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Deriving a Uniform Circular Motion Equation

  • Thread starter ethanh
  • Start date
3
0
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.
 

Answers and Replies

tiny-tim
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Derive the following equation: (mV^2/r)cos(x) = mgsin(x)
Hi ethanh! :smile:

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

Can you show us the whole question? :smile:
 
3
0
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)
 
458
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The only place where I can see you deriving this from is from a banked curve question, so think about the forces acting there.
 
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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