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# Re: [Phys-L] spinning skater

• From: curtis osterhoudt <flutzpah@yahoo.com>
• Date: Sun, 12 May 2013 14:11:27 -0700 (PDT)

This is a question I give in both my algebra-based and calculus-based "general" physics classes. Some students see right away that one's jumping has to exert a torque; others insist that no torque is exerted on the merry-go-round platform itself by the game-player.
Most people answer via the traditional "angular momentum is conserved in this purportedly closed system, and the moment of inertia about the axis of rotation is reducing, so the merry-go-round must spin faster at the end" route, but of course that doesn't get them full points. Nor do most people understand that they need to define carefully what they mean by "the system", even though that's stressed again and again in the lecture and homework problems. Many students draw parallels with the figure skater, not having realized that the figure-skater's arms have to pull at *an angle* to a radial vector in order to be drawn in, and therefore exert a torque on the skater's body, too. This is the difference between the merry-go-round jumper exerting a _force_ due south, and them exerting a force a little west-of-due-south.

The question as worded, though I also include a figure with the "arcs" defined on it:
========

[This question is meant to be entirely conceptual∗ . Probably the best thing to do is to close your eyes and imagine the situation in detail once you've read the problem description.]
You and a friend decide to play a game on an ideal merry-go-round (with frictionless axle), as follows:
1. You stand facing inward on the edge of the merry-go-round while your friend spins it up to some speed. The game starts when you reach the southernmost point of your circular trajectory (marked start in the figure);
2. You ride the merry-go-round all the way around (arc 1) until you reach the starting point, then jump due north (2) to a smaller radius;
3. You ride the smaller radius arc (3) around a full circuit, then jump due north (4) to yet a smaller radius;
4. Repeat the cycle of riding the merry-go-round all the way around to the southernmost point at a given radius, and then jump due north to a smaller radius, until you reach the very middle of the merry-go-round.
By the time you've reached the middle of the merry-go-round, it is spinning faster than when you started. I'd like to know why the merry-go-round is spinning faster than it was. The axle has no friction, and any friction or air resistance would slow the merry-go-round, not speed it up.
You can answer this in terms of Lfinal = Linitial , but that's not really what I'm getting at. What I really want you to think about (and answer) is: Did you exert a torque with all of your jumping around? If so, where and how? If not, then what's going on?
========

From: "LaMontagne, Bob" <RLAMONT@providence.edu>
To: "phys-l@phys-l.org" <phys-l@phys-l.org>
Sent: Sunday, May 12, 2013 11:26 AM
Subject: [Phys-L] spinning skater

I have an issue with the way that Newton's Second Law is presented in many texts and webpages when rotation is involved.

e.g. http://en.wikipedia.org/wiki/Angular_acceleration

An analogy is made between F=ma and T=IA (I'll use I for moment of Inertia and A for Angular acceleration so I don't have to keep writing alpha). Elsewhere, the analogy is made between F=dp/dt and T=dL/dt (L being angular momentum with L=Iw and w being angular velocity).

For most of introductory level physics, one doesn't get into much trouble using the two forms F=ma and F=dp/dt interchangeably because an object can't change its mass at will if mass does not pass in and out of its boundaries.

However, the two forms T=IA and T=dL/dt can give very different results because a rotating object can easily change its moment of inertia even if there is no interaction with the external world. For a spinning skater, T=IA would predict that with no external torque acting that A must be constant - regardless of changes in moment of inertia. But T=dL/dt predicts that the rotation will speed up if the skater's arms are pulled in.

I only bring this up because so many texts and webpages don't emphasize how T=IA assumes that the object is rigid.

As you can probably guess - I'm grading final exams :-)

Bob at PC
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