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Giant Snowball

  1. Jun 24, 2005 #1
    I have a problem here.
    A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straigt down the side. At what point does the skier loses contact with the snowball and fly off at a tangent. That is, at the instant the skier loses contact with the snowball, what angle does the radial line from the center of the snowball to the skier make with the vertical?

    This is the only information I got. I know the radial acceleration stops as soon the skier loses contact so I did a=mg*cos(angle)
    I have no clue how to reach the answer.
    The answer is 48.2°
    Could someone please give me a hint :biggrin:
  2. jcsd
  3. Jun 24, 2005 #2


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    Hint: Find velocity as a function of height using conservation of energy. Use what you know about centripetal forces to find the answer.
    Last edited: Jun 24, 2005
  4. Jun 24, 2005 #3
    Assume the skier leaves the surface at an angle Q with the vertical. Draw FreeBodyDiagram at this point.When the skier leaves the surface , Normal Reaction, N=0, write the expression for centripedal force using FBD.Use conservation of energy to determine 'v' at the leaving point which will be used in centripedal force expression.

  5. Jun 24, 2005 #4
    My work so far is:

    v*v=(2mgR - mgRcos(angle))/m

    Got this using conservation of energy
    Is this what you mean by finding velocity as a function of height, OlderDan?
    The centripetal acceleration should be 0 when the skier loses contact but when I put a=0 I get 1=cos(angle)

    What am I doing wrong here?
  6. Jun 24, 2005 #5
    check your work.
    conserv of energy should give you
    (1/2)mv2 = mgR(1 - cosθ)
    centripetal force from g is:
    (1/R)mv2 = mgcosθ
    mult 1st eq by 2, 2nd eq by R, equate, solve for θ gives point where v is max that can be sustained by the centripetal force after which separation occurs.
  7. Jun 25, 2005 #6
    No.The cetripedal force would not become zero.The normal reaction will become zero because the man leaves the contact at that point.

    At the point of leaving , drawing the FBD and equating the forces:

    [itex]mgcosQ-N=m \frac{v^2}{R}[/itex]

    Now put N=0 , also use conservation of energy for the topmost point and point of leaving to get the value of velocity V in the above expression.

    Now solve for cosQ.

  8. Jun 25, 2005 #7
    Thanks Dr. Brain. I got the right answer. One question though, why do you say that the centripetal force is mgcos(Q)-N
    The way I see it the centripetal acceleration is mgcos(Q) only, when the skier is on top of the snowball Q=0 and the force is mg not 0
    Or am I wrong?
  9. Jun 25, 2005 #8

    Doc Al

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    Staff: Mentor

    You are wrong. Realize that as long as the skier stays on the snowball, there are two forces acting on him: his weight (mg) and the contact force of the snowball (N). Both of those forces have components towards the center of the ball, thus both add to the centripetal force. Only when the skier breaks contact does the normal force go to zero.

    And, at the initial position atop the ball (angle = 0) the net force is zero! If the net force were mg he would fall right through the snowball, without the normal force supporting him. (Just like you would fall right through your chair, if the chair failed to provide a sufficient normal force.) Remember that the initial speed is zero, thus there is no centripetal acceleration at the top.

    (Of course you must assume the initial angle is not exactly zero, but just a little bit off center so that he starts sliding down the slope. Or that he gives himself a little push.)
  10. Jun 25, 2005 #9
    Don't know what I was thinking. I have seen the light. Thanks guys.
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