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Old Nov7-09, 04:24 PM                  #1
richardbsmith

richardbsmith is Offline:
Posts: 6
Centripetal acceleration geometry

This is probably a geometry question more that a physics question. I am trying to prove that in uniform circular motion LaTeX Code: \\Delta VLaTeX Code: / V= sLaTeX Code: / R.

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

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

Probably not explaining this very well.
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Old Nov8-09, 06:11 PM                  #2
Lojzek

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Posts: 227
Re: Centripetal acceleration geometry

Originally Posted by richardbsmith View Post
This is probably a geometry question more that a physics question. I am trying to prove that in uniform circular motion LaTeX Code: \\Delta VLaTeX Code: / V= sLaTeX Code: / R.

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

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

Probably not explaining this very well.
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)
LaTeX Code: \\Delta 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 LaTeX Code: \\Delta VLaTeX Code: / V= sLaTeX Code: / R
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Old Nov8-09, 11:51 PM                  #3
richardbsmith

richardbsmith is Offline:
Posts: 6
Re: Centripetal acceleration geometry

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.

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.
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Old Nov9-09, 09:53 AM                  #4
A.T.

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Posts: 834
Re: Centripetal acceleration geometry

From : http://en.wikipedia.org/wiki/Inscribed_angle_theorem
In geometry, the inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle.
So, to get an inscribed angle of 90° you need a central angle of 180°(=diameter line).
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