Discussion of 370(4)

Agreed with this, any type of torque can be used. I have just sent over Note 370(5), which shows that the use of the angles theta and phi of the spherical polar coordinate system results in a great simplification without loss of generality. This is similar to your calculations with the dumbbell model.

To: EMyrone@aol.com
Sent: 11/02/2017 12:38:30 GMT Standard Time
Subj: Re: 370(4): General Dynamics of the Orbiting Asymmetric Top

Without having read the note:
What about modelling the extra torque by a “rotational potential”

U_rot = Tq_0*phi ?

This gives the torque Tq according to

Tq = partial U_rot / partial phi = Tq_0

that is a consant torque around the Z axis. This can easily be added to the Lagrangian of spinning top with one point fixed for example.


Am 11.02.2017 um 12:51 schrieb EMyrone:

In general the lagrangian is Eq. (2), where the potential energy is a function of r, theta, phi and chi. The problem must be solved with a model for the dependence of U on the Euler angles, so the general solution is found by solving Eqs. (15), (17), 20), (23) and (28) to (30) simultaneously. It is known that this kind of problem can be solved by Maxima on a desktop, so it looks promising. In general the nutations and precessions of the earth, regarded as an asymmetric top, are superimposed on its orbital motion. This kind of approach is completely general, and ought to give patterns reminiscent of the Milankovitch cycles. The earth modelled as a dumbbell has already been solved by Horst Eckardt, and those results might also show some patterns reminiscent of the climate cycles. The extra ingredient in this note is that the potential is assumed to depend on the Euler angles. This assumption means that the nutations and precessions of the earth affect its position in an orbit. If it is assumed as usual that U depends only on r, the problem simplifies. The lagrangian (2) simplifies to that of a symmetric top with one point fixed when r dot = 0 and U = mghcos theta, and when I1 = I2. Here g = – MG / R squared is the acceleration due to gravity, M being the mass of the earth and R its radius. The minus sign denotes attraction. The problem of the gyroscope subjected to an extra potential can also be approached by modelling the potential of the extra force. The extra force will change the nutations and precessions. When the extra force balances the force due to gravity the net potential is zero. So it all comes down to modelling the potential energy.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: