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Calculating Mss of a Rod using Integration

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Let a rod with length L and constant cross-sectional area A have the density
    ρ(x) = ρ((3.x^2 + 2.x.L)/L^2)

    0 ≤ x ≤ L

    where x is the distance from one end of the rod and ρ0 is a real constant.
    (a) Find the total mass M of the rod.
    (b) Find the x-coordinate xc of the centre of mass of the rod.
    (c) Find, in terms of M, the moment of inertia of the rod about the vertical axis through x = 0.


    2. Relevant equations

    M = ∫Aρ(x).dx

    3. The attempt at a solution

    M = ∫Aρ(x).dx

    M= A.ρ∫((3.x^2 + 2.x.L)/L^2).dx

    I am slightly confused where to go from here! Do I use integration by parts?
     
  2. jcsd
  3. Feb 26, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why? That is just the integral of a polynomial. What is the integral of [itex]3x^2 dx[/itex]? What is the integral of [itex]2x dx[/itex]?
     
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