Discussion of 368(2)

Many thanks! Marion and Thornton chose the origins of the fixed and moving frames to be the same, so r double dot does not appear in their lagrangian (see second sentence of Section 10.10). It appears in my lagrangian simply because the MT assumption is not made. It is always possible to represent any rigid body by its centre of mass, and the use of Euler angles is equivalent to the use of spherical polar coordinates. My calculations reveal the meaning of beta dot in UFT270 and give a lot of insight. Therefore we can simply remove the term in r dot in Eq. (10) and recalculate. The constants of motion can be calculated using Euler angles as in MT or using spherical polar coordinates with r double dot removed. So the MT treatment can be extended to finite r double dot, or my treatment can be simplified by omitting r double dot. In ether case the point of the top is not lifted by its own rotation. So another force is needed. The purpose of UFT367 and UFT368 is to demonstrate these points rigorously.

To: EMyrone@aol.com
Sent: 20/01/2017 16:36:00 GMT Standard Time
Subj: Re: 368(2): Motion of a Gyroscope in Spherical Polar Coordinates

I checked the equations, see attachment. What you denoted by h is denoted R therein. In the Lagrange equations of the mass point, no r dot should occur because h is constant, insofar I wonder why it appears in eq.(10) of the note. I added an internal rotational kinetic energy (angle alpha) of the mass point but this decouples from the other coordinates and does not impact the motion, this is probably wrong. I should rather use a dumbbell or something like that.

In (11) ff. you introduced the Lagrangian of the rigid body where psi is an Euler angle. How can you compare this withan expression containing r dot in (15), this should be zero. The constants of motion of the symmetric top are L_phi and L_psi according to M&T. Similar arguments apply for the rest of the note. In M&T no radial coordiante appears.

I guess that these problems arise because Lagrangian theory is made for mass points. It is difficult to etxtend it for rigid bodies.

Horst

Am 19.01.2017 um 14:40 schrieb EMyrone:

This note gives the complete set of equations for the motion of a gyroscope modelled as a symmetric top with one point fixed. These are Eqs. (27) to (29), and Eqs. (32) and (33) for the motion of Z(r), the height of the centre of mass of the gyro above its fixed point. In order for the gyro to lift off the ground, another force is needed in the positive Z(r) direction. Eqs. (32) and (33) can be solved for the motion of r using computer algebra.

368-Lagrange.pdf

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