rotational kinetic energy lost, as calculated with conservation laws, is equal to the the work done by centrifugal force. This means that FE through centrifugal force is not possible with the system in question, which is very general: there is an intrinsic rotational inertia which is assumed to be, say, a framework, none of which moves radially, and a point mass which moves in, out, and around along the arbitrary path given by r(@). The system is started with an initial angular velocity w_0.
The first line are just basic things taken for granted. On the right, all lines are devoted to finding the work done by centrifugal force. The first line on the left finds the angular velocity as a function of angular position using conservation of angular momentum. The next line find the rotational kinetic energy as a function of angular position. The last line on the left uses the work energy theorem and finds the work that must be done to loose a given amount of rotational kinetic energy, as a function of angular position.