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Problem from physics teacher newsletter

  • Thread starter what2wham1
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  • #1
1. A particle is moving around a circle of radius R in the x-0-y plane. During the motion, neither the x nor the y component of the particle's velocity exceeds v. Find the minimum possible period of revolutions.



2. V=2*∏/TR



3. x=y=v
x^2+y^2=2v^2=V^2=sqrt{2}v

T=2∏R/sqrt{2}/v


Is this how you would do it? Can you assume x^2+y^2=2v^2?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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Can you assume x^2+y^2=2v^2?
That has inconsistent units, it cannot be right.

You can use ##v_x^2 + v_y^2 = v^2##, but I don't see why this would be necessary.
##|v_x| \leq |v|## and ##|v_y| \leq |v|## should be obvious. Therefore, you can just use the formula in (2) to solve for T.
 
  • #3
TSny
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You can use ##v_x^2 + v_y^2 = v^2##, but I don't see why this would be necessary.
##|v_x| \leq |v|## and ##|v_y| \leq |v|## should be obvious. Therefore, you can just use the formula in (2) to solve for T.
It's a more interesting problem if the quantity given as ##v## in the problem is not the speed of the particle, but just an upper bound on ##|v_x|## and ##|v_y|##.

The fact that it is a problem out of The Physics Teacher makes me think that it might be this interpretation. But I'm not sure.
 
  • #4
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Ah right, the restriction is for the individual components. Well that makes it even easier, you can consider both components separately.

Edit: No, you cannot. Okay, the problem is interesting, you have to think about it. The calculations are easy.
 
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  • #5
CWatters
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Means the speed isn't constant though.
 
  • #6
haruspex
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The particle moves in a circle. That means you cannot assume the two velocity components are always v in magnitude. That will only happen when the particle is moving at 45 degrees to the axes.
Try polar coords. Express the two velocities in terms of derivative of theta. At any given time, one of them will have magnitude v. You just have to figure out which when, and what it implies for the magnitude of the other.
 
  • #7
rude man
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Hint #1: by symmetry, you only need to consider the first octant, i.e. theta = 0 to pi/4. Theta is the polar angle with the circle centered at the origin of a polar coordinate system.

Hint #2: what is the relationship between theta, vx and vy that defines a circle as opposed to any other shape, e.g. ellipse?

Hint #3: change the problem v to k. You want v to be the actual instantaneous speed. Otherwise you get confused as several posters did already.

Hint #4: the answer requires solving a non-trivial integration.

This is a really neat problem; I was so fascinated I worked it out last night at 2 a.m.!
 
  • #8
rude man
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It's a more interesting problem if the quantity given as ##v## in the problem is not the speed of the particle, but just an upper bound on ##|v_x|## and ##|v_y|##.

The fact that it is a problem out of The Physics Teacher makes me think that it might be this interpretation. But I'm not sure.
I'm quite sure it is, reading the problem carefully.
 
  • #9
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Hint #4: the answer requires solving a non-trivial integration.
You can avoid the integration if you keep the xy-coordinates.
 
  • #10
haruspex
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You can avoid the integration if you keep the xy-coordinates.
Quite so. I realised about one minute after my previous post that switching to polar was counterproductive, but then I lost internet access for two days! Once you've carved up the circle into portions according to which velocity component is the greater, it's very easy.
 

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