Turning in DanceDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/