Two precession rates for a spinning top (forced precession)

[garygary]

When you have a spinning top undergoing forced precession (due to the torque created by its weight trying to tip it), you seem to get two precession rates. Why?

I get this result from using Euler's equations, specifically that the sum of moments tipping the top is (in the case of steady precession):

$$\sum$$ M₀ = $$\psi$$' sin($$\theta$$) (I ($$\psi$$' cos($$\theta$$) + p) - I₀ $$\psi$$' cos($$\theta$$))

Where:
$$\psi$$' = angular velocity of precession
I = mass moment of inertia of top about spin axis
I₀ = mass moment of inertia of top about transverse axis
p = angular velocity of top about its spin axis, its spin
$$\theta$$ = angle of precession

The only moment acting on the top is its weight, which can be written as:
mgl sin($$\theta$$)

Where:
m = mass of top
g = gravity
l = distance from tipping axis to centre of mass of the top

This gives me the quadratic equation:

0 = ($$\psi$$')² cos($$\theta$$) (I - I₀) + I p $$\psi$$' - lmg

Which when solved for $$\psi$$' gives two precession rates. Which one will the top actually precess at, and why? One precession rate is much smaller than the other. If instead of a top it is a cylindrical shape, with I₀>I, both precession rates have the same sign. I'm having trouble working out exactly what this means in reality. So the top can precess stably at both of these precession rates? What makes it go at one rate rather than the other?

In most other examples I can find it is typically assumed that the precession rate is much less than the spin rate, so the squared term of the quadratic is ignored and you only get one answer. This answer is always closest to the larger precession rate that I find above.

This is my first post, so hopefully you can understand my technique and exactly what I'm asking, any help is greatly appreciated.

 

[Cleonis]

When you have a spinning top undergoing forced precession (due to the torque created by its weight trying to tip it), you seem to get two precession rates. Why?

I get this result from using Euler's equations [...]

I wonder whether the two rates you are getting are for precession rate and nutation rate respectively.

Nutation rate is always faster than precession rate. As you refer to, when the spin rate is not much faster than the precession rate things get critical.

The most informative discussion of nutation that I know of is in two articles by Eugene Butikov.
http://faculty.ifmo.ru/butikov/Applets/Precession.html"
http://faculty.ifmo.ru/butikov/Applets/Gyroscope.html"

And possibly the article about http://www.cleonis.nl/physics/phys256/gyroscope_physics.php" on my own website will be helpful to you.

 

[D H]

I wonder whether the two rates you are getting are for precession rate and nutation rate respectively.

The two rates are for steady-state precession. In other words, no nutation is present. The physics of gyroscopic precession is covered in detail in most undergrad classical mechanics texts. The reason you get two precession rates is because the equation for the steady-state precession is a quadratic.

It is usually the slow rate that is observed.

 

[D H]

In most other examples I can find it is typically assumed that the precession rate is much less than the spin rate, so the squared term of the quadratic is ignored and you only get one answer. This answer is always closest to the larger precession rate that I find above.

Dropping the quadratic term yields the smaller precession rate, not the larger one.

Consider the quadratic equation

$$ax^2 - bx + c = 0,\quad a,b,c>0$$

This obviously has two solutions given by

$$x=\frac{b\pm\sqrt{b^2-4ac}}{2a} = \frac{b}{2a}\left(1\pm\sqrt{1-\frac{4ac}{b^2}}\right)$$

When $b^2\gg 4ac$, the radical can be expanded to yield two solutions,

$$\begin{aligned}<br /> x_+ = \frac{b}{2a}\left(1+\sqrt{1-\frac{4ac}{b^2}}\right) \approx \frac b a \\[10pt]<br /> x_- = \frac{b}{2a}\left(1-\sqrt{1-\frac{4ac}{b^2}}\right) \approx \frac c b<br /> \end{aligned}$$

Since $b^2\gg 4ac$, $x_+\gg x_-$.

Another way to look at this: The $x_+$ solution results from ignoring the constant term in the quadratic, leaving $(ax-b)x = 0$. The $x_-$ solution results from ignoring that quadratic term, leaving $-bx+c = 0$.

 

[garygary]

Dropping the quadratic term yields the smaller precession rate, not the larger one.

Yes you're right, I'd made a mistake when calculating it before. And yes, both the rates I get are definitely precession rates.

I asked one of my lecturers about this problem today, and he explained it all to me. Both precession rates exist, and the top can stably precess at either one of them. The difference is however in the energies of both states. The higher precession rate requires significantly more energy in the system, and as such it rarely exists. If it does occur, any damping will reduce its energy and it will try to switch to the lower energy state, with the slower precession rate.

If you imagine a pendulum say, it also has two stable states. The lowest energy stable state is with the pendulum pointing directly down, but it can also be stable with the pendulum perfectly balanced and pointing up. If there is any disturbance the system will try to switch from the high energy state to the lower energy state, ie; the pendulum will unbalance and try to point directly down. A similar thing occurs for the top apparently, which is why the higher precession rate rarely occurs and when it does it is short lived.