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Conservation of energy

  1. Aug 13, 2010 #1
    A ball is rolled down a smooth ramp, that curls upwards.
    Where will the ball lose contact with the surface?

    According to the conservation of energy law, the ball is able to reach its original height, which means the ball should stop at A. However, after the ball passes point B, won't the normal contact force be acting downwards on the ball? I'm unsure whether this will affect the point where the ball loses contact.

    Also, I've tried using Normal force [tex]\leq[/tex] 0 to determine the position where the ball loses contact, but i dont seem to be getting anywhere.

    [PLAIN]http://img15.imageshack.us/img15/4441/17201057.gif [Broken]

    Am I approaching it the correct method by using the normal force, if not what will be more appropriate for this question? Many thanks in advance!


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    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 13, 2010 #2

    uart

    User Avatar
    Science Advisor

    Assuming no friction then the ball will lose contact somewhere between B and A. It will then follow a parabolic trajectory that will at some point re-collide with the track.

    To find where the ball first loses contact you should try to find where the normal (to the curve) component of gravity first exceeds the required centripetal force.
     
  4. Aug 13, 2010 #3
    Let's suppose that the curved line is a circle of radius R. Use an angle [tex]\alpha[/tex] to describe the position of the ball on the circle, so that the point B is [tex]\alpha=0[/tex] and point A is [tex]\alpha=\pi/2[/tex]. Conservation of energy says

    [tex]\frac{v^2}{2}+gR\sin\alpha=gR[/tex]

    (we suppose the ball starts at rest). We derive the centripetal force

    [tex]\frac{mv^2}{R}=2mg(1-\sin\alpha)[/tex]

    The component of the gratitational force normal to the trajectory is

    [tex]mg\sin\alpha[/tex]

    Now we write F=ma:

    [tex]mg\sin\alpha+T=2mg(1-\sin\alpha)[/tex]

    where T is the reaction of the circle on the ball (inwards going). Since T can only be positive, we have

    [tex]3\sin\alpha<2[/tex]

    That is \alpha < 42 degrees (almost half way between B and A).
     
  5. Aug 14, 2010 #4
    thanks, very detailed workings.
     
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