1
$\begingroup$

We know that an increase in height of a spinning top decreases the time it spins for without toppling. Why does this happen?

I suspect that this is connected to the fact that an increase in height of a spinning top increases the rate of precession of the top. However, I am not able to mathematically determine any such connection.

$\endgroup$
1
  • $\begingroup$ Related: physics.stackexchange.com/a/4871 The calculations therein are dependent upon the height of the top. $\endgroup$ – user191954 Jan 27 '19 at 9:34
1
$\begingroup$

Allow me to restate your question in the following sharper form:

Given two spinning tops with the same moment of inertia, but the second top with the center of mass higher above the supporting point, will the two tops behave differently?

I assume that at the start the spinning tops are as vertical as possible, and they start of course at the same spin rate. I also assume the two shapes are matched in such a way that the air resistance for both the tops comes out the same. I assume a normal friction for the point of contact of the spinning top.

After a while the top start tipping over visibly, and from then on things start snowballing.

As the top starts tipping a bit you get a torque from gravity. For the top with the higher center of mass this torque will be larger.

A larger torque has a larger corresponding precession rate. The kinetic energy for the precessing motion comes at the expense of the gravitational potential energy of the spinning top. That is, in order for the precessing motion to get going the center of mass must drop a bit. Pitching over provides that drop. All this means that for the same spin rate the top with the higher center of mass will pitch over a bit more than the top with the lower center of mass.

The point of contact of the spinning top is not an infinitely sharp point, it has a radius. As the top start pitching over the point of contact with the surface that it is resting on is no longer coaxial with the main symmetry axis of the spinning top. This gives rise to additional friction. The more the top pitches over the stronger the adverse effect of the additional friction.

So indeed you do expect that in normal circumstances, normal friction, the top with the higher center of mass will have shorter endurance.


As to why you cannot find a mathematical determination:
A possible explanation for that is that the usual approximation does not take into account that the center of mass must drop a bit in order for the precessing motion to get going. A common approximation is to set up the calculation as if the spinning top starts precessing instead of pitching.

The following two resources provide all the information that is needed to understand gyroscopic precession.
A 2012 answer here on physics.stackexchange, by me, to the question: What determines the direction of precession of a gyroscope?
Paper by Svilen Kostov and Daniel Hammer: 'It has to go down in order to go around'.
Kostov and Hammer discuss how a spinning gyroscope responds to a torque, and they have performed a benchtop experiment to verify the expected behavior

$\endgroup$
4
  • $\begingroup$ Very useful! I almost made a fatal blunder in assuming that the kinetic energy required to precess is matched with a reduction in rotational kinetic energy. $\endgroup$ – crash Jan 27 '19 at 17:06
  • $\begingroup$ "For the top with the higher center of mass this torque will be larger." - why is that? The force is the same and the moment arm is shorter, so the torque will be smaller - what am I missing? $\endgroup$ – NickD Feb 12 '19 at 17:53
  • $\begingroup$ Clearly what I have in mind is that with a higher center of mass the arm is longer. Your question just screams for a picture. What you could do is submit this follow-up question as a new physics.stackexchange question, and that you create and upload a picture of what you have in mind. With a picture babylonian confusion may be avoided. $\endgroup$ – Cleonis Feb 12 '19 at 19:34
  • $\begingroup$ I am not the OP: I just had a question about your answer. But it does seem indeed that we are all picturing different things in our minds. $\endgroup$ – NickD Feb 13 '19 at 0:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.