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Kinetic / Potential Energy Problem

  1. Nov 29, 2011 #1

    x93

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    1. The problem statement, all variables and given/known data
    A small block of mass m starts from rest and slides along a frictionless loop-the-loop as shown in the image. What should be the initial height h, so that m pushes against the top of the track (at a) with a force equal to its weight?

    Ans. h=3R

    prob17a.gif


    2. Relevant equations
    KE=(1/2)mv2
    PE=mgh
    Fc=mv2/r


    3. The attempt at a solution
    PEi=mgh
    KEa=PEi-PEa=mgh-2mgR
    (1/2)m(va)2=mgh-2mgR
    va=√(2gh-4gR)
    Fc=mg
    mg=(2mgh-4mgR)/R
    2h-4R=R
    h=(5/2)R

    I think I attempted the problem in the correct manner, but perhaps made a subtle algebraic mistake since the coefficient is only off by 1/2.
     
  2. jcsd
  3. Nov 29, 2011 #2

    PhanthomJay

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    Your error is in the calculation of Fc, which is not equal to mg. The net centripetal force, Fc, is the sum of all forces acting on the block in the cenripetal direction.
     
  4. Nov 29, 2011 #3

    x93

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    Okay, would the net centripetal force be 2mg? I'm not sure that I understand why. My guess is that the centripetal force is constant all the way around the loop, and at the bottom/start of the loop the centripetal force is equal to the normal force mg. Therefore, the magnitude of the net force at the top of the loop is the weight of the mass, mg, plus the normal force, mg. Is that right?
     
  5. Nov 29, 2011 #4

    PhanthomJay

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    yes...
    no, it is not constant, since v is changing as the object makes its way around the loop.
    no-o.
    Yes, but you have the wrong reasoning. Draw free body diagrams. At the top of the loop, there are 2 forces acting downward (centripetally) on the object. Since it is given that the mass pushes (upward) on the track at the top of the loop with a (normal) force of mg, then by Newton's 3rd Law, the track normal force pushed downward on the object with a force of mg. The object's weight also acts downward on it. The net sum of these 2 forces is ___?___., which is the net centripetal force acting on the object.
     
  6. Nov 29, 2011 #5

    x93

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    Ah, okay, I see. Thanks for your help.
     
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