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Sliding Down a Spherical Surface

  1. Jun 5, 2007 #1
    1. The problem statement, all variables and given/known data

    A person starts from rest at the top of a large frictionless spherical surface, and slides into the water below. At what angle (theta) does the person leave the surface?

    Then the book gives a "hint": (Hint: The person will lose contact with the surface when the Fn (normal force) is zero.)

    2. Relevant equations

    So I was playing with a few quations and with the gravitational force and its vectors and ended up with Cos(theta)=0, but that would be at 90 deg. and I don't think that's right.

    3. The attempt at a solution

    So, gravity would be the only force acting here, right? So gravity would be causing the motion and the centripetal force, etc.?

    It's fairly obvious that if the person had remained in contact witht he surface until hitting the 90 deg. mark, the contact would be lost, but, like I said, it seems contact would be lost before that due to accelration from gravity...I can't quite figure how to intertwine everything to find out the velocity and angle, etc...

    Any help in the right direction would be great, thanks.
  2. jcsd
  3. Jun 5, 2007 #2


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    Staff: Mentor

    Well gravity is pulling straight down always, but as the person moves the vector component pointing normal to the surface decreases. Also the person is accelerating, so v tangent to the spherical surface is increasing, so mv2/r is increasing.

    The point at which contact is lost is when mv2/r = mg(rad), where mg(rad) is the radial component of mg pointing toward the center of the sphere.

    See also - https://www.physicsforums.com/showthread.php?t=100734 - on page 303 (today) of the thread.
    Last edited: Jun 5, 2007
  4. Jun 5, 2007 #3
    is there a picture of this surface?
  5. Jun 6, 2007 #4
    Here's a picture:

    OK, so contact is lost when mv^2/r = cos(theta)mg (cos(theta)mg being the radial component of mg). That makes sense.

    So then v^2/r = cos(theta)g, and v^2 = cos(theta)gr.

    I read through that other problem you linked to, so I assume the answer would be the same here (cos(theta)=2/3), using the equation kp posted in that other thread:

    mgr(top) = 1/2 mv2 + mgrcos(theta) (the angle slides off)

    But I don't quite understand 100% how that equation is set up...
    Last edited: Jun 6, 2007
  6. Jun 6, 2007 #5
    let t = theta

    (PE_o - PE_f) = (KE_f - KE_o)
    mgr - mgr*cos(t) = (1/2)mv^2
    solve for v^2 ...
    mgr[1 - cos(t)] = (1/2)mv^2
    2gr[1 - cos(t)] = v^2

    radial component of weight = centripetal force
    mg*cos(t) = mv^2/r
    substitute for v^2 ...
    mg*cos(t) = (m/r)*2gr[1 - cos(t)]
    cos(t) = 2 - 2cos(t)
    3cos(t) = 2
    cos(t) = 2/3

    t = arccos(2/3) = approx 48.2 degrees
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