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Centre of mass of a rod with varying density

  1. Mar 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A solid metal rod with dimensions 5 x 2 x 1 is placed with one corner at the origin, such that 0≤x≤5, 0≤y≤2 and 0≤z≤1. The rods density is described by ρ(x) = (3x2 +10x) / 25
    a) find the total mass of the rod
    b) find the x-coordinate of the centre of mass of the rod.
    2. Relevant equations
    Total mass, M = ∫ dm = ∫ ρ dV
    X coordinate of COM = (1/M) * ∫xdm = (1/M) * ∫xρ dx


    3. The attempt at a solution
    a) By substituting the density function into the Total mass integral and doing a triple integral over dxdydx, i get:
    M = ∫∫∫ (3x2 +10x) / 25 dx dy dz = ∫∫ [(x3 +5x2) / 25] dy dz with limit of 5 and 0.
    From there I continue and get an answer of M = 20 units
    My problem arises in part b) when I do:
    (1/M) * ∫xdm = (1/M) * ∫xρ dx = (1/M) * ∫ (3x3 +10x2) / 25 dx and use the limts of 5 and 0, i arrive at an X-coordinate for the centre of mass as 1.77. This doesnt make sense to me as surely with the stated denisty function, the mass if increasing as you progress from x=0, resulting in the COM being shifted towards the x=5 end?
    Any help would be appreciated !
     
  2. jcsd
  3. Mar 25, 2015 #2

    SammyS

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    Hello Barnes1995-5-2. Welcome to PF !

    Yes, you are correct to reason that the Center of Mass must lie to the right of the center of the bar.

    You may have to show details of how you evaluated those integrals in order that we can give much help.

    I disagree with your mass calculation.
     
  4. Mar 25, 2015 #3

    NascentOxygen

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    Could you find the mass for 0≤x≤w and equate it to the mass from w≤x≤5?
     
  5. Mar 25, 2015 #4
    This is how i did the mass integral
     

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  6. Mar 25, 2015 #5

    SammyS

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    That's fine.

    How about integral in the numerator for the COM calculation?
     
  7. Mar 25, 2015 #6
    In getting the COM, you forgot to integrate with respect to y. Your answer should be 2x as large.

    Chet
     
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