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Finding the center of mass of a cone

  1. Nov 10, 2005 #1
    I am trying to understand this example of finding the center of mass of a uniform solid cone.
    please refer to the attached figure. We know for obvious reasons that the center of mass will be on the z-axis. I will be referring to the integral that my book used to find the center of mass which is (using the letter S for the integral sign)
    R = (1/M) S r dm = (1/M) S k r dV
    where R is the position vector, M is the total mass, dm is the element of mass, k is the density, and dV denotes the element of volume.
    so for this problem, we have
    Z = k/M S z dx dy dz (again, S z dx dy dz means the integral of z dx dy dz)
    my first question is how do you evaluate this integral?

    Attached Files:

  2. jcsd
  3. Nov 10, 2005 #2
    Consider that the centre of mass (COM hereafter), is the Sum of the product of their masses and their displacement vectors all divided by the sum of the masses.
    Note that a mass at a small change in height:

    m = k * π(z*r/h)2 * dz
    Where (z*r/h) equals the radius of a circle at a height z.

    m*r = π(z*r/h)2 * dz * k * z
    i.e - Mass * Position = (Area * Height * Density) * Position

    So the sum of all the masses time the vectors would be the integral from 0 to h of the above equation (with respect to dz).

    After this you can divide by the total mass.

    I don't like the look of the equation in your book... Try to understand what is going on rather than how to do it. You will be able to apply your techniques to other questions easier if you understand what is going on.

    Sorry mate, that was quite an essay... Post back if I was poor at explaining this. :wink:
  4. Nov 10, 2005 #3


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    Of course, k is simply the mass divided by the volume so "M" will cancel out of this problem what you really have is
    [tex]Z= \frac{\int \int \int z dzdydz}{\int \int \int dzdydx}[/tex]
    You can do that in x-y-z coordinates but because of the circular symmetry it is much easier to do it in cylindrical coordinates.
    If you look at the cylinder from the side, along the y-axis, the cone is bounded by the two lines through (0,0) and (R,h) on one side and (0,0) and (-R,h) (that is, when x= R, z= h). Of course, the first line has equation z= (h/R)x so that x= (R/h)z and the second line is x= (-R/h)z. The plane cutx the cone at that z in the circle x2+ y2= (R2)/h2)z2 so for a given x and z,
    [tex]y= \pm \sqrt{\frac{R^2}{h^2}z^2- x^2}[/tex]
    If you really want to use x-y-z coordinates, the "outer integral" would have limits z= 0 to z= h, and the second integral would have limits x= -(R/h)z to x= (R/h)z, and the inner integral would have limits
    [tex] y= -\sqrt{\frac{R^2}{h^2}z^2- x^2}[/tex]
    [tex] y= \sqrt{\frac{R^2}{h^2}z^2- x^2}[/tex]
    But in polar coordinates, r, the radius of the circle at height z is just that (R/h)z. To cover the entire cone, z runs from 0 to h, r from 0 to (R/h)z, [itex]\theta[/itex] from 0 to [tex]2\pi[/itex]
    Dont forget that the "differential of area" in polar coordinates is [itex]rdrd\theta[/itex] so the "differential of volume" in cylindrical coordinates is [itex]rdrd\thetadz[/itex].
    [tex]Z= \frac{\int_{z=0}^h\int_{r=0}^{\frac{R}{y}z}\int_{\theta=0}^{2\pi}zrd\thetadrdz}{\int_{z=0}^h\int_{r=0}^{\frac{R}{y}z}\int_{\theta=0}^{2\pi}rd\thetadrdz}[/tex]
  5. Nov 20, 2005 #4


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    I have been asked by moneer81 to give more detail. Since it has been long enough that the homework should have been turned in by now,here's the whole thing.

    Draw a picture looking at the cone along the y-axis so that the positive z-axis is up and the positive x-axis is to the right. The side of the cone looks like two lines: one passes through (0,h) and (R, 0): it has equation
    [itex]z= h-\left(\frac{h}{R}\right)x. Since the cone is rotated around the z-axis, that is in fact [itex]z= h- \left(\frac{h}{R}\right)r[/itex] where r is the horizontal distance fromthe z-axis to the cone. In other words [itex]z= h- \left(\frac{h}{R}\right)\sqrt{x^2+ y^2}[/itex].
    Simplifying that gives [itex]x^2+ y^2= \left(\frac{R^2}{h^2}\right)(z- h)^2[itex]. That is, a horizontal circle with center (0,0,z) and radius [itex]\left(\frac{R}{h}\right)(z-h)[/itex]. To cover that circle, for every z, take x ranging from [itex]-\left(\frac{R}{h}\right)(z-h)[/itex] to [itex]\left(\frac{R}{h}\right)(z-h)[/itex], and, for every x and z, y ranging from [itex]-\sqrt{\left(\frac{R^2}{h}^2\right)(z-h)^2- x^2[/itex] to [itex]-\sqrt{\left(\frac{R^2}{h}^2\right)(z-h)^2- x^2[/itex].
    That is, the volume is given by
    [tex]\int_{z=0}^h\int_{x=-\left(\frac{R}{h}\right)(z-h)}^{\left(\frac{R}{h}\right)(z-h)}\int_{y= -\sqrt{\left(\frac{R^2}{h^2}(z-h)^2- x^2}^{\sqrt{\left(\frac{R^2}{h^2}(z-h)^2- x^2} dydxdz[/tex]

    The first integral is easy; just y evaluated between the limits:
    [tex]2\int_{z=0}^h\int_{x=-\left(\frac{R}{h}\right)(z-h)}^{\left(\frac{R}{h}\right)(z-h)}\right(\sqrt{\left(\frac{R^2}{h^2}(z-h)^2- x^2\right)dxdz[/tex]
    The integral with respect to x, since it involve [itex]\sqrt{c^2- x^2}[/itex] cries out for a trig substitution: let [itex]x= \left(\frac{R}{h}\right)(z-h)sin(\theta)[/itex]. Then [itex]dx= \left(\frac{R}{h}\right)(z-h)cos(\theta)d\theta[/itex] and [itex]\sqrt{\left(\frac{R^2}{h^2}\right)(z-h)^2- x^2}= \left(\frac{R}{h}\right)(z-h)\sqrt{1- sin^2(\theta)}= \left(\frac{R}{h}\right)(z-h)cos(\theta)[/itex].
    When [itex]x= \left(\frac{R}{h}\right)(z-h)[/itex] we have [itex]\left(\frac{R}{h}\right)(z-h)= \left(\frac{R}{h}\right)(z-h)sin(\theta)[/itex] or [itex]sin(\theta)= 1[/itex] so the limits of integration are [itex]\theta= -\frac{\pi}{2}[/itex] and [itex]\theta= \frac{\pi}{2}[/itex].
    The remaining integrals are now
    [tex]\left(\frac{2R^2}{h^2}\int_{z= 0}^h\int_{\theta=-\frac{\pi}{2}}^{\frac{\pi}{2}}(z-h)^2 cos^2(\theta)d\theta dz[/tex].

    Since that is an even power of cosine, we us the trig identity [itex]cos^2(\theta)= \frac{1}{2}(1+ cos(2\theta)[/itex] to get
    [tex]\left(\frac{R^2}{h^2}\int_{z= 0}^h\int_{\theta= -\frac{\pi}{2}}^{\frac{\pi}{2}}(z-h)^2 (1+ cos(2\theta))d\thetadz[/tex]
    It should be clear that the integral of [itex]cos(2\theta)[/itex] will be [itex]-\frac{1}{2}sin(2\theta)[/itex] which, evaluated between [itex]-\frac{\pi}{2}[/itex] and [itex]\frac{\pi}{2}[/itex] will give 0. The integral of 1, however, is [itex]\theta[itex] which, evaluated between [itex]-\frac{\pi}{2}[/itex] and [itex]\frac{\pi}{2}[/itex] will give [itex]\pi[/itex]. That is, the final integral is
    [tex]\frac{\piR^2}{h^2}\int_{z=0}^h(z-h)^2 dz[/tex]
    Obviously that gives
    [tex]\frac{1}{3}h^3\frac{\piR^2}{h^2}= \frac{\pi}{3}R^2h[/tex]
    the standard formula for volume of a cone.

    Now, what about that "z" integral. That is:
    [tex]\int_{z=0}^h\int_{x=-\left(\frac{R}{h}\right)(z-h)}^{\left(\frac{R}{h}\right)(z-h)}\int_{y= -\sqrt{\left(\frac{R^2}{h^2}(z-h)^2- x^2}^{\sqrt{\left(\frac{R^2}{h^2}(z-h)^2- x^2}z dydxdz[/tex]
    Since dz is the last ("outer") integral, z is treated as a constant and so (fortunately!) we can just "carry it along" until the final integral which is
    [tex]\frac{\piR^2}{h^2}\int_{z=0}^hz(z-h)^2 dz[/tex]
    Just multiply that out:
    [tex]\frac{\piR^2}{h^2}\int_{z=0}^h(z^3- 2hz^2+ h^2z)dz[/tex]
    which is
    [tex]\frac{\piR^2}{h^2}(\frac{1}{4}h^4- \frac{2}{3}h^4+ \frac{1}{2}h^4)= [/tex]\frac{\piR^2}{h^2}\left(\frac{h^4}{12})= \frac{\pi}{12}R^1h^2[/tex]
    That is
    [tex]Z= \frac{h}{4}[/tex]

    However, as I said before, it is much easier to use the circular symmetry of the cone and do this in cylindrical coordinates. As we saw at the start, for each z, a horizontal cross section is a disk, boundary a circle with equation [itex]z= h- \left(\frac{h}{R}\right)r[/itex] so that [itex]r= \left(\frac{R}{h}\right)(h- z)[/itex]. To cover that disk, r must range from 0 to [itex]\left(\frac{R}{h}\right)(h- z)[/itex] and [itex]\theta[/itex] must range from 0 to [itex]2\pi[/itex], of course. The volume is given by
    [tex]\int_{z=0}^h\int_{r=0}^{\left(\frac{R}{h}\right)(h-z)\int_{\theta= 0}^{2\pi} rd\thetadz[/tex]
    The first integral is easy leaving
    [tex]2\pi\int_{z=0}^h\int_{r=0}^{\left(\frac{R}{h}\right)(h-z) rd\thetadz[/tex]
    The integration with respect to r is also easy leaving
    That is precisely the integral we had before (although much easier than using that trig substitution and so gives [itex]\frac{\pi}{3}R^2h[/itex]
    The integral with z in it,
    [tex]\int_{z=0}^h\int_{r=0}^{\left(\frac{R}{h}\right)(h-z)\int_{\theta= 0}^{2\pi}z rd\thetadz[/tex]
    is, once again, exactly the same until the final integral which is again
    [tex]\frac{\piR^2}{h^2}\int_{z=0}^hz(z-h)^2 dz[/tex]
    which again gives the same result: [itex]\frac{\pi}{12}R^1h^2[/itex]
    and, of course, we get the same result for Z:
    [tex]Z= \frac{h}{4}[/tex]
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