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Deriving a Uniform Circular Motion Equation

  1. Apr 1, 2008 #1
    Derive the following equation: (mV^2/r)cos(x) = mgsin(x)

    I don't know how you would exactly derive the equation but you can simplify it to:

    tan(x)mg = mV^2 /r

    You can also divide by the m and get tan(x)g = V^2/r ...


    Any help is appreciated.
     
  2. jcsd
  3. Apr 1, 2008 #2

    tiny-tim

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    Hi ethanh! :smile:

    hmm … doesn't look right to me … :confused:

    Can you show us the whole question? :smile:
     
  4. Apr 1, 2008 #3
    The teacher just said to derive the equation using Circular Motion Equations,Newton's 2nd Law, etc.

    The equation is:

    m(V^2/r)cos(θ)=mgsin(θ)
     
  5. Apr 1, 2008 #4
    The only place where I can see you deriving this from is from a banked curve question, so think about the forces acting there.
     
  6. Apr 2, 2008 #5

    tiny-tim

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    … draw a diagram … !

    Yeah … Snazzy's right (and I was being dense). :redface:

    Thanks Snazzy! :smile:

    It's a horizontal circle, banked like a cycle track.

    ok, draw a diagram of just one bit of the track, and draw in the two forces and the acceleration (so that's three arrows … and give each one a letter, of course), and the angle.

    Do you know the values of any of these forces or acceleration (without any theta)? :smile:
     
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