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Help! Bowling ball slides off table

  1. Apr 26, 2005 #1
    A bowling ball of radius 6'' slides off the edge of a table of height 20''. How far from the base of the table does it land? The ball is initialy at rest, right on the edge, but you bump into the table and it slides off. Assume no friction. Any help? Thanks!
     
  2. jcsd
  3. Apr 26, 2005 #2
    It depends on purely how fast its rolling
     
  4. Apr 26, 2005 #3

    OlderDan

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    You need to look at the normal force against the ball as it moves over the edge of the table. Without friction there will be no rolling and the line of the force must be through the center of the ball.

    Decompose the weight of the ball into components along a radius of the ball (normal) and tangential to the ball. The tangetial component will accelerate the ball.
     
    Last edited: Apr 26, 2005
  5. Apr 26, 2005 #4
    Okay thank you for the help! Right, the ball does not roll off the edge, it slides. Since the normal force is not constant, how do i figure the final horizontal velocity?
     
  6. Apr 26, 2005 #5
    I would think integrating the normal force over 0 to Pi/2 would work..
     
  7. Apr 26, 2005 #6

    OlderDan

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    More or less, but it will not get to pi/2. What is really needed is to figure out the height where the separation from table occurs, and what the horizontal and vertical velocity components are at that instant. My instinct at this point is that the separation will occur at about pi/4. But I have not worked it out and it's very late where I am.
     
  8. Apr 26, 2005 #7
    How would you figure out the separation point? The force in the x direction is proportional to the force in the y direction, and I think their magnitudes would add up to mg, no?
     
  9. Apr 26, 2005 #8

    OlderDan

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    If you divide the normal force into horizontal and vertical components, the vertical component is initially mg, but it decreases as the ball starts to slip over the edge. If [itex]\theta[/itex] is the angle relative to vertical of the line from the contact point at the edge of the table to the center of the ball, the vertical component will be [itex]Ncos\theta[/itex] and the horizontal component will be [itex]Nsin\theta[/itex], but N will be a decreasing function of [itex]\theta[/itex].

    I don't think this is the easiest way to look at the problem. My orignal suggestion about considering the normal force and decomposing the weight into components was intended as a hint to look at the problem in comparison to things like blocks sliding down planes and roller coasters. How much work does the normal force do on the ball? The intial motion of the ball is circular motion, and the component of gravity in the direction of the normal is a centripetal force. The tangential component produces an increase of angular velocity along the circular path until the centripetal force vanishes.
     
    Last edited: Apr 26, 2005
  10. Apr 26, 2005 #9
    Ooook. So the ball moves in a circle around the corner. that makes sense. I don't think the normal force does work on the ball because it moves perp. to that force. Now i just have to find out when the centripetal force stops?
     
  11. Apr 26, 2005 #10

    OlderDan

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    And if the force does no work, what happens to the energy?
     
  12. Apr 26, 2005 #11
    It stays the same, right?

    add: ok i see how to get the velocity as the ball moves on the edge. But how to find where it leaves the edge?
     
    Last edited: Apr 26, 2005
  13. Apr 26, 2005 #12

    OlderDan

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    When the component of its weight directed toward the point of contact with the table equals the centripetal force, the normal force will be????
     
  14. Apr 26, 2005 #13
    i see what yuo are getting at. The normal force is zero there. So it loses contact. But when calculuating the centripetal force, can i pretend all the mass is located at the center of the ball? All parts of the ball are moving with the same speed, but at different distances from the edge. See what i mean? thanks for all the help so far, it really helped a lot.
     
  15. Apr 27, 2005 #14

    OlderDan

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    You make a good observation, but yes go ahead and treat all the mass as if it were at the center. That's the beauty of the center of mass. The summation of all the external forces acting on the pieces of an extended object result in an acceleration of the center of mass as if there were one mass at the center.

    If you think about it a bit more you will realize that all parts of the ball are moving on circular paths of the same radius. If you put a mark on the top of the ball, what is the path of that mark? If the ball were rotating about the contact point it would be different, but we have a way of taking that into consideration. That's what moment of inertia is all about.
     
  16. Apr 27, 2005 #15
    Great! That makes sense. Thanks again for the help!
     
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