Impulse on a Gyroscope

Question

I've been studying gyroscopes, and I understand why a fast spinning gyroscope doesn't fall. However, I don't understand why a slow spinning gyroscope does fall, or why a large enough impulse to a fast spinning gyroscope will cause it to fall (ie if I push a gyroscope down with my hand, it will fall). If a gyroscope is spinning, it must have angular momentum outward or inward, along the radial axis, due to the spinning wheel/disk. If a slow spinning gyroscope (or a fast spinning one experiencing a large impulse) falls, this angular momentum must change direction. Only torques cause changes in angular momentum, which means this gyroscope must be experiencing a torque perpendicular to the angular momentum vector of the spinning wheel. However, I can't see where such a torque might be coming from; none of the torques that I've considered act in this direction.

Answer 1

When you are working from the frame of reference of the center of mass of the gyroscope , which is at the center of the disc, the torque due to gravitational force is zero and when the gyroscope is fully upright even the torque due to the normal force by the ground is zero.This is the case for ideal conditions(smooth floor,no irregularities,no dissipative forces such as friction).

In real world conditions the normal force vector moves around the base of the gyroscope due to irregularities in the surface on which it is kept.So a small torque is produced which slowly tilts the axis of rotation.This torque increases in magnitude as the axis makes a larger and larger angle with the vertical till the gyroscope finally topples.Smaller the angular velocity, lesser is the resistance of the body towards this torque.

Hope this helps!If you need me to give a diagram please tell me.

Answer 2

First: you describe two cases, distinguishing between them, but making that distinction isn't necessary.

The two scenarios you describe:

• A slow spinning gyroscope

• A fast spinning gyroscope that you push by hand, exerting a sufficiently strong force to make it topple over.

Here is a circumstance where the slow spinning gyroscope would remain close to upright: when there is only a very small gravitational force, such as when positioned on the surface of an asteroid (so that the gravitational acceleration is only centimeters per second squared.)

Conversely, if you would place a gyroscope in an environment with a far stronger gravity than on Earth then a spin rate that would be sufficient on Earth to remain upright (for quite a while) would be insufficient in that high G case.

The point of view that unites the two cases is the following: it's about the ratio of capacity to remain upright and the magnitude of the force that tends to topple it over.

For every spinning gyroscope there is an amount of force that is sufficient to make it topple over immediately. It's a rule of proportion: the faster the gyroscope spins, the larger the required force to make it topple over immediately.

With that distinction out of the way I move to a more general discussion.

I assume you are thinking about the case where the initial orientation of the gyroscope spin axis is as vertical as you can possibly get it. It's as if you are trying to balance a pencil on its tip. If the initial orientation is very close to perfectly vertical then maybe you get a bit of "hesitation" in tipping over.

So let's say the gyroscope is initially very close to perfectly vertical, so close to perfecly vertical that even if it would not be spinning you would see a slight "hesitation" to topple over. No matter how close to perfectly vertical, in the end the spin axis will start to move away from vertical. (In the case of the actual Earth, the Earth's rotation underneath the gyroscope will make the local surface rotate relative to the gyroscope's axis of rotation.)

Once the gyroscope spin axis is no longer vertical the center of mass is no longer straight above the point of support, so you get a torgue from gravity that tends to topple the gyroscope. In response to the exerted torque a precessing motion commences.

The ability to oppose being toppled over is proportional to the spin rate. When the spin rate is very fast a slow precession rate is sufficient to prevent being toppled over. The slower the spin rate, the faster the required precession rate.

So if you would try this setup multiple times, each time with a slower spin rate you will arrive at a spin rate where there is not enough time and not enough kinetic energy available for the gyroscope to transition to a precessing state.

The precessing motion itself has a corresponding kinetic energy and a corresponding angular momentum, and those two have to be obtained from somewhere else.

When the spin rate is close to critically slow then you really have to scrape the barrel to obtain the necessary energy and momentum. If you have a gyroscope I encourage you to try and find that critical case, and watch it many times.

The slower the spin rate, the faster the required precession rate. So in the critical region the gyroscope can barely get up to the required precession rate. By the time the precession rate is large enough the gyroscope has toppled over through a siginificant angle. In fact, the source of the kinetic energy corresponding to the precessing motion is the move of toppling over. As the gyroscope topples through some angle some gravitational potential energy is converted to kinetic energy. Likewise the angular momentum corresponding to the precessing motion comes from the change of angular momentum as the gyroscope wheel topples through some angle.

The transition to precessing motion is not instantaneous. I have discussed the mechanics of that transition in a 2012 answer here on Stackexchange, to a question titled What determines the direction of precession of a gyroscope?

[Later edit]
The following was initially a comment, but it is too long for a comment.
The following may or may not be relevant to your case. There is a widespread overgeneralization that goes something like this:

"When you exert a torque on a spinning gyroscope wheel then that gyroscope wheel will respond with an infinite opposition to yielding to the applied torque. Instead the gyroscope wheel will move around an axis given by the vector cross product of the spin axis and the axis of the applied torque."

In fact: the gyroscope does yield to the applied torque. It's a matter of proportion. The faster the spin rate the stronger the response of the gyroscope wheel to changing axis orientation. In demonstrations is it common, of course, to push to a very high spin rate; that gives a vivid demonstration. When the spin rate is very high the gyroscope will yield only a little bit, imperceptible to the unaided eye. So at a very high spin rate it's an acceptable approximation to describe the gyroscopic effect as turning around an axis that is perpendicular. This is valid (in approximation) only for high spin rate; it doens't generalize.

That is why it is crucially important to try a range of spin rates, down to spin rates where the gyroscope is just barely able to get to the required precession rate, the case that I referred as 'the critical case'.

If you understand all of the motion that you see in the critical case then you are in a position to understand the motion of the gyroscope in all cases.