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At what angle does the normal force go to zero?

  1. Nov 26, 2015 #1
    1. The problem statement, all variables and given/known data
    An ice cube is placed on top of an overturned spherical bowl of radius r, as indicated in the figure. If the ice cube slides downward from rest at the top of the bowl, at what angle θdoes it separate from the bowl? In other words, at what angle does the normal force between the ice cube and the bowl go to zero?
    08-27.gif

    2. Relevant equations
    ∑F=ma
    ΔK=-ΔU

    3. The attempt at a solution
    I drew a free body diagram, with the force on the ice cube from the earth pointed straight down and the normal force perpendicular to the surface of the globe. I then summed the forces.
    ΣFy=Wy-N=mac
    Based on trig, I know that Wy= cosθW = cosθmg.
    Then: ac= Vt2/r and the normal force is 0. Substituting all this in I get:
    cosθmg=mVt2/r
    cosθ=Vt2/gr

    So the next step would be to find for Vt.
    ΔK =-ΔU
    Kf =1/2mv2
    KO = 0
    Uf = 0
    UO = mgr
    1/2mv2 = mgr
    v2 = 2gr

    Plugging in this to the above:
    cosθ=2gr/gr
    cosθ=2
    which isn't the answer.

    I think I am making a mistake possibly in Uf, but I am not sure what exactly it would be. Would it be mgh, where h is the height it falls off? How would I find h then?
     
  2. jcsd
  3. Nov 26, 2015 #2
    The drop should be less than r.
     
    Last edited: Nov 26, 2015
  4. Nov 26, 2015 #3
    Till this point it's correct.

    This equation generally is correct, but it gives you the velocity of the ice cube, when it hits the ground (as the entire potential energy is converted into kinetic energy). You have to find a connectedness between the velocity of the ice cube and the height above ground depending on the angle θ.

    So instead of

    you need a function

    1/2⋅m⋅v2 = m⋅g⋅Δh(r,θ)

    How could you express Δh in terms of radius r and angle θ?
     
  5. Nov 26, 2015 #4
    I made a right triangle, and I found that r-h=rcosθ, so then would h = r-rcosθ?
     
  6. Nov 26, 2015 #5
    I attached a picture of the drawing I made.
     

    Attached Files:

  7. Nov 26, 2015 #6
    Yes, that's it.
     
  8. Nov 26, 2015 #7
    Looks good.
     
  9. Nov 26, 2015 #8
    This makes so much more sense, thank you!!
     
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