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Cylinder rolling down a path.

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data

    A cylinder rolls without slipping down a hill. It is released from height h. What is its speed when it come down? The cylinder mass may be completely concentrated on the radius R, which is the radius of the cylinder.

    http://i.imgur.com/Ge3x1nu.png

    3. The attempt at a solution

    The answer is supposed to be v=√(gh) but my calculations give;

    At the top the potential energy is E=mgh and at the end(h=0) all energy has become kinetic energy since no friction/air drag is acting. Thus mgh=(1/2)mv^2 <--->v=√(2gh). Why is this wrong??
     
  2. jcsd
  3. Jun 13, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    The cylinder is not a point-mass - some of the kinetic energy is required to have it rolling.
     
  4. Jun 13, 2013 #3
    I am trying but I am not following. The only acting force is the downward gravitational force mg and normal force from the ground. So if energy isn't the same at the top and bottom, where has it gone.

    I also don't understand that kinetic energy is required to have it rolling. The kinetic energy at the top is zero and keeps increasing as h decreases(same rate inversely right?).
     
  5. Jun 13, 2013 #4
    Garvity and Normal the only acting forces, then what makes the cyllinder roll?
    HINT: There is one more force that you are missing, draw the diagram and you will get it!
     
  6. Jun 13, 2013 #5
    Consider the moment of inertia of the cylinder
     
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