1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Center of mass of pyramid

  1. Oct 13, 2004 #1
    I know how to find the center of mass of a 2 dimensional object like a piece of plywood or something like that, but when it comes to 3-D objects i'm clueless.
    All I know is Mx=x1m1+x2m2+...xnmn

    here is my problem, i don't know if i should split the pyramid into 4ths or not.

    The Great Pyramid of Cheops at El Gizeh, Egypt, had a height H = 144.9 m before its topmost stone fell. Its base is a square with edge length L = 233 m. Its volume V is equal L2H/3. Assuming that it has uniform density p(rho) = 1.8 x 103 kg/m3.

    (a) What is the original height of its center of mass above the base?
    (b) What is the work required to lift all the blocks into place from the base level?
     
  2. jcsd
  3. Oct 13, 2004 #2

    Chronos

    User Avatar
    Science Advisor
    Gold Member
    2015 Award

    Find the center of mass for one face [treating it as a two dimensional triangle]. Project a line through and perpendicular to the center of the face. Find and project a line perpendicular to the center of mass of the base [a square]. Align the base of the triangle with the pyramid base and rotate it to the correct angle. The center of mass is where the two projected lines intersect [don't forget to plot the coordinates in three dimensions - x,y,z].
     
  4. Oct 13, 2004 #3

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    If B is the area of the base of the pyramid then the area of a cross section y units above the base is

    [tex]A(y) = B \left( \frac {h-y}{h} \right)^2[/tex]

    where h is the height of the pyramid. You can use this to find both the volume of the pyramid and the center of mass:

    [tex]V = \int_0^h A(y) dy = \frac {1}{3} B h[/tex]

    and

    [tex]\bar y = \frac {\int_0^h y A(y) dy}{V}[/tex]

    but I'll leave the second integral for you to do. :smile:
     
  5. Oct 17, 2004 #4
    the work?

    As to part B of this problem. Would the work be the integral of mgh? where after you integrate it would be mg*h^2/2 ?
     
  6. Oct 17, 2004 #5

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    The work would be the integral of [itex]\rho g y A(y)[/itex] which should look familiar to you.
     
  7. Oct 17, 2004 #6
    Recall for the laminar ones you were describing:

    [tex]\overline{x} = \frac{1}{A}\int{x}{f(x)}dx[/tex]

    [tex]\overline{y} = \frac{1}{2A}\int{[f(x)]^{2}dx[/tex]
     
  8. Oct 17, 2004 #7
    What if you want to find the center of mass of a triangle?

    By symmetry, locate the y-coordinate of the center of mass of an equilateral triangle of side length l=95cm located with one vertex on the y axis and the others at (-l/2,0) and (l/2,0).

    does this also involve integrating?
     
  9. Feb 8, 2005 #8
    I have a question about this. Wonder if I can bring this thread back.

    - harsh
     
  10. Feb 8, 2005 #9
    Ok, phew. So, here is my problem when I try to figure out the center of mass of a pyramid. My answer is outside the damn pyramid, which doesnt really make sense at all.

    The pyramid has uniform mass density, so rho = M(tot)/ Volume

    Volume = 1/3 b h

    Now, when I try to integrate, I get my coordinates to be outside the damn pyramid. Any ideas? Where am I going wrong?

    - harsh
     
  11. Feb 8, 2005 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, how did you do this integral then?
     
  12. Feb 8, 2005 #11
    Well, okay, so here goes:

    Say the base is a square with length s. And I put the pyramid at the origin. My limits, whether they go from 0 to S or -s/2 to s/2, should give me the same result. Anyways

    Triple integral, with limits 0->h, 0->s, 0->s, with respect to dx, dy, dz respectively. Then, since rho is constant, (I can take out 3Mtot/s^2*h) out of the triple integral.
    Since its also independent of z and y, (for the x coordinate of the center of mass), you can solve those integrals readily. So far, after cancellations and doing the integral, I get 3s/2. Is that correct?

    - harsh
     
  13. Feb 8, 2005 #12

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    What triple integral are you talking about...?Tide has come up with a simple integral.Why didn't u take his advice...?

    Daniel.
     
  14. Feb 8, 2005 #13
    I am using the generic formula for the center of mass:

    x(cm) = Triple integral, with the limits, of ((rho)(x)(dV)) / mass

    Same thing with y and z, except inside, you replace with y and z respectively. I actually looked at that, but didnt understand the A(y) part of it.

    - harsh
     
  15. Feb 8, 2005 #14

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Well,Tide explains what it means.It's the area of the cross-section through the pyramid made at the height "y".
    Just make a drawing and convince yourself it is true.

    Daniel.
     
  16. Feb 8, 2005 #15
    Well, I will try to figure it out myself. Thanks for your help

    - harsh
     
  17. Feb 8, 2005 #16

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    First off: Your limits are totally wrong!
    You are describing a box, not a pyramid.
    Secondly, for your box:
    [tex]\hat{x}=\frac{\int_{0}^{h}\int_{0}^{s}\int_{0}^{s}\rho{x}dxdydz}{\rho{h}s^{2}}=\frac{\rho{h}\frac{s^{3}}{2}}{\rho{h}s^{2}}=\frac{s}{2}[/tex]
     
  18. Feb 8, 2005 #17
    I have a question. In the denominator, shouldnt it be (rho)*1/3*s^2*h ? Since its rho * volume?

    I see what you are saying about my limits, I guess they do describe a box. I think I need to change my x and y limits, maybe some sort of a plane?

    - harsh
     
  19. Feb 8, 2005 #18

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Since the geometry is really simple, you can change your variables as follows:
    Regard the pyramid as the set of all straight lines connecting a point in the (x,y)-plane with the vertex (s/2,s/2,h).
    That is, by letting u,v be coordinates of a point in the (x,y)-plane (ranging from 0 to s), while w measures where on a particlar line segment you are (ranging from 0 to 1), you have:
    [tex](x,y,z)=((u,v,0)-(\frac{s}{2},\frac{s}{2},h))w+(\frac{s}{2},\frac{s}{2},h),0\leq{u,v}\leq{s},0\leq{w}\leq{1}[/tex]
    or:
    [tex]x=(u-\frac{s}{2})w+\frac{s}{2}[/tex]
    [tex]y=(v-\frac{s}{2})w+\frac{s}{2}[/tex]
    [tex]z=h(1-w)[/tex]
    Hence, the Jacobian is:
    [tex]\frac{\partial(x,y,z)}{\partial(u,v,w)}=w^{2}h[/tex]
     
  20. Feb 8, 2005 #19

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    I used the volume of a BOX, since a box it was..
     
  21. Feb 8, 2005 #20
    I sort of see what you did. You are basically coming up with an equation of line that has intercepts at the summit of the pyramid and the base (being the x-y plane) ? Is that the correct way to think about this?

    Damn, I should have thought of coordinate change. Thanks for all your help. Please let me know if the way I am thinking of your coordinate change is okay or not.
    - harsh
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?