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Thread: Does center of mass of symmetrical figure split figure into equal volumes?

  1. #1
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    Does center of mass of symmetrical figure split figure into equal volumes?

    The question is as follows:

    a: Find the center of mass of a solid of constant density bounded below by the paraboloid $\displaystyle z^2 =x^2 + y^2 $
    and above by the plane z=4
    b. Find the plane, z=c, that divides the solid into two parts of equal
    volume. This plane does not pass through the center of mass.

    I already found part A to be (0,0,8/3). My question is how come the plane z=8/3 doesn't split the figure into 2 equal volumes? It's the center of mass point in the z direction AND the density is uniform. Because D=M/V $\displaystyle \Rightarrow $ and D is constant, [tex] V=cM [tex].

    To make matters more confusing (to me), my math textbook (in the solution) simply sets M/2 (which is 4pi) equal to the integral with upper limit limit of z as C. Isnt the center of mass supposed to be the point where it in the center of the weighted distribution of the mass? I attached a picture of the textbooks solution to better illustrate my point. The problem I am confused about is 25 b).Does center of mass of symmetrical figure split figure into equal volumes?-math-solution.jpg

    Thank You.
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  2. #2
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    Re: Does center of mass of symmetrical figure split figure into equal volumes?

    Hey toesockshoe.

    With regard to the evaluation side of things have you tried doing the integral for the limit involving z = [0,8/3] to calculate the answer of the volume in that region?
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    MHF Contributor ebaines's Avatar
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    Re: Does center of mass of symmetrical figure split figure into equal volumes?

    Quote Originally Posted by toesockshoe View Post
    My question is how come the plane z=8/3 doesn't split the figure into 2 equal volumes? It's the center of mass point in the z direction AND the density is uniform.
    The center of mass of an object is the point where the distance to the center of mass of what's above times the mass above = the distance to the center of mass of what's below times the mass below. Having those two masses being equal is not sufficient - the moment arm of those two masses must be considered as well. Here's an analogy that might help: consider a barbell with two masses on either end of a long weightless bar, where one mass = M and the other = 2M. Where is the center of mass of the barbell? It turns out to be at 2/3 of the way along the bar from the small mass to the large mass. That's where you could hold the bar and it would be balanced. I'm sure intuitively you can sense this. But the point where the total mass of the barbell is split in two is somewhere inside the larger mass. So here's an obvious case where the center of mass is not at the same point as where you have equal mass to the left and to the right.

    Quote Originally Posted by toesockshoe
    To make matters more confusing (to me), my math textbook (in the solution) simply sets M/2 (which is 4pi) equal to the integral with upper limit limit of z as C. Isnt the center of mass supposed to be the point where it in the center of the weighted distribution of the mass?
    The integral determines only the volume of the paraboloid from 0 to C - not the "weighted distribution" of the volume. Since the density is uniform this is equivalent to calculating the mass from 0 to C.
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