Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-L] spinning skater



Oh, oops. My mistake! Of course, it can't matter HOW it's done. Conservation of angular momentum insures that the final angular velocity will be larger. I neglected to take into account the forces on landing!

If you jump due north in the rotating frame, there is no torque on take off, but there IS a positive (i.e., accelerative) torque on landing. If you jump due north in the ground frame then there is a large positive torque on take off and a smaller negative torque on landing. So the easiest way to understand the result DOES seem to be to "jump due north" in the rotating frame in which case ALL torques are positive.

John Mallinckrodt
Cal Poly Pomona

On May 12, 2013, at 3:37 PM, John Mallinckrodt wrote:

Curtis,

It seems to me that there is a critical ambiguity here.

If you "jump due north" in your own rotating frame by pushing off on the merry-go-round with a force that is directed due south then there is no torque on the merry-go-round and it's angular velocity won't change. As long as the jump isn't too big, you will end up at a smaller radius AND at an advanced angular position on the merry-go-round.

If, on the other hand you "jump due north" as seen by observers on the ground, then you must push off with a component of force in the tangential direction of motion in order to entirely eliminate your own tangential motion in that frame. As long as the jump isn't too big, you will end up at a smaller radius AND at a retarded angular position on the merry go round.

It sounds like you intend for students to make the second interpretation.

John Mallinckrodt

On May 12, 2013, at 2:11 PM, curtis osterhoudt wrote:

[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?

_______________________________________________
Forum for Physics Educators
Phys-l@phys-l.org
http://www.phys-l.org/mailman/listinfo/phys-l