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Sliding down a sphere of ice; when do you lose contact?

  1. Jun 7, 2014 #1

    Nathanael

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    1. The problem statement, all variables and given/known data
    "A boy is initially seated on the top of a hemispherical ice mound of radius R = 13.8 m. He begins to slide down the ice, with a negligible initial speed. Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?"

    2. The attempt at a solution
    I don't have much of a solution attempt. I know that the boy's speed at any given height is known. The direction should be tangent to the circle until contact is lost. I've thought about it a few ways but none seem correct.

    I don't know why this seemingly simple problem is so tricky for me.

    Any hints will be appreciated (full solutions will not be appreciated)
     
  2. jcsd
  3. Jun 7, 2014 #2

    SammyS

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    How would you solve for the boy's motion if he were constrained to maintain contact with the surface ?
     
  4. Jun 7, 2014 #3

    Nathanael

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    [itex]V_x=cos(θ)\sqrt{2gh}[/itex]
    [itex]V_y=-sin(θ)\sqrt{2gh}[/itex]

    [itex]V_x[/itex] is horizontal velocity [itex]V_y[/itex] is vertical velocity


    This only applies to a quarter of a circle (from 90° to 0°) but that's all that's relevant (as far as I can tell).
     
  5. Jun 7, 2014 #4
    You need to draw a free body diagram, and then you need to do a force balance on the boy, resolving the forces into components in the radial and tangential directions. The radial force balance will give you your answer.

    Chet
     
  6. Jun 7, 2014 #5

    Nathanael

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    But I don't know what radial force is required to move in the circle because the velocity is not constant, and so [itex]F_c=\frac{mv^2}{R}[/itex] doesn't apply.
     
  7. Jun 7, 2014 #6
    What forces are acting on the boy as it slips down the sphere ?

    Consider an instant where the line joining the boy with the center of the sphere makes an angle ##\theta## with the vertical .What is the component of force in radial direction ? Write force equation in radial direction.
     
  8. Jun 7, 2014 #7

    Nathanael

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    There would be the force of gravity and the normal force.

    The component of gravity in the radial direction would be equal to [itex]mgsin(θ)[/itex] which would equal the normal force.


    Edit: My θ was to the horizontal. I just realized you said "an angle θ with the vertical" so for your θ it would be [itex]mgcos(θ)[/itex]
     
  9. Jun 7, 2014 #8
    Yes.I would suggest you to work with the vertical angle rather than horizontal.

    If the two forces were equal there would be no acceleration towards the center .Do you think this is the case ?
     
    Last edited: Jun 7, 2014
  10. Jun 7, 2014 #9

    SammyS

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    That is the radial force needed: [itex]\displaystyle \ F_c=\frac{mv^2}{R}[/itex]

    Tanya suggests what supplies that force.
     
  11. Jun 7, 2014 #10

    Nathanael

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    Ahh, yes, I wasn't really thinking... But I just run into the same problem as before: I don't know the necessary net radial force to move in a circle at a non-constant speed

    How? I was taught that equation applies ONLY with constant speed (and constant radius of course)
     
  12. Jun 7, 2014 #11
    1) What is the component of mg towards the center ?
    2) What is the component of N towards the center ?
    3) How do you write the sum of the two ?
    4) The net unbalanced force calculated in part 3) is what supplies the centripetal acceleration i.e you equate it to mv2/R .
     
  13. Jun 7, 2014 #12

    Nathanael

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    Yes I've understood this but my problem has been that the equation [itex]\frac{mv^2}{R}[/itex] should NOT apply.... Because that is the equation for acceleration of circular motion at a CONSTANT velocity, which is not the case in this situation
     
  14. Jun 7, 2014 #13

    SammyS

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    Actually it always applies for the radial component of acceleration. For uniform circular acceleration, the tangential component is zero, so the magnitude of the acceleration is the centripetal acceleration.

    The formula for radial acceleration turns out to be very useful for finding the radius of curvature for particle motion in 3 dimensions (or simply curves in 3-D)), such as encountered in multi-variable calculus.
     
  15. Jun 7, 2014 #14

    Nathanael

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    Interesting, I never knew this!

    So we have:

    [itex]mgsin(θ)=\frac{mv^2}{R}[/itex]

    and:

    [itex]\frac{mv^2}{2}=mg(R-h)[/itex]

    and:

    [itex]h=Rsin(θ)[/itex]

    Putting it all together you get:

    [itex]sin(θ)=2/3[/itex]

    Which means:

    [itex]h=\frac{2R}{3}=9.2[/itex]

    Which is the correct answer.


    Thank you all very much
     
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