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Homework Help: Particle in a bowl

  1. Jul 3, 2005 #1

    Päällikkö

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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?
     
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  3. Jul 3, 2005 #2

    siddharth

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    I got
    [tex]
    d = \frac {-v^2 + \sqrt {v^4 + 4g^2r^2}}{2g}
    [/tex]
    I think that's the same as what you got.
    By putting v=0, I get d=R. Which is a reasonable answer.

    I got a quadratic in d. That's where the minus sign came from.
     
  4. Jul 3, 2005 #3

    Päällikkö

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    Yep, it is the same.
    I got a quadratic too, from which I get two solutions (the plus/minus-sign). I'm uncertain why the other solution is wrong.
     
  5. Jul 3, 2005 #4

    siddharth

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    I took the value of d to be from the center of the bowl to the projection of the particle on the vertical axis.

    This value must be positive.If I take one of the solutions of the quadratic, then the value of d will always be negative.

    Because this is not possible (ie, the value of d must always be positive as the particle moves only in the bowl), I rejected the value and took the other value.
     
  6. Jul 3, 2005 #5

    Päällikkö

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    The outcome cannot be predicted before the final solution(s), or can it?

    Is taking a derivative the only "proper" way to figure out which solution is correct?
     
  7. Jul 3, 2005 #6

    siddharth

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    'd' must be positive because the way I measure it, a negative value of 'd' would mean that the particle is above the hemi-spherical bowl (Where there is no bowl!).Thus, in the context of the question, this answer can be neglected.
     
  8. Jul 3, 2005 #7

    Päällikkö

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    Yes, I know it. But without a calculator or taking derivative, is there a way to figure out which of the solutions is right?
     
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