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Potential energy of a pyramid 
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#1
Jan3012, 04:29 PM

P: 29

1. The problem statement, all variables and given/known data
The potential energy of a mass element dm at a height a above the earth's surface is dV = (dM)gz. Compute the potential energy in a pyramid of height h, square base b x b, and mass density p. 2. Relevant equations dV = (dM)gz V = 1/3 Bh  pyramid volume 3. The attempt at a solution I have drawn out the dimensions and tried changed the form of the potential differential from dV = (dM)gz to dV = [1/3 (b^2)h]pgdm but I'm pretty sure it wouldn't be dm and dm transforms when you consider the entire potential of the volume with respect to the mass element. Even if it transforms to something like dh however, I would still be a factor of two off. Thanks to anybody that helps. 


#2
Jan3012, 04:36 PM

P: 552

The pyramid stands over its base? I think you should find the center of mass. Then if its stands over the base, consider the height at which the center of mass is located, and compute the potential for the point mass located at that height.
Do you know how to get the center of mass? 


#3
Jan3112, 10:47 AM

P: 29

Alright, I was able to get the center of mass in an understandable way which came out to h/4. My question which be how that relates to the equation, is the center of mass the integral of the dm mass elements and you still have to take into consideration of the entire volume of the pyramid for the z term relative to the mass density p? From that I did the integration and got V = (1/12)(b^2)(h^2)pg.



#4
Jan3112, 11:36 AM

P: 552

Potential energy of a pyramid
You got the center of mass, you know the total mass, then the potential energy is just V=mgh, with h=(height of the pyramid)/4
I'm supposing here that the pyramid is standing over its base. The density will just affect in the location of the center of mass, if it's not constant, then you must have it in consideration when making the integrals. The potential energy is given just by the position of the center of mass. 


#5
Jan3112, 12:03 PM

P: 29

Looking at the equation what I'm seeing is that the equation does reduce down to V=mgh because what I see is that (1/3)(b^2)h*p where you multiple the volume and mass density would give you the total mass of the pyramid so that is why the volume of the pyramid is apart of the equation. Is this crrect?



#6
Jan3112, 12:21 PM

P: 552

I'm not sure what you mean. I'll try to be more clear, excuse me if I'm not, I have some difficulties to express my self in english.
This: dV=dmgz, when integrating will give vv_{0}=mgz, which gives the potential energy. In the other hand, Volume*density=mass, density is mass/volume. That is for a uniform mass density, it could be that the distribution of mas being not uniform, then you should have that in consideration when finding the center of mass, inside the integral. The answer that you found is fine for an uniform mass distribution (the density being constant). The center of mass is given by: [tex]r_{cm}=\frac{1}{M} \int_V \rho (r) rdV[/tex] Here M represents the total mass, rho gives the mass density distribution, r is the position, and the integration shields over the entire volume. http://en.wikipedia.org/wiki/Center_of_mass Now, the potential energy actually depends on the frame of reference (you can choose in the integration the V_{0}. You can choose the point from which you are choosing to measure the height. I supposed that the exercise asks you to measure the potential energy for the pyramid standing over its base on the surface of the earth, with respect to the surface of the earth. Then what you need to find is the height at which the center of mass is located. Once you got the center of mass, you forget about the pyramid, and just compute the potential energy for a mass point with the mass of the pyramid, located at the center of mass. I don't know if I'm being clear, but don't doubt in asking again and insisting with your doubts, I'll do my best. BTW, if you don't know the mass, but you know rho, and rho is not uniform, you can compute the total mass this way: [tex]M=\int_V \rho (r)dV[/tex] 


#7
Jan3112, 01:01 PM

P: 29

After looking at the next problem I need to do I saw it was based off of this formula that I found on the other post and using it to find the amount of person years to build the pyramid where formula I found gives the correct answer so it's taking that into consideration I guess.
Thanks alot for the help! 


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