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A Compound Pendulum's Period

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a light rod of negligible mass and length "L" pivoted on a frictionless horizontal bearing at a point "O." Attached to the end of the rod is a mass "m." Also, a second mass "M" of equal size (i.e., m=M) is attached to the rod (0.2L from the lower end). What is the period of this pendulum in the small angle approximation?

    2. Relevant equations

    T=2pi(I/Hgm)^0.5
    where H=the length from the center of mass to the point of rotation
    rcm=(r1m+r2M)/(m+M)
    Icm=m(L1)^2+M(L2)^2
    I=Icm+m(d)^2

    3. The attempt at a solution
    rcm=(r1m+r2M)/(m+M)
    rcm=[(L+0.8L)m)]/2m
    rcm=0.9L (from the point of rotation)

    Icm=mL^2+mL^2
    Icm=m[(0.1L)^2+(-0.1L)^2]
    Icm=0.02mL^2

    I=Icm+md^2
    I=0.02mL^2+m(0.9L)^2
    I=0.83mL^2

    T=2pi(I/Hgm)^0.5
    T=2pi(0.83mL2/0.9Lgm)^0.5
    T=2pi(83L/90g)^0.5

    I'm really unsure as to what I'm doing wrong so it'd be great if someone could point out what I'm doing wrong--perhaps I've mistaken what one of the variable is supposed to represent? Thanks!

    P.S. if it helps, the answer is 2pi(41L/45g)^0.5, I'd just love to know how to get to that.

    OK so another update. I just realized I'm one digit off--because the answer is 82/90, not 83/90. I think the error must be with finding the Icm, because everything would simplify properly if Icm=0.01 instead of 0.02. I'm just confused as to why Icm would only be with regards to one mass instead of both...

    FIGURED IT OUT:
    No parallel axis theorem and now the period equation makes sense because we can put in 2m instead of m-->
    Solution:

    rcm=(r1m+r2M)/(m+M)
    rcm=[(L+0.8L)m)]/2m
    rcm=0.9L (from the point of rotation)

    Icm=mr^2+mr^2
    Icm=m(1L)^2+m(0.8L)^2
    Icm=1.64L^2

    T=2pi(I/Hgm)^0.5
    T=2pi(1.64mL^2/0.9Lg2m)^0.5
    T=2pi(82L/90g)^0.5
    T=2pi(41L/45g)^0.5
     
    Last edited: Jan 12, 2010
  2. jcsd
  3. Jan 12, 2010 #2

    ideasrule

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    Homework Helper

    Is there a diagram that goes with this? Where's O?
     
  4. Jan 16, 2010 #3
    It is a pendulum, thus O is at the opposite end of the rod.
     
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