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Pendulum's Max Speed

  1. Mar 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Rank each pendulum on the basis of its maximum speed.
    [​IMG]

    2. Relevant equations

    m*g*[tex]\Delta[/tex] = [tex]1/2[/tex]*m*v^{2}



    3. The attempt at a solution


    At first I thought that all the Max speeds were the same, but I was wrong.

    I know that the kinetic energy is equal to the change in potential energy since I solve the first 2 parts correctly.


    If I follow the equation, I know the mass will cancel out. So am I solving for v^{2}?


    So if I simplify the equation is it:

    v = \sqrt{2*g*\Deltah}

    Then would i just apply it to all the situations?

    Thanks

    Sorry for the bad format for the equations.
     
  2. jcsd
  3. Mar 6, 2008 #2

    dynamicsolo

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    Homework Helper

    Yes, you have the important part right here. We assume that the mechanical energy, E = K + U (kinetic+potential energy) is conserved in the pendulum swing, because gravity is a conservative force and we ignore air resistance and friction in the bearing the pendulum swings on.

    So when the pendulum is initially released, all of the mechanical energy is in gravitational potential energy (relative to the bottom of the swing), all of which is converted into kinetic energy when the pendulum reaches the bottom. This lets you write the equation above.

    You solved this for v^2 = 2gh correctly. So this tells you that in an entirely gravitational process of this sort, the mass is irrelevant and that the speed at the bottom of the swing will depend only on the local gravitational acceleration and the height of release of the pendulum bob. (Thus, in sorting your examples, some will be of equal rank!)
     
  4. Mar 6, 2008 #3
    Thanks! I just needed to be sure.
     
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