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Center of mass of a rod question.

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A rod of length 24.5 cm has linear density (mass-per-length) given by the following equation, where x is the distance from one end.

    λ = 50.0 g/m + 20.5x g/m2

    (a) What is its mass?

    (b) How far from the x = 0 end is its center of mass?





    2. Relevant equations

    A. lambda=50.0 g/m +20.5x g/m2


    3. The attempt at a solution

    I have been having a tough time trying to figure out part A. I am pretty lost, at first i thought you could put in .245 into the equation of the linear density to find the density and then multiply that by .245 but i am positive that this is wrong.

    I am not really good with the definition of center of mass
    Xcm=1/M Sigma(i) mixi

    if someone could also explain that it would be much appreciated.
     
  2. jcsd
  3. Jan 13, 2009 #2
    g/m2 ?
    Looks like this unit should be g/m
     
  4. Jan 13, 2009 #3

    LowlyPion

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    I will presume that your density distribution with x would be 50 + 25x²

    Since you have a formula for the distribution of the mass then your summation will look like an integral then won't it?

    Hence you will have x*δm elements where δm at any x is given by 50 + 25x²

    Looks like this suggests 50x + 25x³ integrated from 0 to .245 m.

    Of course you also need to integrate the volume of the object to determine the overall mass for your 1/M.
     
  5. Jan 13, 2009 #4

    LowlyPion

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    Sorry, that should be 20.5 not 25 in the previous post. I misread it I see.

    The idea is the same of course.
     
  6. Jan 13, 2009 #5
    thanks, could you also help me with how i would go about getting part b?

    I think you misread the question, i did what you suggested and took the integral of the equation of the equation from 0 to .245 m and got the correct answer. Thank you for your help again.
     
  7. Jan 13, 2009 #6

    LowlyPion

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    Actually I provided the solution for b) already. You need the total mass from a) as I already outlined that you divide into the integral of the moment summations.
     
  8. Jan 15, 2009 #7
    oh sorry, i misread. Thanks for the help! I was really confused thanks for clearing it up.
     
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