Precession Tilt Angle: Exploring Physics of Spinning Top

[dangus]

Im trying to understand the physics of the precession of a spinning top. What determines the tilt angle (the angle the top is tilted off the vertical axis)?

 

[Dr_Morbius]

http://www.lmgtfy.com/?q=precession

 

[K^2]

What determines the tilt angle?

If there is no friction, tilt angle is determined by initial condition only and remains constant. If there is friction, it decays.

 

[dangus]

If there is no friction, tilt angle is determined by initial condition only and remains constant. If there is friction, it decays.

Thank you for the quick reply. I have a couple quick follow up question: If I have a spinning top in a frictionless environment set at a tilt angle θ and I wished to apply a force to increase this tilt angle how would I do that?

Instinctively I would want to apply an additional force downwards in the direction of gravity. It is my understanding, however, that the downwards torquing force of gravity is what creates the precession motion in the first place and that changing its magnitude (by adding an additional force downwards) would change the precession frequency not the tilt angle. Is this correct? How then should I apply a force to change the tilt angle?

Im also curious about precession and how it relates to the motion of a spinning proton in a magnetic field. Since the tilt angle is determined by initial conditions, what determines the tilt angle of a precessing proton?

Thank you

 

[K^2]

Yes, your understanding is correct. To change the tilt angle, you must apply force along or against the direction of prrecession. In other words, a force that's perpendicular both to gravity and to the axis of rotation. To see which increases the tilt angle and which decreases, find direction of the torque.

There is another very creative way to change tilt angle. You can shake the support surface in resonance with precession frequency. It will effectively do the same thing as above, but in some situations, can be a lot easier to achieve mechanically.