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Period of a pendulum with two masses.

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data
    We have a light rigid pendulum with length ##l##. A mass ##M## is placed at the end and a mass ##m## is placed a distance ##x## from the pivot. What is the period of the pendulum?

    2. Relevant equations


    3. The attempt at a solution

    Reduce the problem to a single mass situation using the center of mass. From the pivot, I get the center of mass ##r = \frac{mx+Ml}{m+M}##

    Using the small angle approximation we end up with ##\frac{d^2\theta}{dt^2} + \frac{g(M+m)}{mx + Ml} \theta = 0## and hence ## T = 2\pi \sqrt{\frac{mx + Ml}{g(M+m)}}##

    I suspect this is incorrect as the second part of the questions asks for the value of ##x## which has the greatest impact on ##T## which cannot be obtained as ##\frac{dT}{dx} = 0 ## has no solutions in ##0<x<l##. Help is much appreciated.
     
  2. jcsd
  3. Mar 29, 2015 #2

    TSny

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    Problems like this are usually given after a discussion of the "physical pendulum" as opposed to the "simple pendulum". So, you might check your notes or text to see if a general formula for the period of a physical pendulum is given. Or see http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html
     
  4. Mar 29, 2015 #3

    Matternot

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    The 2 masses cannot be reduced to a single mass at the centre of mass. This is because the inertia of the system is dependant of the moment of inertia of the 2 masses (Which is different to that of 1 mass at the centre of mass).

    If you haven't seen this before, take a look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

    If you are yet to visit moment of inertia, consider the potential and kinetic energy of the system (V+T)=E. By differentiating with respect to time and setting dE/dt to 0, the equation of motion can be obtained.
     
  5. Mar 29, 2015 #4
    Ah yes of course! Many thanks for the help guys!
     
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