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A person slides downa spherical, frictionless surface.

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A person starts from rest at the top of a large, frictionless, spherical surface, and slides into the water below. At what angle θ does the person leave the surface? (Hint: When the person leaves the surface, the normal force is zero.)


    2. Relevant equations
    Don't Know


    3. The attempt at a solution
    I honestly would love to give this an attempt, but I just don't know where to start. It's a problem in my Energy and Power chapter, so I would assume that at the top E(total)=mgh. I just don't know where the angle comes into play. The hint leads me to believe that it would be at an angle of 90 because the entirety of the persons weight force is tangent to the circle, but this answer was wrong.
     
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  3. Oct 12, 2012 #2

    TSny

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    To get started: Draw the person at an arbitrary angle on the sphere. Set up Newton's 2nd law and see if you can get a relation between the normal force and the speed of the person at that point. (Energy will come in later.)
     
  4. Oct 12, 2012 #3
    Ok I drew him at the top, therefore he has to be at rest. And I have the normal force and weight force acting on him
     
  5. Oct 12, 2012 #4

    TSny

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    The top is too special. You want a general relation between the normal force and the speed. So, think about an arbitrary point at some angle θ from the vertical.
     
  6. Oct 12, 2012 #5
    That's what the question says.
     
  7. Oct 12, 2012 #6

    TSny

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    Yes, but he doesn't stay at the top. We need to know what happens as he slides down along the surface.
     
  8. Oct 12, 2012 #7
    Oh, ok. So I'll just pick 60 degrees then. So there is the Normal force and two compnents of the weight force.
     
  9. Oct 12, 2012 #8

    TSny

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    We don't want to favor any particular angle. Work with symbols and let θ denote any (unspecified) value of the angle. Likewise, just let v represent the speed at that point.

    The problem gives you a hint that you should consider the magnitude of the normal force. Is there a way to get an expression for the normal force in terms of the angle θ and the speed v?
     
  10. Oct 12, 2012 #9
    Well I'm thinking centripital acceleration ac=v^2/r. And as far as theta, y=r-rcos(theta), where theta is subtended between r and y. I'm still not sure where the Normal force fits in.
     
  11. Oct 12, 2012 #10

    TSny

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    Great!
    How do you relate forces to acceleration?
     
  12. Oct 12, 2012 #11
    F=m(v^2/r)
     
  13. Oct 12, 2012 #12

    TSny

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    How would you write out the left side of the equation in terms of all of the forces acting on the guy?
     
  14. Oct 12, 2012 #13
    OH wait, would the point where he falls off be where Fc=Weight Force? And KE + PE @ the moment he falls off equals the PE at the start point. Then i can solve for v^2 relate that to the weight force which in this case would mgCos(theta)
     
  15. Oct 12, 2012 #14

    TSny

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    No, let's let the equations tell us the answer. When you write down the million dollar equation, F = m(v^2/r), what does F represent?
     
  16. Oct 12, 2012 #15
    Centripital force?
     
  17. Oct 12, 2012 #16

    TSny

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    Oh, sorry. I think you might actually have it. When you first said "weight force" I was thinking mg. But later you said "weight force" is mgcos(theta), so you were actually thinking about the component of the weight acting toward the center of the sphere. That's right. So, Sounds good to me, now. Work it out and see what you get.
     
  18. Oct 12, 2012 #17
    Yea it came out right. Thanks. There's another probelm I'm stumped on, you think you could help. I'm gonna make it into a new thread.
     
  19. Oct 12, 2012 #18

    TSny

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    "Centripetal force" is just the name we give to the net force acting toward the center of the sphere. It's always due to whatever physical forces are acting on the object. You have identified mgcos(theta) as acting toward the center. Should you also include the normal force as contributing to the centripetal force?
     
  20. Oct 12, 2012 #19
    No, because the normal force acts on the slider away from the center of the circle, while the centripital force acts on the slider towards the center of the circle
     
  21. Oct 12, 2012 #20

    TSny

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    Acting away from the center is like acting negatively toward the center. So, while sliding down, the centripetal force would be mgcos(theta) - Fn where Fn is the normal force. But you are looking for the place where he just leaves the surface, and that's where Fn = 0 as stated in the hint.
     
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