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Tough question on hoops and beads

  1. Oct 19, 2006 #1
    Two beads of mass m are positioned at the top of a frictionless hoop of mass M and radius R, which stands vertically on the ground. The beads are given tiny kicks, and they slide down the hoop, one to the right and the other to the left. What is the largest value of m/M for which the hoop will never rise up off the ground?

    Two questions, why would the hoop even rise up, and two how exactly do you go about doing this, I figure out most of the FBD but I don't even know if that's relevant to the solution.
     
  2. jcsd
  3. Oct 19, 2006 #2

    OlderDan

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    This is a centripetal acceleration problem. As the beads slide down the hoop they reach a point where the gravity component acting toward the center of the circular motion is no longer sufficient to provide the required centripetal force. If the beads were not attached to the hoop at that point they would fly off. Since they are attached, the hoop pulls them inward. At the point where this begins to happen, the net force on the two beads by the hoop is downward. By Newton’s third law, the beads are exerting an upward force on the hoop. If the hoop had no mass, it would leave the ground as soon as the beads reach the point where they tend to fly off. If the hoop has mass, the beads have to get going faster to lift the hoop. If the hoop has enough mass, it will never leave the ground.

    You need to consider the FBD for one bead in detail and pay careful attention to the directions of the forces. The other bead does exactly the same thing on the opposite side, so the horizontal forces of the beads on the hoop cancel. The vertical forces add. Then examine the FBD for the hoop.
     
    Last edited: Oct 19, 2006
  4. Nov 20, 2010 #3
    I'm working on a similar problem. Is the upward force on each bead equal to the mass of the bead times gravity?
     
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