I suppose I could've used the "Fast particle in a bowl" thread, but as this is a different problem, I decided not to.(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Particle in a bowl

**Physics Forums | Science Articles, Homework Help, Discussion**