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Math Help - [SOLVED] Time period of simple pendulum(with a catch!)?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Time period of simple pendulum(with a catch!)?

    Find the time period of a simple pendulum if the mass of the bob is m and mass of the rod is M.

    Answer:
    Spoiler:
    T = 2\pi\sqrt{\left(\frac{m + \frac{M}{3}}{m + \frac{M}{2}}\left(\frac{L}{g}\right)\right)}


    Anyone?
    Last edited by fardeen_gen; June 1st 2009 at 10:50 PM. Reason: Proper brackets
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Compound Pendulum

    Hello fardeen_gen
    Quote Originally Posted by fardeen_gen View Post
    Find the time period of a simple pendulum if the mass of the bob is m and mass of the rod is M.

    Answer:
    Spoiler:
    T = 2\pi\sqrt{\left(\frac{m + \frac{M}{3}}{m + \frac{M}{2}}\frac{L}{g}\right)}


    Anyone?
    You need the formula for the period of oscillation of a compound pendulum, which is:

    T = 2\pi\sqrt{\frac{I}{mgh}}

    where I is the moment of inertia about the axis of rotation
    m is the mass
    h is the distance of the centre of mass from the axis of rotation

    The moment of inertia of a rod of mass M, length L about one end is \tfrac13ML^2, and the moment of inertia of the bob is mL^2, so I= \tfrac13ML^2+mL^2

    Total mass = M+m

    and, taking moments about the axis of rotation:

    (m+M)h = \tfrac12ML+mL

    \Rightarrow T =2\pi\sqrt{\frac{\tfrac13ML^2+mL^2}{g(\tfrac12ML+m  L)}}

    =2\pi\sqrt{\frac{(\tfrac13M+m)L}{g(\tfrac12M+m)}}

    (I think you have some brackets missing in your answer!)

    Grandad
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