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I suppose I could've used the "Fast particle in a bowl" thread, but as this is a different problem, I decided not to.
If you set a particle to move in a frictionless bowl (radius R) at the velocity v, how far is the orbit from the equator?
I got the following equations:
[tex]tan\alpha = \frac{v^2}{gr}[/tex]
[tex]r = Rsin\alpha[/tex]
[tex]d = Rcos\alpha[/tex]
where g is the constant 9,81[tex]\frac{m}{s^2}[/tex] and r the radius of the orbit.
I ended up with [tex]d = \frac{\frac{v^2}{gR}-\sqrt{(\frac{v^2}{gR})^2+4}}{-2}R[/tex], which seems a little complicated but gives reasonable answers. I haven't yet figured out why it only works with - in front of the square root (I got both plus and minus when I solved the equations). Is my solution correct?
If you set a particle to move in a frictionless bowl (radius R) at the velocity v, how far is the orbit from the equator?
I got the following equations:
[tex]tan\alpha = \frac{v^2}{gr}[/tex]
[tex]r = Rsin\alpha[/tex]
[tex]d = Rcos\alpha[/tex]
where g is the constant 9,81[tex]\frac{m}{s^2}[/tex] and r the radius of the orbit.
I ended up with [tex]d = \frac{\frac{v^2}{gR}-\sqrt{(\frac{v^2}{gR})^2+4}}{-2}R[/tex], which seems a little complicated but gives reasonable answers. I haven't yet figured out why it only works with - in front of the square root (I got both plus and minus when I solved the equations). Is my solution correct?