Conservation Of Angular Momentum (COAM)

[Merlin]

The magnitude of the AM is so low as to be inconsequential in the physics of loop propagation. That is, there are factors in the physics of the loop that swamp the influence of AM in the loop, so much so that if we ignore the loop's AM, we don't significantly alter the understanding of how the loop works.

Sorry to disagree Graeme

Without AM one cannot understand the underlying mechanism of pullback. One can comment what one sees, but one cannot explain (and certainly not simulate) the behavior of the loop in that case.

Merlin

[Graeme H]

One can but he won’t.

[gordonjudd]

FYI, if anybody is interested I’ve done the full calculation for angular momentum of the loop in the fixed frame.

Walter,
I am very interested to see how you are calculating the angular momentum of the loop in an earth frame. What did you use for the fixed rotation center in that calculation to conclude that the angular momentum of the moving mass in the loop measured in an earth frame was pi/2 times the AM calculated in a moving frame?

Thanks for giving us the details of how you are calculating the AM of the moving mass of the loop in an earth frame. That has to be a complicated problem.

Gordy

[Walter]

Gordy,

I started with the loop centric frame and then I applied the transformation between the loop centric frame and the fixed frame.

[gordonjudd]

I started with the loop centric frame and then I applied the transformation between the loop centric frame and the fixed frame.

Walter,
What did that transformation involve?

Did it just result in adding v_loop to the x velocity of the points going around the loop in the loop centric frame to get the x velocity in the earth frame while the y velocity remained the same in both frames?

Also does that transformation give a fixed rotation center at the center of the semi-circle that is the same for both frames?

Gordy

[Walter]

Gordy,

Are you trying to do an integral with respect to time? Don’t do that. It’s nasty.

Use the equations for motion that Perkins derived.

[gordonjudd]

Are you trying to do an integral with respect to time? Don’t do that. It’s nasty.

Walter,
No. The L=rxp calculation around the loop is done for one instant of time.

Use the equations for motion that Perkins derived.

From eqn 1. in his note on "WHAT ARE THE DRAG FORCES ON THE LOOP AND HOW DO THEY SLOW ITS FALL?" I understand he is getting a velocity profile going around the loop in an earth frame that is equal to:

velocity_around_loop_in_earth_frame.jpg (7.18 KiB) Viewed 362 times

If I do the L=rxp calculation for that velocity profile in an earth frame I get the same AM value as the one I calculated for a moving frame.

Are you using a different approach other than L=rxp to get your pi/2 difference?

Gordy

[Walter]

Those are the individual components of velocity. Are you using the magnitude of Vp to calculate p?

[gordonjudd]

Are you using the magnitude of Vp to calculate p

Walter,
No. As you know, momentum is a vector quantity so you have to use velocity vectors to compute its x and y components.

Are you using a L=rxp approach to compute the angular momentum around the loop?
Gordy

[Walter]

We can add the components of two vectors in a Cartesian co-ordinate system, for example if I add the vector (3, 0) to (0, 4) I get the vector (3, 4). I think that’s straightforward, but the magnitude of the resulting vector is 5 = (sqrt(3^2 +4^2)) not 7 = (3+4). If my velocity consists of the components of 3 m/s North and 4 m/s East then I’m going 5 m/s sort of Northeast ish.