# What fundamental force keeps a tilted top steady?

1. Dec 31, 2011

### kmarinas86

A tilted top that would otherwise topple over can somehow not do so if it spins on an axis fast enough, but only if that axis also pivots fast enough circumferentially with respect to a "principal axis". Often conservation of angular momentum is used to predict this. But the rotational pull away from the principal axis caused by the force of gravity on a tilted top must clearly have an opposing torque that is not this gravitational force. The force that keeps the top up must be rotational *toward* the principal axis. No force of nature we presently know about, whether gravitiational, electric, magnetic, or nuclear, seems to possess any capacity for some innate attraction towards an axis simply by spinning mass in such a manner, though it would appear to me that of all the four just mentioned, only the magnetic, with its inherent or "irreducibly axial" nature, appears to resemble this phenomenon, and especially so because stability of a top requires that the spin of the top and the movement about the principal axis be at least within 90 degrees of alignment, similar to the rotating currents in the loop model for electromagnetism.

Last edited: Jan 1, 2012
2. Jan 1, 2012

### Staff: Mentor

No it doesn't. Why do you think this? (Don't forget that angular momentum and torque are vector quantities.)
No "force" is needed to keep the top up. The torque on the spinning top changes the angular momentum of the top accordingly. It's not in rotational equilibrium.

(It's like asking what force counters gravity to keep the moon from falling. Obviously no such force is necessary, since the moon is accelerating.)

3. Jan 3, 2012

### kmarinas86

Net torque leads to a rate change in angular momentum. If the rotational speed is fixed, then the torques must be equal and opposite. Such opposing torque vectors must be anti-parallel.

If I have three principal coordinates, say r, phi, and h, I have three types of surfaces whose normals are projected across the domain of these coordinates, these being phi-h ("tube-shaped") surfaces, h-r ("flat vertical") surfaces, and r-phi ("flat horizontal") surfaces, respectively.

The torque on the top due to gravity and the ground induces a rotational derivative within an h-r surface whose normal is extended across the phi domain. The torque vector itself is normal to that same h-r surface. Where do I find the torque that is ANTI-PARALLEL to THIS?

I could not find it:

When the top, already spinning, is made to precess, one should not be adding a rotation bounded inside an r-phi surface, because like you say, the angular momentum about the principal axis is conserved. The net torque on the principal axis is zero.

Well it can't be parallel to the principal axis anyway, so moving on:

How does constant integral of torque (angular momentum) in the phi-h surfaces cancel a torque (a derivative of angular momentum) in an h-r surface?

Let me note that a constant (angular momentum) has no torque! It's not found by motions within the phi-h surfaces!

From the point of view of the principal axis (where the r domain is seen as an "altitude" domain), the top is basically flipping end-over-end at some diagonal angle existing in any of the h-r surfaces at any given time, traversing phi coordinates.

Yes, I can see that as being a change of angular momentum per change in time where in this case:

* The angular momentum vector "clocks" through the phi-domain, while
* This vector is always at tangent to an h-r surface for each phi coordinate.

It also turns out that the sweeping velocity during precession is the same direction as the torque vector produced by gravity. However, torque causes acceleration circumferential with respect to its vector. It's not supposed to cause acceleration parallel to it!

Why should a combination of:
* top spin angular momentum (whose vector is on the top's axis of symmetry, embedded in a r-h plane), and
* some angular velocity about the principal axis (whose vector is on the principal axis, at r=0)
.....somehow counter act a torque (whose vector that is perpendicular to both the top's axis of symmetry and the principal axis, normal to an r-h plane)?
Does the does the product of the angular momentum and some angular frequency always produce torque this way? I mean, is it a source of some kind rotational force? Could this expressed as some kind of continuum field? If so, what is there?

Last edited: Jan 3, 2012
4. Jan 3, 2012

### Staff: Mentor

If the angular velocity were fixed, then you would be correct. But it's not. There is an unbalanced torque continually acting perpendicular to the angular velocity of the spinning top, causing it to change direction (precess).

5. Jan 3, 2012

### kmarinas86

Why does an unbalanced torque whose vector is in a horizontal plane, get counteracted by a vector angular velocity $\overrightarrow{\omega}_p$ that exists at the principal axis?

And if you did have an accumulation of angular momentum at the principal axis or the top spin axis, what does this accumulation of spin do to accumulate a resistance to a change in angular velocity, orthogonal to these two (somewhat aligned) vectors, that would otherwise be caused by "gravity+fulcrum"?

How exactly do you combine an angular velocity of a precession vector and a top spin angular momentum vector into a torque? The cross product right? That would give us a vector perpendicular to both. What is the physical meaning of this cross product? Why would it in anyway act to resist the top spin angular momentum from falling wayside from the principal axis?

Is there something material inside that transfers a force at a right angle relative to product of the angular velocities $\overrightarrow{\omega_p}$ and $\overrightarrow{\omega}_s$ and the mass moment $mr$ in the direction from $\overrightarrow{\omega_s}$ to $\overrightarrow{\omega}_p$ to balance out the torque cause by "gravity+fulcrum"?

Last edited: Jan 3, 2012