Why does a gyroscope accelerate?

AlwaysCurious

A gyroscope has two forces acting on it (is this wrong?): the normal force and its weight. Assuming they cancel, the total acceleration of the center of mass should be zero. How then, does the center of mass spin in a circle? Is there another force (friction?) present? If I released a gyroscope on ice, would it not spin?

Simon Bridge

...but the acceleration of the center of mass is not zero - what does that tell you about the how gravity gets balanced?

There is plenty written on this online - is there something you don't follow from those descriptions? eg.
http://science.howstuffworks.com/gyroscope1.htm
http://www.physicsforums.com/archive...p/t-75134.html
http://www.physicsforums.com/archive...p/t-50897.html
http://www.physicsforums.com/archive...p/t-21495.html

AlwaysCurious

This is the question I asked: the normal force acts entirely vertically, and so does the weight: because, during uniform precession, the center of mass does not accelerate downwards (or up), the two forces have to cancel. However, the center of mass accelerates, thus ƩF cannot be zero. What is the other force?

I understand the torque and the dL/dt business explaining the rotation. I still, however, cannot understand why the center of mass accelerates, because as said above, one side of the argument says ƩF =0 while another says ma ≠0.

Simon Bridge

... I had hoped that by focussing your attention on only part of your question, you will see that it contains the answer. I also hoped you would look through the links I gave you and attempt the question I asked you.

The normal and gravitational forces only cancel when they both act through the center of mass - which is when the gyroscope is upright. What is the rate of precession in that case?

If you tilt the gyro without spinning it up first, what does it do?
What do you conclude about the balance of the normal and gravity forces in that case?

AlwaysCurious

As a gyroscope precesses uniformly, the vertical component of the center of mass' acceleration is zero, but the center of mass does accelerate. My guess (which I think is correct) is that the table has to force the gyroscope inwards via friction.

If there was no friction, the gyroscope's center of mass would then move in a straight line over time, which seems to be the case in this video: http://www.youtube.com/watch?v=DgZFtCvwTZQ.

Am I correct?

Simon Bridge

I don't think so - how is the direction of the linear motion determined?

Precession in terms of the unbalanced forces.
http://www.youtube.com/watch?v=ty9QSiVC2g0

Notice that the hung gyro preceses (tries to) about it's center of mass.
The usual one has an extra horizontal reaction - friction - keeping the precession about the end-point. When the gyro is tipped over, the normal force no longer balances the weight - and the weight produces a torque ... but if it just fell over, angular momentum would not be conserved.

But you don't seem interested in following advise so I'll leave this for someone else.

TSny

Here is a short video which demonstrates precession in the case of no friction at the supporting surface. At this site there are also some flash demos.

AlwaysCurious

Thank you, TSny! This is what I was wondering (answered in the first link) - without friction from the surface, the center of mass experiences zero acceleration, so the whole gyroscope must turn about the center of mass, and not its base.

TSny

Yes, without friction at the base of support there can be no horizontal component of acceleration of the cm (so the cm cannot swing around in a horizontal circle). There could still be "nutation" in which the cm has a varying vertical component of acceleration (the normal force will be varying in magnitude).

Here's an interesting video I just found on nutation that includes some formidable mathematical references (as well as some Beethoven). Here we do have friction at the base, but it still gives a nice look at nutation.

AlwaysCurious

Thank you again - although I know very little about the math (my book mentioned elliptic integrals but I skimmed over it), it was cool to see nutation, which I had not seen before.