1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.


    The lagranian is


    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


    When i have solved similiar problems i have used that


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


    But i dont think this works now.

    Any ideas on how to continue?

    Attached Files:

  2. jcsd
  3. May 6, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The new velocity of the second rod's C.M must satisfy:
    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:

    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:
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook