Derivation of Acceleration Formula in Uniform Circular Function

[Mao]

This has been one of my biggest hurdles in VCE Physics, as I was not able to derive the formula. I have been told that truetears has posted a quite elegant solution involving similar triangles, though I cannot find it. [If someone could link/copy it, that'll be great =) ]

Anyhow, bored in my lecture today, here's a derivation that is slightly beyond the level of Specialist but some of you still may enjoy.


Let be the radius vector pointing away from the centre with magnitude , and is the angle it makes with the horizontal. Let be a unit vector pointing in the direction of the horizontal axis, and be a unit vector pointing in the direction of the vertical axis.


Let be the unit vector of ,

Let be the velocity vector with the magnitude . Assuming the motion is observed from a point such that the object is traveling in a counter-clockwise direction. Since is perpendicular to the radius vector, it makes an angle of with the horizontal axis.





Using the arc-length







Let the acceleration vector be



Magnitude of acceleration:

The direction of acceleration () is , opposite the radius vector, i.e. in towards the centre.

[the_head]

I am proud to say I understand none of that

[/0]

Nice proof, here is the visual proof
http://www.youtube.com/watch?v=TNX-Z6XR3gA

[kamil9876]

I remember seeing a proof that doesnt use the rigours of calculus but rather infinitesimal geometry in this cheap physics book we used in year 12 (Jacaranda(yuck, i know)). Didn't watch that video but skimmed through it and it looked similair to what was in the book. The fathers of calculus often used infinitesimal and geometry based arguments to prove stuff. Not long ago actually I thought of a similair proof to Mao's, although a bit different and more simple, however i didn't bother to finish it off(my intuition sort of told me it should work and so in the spirit of mathematics I was too lazy to do the algebra).

again:

arc length: Assuming the particle starts at (r,0)






 







which is what mao got. Note: we couldve worked out the magnitude by directly applying the magnitude formula for in equation 1, however this would be incomplete as it wouldnt specify direction like Mao's proof did.

Edit: What an epic 50th post hahaha. Latex syntax errors galore.

[TrueTears]

(sorry for bad paint so messy)



so, basically in the image on the left the 2 radius form an isosceles triangle with a chord AB and forms an angle alpha

the figure on the left is just 'taken' out of the circle, assuming uniform circular motion v1 and v2 are congruent, so AB = BC. so it also forms an isosceles triangle with angle alpha.

So the distance of the chord AB can be calculated simply by x = vt

the BC is just simply the change in velocities, , or -v1 + v2

Since they are similar triangles ratio of their sides must be constant









but

so

[kamil9876]

yep, that's the one in our textbook