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Find distance when slipping down the roof of a sky dome stadium

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Mike Harris stands at the very top of the sky dome stadium roof to pitch the opening ball for the Blue Jays game. He slips a bit left of centre and slides down along the frictionless roof surface. For what distance, measured along the curve, will he slide before leaving the roof? Assume the roof has a circular cross section of radius 125m.


    2. Relevant equations
    FUN=ma
    ETi=ETf
    a=(mv2)/r


    3. The attempt at a solution
    I think, at the point before Mike Harris falls off the roof, FN=0
    So I wonder if this question would involve centripetal forces because the roof is shaped like a circle.

    I have attached my diagram and here's how I decided to tackle the question.

    FUN=ma
    Fg[itex]\bot[/itex]-FN=(mv2)/r
    mgcosθ-0=[m(2gr(1-cosθ))]/r
    cosθ=2(1-cosθ)
    cosθ=2-2cosθ
    cosθ=2/3
    θ=48°

    I found the angle, don't quite know how to proceed
     

    Attached Files:

    Last edited: Apr 10, 2013
  2. jcsd
  3. Apr 10, 2013 #2

    mukundpa

    User Avatar
    Homework Helper

    You are correct for the angle and can use

    angle (in rad) = arc length / radius
     
  4. Apr 10, 2013 #3
    OOH, so the distance the question is called the arc length? Didn't know that!

    So, if I used that formula:

    angle (in rad) x radius = arc length
    arc length = (48 x ∏/180) x 125
    arc length = 105m

    So the total distance Mike Harris travels on the sky dome rooftop is 105m? :smile:
     
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