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Moments of Inertia of non-uniform rod

  1. Jan 4, 2014 #1
    1. The problem statement, all variables and given/known data
    a rod of mass M and length L is supported by a smooth horizontal floor and leans against a smooth vertical wall, the mass density increases linearly with p=kr where r is the distance from the wall and k is a positive constant.
    a.finf the moment of inertia of the rod with respect to the centre of mass.
    b.the beam is released from a position of rest, at 60 degrees to the downward vertical, find the energy conservation equation of the beam.



    2. Relevant equations

    I=∫r^2 dm


    3. The attempt at a solution

    for part a, I considered a small length of rod, dl, and its mass as dm=(m/l)*dl

    and using the fact that density ,p=m/l , m=krl

    giving dm=kr dl , heres where I got stuck - I assumed that since dl and dr would be proportional you could simply replace dl with dr in this equation and integrate for mass and moment of inertia etc. doing it this way I got an answer for moment of inertia = (1/2)ML^2 which looks OK. But I am very uncertain about substituting dl for dr.

    for part b I'm not completely sure how to even approach this, it would be wonderful if someone could verify my approach and perhaps give me a hint for part b.

    thanks in advance,
     
  2. jcsd
  3. Jan 4, 2014 #2

    SammyS

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    Of course it's fine to use dl for dr.

    What is the location of the center of mass ?
     
  4. Jan 4, 2014 #3
    I worked out that COM is (KL^3/3M) along the rod
     
  5. Jan 4, 2014 #4
    I worked out that COM is (KL^3/3M) along the rod
     
  6. Jan 4, 2014 #5

    SammyS

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    That can be expressed solely in terms of L, without M or k .

    What is the mass of the rod, M, in terms of L and k ?
     
  7. Jan 4, 2014 #6
    the mass M=1/2 * KL^2
     
    Last edited: Jan 4, 2014
  8. Jan 4, 2014 #7

    SammyS

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    No. If you replace dr with dl in the integral, then you need to replace r with l for the density in that integral.
     
  9. Jan 4, 2014 #8

    sorry, I was reading that of my old piece of working, I have the correct answer up now :)
     
  10. Jan 4, 2014 #9
    BUT to be honest with you I'm getting myself more and more confused with this question, I''m not really sure where I'm going anymore with it. I Do you know the correct answer?
     
  11. Jan 4, 2014 #10

    SammyS

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    It's easier for people to follow a thread at a later date if you don't use the "Edit" feature to change a post after it's been responded to. (Some typo's may be edited after the fact to make things clear.)

    You previously had M=KLr .



    Now substitute the expression, 1/2 * KL^2, for M into your expression for the location of the center (centre) of mass.
     
  12. Jan 4, 2014 #11
    Oh sorry I didn't realise.
    yep, did that and managed to get L/3 for COM, and managed the rest of the question! Thanks for your help
     
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