Deriving a Uniform Circular Motion Equation

[ethanh]

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.

 

[tiny-tim]

Derive the following equation: (mV^2/r)cos(x) = mgsin(x)

Hi ethanh!

hmm … doesn't look right to me …

Can you show us the whole question?

 

[ethanh]

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)

 

[Snazzy]

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]

… draw a diagram … !

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

Thanks Snazzy!

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)?