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Centripetal acceleration: angle at which one will leave a dome

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data
    see attached.


    2. Relevant equations
    centripedal acceleration: a=v^2/r


    3. The attempt at a solution
    Okay I found the solution for part 1.

    http://www.vic.com/~syost/baylor/phy1422s07/final-answers2.pdf [Broken]

    question 2. I know the change in energy part. But I don't really understand how

    "Newton’s law states that the centripetal force must be the net radial force
    on the skater, which is the normal force FN directed radially
    outward, and the radial component of his weight, mg cos θ
    directed inward."

    and the equation that follows.

    I think I can do part 2 and 3 by vectors and projectile motion once the 1st part is solved.
     

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    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 19, 2012 #2
    The law states F = ma, and that is also true for any projection of the vectors involved. If we take the radial projection, we end up with the quoted statement.
     
  4. Aug 19, 2012 #3
    okay thanks now I understand. Thanks so much for always coming to my rescue!
     
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