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Conservation of Energy?

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Two heavy balls of equal mass M are attached to the long but light metallic rod standing
    on the floor. The rod with the balls falls to the floor. Find the velocity of each ball at the moment
    when the rod hits the ground. Neglect mass of the rod and the friction between the balls and the
    floor.


    2. Relevant equations

    E1=Ef
    m1v1 + m2v2 = m1v1 + m2v2
    E1 = mgh


    3. The attempt at a solution

    Mass 2 initial energy = mgh, mass 1(on floor) = 0
    When rod falls Mass 2 = 0; mass 1 = 0.5mv^2
    so mgh = 0.5mv^2

    V=sqrt(2gh) ??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2009 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Can you explain the problem better? Is there a diagram?

    AM
     
  4. Oct 24, 2009 #3
    You have to take into consideration all the energies that are present in this problem. When the dumbbell is at a certain height, it has potential energy. When the Dumbbell is acceleration towards the ground, it has kinetic energy. There you have your answer. As long as you set your potential energy equal to zero at the ground just make your before and after energies equal each other, mgh = (1/2)mv^2 and just solve for v.

    Edit: Hmmm, I'm looking at your question again and maybe you didn't describe it well enough. Does the dumbbell hit the floor horizontally?
     
  5. Oct 24, 2009 #4
    Yes the rod hits horizontally. Do I have to consider the normal force on the mass at the floor.

    The Diagram is basically a vertical rod with one end on the floor and the other in the air. both ends have spherical masses attached to them.
     
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