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Find the center of mass of the solid

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the center of mass of the solid figure similar to a cone pointing upward with slope = 1
    Note: the density varies with z^2 and the edge has a slope of 1. From symmetry we see that both Xc and Yc are equal to zero. Find the center of mass in the z direction as a function of h by doing the appropriate integral.

    2. Relevant equations

    p(vector r) = z^2 z^
    slope = 1

    3. The attempt at a solution
    I'm thinking about using cylindrical coordinates

    @ radius = sqrt of (1 - h^2)

    @ Z = 1/V ∫ z dV (lower limit V)

    @ V = 1/3*pi*r^2*h

    @ dV = r⊥dr⊥dθ dz.

    @ ∫ z dV (lower limit V) = ∫ (∫ (∫zr⊥dr⊥) dθ) dz

    Now im stuck there i dont know if im doing it right or wrong.
    Any help or idea ?
     

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    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    gneill

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    Staff: Mentor

    If the sides are straight and with slope 1, wouldn't the radius at height z be (h - z) ?

    Since the density varies with height you'll need to find the mass of the object via an integration; you can't just use the volume of the object as a stand-in for mass.
     
  4. Mar 10, 2012 #3
    "wouldn't the radius at height z be (h - z) ?"
    is that the radius at the base depends on the change in height ? and it is equal h - z ?
    if so, would the integral R = 1/M ∫p(r)rdV solve this problem?
     
  5. Mar 10, 2012 #4

    gneill

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    Staff: Mentor

    The radius at the base is given by (h - z) when z=0. That is, the radius at the base is h. As z increases the radius grows smaller. When z=h you've reached the apex of the object and the radius is zero there.

    To find the center of mass you want the weighted sum of mass elements, dm, as you go up the z-axis, divided by the overall mass of the object. You need to determine an expression for an appropriate dm.
     
  6. Mar 11, 2012 #5
    So my interation will be like this, correct me if im wrong
    r⊥ = h-z
    r sqr(z^2+(h-z)^2)
    M = integral from 0 to h of pi(h-z)^2*z^2dz
    *
    components:
    x=0
    y=0
    z = (1/M)*∫ (from 0->h) ∫(from0->2pi)∫(from 0->h)(sqr(z^2+(h-Z)^2))drdθdz.
     
  7. Mar 11, 2012 #6

    gneill

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    Staff: Mentor

    What does this represent?
    Okay, so the overall mass is the sum of the dm's, which are individually disks of radius (h-z) and thickness dz with density z2. That looks okay (although technically you're told that density varies as z2, so you should write ρ = k*z2).
    You really only need a single integral over the dm's if you take each dm as a disk; you know that the center of mass of a disk is at its center and you know the mass of each disk via its radius and density.
     
  8. Mar 11, 2012 #7
    thank you so much, now I get it :D
     
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