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Forces during the motion of an object in a vertical circle

  1. Apr 14, 2015 #1
    ImageUploadedByPhysics Forums1429010949.827634.jpg ImageUploadedByPhysics Forums1429010959.628482.jpg

    At D, by resolving and equating to centripetal force we know that R + W = mv^2/r, however, why do we assume R=0? Based on what?

    I know my post is short and usually a longer solution is required but I do not have a problem with the maths in the attached, it's just the concept of letting R=0 that I need clarified.

    Thanks
     
  2. jcsd
  3. Apr 14, 2015 #2

    PhanthomJay

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    Based on the fact that normal forces are contact forces perpendicular to the tangent of the curve that in this instance must push on the object. If it is less than 0, contact is lost. So it must push with a value greater than 0. Problem is looking for minimums.
     
  4. Apr 14, 2015 #3
    What happens at exactly R=0? It remains in a circular motion?
     
  5. Apr 14, 2015 #4

    PhanthomJay

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    Yes, gravity only supplies the centripetal force without contact at that point D. The pellet remains at a distance 2 feet from the center of the circle without 'dropping', even though contact is momentarily lost, and with a speed of [itex]\sqrt{rg}[/itex]. It behaves like a satellite in orbit at that specific point, no contact necessary at that point. If the speed was less than that, then continuous circular motion could no be maintained.
     
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