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Kleppner/Kolenkow: Two Points Around a Circle

  1. Jan 18, 2016 #1
    • Moved from a technical forum, so homework template missing.
    Hey, all. I've decided to go back and work on some old K&K problems that I didn't finish last time. Here's a neat one that's been giving me trouble.

    2ab1cfa31f.png

    I hadn't attempted problem a yet (I admittedly completely overlooked it by accident!). For b, I had a difficult time finding a good starting point. I initially took ##B##'s frame of reference to be stationary and decided that would mean ##A## would be the only object moving.

    For this case, I centered the circle at the origin ##(0,0)##. Let ##A## and ##B## be two points on the circle. From ##B##'s frame, ##A## travels around the circle. ##A##'s position is given by the vector ##\vec{R}##, drawn from ##B##. ##\vec{R}## is a chord on the circle. ##\theta## is the angle made by ##\vec{R}## and the line ##BO = L = AO##, where ##O## is the origin ##(0,0)##.

    Because I am finding the velocity of ##A## from ##B##'s point of view, the velocity vector will not be purely tangential to the circle. So, in polar coordinates, I get:
    [tex] \vec{v_A} = -v \sin(\theta) \hat{r} + v \cos(\theta) \hat{\theta} [/tex]

    After discussing this with a friend who has completed his undergrad, he informed me that I was wrong. From ##B##'s point of view, both ##A## AND the circle are moving. If this is the case, I am having a difficult time finding a starting point. So, help is welcome as always!
     
  2. jcsd
  3. Jan 18, 2016 #2

    A.T.

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    You should. It asks for the general solution, which you can then apply in b).
     
  4. Jan 18, 2016 #3
    Thanks, I'll take a look at it.
     
  5. Jan 18, 2016 #4
    For part a, I've found

    [tex] B = A - R [/tex]
    [tex] v_B = v_A - \frac{dR}{dt} \qquad \to \qquad v_A = v_B + \frac{dR}{dt} [/tex]
     
    Last edited: Jan 18, 2016
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