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Pendulum question

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A pendulum consists of mass M hanging at the bottom of a mass-less rod of length L with a frictionless pivot at its top. A bullet of mass m and velocity v collides with M and gets embedded. What is the smallest velocity v of the bullet sufficient to cause the pendulum with the bullet to clear the top of the arc.


    2. Relevant equations

    [tex]mv=(m+M)v'[/tex]
    [tex]KE=\frac{1}{2}(m+M)v'^2[/tex]
    [tex]PE=(m+M)gh[/tex]


    3. The attempt at a solution

    First I found out the velocity of the moving block.

    [tex]mv=(m+M)v'[/tex]
    [tex]\frac{mv}{(m+M)}=v'[/tex]

    and [tex]KE_{bottom}=PE_{top}[/tex]

    [tex]\frac{1}{2}(m+M)v'^2=(m+M)g(2L)[/tex]

    plug in [tex]v'[/tex]

    [tex]\frac{1}{2}(m+M)\frac{{m^2}{v^2}}{(m+M)^2}=(m+M)2gL[/tex]

    cancel out [tex](m+M)[/tex]

    [tex]\frac{1}{2}\frac{{m^2}{v^2}}{(m+M)}=(m+M)2gL[/tex]

    cross multiply

    [tex]{m^2}{v^2}={(m+M)^2}2gL[/tex]

    divide by [tex]m^2[/tex]

    [tex]v^2=\frac{{(m+M)^2}2gL}{m^2}}[/tex]

    [tex]v=\frac{(m+M)\sqrt{4gL}}{m}[/tex]

    [tex]v=\frac{(m+M)\sqrt{4}\sqrt{gL}}{m}[/tex]

    [tex]v=\frac{(m+M)2\sqrt{gL}}{m}[/tex]
     
    Last edited: Nov 3, 2009
  2. jcsd
  3. Nov 2, 2009 #2

    rl.bhat

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    Homework Helper

    At the top of the arc, the velocity of M is not equal to zero because it clears the top.
     
  4. Nov 2, 2009 #3
    but is it more or less correct?
     
  5. Nov 3, 2009 #4

    rl.bhat

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    Homework Helper

    The total energy at the top is
    (m + M)*2gL + 1/2*(m + M)*v"^2
    At the top acceleration v"^2/L = g. Or v"^2 = Lg. Sustitute this value in the above expression.
    One more mistake.
    Check the step from v^2 to v.
     
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