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Finding the angle of a falling person on a sphere

  1. Sep 3, 2007 #1
    1. The problem statement, all variables and given/known data

    A sphere is on the ground, with the sphere having a radius R. A person (skier) is standing on top of the sphere. neglecting friction, find the angle at which the person will slide off the sphere if he moves from rest to a constant velocity of v.


    2. Relevant equations

    F = ma
    F = mv^2/r (centripetal force)
    U + K = U + K (conservation of energy)

    3. The attempt at a solution

    1/2mv^2 + mgh = 1/2mv^2 + mgh
    0J + m(9.8m/s^2)(2R) = 1/2mv^2 + m(9.8m/s^2)(R + Rcos (angle))

    So that simplifies to v^2/2 = 9.8 m/s(R) - 9.8m/s(Rcos (angle))

    Drawing a free body diagram, I know that the centripetal force is equal to the Normal foce*cos(angle):

    mv^2/r = mgcos(angle)
    v^2/r = 9.8m/s (cos (angle)
    v^2 = 9.8 m/s (Rcos(angle))

    Substitute that into v^2/2 = 9.8 m/s(R) - 9.8m/s(Rcos (angle)) and get:

    9.8m/s Rcos (angle)) = 19.6 m/s(R) - 19.6m/s (Rcos (angle))

    That reduces to:

    cos (angle) = 2/3
    so the angle must be 48.2 degrees

    My teacher said it was wrong, can anyone assist me?
  2. jcsd
  3. Sep 3, 2007 #2


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    I a person is sking down a sphere, there is no "centripetal force"- there is no force on the person toward the center of the sphere. The only force is that of gravity, straight down.
  4. Sep 3, 2007 #3
    A centripetal force is included in the problem. I didn't type the whole problem out.
  5. Sep 3, 2007 #4


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    The centripetal force is the force towards the center due to gravity and the normal... if the skier is moving along the sphere:

    the net force towards the center = mv^2/r

    component of gravity towards the center - normal force = mv^2/r

    When the component of gravity towards the center becomes less than mv^2/r it is impossible for the skier to move along the sphere... force towards the center at this point cannot be mv^2/r because the normal force can't be negative...
  6. Sep 4, 2007 #5
    I got it now! Turns out my teacher made an error. :smile:

    Thanks everyone
  7. Sep 4, 2007 #6


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    Cool. Looking at your solution now, I see it is exactly right.
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