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Centripetal acceleration geometry

  1. Nov 7, 2009 #1
    This is probably a geometry question more that a physics question. I am trying to prove that in uniform circular motion [tex]\Delta[/tex] V[tex]/[/tex]V= s[tex]/[/tex]R.

    I am basically trying to show that S forms a right triangle with [tex]\Delta[/tex]V, when [tex]V{1}[/tex] is added to [tex]V{2}[/tex] as a vector. (This is to demonstrate that the triangles are similar.)

    I understand that the angle formed by S and [tex]\Delta[/tex]V is a right angle because it obviously inscribes the diameter. I just cannot seem to find a satisfactory proof that [tex]\Delta[/tex]V must necessarily intersect the circle at the diameter.

    Probably not explaining this very well.
  2. jcsd
  3. Nov 8, 2009 #2
    If you are solving geometry problems with both distances and velocities involved, then you
    are probably making a mistake: remember that they have different units! (unless you study relativistic theory)
    For the mentioned problem you should use formulas:
    s=R*fi (fi is angle of the part of orbit traveled in radians)
    [tex]\Delta[/tex]V=V*sin(fi') (fi' is the angle between the old and new velocity vector)

    Prove that fi=fi' and use sin(fi)=fi (for small angles) and you will get [tex]\Delta[/tex] V[tex]/[/tex]V= s[tex]/[/tex]R
  4. Nov 8, 2009 #3
    Thank you so much for responding. I think though my question which started with uniform motion and delta V, is now simply a geometry question.

    I will try to put up a drawing of what I so pitifully tried to explain.


    Here is another image with a different angle and size of the tangents. uniformcircularmotion2.png

    I have tried several approaches, but I cannot prove that the angle formed from tangent 1 to tangent 2 to the circle must inscribe a 180 degree arc and must be a right angle.
  5. Nov 9, 2009 #4


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    From : http://en.wikipedia.org/wiki/Inscribed_angle_theorem
    So, to get an inscribed angle of 90° you need a central angle of 180°(=diameter line).
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