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Pendulum Pulley System

  1. Feb 17, 2009 #1
    This is not a homework problem. It was a problem posed by my professor at the end of the semester but had me puzzled for a while. I was able to satisfactorily solve it for the case when gravity is not present but I was not so successful with gravity.

    The question is: to determine the trajectory of the mass m1, in terms of
    -mass (m2),
    -initial length (initial radius r0) of the pendulum and
    -the initial angle (theta0) that the string makes with respect to the side of the table and
    -the initial distance of m2 from the pulley nearest to it (x0).

    Solution would be appreciated. Or, explanation of why a college freshman can not solve such a problem would also do.

    Thank you.
    Last edited: Feb 18, 2009
  2. jcsd
  3. Feb 18, 2009 #2


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    Welcome to PF!

    Hi H.G.! Welcome to PF! :smile:

    Just use conservation of energy

    (and remember the speed of m2 is the same as the rate of increase in length of the hanging part of the string) :wink:
  4. Feb 18, 2009 #3
    Thanks Tiny Tim.

    --- I had this question posed to me few years back. Sorry if I am not clear on the problems I ran into.

    (BTW: Pendulum - m1 - is not attached to the pulley but is descending to the ground as it pulls m2 with it while at the same time swinging)
    Last edited: Feb 18, 2009
  5. Feb 18, 2009 #4


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    Just call the angle θ, and express the velocity in terms of r' and rθ' :smile:
  6. Feb 18, 2009 #5
    so the velocity of m2 would be dr/dt.

    what is the velocity of m1? it can not be dr/dt because it is also swinging. What value of velocity would u use for the calculation of its (m1's) kinetic energy?

    P.S> tiny_tim: *I appreciate the quick response*
    p.s.s> I will post my solution for the case when there is no gravity
    Last edited: Feb 18, 2009
  7. Feb 18, 2009 #6


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    You must learn this … radial speed = dr/dt, tangential speed = r dθ/dt. :smile:
  8. Feb 18, 2009 #7

    so, we have non-linear differential equation in two variables (not counting t)..

    [tex]\Delta[/tex]U + [tex]\Delta[/tex]KE = 0;

    m1 gr( cos ([tex]\theta[/tex]) - cos ([tex]\theta0[/tex]) ) + 0 + 0.5*(m1[(dr/dt)[tex]^{2}[/tex] + (rd[tex]\theta[/tex]/dt)[tex]^{2}[/tex] ] + m2* (dr/dt)[tex]^{2}[/tex]) = 0;

    0 = k1 + k2 * cos([tex]\theta[/tex]) + k3 * r *[tex]^{'}[/tex][tex]^{2}[/tex][tex]\theta[/tex] + k4 *(r[tex]^{'}[/tex])[tex]^{2}[/tex] ;

    What other relationship does exist that would turn this into a simultaneous equation?
    Last edited: Feb 18, 2009
  9. Feb 18, 2009 #8


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    Hi hgetnet! :smile:
    Yes, we need one more equation …

    which we can get by taking components of force and acceleration for the swinging mass either in the radial or the tangential direction …

    though I must admit, having tried it, that I can't actually see any simple solution to the combined equations. :redface:

    hmm … perhaps that's why your professor posed it at the end of the semester? :smile:
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