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Coin-vertical displacement question and other

  1. Jan 25, 2010 #1
    Coin--vertical displacement question and other

    1. The problem statement, all variables and given/known data

    A Perfectly hemispherical dome with a 10. meter radius is treated with a frictionless coating. A marble and a coin are released from the top of the dome simultaneously. We will assume there is no air friction. The coin will slide, not roll. The coin will hit the ground first because it has more kinetic energy.



    2. Questions
    Show mathematically whether or not the coin will stay in contact with the dome until it reaches the ground. If it does not, what will the coin's vertical displacement be at the instant it loses contact with the dome, and how far from the base of the dome will the coin hit the ground? (be clear and include explanations where necessary)

    3. Attempt at the solution


    This part is where I am confused.
    I know that when the normal force is greater then the weight (or gravitational force) the coin will lose contact with the dome.

    I am lost on how to show it mathematically and where to start.

    Please Help!
     
  2. jcsd
  3. Jan 26, 2010 #2

    tiny-tim

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    hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello hello Elo21 ! :smile:
    Nooo :redface:

    the coin will lose contact when the normal force is zero. :wink:
     
  4. Jan 26, 2010 #3
    Re: Coin--vertical displacement question and other

    Okay.

    So to solve it I would have to find where on the surface Fn=0, like the height from the radius/ diameter of the circle. Then using that use an equation to find the angle that it comes off at?

    Sorry if I do not make sense. :(
     
  5. Jan 26, 2010 #4

    tiny-tim

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    Yes, find the height or the angle where FN = 0. :smile:
     
  6. Jan 26, 2010 #5
    Re: Coin--vertical displacement question and other

    so then...

    if i go about finding the height using energy...

    Total E= mgh+1/2mv2

    and GPE equals mgh but becuase of the radius I can assume it equals mgr

    GPE= total E= mgr

    mgr=mgh+ 1/2mv2

    Cancel m so I have

    gr=gh+1/2v2

    I know all variables now except for h and v and I want to solve for v.
    So what would I put in for v to find h?

    Once I know that I can do the rest... I just can't figure it out!

    THANK YOU!

    Sorry if that did not make sense...I am bad at explain things. :)
     
  7. Jan 27, 2010 #6

    tiny-tim

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    (just got up :zzz: …)
    No, that would mean h is constant.

    You need v as a function of θ, so use h = r(1 - cosθ).

    Try again. :smile:
     
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