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Normal Force of a Puck on a Sphere

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a small frictionless puck perched at the top of a fixed sphere of radius R. If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere?


    2. Relevant equations



    3. The attempt at a solution
    I recognize that when the normal force on the puck exerted by the sphere is zero, the puck is no longer on the sphere. Therefore, I seek to express the normal force as a function of height, or of the angle measured down from the vertical, from which the height is easily calculated.

    I can't find a way to write the normal force. What would the normal force be, at a point? Looking at a solution, someone wrote:

    N = mgcosø + (mv^2)/R

    where ø is the angle measured down from the vertical, so that it could also be written

    N = (mgh + mv^2)/R

    Where does this come from? I don't see the truth in this expression of the normal force.
     
  2. jcsd
  3. Jan 21, 2010 #2

    tiny-tim

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    Hi Oijl! :wink:

    It comes from good ol' Newton's second law … F = ma … in the radial direction.

    So long as the puck remains on the sphere, the radial acceleration is obviously the standard centripetal acceleration (that's just geometry).

    So put all the forces on the LHS (that'll be N and the radial component of mg), and m times centripetal acceleration on the RHS. :smile:
     
  4. Jan 21, 2010 #3
    Hey, that makes perfect sense! Thanks.


    BUT, oh no! When I solve for N, using mgR = (1/2)mv^2 + mgRcosø (which is E = T + U) to replace mv^2 in the equation with N, I get

    N = mg(2-cosø)

    So N = 0 when cosø = 2, which doesn't happen.

    I have access to the solution, and the way they do it, they get down to N = mg(2-cosø) as well, but the next line they have written that, for mg(2-cosø)=0,

    ø = arccos(1/2)

    How can this be? It's algebra that I must be doing wrong, I believe, although for my life I cannot see it.
     
  5. Jan 22, 2010 #4

    tiny-tim

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    Hi Oijl! :smile:

    (just got up :zzz: …)
    (try using the X2 tag just above the Reply box :wink:)

    mv2 = 2mg(1 - cosø) is right, but "the solution" is wrong.

    You didn't do the F = ma that I suggested, with forces on one side and centripetal acceleration on the other.

    As a result, you got a minus sign in the wrong place (and so, presumably, did "the solution").

    This is a fundamental mistake that people keep making, they treat centripetal acceleration as if it was a force, so they get it on the wrong side. :rolleyes:

    Try again, and this time write it all out properly! :smile:
     
  6. Jan 22, 2010 #5
    F = ma
    a = -(v^2)/R
    F = N - mgcosø......?

    Is that the radial component of the weight, mgcosø?

    If it is, then, using
    N - mgcosø = -((v^2)m)/R,
    I get an answer that N = 0 when ø = arccos(2/3).
     
  7. Jan 22, 2010 #6
    Well, you know, I got it anyway. I used a differently-defined angle, it was easier to see what the radial component of the weight was if I measured the angle up from the horizontal. Difference was I used sin instead of cos.

    I got an answer of one-third the radius of the sphere for being the vertical distance through which the puck descends before it leaves the surface of the sphere.

    Thank you for your help.
     
  8. Sep 17, 2011 #7
    I think you were right the first time Oijl--angle=arccos(2/3)
     
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