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Point mass in a (non-hyperbolic) paraboloid.

  1. Oct 14, 2013 #1
    Need a 2nd opinion on my solution.

    1. The problem statement, all variables and given/known data
    A point mass moves frictionlessly in a circle inside a parabolic cup, with the radius at the top being R. The particle's position vector makes an angle theta wrt the center of symmetry (generatrix going from -z to +z).


    2. Relevant equations
    F=ma


    3. The attempt at a solution
    Confined to a certain height implies [tex]mg = N\cos \theta[/tex]

    Circular motion at an arbitrary radius r<R gives
    [tex]m\frac{v^2}{r}=N\sin\theta[/tex]

    which yields
    [tex]\frac{v^2}{rg} = tan \theta[/tex]

    In my attempt to leave this as a function only of the given parameters (R,g and theta), I use the equation for a paraboloid:

    [tex]z=x^2 + y^2[/tex]

    and the fact that

    [tex]r=sqrt(z)\\
    \frac{r}{z}=\tan \theta
    [/tex]

    to get

    [tex]v=\sqrt(g)[/tex]

    This doesn't make sense to me on physical grounds. My argument: the speed of the particle in a paraboloid should depend on the angle to enable the requirement that at higher values of r<R, a greater y-component of the normal force is required to remain at that height. Also, units are off by a factor of [L]^(1/2).

    Just what am I doing wrong?
     
    Last edited: Oct 14, 2013
  2. jcsd
  3. Oct 14, 2013 #2

    tiny-tim

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    Hi Lavabug! :smile:
    Try [itex]Cz=x^2 + y^2[/itex] :wink:
     
  4. Oct 14, 2013 #3
    Not sure I follow, is this not the proper equation for a paraboloid?
     
  5. Oct 14, 2013 #4

    tiny-tim

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  6. Oct 15, 2013 #5
    I did, I suppose my z should really be z/R then, that's the only way I can get v=sqrt(Rg).

    However is my physical solution correct? Something about v being independent of the angle makes me worried. Common sense tells me I'd need a greater (and eventually divergent) velocity the higher I wish to make the mass orbit higher and higher (as theta approaches 90º and the normal force starts to vanish). How does my solution account for this?
     
  7. Oct 15, 2013 #6

    tiny-tim

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    i didn't notice before, your [itex]\frac{r}{z}=\tan \theta[/itex] is wrong :redface:
     
  8. Oct 15, 2013 #7
  9. Oct 15, 2013 #8

    tiny-tim

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    r/z is radius /height

    if a point on the rim is P, and if the bottom of the cup is B, then r/z is the slope of the line BP, which goes through the cup: you want the slope of the line that touches the cup :wink:
     
  10. Oct 15, 2013 #9
    Clarify please? I take it my substitution is not right, the x's and y's in my expressions for r and z are not the same. But I don't know how I would leave this independent of anything but R, theta and g. I don't see any convenient substitution that would make it work.
     
  11. Oct 15, 2013 #10

    tiny-tim

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    if z = f(x) = Cx2 is your parabola (that generates the paraboloid z = C(x2 + y2 )),

    then the slope is dz/dx :wink:
     
  12. Oct 15, 2013 #11
    Why would I need the slope of this segment? I don't see what you're trying to get at. This is supposed to be a problem that should be done in under a minute...
     
  13. Oct 15, 2013 #12

    tiny-tim

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    oh wait, your θ isn't defined in the usual way, it's …
    so your tanθ = r/z is correct after all (sorry),

    but your mg = Ncosθ is wrong, you need cos of the slope (tan-1 dr/dx)
     
  14. Oct 15, 2013 #13
    I should've included a picture:
    http://s9.postimg.org/5vouf3fni/IMG_20131015_153139_269.jpg
    I just realized that the normal force in a parabolic cup is not necessarily parallel to the position vector from the center. The angle between the normal force and position vector is 90-theta. So my trigonometry is off.
     
  15. Oct 15, 2013 #14

    tiny-tim

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    how do you get 90° - θ ? :confused:

    (and i was assuming the "position vector" was measured from the bottom of the cup, not from the centre of the top of the cup)
     
  16. Oct 15, 2013 #15
    Sorry, brainfart.

    Have a look at the pic. The position vector starts at the top-center of the paraboloid. A paraboloid of finite length is just like an ellipsoid that was twice as long chopped in half right? All I am given is the top (max) radius R.

    The angle the position vector makes wrt the z axis does not coincide with the one the normal force makes at every point, since a paraboloid is not isotropic in theta.

    What I need is the angle the normal force makes wrt the vertical, in order to get it's correct projections. But since this is a paraboloid and not a semisphere, it's not the same as the position vector angle, so I'm screwed? There must be a way to do this...
     
  17. Oct 15, 2013 #16

    tiny-tim

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    nooo!

    a parabola of finite length is not an ellipse chopped in half
    yes, you can use the equation z = Cx2, and then the slope (tan of the angle) is dz/dx

    (and the normal is perpendicular to that)
     
  18. Oct 15, 2013 #17
    The normal FORCE from the wall is not perpendicular to the slope you're describing, which is just the slope of the position vector yes?
     
  19. Oct 15, 2013 #18

    tiny-tim

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    the normal force is always along the normal, ie the direction perpendicular to the surface

    the slope i'm describing is the slope of the cup (the slope of the tangent plane to the cup, ie the slope of the tangent to the generating parabola)
     
  20. Oct 15, 2013 #19
    Ok, but any segment normal to that tangent line (ie the normal force) is NOT parallel to the line from the origin to the mass, in general.

    It would be if it was a revolved semicircle/semisphere, but not a paraboloid/egg shaped surface, am I right?
     
  21. Oct 15, 2013 #20

    tiny-tim

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    yes, you're right: the slope of the normal is not θ, the slope of the line from the origin
     
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