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Circular loop-de-loop physics

  1. Mar 30, 2007 #1
    How does one describe the trajectory of a ball which has enough velocity to climb the first 1/4 of a circular loop-de-loop, but not enough to smoothly descend the last 1/4 of the loop? E. g., in this situation what discontinuity does the ball path experience in curvature or its derivative?
     
  2. jcsd
  3. Mar 30, 2007 #2
    If the ball has only sufficient speed to climb the first 1/4 of the loop, then obviously it will not make it ot the last 1/4 of the loop.
     
  4. Mar 30, 2007 #3
    nice name you have for this track, loop de loop. lol
     
  5. Mar 31, 2007 #4
    Geometrodynamics will throw one for a loop.:bugeye:

    e(ho0n3,

    Please read the problem again. The ball may have enough velocity to stay on the track approaching the 3/4 mark.
     
  6. Mar 31, 2007 #5
    I'm a bit confused as well, if it got to the 3/4 mark (lets call this 3 o clock), the rest is easy as it has both gravity and the support of the track for the rest of the way. Now if it had problems beween say 10:30 and 1:30, I could understand.
     
  7. Mar 31, 2007 #6

    Doc Al

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    Staff: Mentor

    Your description of the problem is confusing; why don't you try again. What do you mean by first 1/4 of the loop? Last 1/4? 3/4 mark? Describing the path in terms of angle might be clearer: 0 degrees is at the bottom; 180 at the top; etc.

    In any case, unless the ball has sufficient speed to maintain contact throughout the loop, it will at some point lose contact and become a projectile in free fall. If it maintains contact at the top part of the loop then it will maintain contact to the bottom again, ignoring losses.
     
  8. Mar 31, 2007 #7
    OK. The ball has at least enough velocity to stay on track past 90o from the bottom, but at most not enough to stay on until 270o. How can one generally describe its trajectory after falling off the track under such conditions?

    I guess there are two components to this "free fall" trajectory: the momentum due to its angular velocity when leaving the track, and that due to gravitational potential from the point of departure. I thought there might be some discontinuity in the path of the ball at this point.
     
  9. Apr 1, 2007 #8

    Doc Al

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    Again, I do not understand. 0 degrees is at the bottom; 180 degrees is at the top. I can understand having enough speed to make it past 90 degrees, but not enough to make it to the top (180 degrees); but if it does make it to 180, it will surely make it past 270 degrees as well (ignoring losses).
     
  10. Apr 1, 2007 #9
    Therein may lie the fault to my initial reasoning, Doc Al (I mentally added friction to the situation). What are the governing equations to describe the frictionless ball falling off between 90o and 180o?
     
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