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Homework Help: Help quite hard pendulum problem

  1. May 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A very light rigid rod with a length of 0.062 m extends straight out from one end of a metre stick. The combination is suspended from a pivot at the upper end of the rod as shown in the following figure. The combination is then pulled out by a small angle and released.

    2. Relevant equations

    3. The attempt at a solution

    well using parallel axis Theorem

    I= 1/12mL2+md2




    = 2∏√((1/12)(m)(L)2+md2)/mgd

    m's cancel


    but what should d be??

    is it a half L, I'm confused:confused:

    is this the right method as well?
  2. jcsd
  3. May 19, 2012 #2


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    hi charmedbeauty! :smile:
    looks ok :smile:

    you're using the parallel axis theorem, and your axis through the c.o.m. is perpendicular to the rod, so d is the perpendicular distance from that axis to the pivot, ie the distance PC from the pivot P to the c.o.m. C

    (if you're wondering how the moment of inertia can be the same whatever the angle θ of the stick, just do a bit of geometry using the cosine formula for two points A and B on the stick equidistant from C … you'll find that PA2 + PB2 is independent of θ :wink:)
  4. May 19, 2012 #3
    so say I have the pivot point which is connected to a rod which in turn is connected to a ruler then d would be the length of the the rod+half that of the ruler?? i.e. 0.5+0.062=d


    thats the pic there with all details given in question.
  5. May 19, 2012 #4


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    (oh it's straight!)

    yes :smile:
  6. May 19, 2012 #5
    Thanks a bunch tiny-tim:approve:
  7. May 19, 2012 #6
    and just to double check L is the length of the rod (0.062) and not the length of the ruler?
  8. May 19, 2012 #7


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    no, L is in 1/12 mL2

    L is the length of the thing that has mass m (in this case, the metre stick)
  9. May 19, 2012 #8
    oh right so I can just leave it out since it is 1 metre. cool.
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