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Author Topic: Minimum critical spin  (Read 440 times)

ta0

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Minimum critical spin
« on: June 08, 2014, 11:59:22 PM »

EDIT: I split and combined these posts as they are mathematically involved.

[NERD alert]

I was browsing a technical paper on the mechanics of the spinning top ("Revisiting the Spinning Top" by C. Provatidis, pdf available here) when something applicable to us caught my eye. The paper has some calculations on the oscillations with respect to the vertical ("nutation") for different rates of spin.  Basically, a fast top will be very stable, keeping a constant angle, while a slow top will oscillate widely.  We all experience this: when a top slows down it reaches a point when it starts to wobble frantically and it is not playable anymore.

From the paper, a top is spinning "fast" if the energy of rotation is much greater than the maximum possible change in gravitational energy (i.e. between standing up and hanging from the tip). 
This translates to 1/2 I w2 >> 2 M g L , or:

w2 >> 4 M g L / I , with w in rad/sec

Using the numbers from the previous example for the Spintastics Trompo Bearing, the critical spin velocity would be:

w >> sqrt(4 x 0.076 kg x 9.81 m/s2 x 0.049 m / 0.0000326 kg m2) x 60 / (2 x 3.1416) = 640 RPM

This result is pretty reasonable.  From measurements I have taken in the past, a top the size of the TG becomes unplayable between 1500 and 1000 RPM.

EDIT: I found another paper mentioning a slightly different equation ("The heavy top: a geometric treatment," by Lewis D. et al, 1992, pdf available here

« Last Edit: July 15, 2016, 10:42:13 AM by ta0 »
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Jeremy McCreary

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Minimum critical spin
« Reply #1 on: July 15, 2016, 05:08:49 AM »

I believe the formula Jeremy posted and a different one I once posted give an estimate on the transition speed between precession at a more or less uniform angle and the start of large oscillations (nutation). If I recall correctly, it starts when the energy of spin is of the magnitude of the gravitational energy.

I have seen a few tops rise from nearly touching the floor to standing up and sleeping.
The formula I gave really does estimate the minimum spin rate sc for stable sleeping, but some of the underlying assumptions might be questioned. A more general formula is

sc2 = 4 (I1 - I3) M g h / I32,

where I1 is the transverse moment of inertia about the contact point, not the center of mass; I3 is the axial moment of inertia; M is the top's mass; g is the acceleration of gravity; and h is the axial distance from the contact point to the top's center of mass.

A similar formula gives the minimum spin rate scp for stable steady precession at constant inclination angle a:

scp2 = 4 (I1 - I3) M g h cos(a) / I32,

These formulas skirt a thorny issue having to do with the fact that the precession rate p about the vertical isn't necessarily zero in a sleeping top. They work only when I1 > I3, but that's usually the case for realistic values of h.

The spin rate at which axial kinetic energy equals gravitational potential energy (GPE) is different from both of these.

Assume steady precession at constant a. Relative to the contact point, the GPE is

V = M g h cos(a).

The kinetic energy about the top's symmetry axis (subscript 3) is

T3 = ½ I3 w32,

where w3 is the total angular speed about the symmetry axis given by

w3 = s + p cos(a).

The total angular speed w3eq when T3 = V is then

w3eq2 = 2 M g h cos(a) / I3

Unfortunately, one must specify the precession rate p in order to extract the corresponding spin rate from w3eq. The usual formulas for estimating p won't do here, as they all assume that

w3 >> w3eq.

Nor is it kosher to assume that

s >> p

in this low-speed regime, where the usual formulas surrounding nutation also break down.
« Last Edit: July 15, 2016, 10:43:56 AM by ta0 »
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Jeremy McCreary

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Re: Minimum critical spin
« Reply #2 on: July 15, 2016, 02:04:59 PM »

The w in ta0's formula at the top of this thread refers to the total angular speed about the top's symmetry axis. In my post above, I used w3 for the same thing.

The spin rate s about the symmetry axis is somewhat different:

s = w - p cos(a),

where p is the precession rate about the vertical and a is the precession angle.

The distinction isn't all that important at the start of a good spin, when s is high and p is typically low, but can become important in a top precessing rapidly at low spin rate. The situation is further complicated by the fact that p is undefined and not necessarily zero in a sleeper.

To get the minimum spin rate smin for stable steady precession at constant nonzero inclination a without having to worry about the precession rate explicitly, you have to start from Equation (20a) in Provatidis rather than from Equation (22). You then end up with

smin2 = 4 (I1 - I3) M g h cos(a) / I32

instead of

w2 = 4 I1 M g h cos(a) / I32.

Among other sources, Jerry Ginsberg's Advanced Engineering Dynamics at http://www.pwut.ac.ir/FA/Colleges/Coll1/Files/Advanced%20Dynamics.pdf supports this formula for smin for steady precession.

The derivation for a stable sleeper (a = 0) has to take a different route to avoid dividing Provatidis' Equation (20a) through by zero, but the result is what you'd get by plugging a = 0 into smin  formula for steady precession -- i.e.,

smin2 = 4 (I1 - I3) M g h / I32

Ginsberg gives this formula as well.

Importantly, none of the formulas above require a "fast top" with much more kinetic than potential energy. That assumption becomes necessary only when you try to go on to derive simple closed expressions for the precession and nutation rates.
« Last Edit: July 15, 2016, 02:15:46 PM by Jeremy McCreary »
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