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Motion of a particle in a verticle ring.

  1. Oct 6, 2012 #1
    See attachment

    the green spot shows the initial position of a particle
    the blue spot shows the position at which the particle loses contact with the ring.

    Intuitively one can easily deduce that the particle would indeed loose contact with the ring.

    Is there a way to prove this mathematically ?

    What exactly causes the particle to loose contact ?
    Is it because its Kinetic energy becomes Zero (ie all of it has been converted to P.E)

    Attached Files:

  2. jcsd
  3. Oct 6, 2012 #2
    For the particle to undergo circular motion. it must undergo a centripetal force(what is the centripetal force in this case?) and if the force is not large enough it would break contact. So the statement that it would break contact anyway can be wrong regarding the different situations. So what we need to do now is to find the mathematical expression for this force then equate it with the expression of centripetal force.
  4. Oct 6, 2012 #3
    The centripetal force in this case is the sum of two forces :

    R (normal reaction from the surface of the ring)
    W(the radial component of the weight )

    Now, which force when not large enuf would cause the particle to stop the circular motion ?
  5. Oct 6, 2012 #4
    Sorry for my English as I am taught in my native language.
    Yes here the centripetal force is the resultant force of the two. But let's consider an extreme case that the particle breaks contact just at the top of the ring. Then what is the minimum centripetal force required? Using the equation we can calculate the minimum velocity required at that point. And it would be easy to extrapolate it to the situation when the particle breaks loose at any given point.
  6. Oct 6, 2012 #5
    Dude if u draw the fbd of the particle at some general theta with the vertical/horizontal(whichever u prefer); try equating the the normal contact force component to zero AND resolving the component of mg along the radius of the circle, that component will provide the centripetal force at that point.... If u r interested to find theta then use W(all) = change in KE
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