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Homework Help: Mechanics - pendulum

  1. May 6, 2005 #1
    a double pendulum made up of two rods...look at the image below.

    The lower rod is struck at a distance a from the point connecting the rods (straight arrow in the image). Before that both rods are at the equilibrium postion and have angular velocity w. Determine a so that the rods have angular velocity w and -w after the lower rod have been struck.

    solution:

    The lagranian is

    [tex]

    L = 1/6*(m_1 + 3m_2)l_1^2\dot{\theta_1}^2 + 1/6*m_2l_2^2\dot{\theta_2}^2 + 1/2*m_2l_1l_2cos(\theta_1 -\theta_2})\dot{\theta_1}\dot{\theta_2} + 1/2*(m_1 + 2m_2)gl_1cos\theta_1 + 1/2*m_2gl_2cos\theta_2

    [/tex]

    When i have solved similiar problems i have used that

    [tex]

    (\frac {dT} {d\dot{q}})_f - (\frac {dT} {d\dot{q}})_i = F_x

    [/tex]

    But i dont think this works now.

    Any ideas on how to continue?
     

    Attached Files:

  2. jcsd
  3. May 6, 2005 #2

    arildno

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    The new velocity of the second rod's C.M must satisfy:
    [tex]\omega\vec{k}\times{l_{1}\vec{i}_{r}-\omega\vec{k}\times{\frac{l_{2}}{2}}\vec{i}_{r}=\vec{v}_{c.m.2}[/tex]
    where [tex]\vec{i}_{r}[/tex] is the unit vector down along the pendulum system.

    Note that if there had been a net impulse couple acting in the joint, i.e, rod 1 imparting an impulse to rod 2, and rod 2 imparting an equal, but oppositely directed impulse on rod 1, then rod 1 would have experienced a change in its angular velocity.
    Thus, no such impulse couple was present.
    But, therefore, the impulse [tex]\vec{I}[/tex] striking at "a" is solely responsible for the perceived change in the momentum of rod 2, that is:
    [tex]\vec{I}=m_{2}(\omega\vec{k}\times(l_{1}-\frac{l_{2}}{2})\vec{i}_{r}-\omega\vec{k}\times(l_{1}+\frac{l_{2}}{2})\vec{i}_{r})=-m_{2}\omega{l}_{2}\vec{k}\times\vec{i}_{r}[/tex]

    But, this must be consistent with the change in angular momentum rod 2 experience as a result of [tex]\vec{I}[/tex] striking at "a":
    Measured, from the C.M of rod 2, we must have:
    [tex]a(-\vec{i}_{r})\times\vec{I}=-\mathcal{I}_{C.M}2\omega\vec{k}[/tex]
    where the moment of inertia with respect to the C.M fulfills: [tex]\mathcal{I}_{C.M}=\frac{m_{2}l_{2}^{2}}{12}[/tex]

    Solving for "a", we get [tex]a=-\frac{l_{2}}{6}[/tex], i.e, it is below the center of mass (the distance from the joint is therefore [tex]\frac{2}{3}l_{2}[/tex]
     
    Last edited: May 6, 2005
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