Turning in Dance

Date: Thu, 30 May 1996 21:31:19 -0700 (PDT) 
From: Wendy Fan
Subject: turning in dance

I am interested in finding out how a dancer is able to sustain a turn. 
What roles do angular velocity and acceleration play? What could I 
do with information gathered about angular velocity, angular 
acceleration, centripetal force, and centripetal acceleration? 

I plan to begin experimenting with the idea of a dancer turning by 
using a top as an example. I was wondering if information on how 
a top turns is available. 

Thank you.  
Wendy Fan

Date: Fri, 31 May 1996 19:23:45 -0400 (EDT) 
From: Dr. Tom
Subject: Re: turning in dance

I'm not positive how to answer this because I don't know your 
background in mathematics and physics, so I'll start with some 
simple stuff, and then get more complex. You can stop reading 
when it stops making sense.

The spinning of a dancer is completely described by physics, but 
the physics can get quite complex. There are many differences 
between a dancer and the "top" from the point of view of a physics 
problem. 

For example, the dancer's toe on the ground is not frictionless. The 
movement through the air is not frictionless, since there's air drag. 
The dancer's body and clothes are not completely rigid during the 
turn, and, due to the friction, the dancer's body may not be turning 
exactly around its center of mass.

However, for a first approximation, a top is not a bad model. 
Depending on the amount of resin on the tip of the toe-shoe, there 
may be very little friction, and turning speeds are not too high, so 
air friction may be low. (The more billowy the dress, or the more 
resin there is, the less like a top the dancer will be. An ice skater in 
a skin-tight outfit will be much more like a top than the best 
ballerina.) 

A top with no friction will spin forever. If it is perfectly 
symmetric, and starts straight up, it will stay that way forever. If it 
starts tilted, after a while it will end up "precessing" - the axis of 
the top will turn around the tip touching the ground. When it's first 
released, it will have some "nutation" that will die out - a sort of 
bouncing around the "perfect" precession path. I learned some of 
the details of this in my first year of college physics, but I really 
didn't understand why until my third year when we re-solved all 
the problems with some better physical intuition and better 
mathematical tools. 

An important thing that a dancer can do is change the moment of 
inertia by bringing the hands/arms closer to the body during the 
spin. As this happens, the moment of inertia decreases, and the 
angular velocity has to increase to keep the energy the same (again, 
ignoring friction). This works great for skaters who start spinning 
with their arms out, and when they bring them in, their spinning 
speeds up like crazy.

One cool thing about physics that's not too hard to understand is 
that there is a more-or-less perfect analogy between the equations 
of linear motion and the equations of rotary motion. If you take 
almost any equation of linear motion, and substitute angular 
velocity for velocity, angular momentum for momentum, moment 
of inertia for mass, angular acceleration for acceleration, torque 
for moment of inertia, and so on, the exact same equations hold. 

So F = ma (force = mass times acceleration) 

becomes:

T = I*alpha (torque = moment of inertia times angular acceleration) 

Or E = 1/2*mv^2 (kinetic energy = half the mass times velocity squared) 

becomes:

E = 1/2*Iw^2 (k. energy = half mom. of inertia * ang. velocity squared) 

This last equation explains the increased speed when the dancer 
pulls the arms in. The energy has to stay the same (assuming no 
friction), so if I is decreased, w must increase to balance it. And the 
equation shows by exactly how much.

Other weird things happen, too. A spinning dancer or skater will 
get dizzy without "spotting", so a good dancer will lock his/her 
eyes on one thing, turn the neck to keep spotting, and then whip it 
around to lock again. The mass of the head is not huge, but it's not 
insignificant, either, and a spotting dancer will therefore have an 
irregular spinning rate.

You can learn about spinning tops in any good physics book on 
mechanics. For a thorough understanding, you'll need a thorough 
understanding of physics, however.

I hope this helps.

-Doctor Tom, The Math Forum

Date: 08/03/2001 at 02:02:35
From: Ard Jonker
Subject: Gyration radius

The key issue here is 'moment of inertia'. Is this the same as 
'gyration radius'? 

Date: 08/03/2001 at 10:01:36
From: Doctor Rick
Subject: Re: Gyration radius

Hi, Ard.

I had no difficulty finding a definition of "radius of gyration" in 
terms of moment of inertia by searching the Web, using Google 
(www.google.com) and searching on the phrases

  "radius of gyration" "moment of inertia"

One such definition can be found here:

  http://www.xrefer.com/entry/644217   

I quote:

radius of gyration
Symbol R0 or k.

For a rigid body of mass M, rotating about a principal axis, the 
radius at which a point mass M would have to be placed to have the 
same moment of inertia, I. Thus the radius of gyration, given by 
k = [sqroot](I/M), depends on the body's shape, not its density.

The New Penguin Dictionary of Science, copyright M. J. Clugston 1998

I think that says it quite clearly. The two are not the same, but they 
are closely related. The radius of gyration is a distance, as the name 
suggests, while the moment of inertia has units of mass times distance 
squared. You can compute one from the other if you also know the mass 
of the object.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/