A Physics Puzzle: Double Circular Motion

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Describe a simple way to force a particle to move along two different circular trajectories at the same time. (Two circles that are not concentric) Hint: The trajectories do not have to be measured from the same frame of reference.

It need not be an elementary particle (could be even a small ball)
 
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  • #2
BobG
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Describe a simple way to force a particle to move along two different circular trajectories at the same time. (Two circles that are not concentric) Hint: The trajectories do not have to be measured from the same frame of reference.

It need not be an elementary particle (could be even a small ball)

Or a planet?

http://faculty.fullerton.edu/cmcconnell/Planets.html#7a
 
  • #4
drizzle
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A gum stuck on a tire of a car that keeps driving around a roundabout! :biggrin:
 
  • #5
berkeman
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Haha i'm sorry but i'm afraid this is not correct. :tongue2:

Why is it not correct? Because he didn't say that the particle is sitting on the surface of the planet, and is not located at either pole?
 
  • #6
drizzle
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He probably should've just said 'me' or 'you Mr. AlexChandler'. :biggrin:
 
  • #7
BobG
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Why is it not correct? Because he didn't say that the particle is sitting on the surface of the planet, and is not located at either pole?

Well, it is true that planets don't orbit the Earth via epicycles. In fact, they orbit the Sun in ellipses. In fact, it would be highly unlikely to find any particle that moved freely in a perfectly circular path.

Perfectly circular motion usually requires the particle to be fixed to the point it's circling.

But the link is an example of double circular motion, even if it doesn't reflect the real motion of the planets (nor was it intended to, since the emphasis was on the history of celestial mechanics).

And it would be simple to rig a device where the axes of a spinning gyroscope were fitted into a forked frame and the frame rotated about a fixed point.
 
  • #8
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A gum stuck on a tire of a car that keeps driving around a roundabout! :biggrin:

Hehe also a nice idea, but i'm afraid only the trajectory around the tire is circular. The trajectory around the roundabout is more like a cycloid.
 
  • #9
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Me on a merry go round spinning a ball on a string?

You're being picky. Strictly speaking you'll never get a perfect circle.

The above are all valid.
 
  • #10
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But the link is an example of double circular motion, even if it doesn't reflect the real motion of the planets (nor was it intended to, since the emphasis was on the history of celestial mechanics).

And it would be simple to rig a device where the axes of a spinning gyroscope were fitted into a forked frame and the frame rotated about a fixed point.

But even if the epicycles were possible, this would not satisfy the puzzle. The epicycles would be circular trajectories measured in the frame of reference co-moving with the center of the epicycles, but the overall trajectory described from the frame of reference of the sun would not be circular at all.

Could you elaborate on the final idea you presented? I think it will also not work for a similar reason, but I do not fully understand what is meant by "fitted into a forked frame".
 
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  • #11
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I think the puzzle has been somewhat misunderstood. The point is to find a forced motion such that two distinct frames of reference could accurately describe the trajectory of the particle as circular, and the two circular trajectories must not have the same point in space as a center.

The trajectories do not have to be full circles, as long as the particle moves along two distinct circular arcs for some time.
 
  • #12
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Me on a merry go round spinning a ball on a string?

You're being picky. Strictly speaking you'll never get a perfect circle.

The above are all valid.

I am not being picky :tongue: but the above mentioned motions do not satisfy the puzzle. I apologize if it is unclear what I am asking for, but I tried to be as exact as possible in my wording.

For example, you on the merry go round: In your frame of reference, the trajectory of the ball is absolutely a circle. But from the frame of reference of the center of the merry go round, the trajectory is very complicated, actually it is the same as thing as the epicycles described before. It is not circular. There is only one circular trajectory in each of these examples.
 
  • #13
drizzle
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[PLAIN]http://i2.squidoocdn.com/resize/squidoo_images/250/draft_lens17687361module148542008photo_1298768215drawing1.jpg [Broken]

A motion of a dropped coin led me to this example. Imagine if there's a small body in the contact point between the coin and the surface of the floor, won't that body be moving two different circular trajectories!... Actually the body should be fixed while the two references move relatively to each other... URGH! I'd like to know the answer to this question, thanks for keeping me curious. :grumpy:
 
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  • #14
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For example, you on the merry go round: In your frame of reference, the trajectory of the ball is absolutely a circle. But from the frame of reference of the center of the merry go round, the trajectory is very complicated, actually it is the same as thing as the epicycles described before. It is not circular. There is only one circular trajectory in each of these examples.
For example, if the ball is being spun about a horizontal axis even as the merry-go-round is rotating about a vertical axis. The center of the first frame of reference is the tip of of JaredJames' fingers holding the string. The center of the second frame is a point moving sinusoidally up and down through the axis of the merry-go-round in time with the rise and fall of the ball.
 
  • #15
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For example, if the ball is being spun about a horizontal axis even as the merry-go-round is rotating about a vertical axis. The center of the first frame of reference is the tip of of JaredJames' fingers holding the string. The center of the second frame is a point moving sinusoidally up and down through the axis of the merry-go-round in time with the rise and fall of the ball.

Actually this does work. Great thinking! I thought at first that it would not work for the following reason: The oscillating frame of reference at the center of the merry go round will see that the ball moves back and forth along the circular trajectory. But actually if the merry go round is moving fast enough, the ball will only slow down in the FOR of the center of the merry go round, so the trajectory will be circular as measured from two distinct reference frames with two distinct centers. Great job! In fact I think that this answer is better than the one I had in mind. And in fact this would satisfy the puzzle even if the ball did travel back and forth along the trajectory. I am quite impressed.
 
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Well done... uh... Jimmy.
 
  • #17
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My solution to the problem is illustrated in the attached picture. It is just a simple machine with a roller an an elbow joint. View the small object attached to the center of the elbow joint from the frame or reference of both points O and P. Point O is held fixed with respect to the piece of paper, and point P is moving at a constant velocity to the right.
 

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  • #18
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How does that answer the question?

If I'm reading it right, it only has one rotational path of a small arc. Where's the second?
Describe a simple way to force a particle to move along two different circular trajectories at the same time. (Two circles that are not concentric) Hint: The trajectories do not have to be measured from the same frame of reference.
 
  • #19
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How does that answer the question?

If I'm reading it right, it only has one rotational path of a small arc. Where's the second?

Viewed from the frame of reference of O, the object travels along a clockwise circular trajectory. Viewed from the frame of reference of P, the object travels along a counterclockwise circular trajectory. It must do so. It remains a fixed distance from the point P. This is the definition of a circle.
 
  • #20
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Viewed from the frame of reference of O, the object travels along a clockwise circular trajectory. Viewed from the frame of reference of P, the object travels along a counterclockwise circular trajectory. It must do so. It remains a fixed distance from the point P. This is the definition of a circle.

EDIT: I see it.
 
  • #21
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In each reference frame the object moves along a trajectory that is a fixed distance from the origin of that reference frame. Also all motion takes place on the plane of the paper. Therefore the trajectory as measured from each reference frame is circular.

Yeah, I saw. See last post.

Can't say I'm impressed. Prefer the other answers (especially the planet one).
 
  • #22
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Yeah, I saw. See last post.

Can't say I'm impressed. Prefer the other answers (especially the planet one).

Well actually the planet one is incorrect, unless you can point out some frame of reference that makes it correct.

But I do agree that the merry-go-round solution is quite nice. I prefer it over my solution.
 
  • #23
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Well actually the planet one is incorrect, unless you can point out some frame of reference that makes it correct.

But I do agree that the merry-go-round solution is quite nice. I prefer it over my solution.

The planet one is no different to the merry-go-round one, just imagine the planet going vertically instead of horizontally.

Regardless, it meets the criteria you laid out.
 
  • #24
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The planet one is no different to the merry-go-round one, just imagine the planet going vertically instead of horizontally.

Regardless, it meets the criteria you laid out.

Yes I suppose that works then if the axis is oriented properly. Good thinking.
 
  • #25
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Well actually the planet one is incorrect, unless you can point out some frame of reference that makes it correct.

But I do agree that the merry-go-round solution is quite nice. I prefer it over my solution.
Which one you like better is simply a matter of taste. I think your solution is quite clever too. I knew the solution would involve a moving axis because three points determine a circle.
 

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