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[Iacopo]

It seems that the very high speed compensates for the high center of mass, so that these tops spin stable.
Also I find interesting the way they spin in tilted position.  They call to mind a kind of gyroscopes.

[ta0]

That is great!

These tops remind me Philippe Dyon's ball top. I generally start it at a large angle to play the inversion game, but it can be started vertically and it will spin straight for about 10 seconds. It probably doesn't spin at more than 6000 RPMs. It is also taller: 17 cm at the top of the ball, 22 cm overall. The tip is relatively pointy (well, small radius).

I have wondered for a long time how much speed would it be needed to spin a pencil vertically. I guess I could try to calculate it with the critical speed formula. By the way, these tops would be ideal to check that expression.

[Softspin]

I see that a bigger eraser might spin that pencil... Those are beautiful the two large white ones seemed to be in a mating ritual. Great workmanship

[the Earl of Whirl]

I love it!  What a great video!!  I can't wait to watch it again!!!

Thanks for posting that.

[Iacopo]

To see a pencil spinning like a top would be amazing.   

[Russpin]

I have wondered for a long time how much speed would it be needed to spin a pencil vertically. I guess I could try to calculate it with the critical speed formula. By the way, these tops would be ideal to check that expression.

The criterion for a sleeping top is S >= 2*sqrt(I1*m*g*h)/I3
where:
 S is the spin angular velocity of the top in rad/s
I1 is the transverse moment of inertia about the contact point
I3 is moment of inertia about the symmetry axis
m is mass of top
g is acceleration of  gravity
h is distance from contact point to the center of gravity of top.


Assuming the pencil is a uniform cylinder of length a and diameter b then it can be shown that:

S >= sqrt(128*a*g*(a*a/3 + b*b/16))/(b*b)

if you let a = 20 cm and b =  1 cm and g = 980 cm/(s*s)

then S >= 18293.8 rad/s or 174692.6 rpm

I had this problem in college from the book “Analytical Mechanics” by Grant R. Fowles. 3rd edition
I remember this problem because I got a different answer than what the book gave. I just found online the 7th edition of this book and found they have corrected the answer to this problem!

[ta0]

Ha! I cannot believe that exact example is used in a textbook!   

A 1 cm diameter pencil is pretty thick, so the speed would actually be somewhat larger.
Mm, 200,000 RPM: that would make for a great youtube video!   

[Jack]

deliciousness @-@

[Jeremy McCreary]

Iacopo: Wonderful video. Thanks for sharing it.

I've been studying the behavior of high-CM tops using LEGO for a while now. The LEGO tops are only a few grams in total mass, but they show all the behaviors seen in the video, though at release speeds of ~5,000 to 9,600 RPM in my case. In particular, I see smooth descents like those at 5:18-6:12 and 15:45-17:02 and bucking descents (weak example at 10:05-10:35). The latter usually start out smooth, but the bucking keeps getting worse once it appears.

The smooth descents look to me to involve pure precession without nutation, while the bucking descents appear to develop nutation after spinning down a while. Reaching the critical spin rate for steady precession may have something to do with the onset of nutation, but tip radius of curvature is also involved.

The persistence of these behaviors over several orders of magnitude in mass confirms that they're largely driven by mass distribution. In my case, the ratio of axial moment of inertia (AMI) to transverse moment of inertia about the tip (TMI) seems to be an important controlling parameter. However, tip radius of curvature also plays an important role: The greater the radius, the slower the rotor's vertical descent, and the smaller the chance of seeing bucking on the way down.

I'll try to post video of some experiments with high-CM LEGO tops in the next few days. One nice thing about LEGO: I can change CM height, AMI, and tip radius of curvature at will. The TMI is hard to isolate, as it's inextricably tied to CM height by the parallel axis theorem (see Wikipedia).

ta0: Do you think that the precession in the video is still slow precession in the physics sense?

[Iacopo]

The criterion for a sleeping top is S >= 2*sqrt(I1*m*g*h)/I3

I see mass compares in the equation, this surprises me a bit.
If I have two tops, with identical design, identical dimensions and identical distribution of weight, but different weight, (for example one made of aluminum and the other of brass), does the brass one have to spin faster than the aluminum one, to stay in sleeping position ?

Intuitively I agree with Jeremy:

the ratio of axial moment of inertia (AMI) to transverse moment of inertia about the tip (TMI) seems to be an important controlling parameter. However, tip radius of curvature also plays an important role

[ta0]

ta0: Do you think that the precession in the video is still slow precession in the physics sense?

When the top is leaning a lot, I think it is clear that most of the energy and rotational momentum is due to the precession and little is due to the spin around the axis of the top, so in that sense I would say it is fast precession. In fact, Philippe's top is very easy to invert to have it end up spinning around the axis in the opposite direction (relative to the top), so the spin has to completely stop at some point (like a tippe top).

I need to record Philippe's top in slow motion.

The criterion for a sleeping top is S >= 2*sqrt(I1*m*g*h)/I3

I see mass compares in the equation, this surprises me a bit.
If I have two tops, with identical design, identical dimensions and identical distribution of weight, but different weight, (for example one made of aluminum and the other of brass), does the brass one have to spin faster than the aluminum one, to stay in sleeping position ?

The mass cancels out. The moments of inertia are proportional to the mass and the equation has square root (I₁*m) divided by I₃

[Iacopo]

The mass cancels out. The moments of inertia are proportional to the mass and the equation has square root (I₁*m) divided by I₃

Ah, ok.  Now I see it. 

[Jeremy McCreary]

The mass cancels out. The moments of inertia are proportional to the mass and the equation has square root (I₁*m) divided by I₃

Ah, ok.  Now I see it.

The fact that mass cancels completely out of the critical speed formula is a crucial point that I have yet to see mentioned in any textbook or journal article. See http://www.ta0.com/forum/index.php/topic,1496.msg46533.html#msg46533 for an attempt to sort out which mass properties control what in top behavior.

BTW, mass also cancels out of the usual approximate expressions for nutation rate and slow and fast precession rate. Hence these are also strictly matters of geometry.

[Jeremy McCreary]

As promised, just uploaded a video of top experiments inspired by this thread. Tops of increasing CM height start appearing right away. Release speeds are typically in the 5,500-6,500 RPM range, but some may go as high as 7,600 RPM by overhand twirl.



Also uploaded a 2nd copy with edge-detection (outline effect) applied to emphasize top behavior. It's interesting to sync the two versions in side-by-side windows.



See either video description for an explanation of the experimental sequence and some things to watch for. (NB: The paragraphs will get run together on a phone.)

I've been studying these tops for several months now but am still puzzled by some of the behaviors observed. This much is clear: (i) Mass distribution and tip radius of curvature both have profound effects on spin-down behavior. (ii) During precession, the vertical precession axis passes through the top's CM, not the tip.

The same behaviors are still present at motorized release speeds of up to 9,600 RPM, but spin times will be longer, as will any sleeping phases right after release.

[Iacopo]

The second video has a nice graphic effect.