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Circular motion with respect to south pole

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The south pole of the circular motion is labeled "O" and it's the center of the xOy axes. The point leaves "O" to the right increasing it's positional vector([tex]\bar{}r[/tex]) dimension and the [tex]\alpha[/tex] angle (the angle is between the positional vector and the Ox axis). The force that keeps the material point on the circular motion is always pointed towards "O". The north pole is labeled "A" and at that specific point the velocity is v[tex]_{}0[/tex]. The radius of the circle is R. Find the force F and the period T.

    2. Relevant equations
    r=r([tex]\alpha[/tex])
    F=ma
    v(subscript-alpha)=r(first derivative of alpha)
    r^2*(first derivative of alpha)=c

    3. The attempt at a solution
    If the centripetal force applies i think F=-m(c^2/r^4)r([tex]\alpha[/tex])
    If i know the force i can compute the T as T=-(2*pi*m*[tex]\vec{}r[/tex])/F
    Can someone please explain if this is right and if not i am open to suggestions :) Thanks
    Sorry for not using the subscripts, but the preview looked kinda weird.
     
  2. jcsd
  3. Oct 9, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi duxy! Welcome to PF! :smile:

    (what is c? and where does your r4 come from? :confused:)

    Can I check that I understand the question correctly?

    An object moves under no constraint (and no gravity), but subject to a force always directed towards O: find how the force depends on position P if the body moves in a circle through O.

    Is that correct?

    If so, then good ol' Newton's second law means that the component of acceleration perpendicular to OP must always be zero.

    Hint: find the radial and tangential acceleration relative to the centre of the circle (because that's fairly easy! :wink:), and from that find the components along and perpendicular to OP. :smile:
     
  4. Oct 9, 2009 #3
    Hey, thanks for the reply and the welcome. About that c it's supposed to be a constant i haven't really figured it out yet. About Newton's second law, the component of acceleration perpendicular to OP is always zero when O is the center and P is a point on the circle, in that case the radial acceleration is always perpendicular on the linear velocity, but if O is at the south point of the circle the tangent line that goes through P is not perpendicular to OP anymore(i think).

    On the other hand i guess that when P is a 2R distance away from O the acceleration components have the same value and direction.

    Are you implying to find the components along and perpendicular to OP geometrically or is there some kind of connection between them in this particular case?

    Thank you
     
  5. Oct 9, 2009 #4

    tiny-tim

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    Sorry, I don't follow any of that. :confused:

    First use geometry to find the acceleration.

    Use the centre of the circle as I suggested.
     
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