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Uniform Circular Motion equations

  1. Apr 5, 2005 #1
    Greetings!

    I'm looking for some help, so here is my predicament:

    In my Physics class we did a lab in which we found two relationships involved with the velocity of an object in uniform circular motion. The relationships were that velocity squared is proportional to not only mass, but radius as well. From that we somehow manifest two equations A=V^2/R and F=MV^2/R; where A is centripetal acceleration, V is the velocity of the object, R is the radius, and F is centripetal force. This isn't a question on any of my assignments, but I'm looking to understand this better. As such, I was wondering if anyone could explain how these equations came about.

    Thanks in advance for any help.

    -Michael
     
  2. jcsd
  3. Apr 5, 2005 #2
    What exactly do you mean ?

    The proof for the centripetal (or normal) acceleration in a circular motion can be executed by using congruent (apparently, they are called similar triangles :rofl: , whatever...) triangles...What does your textbook say ?

    regards
    marlon
     
    Last edited: Apr 5, 2005
  4. Apr 5, 2005 #3
  5. Apr 5, 2005 #4

    dextercioby

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    HERE is the actual proof.

    Daniel.

    P.S.To Marlon,yes,those triangles are similar triangles and not congruent... :wink:

    P.P.S.This was not meant as an article,just as an exposure.Sorry about the mistyping,it was really late.
     

    Attached Files:

  6. Apr 5, 2005 #5
    Ah, see, my confusion in this matter came directly from the fact that I couldn't understand how we came upon our equation. After visiting my teacher today, though, it has become more apparent as to what was really going on. Basically, the previous equation I was familiar with was Fnet=MA, or Newton's second law. From our lab we found the previously mentioned relationships and somehow determined that A, from newton's second law, is equal to V^2/R. Given that, we could say that in any system of uniform circular motion, Fnet=Fcentripetal=MV^2/R. My confusion still lies in how exactly we decided that A=V^2/R. Can this somehow be concluded based on that triangle proof, or do I only need to know the aforementioned relationships? If someone could walk me through how exactly it was deduced that A=V^2/R, that would be greatly appreciated. Here's a little clarification: I'm looking to understand why the V is squared, why it's over R, and why those together are equal to A.

    Thanks again for any help in this matter.

    -Michael
     
  7. Apr 5, 2005 #6

    dextercioby

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    I did the most rigurous proof one can give.So let's wait for Halls or someone else to approve it...

    I hope u're familiar to differential & integral calculus...

    Daniel.
     
  8. Apr 6, 2005 #7

    This little text that you copied is not the most common proof of what was asked here. You are using polar coordinates in order to proof the normal acceleration. That is not necessary. The similar triangles will do...

    regards
    marlon
     
  9. Apr 6, 2005 #8
    Well if we look at the link that i provided, just check out the left triangle. The clue is that the two triangles (denoted by (r,r,s) and (v1,v2,delta v)) are similar. It is important to realize that the delta v is perpendicular to s. Then using the properties of similar triangles : we write : S/R = (delta v)/v

    Then we know also that : s = v * (delta t)

    the acceleration is a = (delta v)/(delta t)

    From the previous expressions :
    delta v = (s/R)*v
    delta t = s/v
    Now calculate the quotient :

    So a = (delta v)/(delta t) = v²/R



    ps : this is a more clear version of what dexter wanted to give you :
    http://www.see.ed.ac.uk/~gph/teaching/eng1/Lectures/Rotation.pdf

    regards
    marlon
     
  10. Apr 6, 2005 #9
  11. Apr 6, 2005 #10

    dextercioby

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    Heh,it took me two hours to write it.:tongue2:

    Yes,just because it is circular motion and polar plane coordinates are used for a circular symmetry,it was preferrable to use them.

    It can't get any more rigurous than that.

    Daniel.
     
  12. Apr 6, 2005 #11

    dextercioby

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    Sides,it was how a physicist would do it.From the general to the particular...

    Sides,this is a perfectly solvable 1-body problem.Why use approximations (arch=chord) when it's a simple circle...?

    Daniel.

    P.S.Writing that .pdf took a lot more time & is was definitely more fun than your searching the internet for "proofs"...
     
  13. Apr 6, 2005 #12
    dexter, you didn't writ it, you just copied this text... :rolleyes:

    regards
    marlon
     
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