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Potential energy, a pyramid, and calculus

  1. Oct 31, 2006 #1
    I have a question involving calculus and Mechanics. It's from AP french, question 10-12.

    There's a pyramid with a square base of side, b, length 230m, and height, h, 150m. The density, p, of the pyramid is 2500kg/m^3. We are asked to find the total potential energy, PE of the pyramid taking the zero potential height as the base of the pyramid.

    Here's what I did. I realized that the problem is to take a slice of volume, find it's mass and then multiply by gravity and the height of the slice. Then sum all the slices from height 0 to height 150.

    It seems that mass is dependant on volume, which is dependant on height and base length. And potential energy is dependant on mass and height.

    m = p * tripleint(dV)
    = p * tripleint(dA dh)

    PE = mg int(dh)


    PE = p * tripleint(dA dh)g * int(dh)
    = pg * int(tripleint(dA dh)dh)

    Then I saw by inspection that tripleint(dA dh) = some function V(h), the volume of the pyramid in terms of height.

    So then subbing in V(h), we find

    PE = pg * int(V(h)dh)

    Then I noted from a thought experiment of cutting a pyramid out of a box, that the volume of pyramid as a function of height, V(h), is,

    V(h) = (1/3)(b^2)h, so I rewrote potential energy by taking out (1/3)(b^2) as constants and keeping 'h' under the integral as:

    PE = (1/3) * pg * (b^2) * int(h dh), with the 'h' bounds from 0 to 150m.

    After integrating I get:

    (1/6) * pg * (b^2) * (h^2), with h going from 0 to 150.

    Plugging all the given values for p, g, b, and h I find:

    (1/6)(2500)(9.8)(230^2)(150^2 - 0^2)

    = 4.860 x 10^12 Joules.

    Unfortunately, this answer is off by a factor of 1/2!

    The answer given in the back of the book is '2.4 x 10^12 Joules'.
    (If you're not familiar with AP French, his answers are always approximations, so the other digits besides '2.4' are not known, and could be non-zero).

    Why am I off by a factor of two? I'm really truly not seeing why my method is incorrect.

    Last edited: Oct 31, 2006
  2. jcsd
  3. Oct 31, 2006 #2


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    So that I can just give you a quick reply, I am not going to look for your factor of 2, but here is what I would do, very similar to what you did.

    Cut the pyramid into slices of thickness dh. The volume of a slice is dV = s^2dh where s(h) is a linear function of h that is b when h = 0 and 0 when h = 150. All you have to do is integrate rho*g*h*s^2*dh from h = 0 to 150
    Last edited: Oct 31, 2006
  4. Oct 31, 2006 #3
    I went to see my prof today for help, and he didn't actually go through the problem with me, but I think the gist of what he was saying is that the way that I was approaching the problem by integrating dV to find the mass, would be fine, if I realized that the height I was dealing with is no longer 150. It's instead the height to the center of mass of the pyramid.

    This sounds about right? (If this were the case, then I think that the center of mass of the pyramid would be at h/sqrt(2).)
    Is this right?
  5. Oct 31, 2006 #4


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    The integral I described (and the one I think you were trying to do in the first place) is exactly the integral that needs to be done to find the center of mass. If you knew where the center of mass was (call the height H) then the GPE would just be mgH. Looking back at your first post, it appears to me that where you have int(dh) you should just have h, but the h belongs inside the triple integral you are doing. You should have

    PE = p * tripleint(dAhdh)g
    instead of
    PE = p * tripleint(dA dh)g * int(dh)

    The integral of dA is the s(h)^2 I was talking about

    s(h) = 230 - (230/150)h

    PE = pg * int(s^2hdh)

    Just expand the s^2 and do the integral.
  6. Nov 1, 2006 #5


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    You give density as
    You don't have two different integrals. Imagine a thin layer at height y above the base. It has the shape of a square, of side length x, say, and so area x2. Since the sides are planes, x is a linear function of y: x= ay+ b. At the base, where y= 0, x= b= 230. At the top, where y= 150, x= a(150)+ 230, so a= -230/150= -23/15. At each height y, the square has area (230- (23/15)x)2. Taking thickness dy, the thin layer has volume (230- (23/15)x)2dy and so mass equal to 2500(230- (23/15)x)2dy kg.
    Of course, at height y, the potential energy is mgy so here it is (9.81)(2500)(230- (23/15)x)2dy. That integral
    [tex](9.81)(2500)\int_0^{150}(230- (23/15)x)^2dy[/tex]
    gives the total potential energy.
  7. Nov 1, 2006 #6
    Ohhhhhh. I get it.
    I see the problem now. Thank you both for helping me to actually understand this problem. I can do the problem now! :)

    My troubles came from 1) not properly understanding how to set up the integral, and 2) Not understanding how to find the appropriate s(h).
    Last edited: Nov 1, 2006
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