acceleration

[jimmy p]

My Physics teacher wanted me to understand how to derive a=v^2/r using the equations v=d/t, \omega=\theta/t and arc length=r\theta but when it came down to it, i had brain freeze..and when my teacher looked at it, he had brain freeze. This isnt part of my syllabus but i was wondering how you derive the equation (so it isnt homework!)

oh yeah, i was given the hint to work it first into the equation v=r\omega and then work from there.

Cool, i just used that LaTex stuff! that was quite challenging!

thanx

Jimmy P

[Doc Al]

Originally posted by jimmy p
My Physics teacher wanted me to understand how to derive a=v^2/r using the equations ...
Perhaps these threads may help:

http://www.physicsforums.com/showthread.php?s=&threadid=8924&highlight=acceleration

http://www.physicsforums.com/showthread.php?s=&threadid=8918&highlight=centripetal+acceleration

[jimmy p]

OK unfortuantely i didnt find that too useful. That is kind of what we tried to do but it failed miserably, because we ended up getting T on both sides of the equation. i could get to v=r\omega (if that is even helpful) and couldnt get any further

[FZ+]

Can you use calculus?

[Doc Al]

Originally posted by jimmy p
OK unfortuantely i didnt find that too useful. That is kind of what we tried to do but it failed miserably, because we ended up getting T on both sides of the equation. i could get to v=r\omega (if that is even helpful) and couldnt get any further
Well, those two threads show how to derive the formula for centripetal acceleration. I know of no other way.

Using the equations you started with, you can certainly get to v=r\omega. But you won't get any further without using the strategy explained in those threads. The "trick" is to find \Delta V---the difference in the velocity vectors---between two points on the circle separated by \Delta\theta.

Why don't you show what you've done and perhaps we can fix it.

[krab]

You have everything EXCEPT a formula for \inline{\Delta v}. This is the real physics; you have to think about it. The rest is just formula manipulation. I remember when I was in highschool, I thought about this very thing for days, because I had difficulty coming to terms with how vectors add; I did not understand \inline{a=\Delta v/\Delta t} as a VECTOR equation. But I put in the effort and was rewarded with a career in physics.

[Moose352]

I haven't looked at either of those links, but I remember first deriving the formula by creating a proportion between the triangle the velocity and acceleration vectors created, and the triangle that the 2 radii and the chord created.

[jimmy p]

cant anyone actually give the answer? i promise that it isnt homework and that i have tried!!

[dodger]

Try this.
Draw a vector for r with a vector for v on the end (will be 90 degrees from this), then a little time later examine what would happen. The arclength s= r theta is an approximation from geometry.

I havent done this yet myself, but i am working on it

[dodger]

for the algebraic way you are looking for you go from v = d/t and find d from the other eqns and get v = r omega
Then with acceleration, a = v / t, substitue the v and you get r omega / t, v=r/t so a = v omega, the trick is a = v^2 omega / v = v^2 omega / ( r omega ) and cancel.

BUT this gives you nothing (not exactly true). Calculus method and geometry - what i siad to do in last post, work the same and give you a good in sight in to what happend.
This is related to how Feynman got to grips with QED.

[dodger]

Although with the geometrical method i cant avoid the final 'trick' bit to go from a = r omega^ 2 to v^2 / r
Doh

[Doc Al]

Originally posted by jimmy p
cant anyone actually give the answer?
We have. Several times over. You just don't seem to want to accept it! [:)]

[jimmy p]

lol sorry, i wasnt concentrating the other night when i wrote that, thanx guys!