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Homework Help: Centroid ( center of mass ) of cardioid revisited

  1. Jun 13, 2008 #1
    Centroid ("center of mass") of cardioid... revisited

    This is actually a variation of the https://www.physicsforums.com/showthread.php?t=119622".

    As it happens, the cardioid I'm working with is identical to jbusc's:

    [tex]r = 1 + \cos(\theta)[/tex] and [tex]\delta \equiv 1[/tex]

    The task is to find the centroid of the above cardioid using the polar versions of the familiar formulas for:

    Mass of the lamina: [tex]m = \int_{}^{}\int_{R}^{}[\delta \ast r]\ drd\theta[/tex]

    Centroid radial coordinate: [tex]\overline{r} = \frac{1}{m} \ast \int_{}^{}\int_{R}^{}[r \ast \cos(\theta) \ast \delta \ast r]\ drd\theta[/tex]

    Centroid angular coordinate: [tex]\overline{\theta} = \frac{1}{m} \ast \int_{}^{}\int_{R}^{}[r \ast \sin(\theta) \ast \delta \ast r]\ drd\theta[/tex]

    ... but I need to find the centroid coordinates for the portion of the cardioid in the first-quadrant only.

    No sweat, right? I've done problems like this--although never with restrictions on the quadrants--before in both cartesian and polar systems, how hard can it be?

    So given the information above, I began by calculating the mass of the cardioid with the appropriate formula (given above):

    [tex]m = \int_{}^{}\int_{R}^{}[\delta \ast r]\ drd\theta \Rightarrow \int_{0}^{\frac{\pi}{2}}\int_{0}^{1 + \cos(\theta)}[r]\ drd\theta = \frac{3\pi + 8}{8}[/tex]

    [tex]\overline{r} = \frac{1}{m} \ast \int_{}^{}\int_{R}^{}[r \ast \cos(\theta) \ast \delta \ast r]\ drd\theta \Rightarrow \frac{8}{3\pi + 8} \ast \int_{0}^{\frac{\pi}{2}}\int_{0}^{1 + \cos(\theta)}[r \ast r \ast \cos(\theta)]\ drd\theta = \frac{5\pi + 16}{6\pi + 16} \approx 0.91[/tex]

    [tex]\overline{\theta} = \frac{1}{m} \ast \int_{}^{}\int_{R}^{}[r \ast \sin(\theta) \ast \delta \ast r]\ drd\theta \Rightarrow \frac{8}{3\pi + 8} \ast \int_{0}^{\frac{\pi}{2}}\int_{0}^{1 + \cos(\theta)}[r \ast r \ast \sin(\theta)]\ drd\theta = \frac{10}{3\pi + 8} \approx 32.882^\circ[/tex]


    [tex](\overline{r}, \overline{\theta}) \Rightarrow (0.91, 32.882^\circ)[/tex]

    So that's what I have so far, but when I use a program like MathCAD 2001 Professional to generate a graph, the result doesn't look like it would balance on the head of a pin. Generating the https://www.physicsforums.com/attachment.php?attachmentid=14369&stc=1&d=1213362232" of the cardioid is simple enough; then, I added the graph of a circle with the radius of the radial centroid coordinate [tex]\overline{r}[/tex] and used the trace option on the graph to view the intersection of a line at the proper angle--or close to it--with the circle of the proper radius and it just doesn't look like a "center of mass" to me. It looks as iff you tried to balance the lamina on the tip of needle you were holding, you simply end up with a sore foot--depending on what the lamina was made out of and how big it was.

    So does this point look right to anyone? If I had the resources, I would test it with an experiment, but I don't. I know this may all be elementary, and I'm reasonably certain that my math is correct, but some of the answers I've been getting since I began with this problem have caused me to doubt my results--even when they agree with the text. For instance:

    Find the centroid of the region inside the circle "[tex]r = 2 \ast \sin(\theta)[/tex]" and outside the circle [tex]r

    = 1[/tex]. Use [tex]\delta \equiv y[/tex].

    So, the usual calculations go something like this:

    [tex]m = \int_{}^{}\int_{R}^{}[\delta \ast r]\ drd\theta \Rightarrow \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{1}^{2 \ast \sin(\theta)}[r \ast r \ast \sin(\theta)]\ drd\theta = \frac{8\pi + 3\sqrt{3}}{12}[/tex]

    [tex]\overline{r} = \frac{1}{m} \ast \int_{}^{}\int_{R}^{}[r \ast \cos(\theta) \ast \delta \ast r]\ drd\theta \Rightarrow \frac{12}{8\pi + 3\sqrt{3}} \ast \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{1}^{2 \ast \sin(\theta)}[r \ast r \ast r \ast \sin(\theta) \ast \cos(\theta)]\ drd\theta = 0[/tex]

    [tex]\overline{\theta} = \frac{1}{m} \ast \int_{}^{}\int_{R}^{}[r \ast \cos(\theta) \ast \delta \ast r]\ drd\theta \Rightarrow \frac{12}{8\pi + 3\sqrt{3}} \ast \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{1}^{2 \ast \sin(\theta)}[r \ast r \ast r \ast \sin(\theta) \ast \sin(\theta)]\ drd\theta = \frac{36\pi + 33\sqrt{3}}{32\pi + 12\sqrt{3}} \approx 80.409^\circ[/tex]


    [tex](\overline{r}, \overline{\theta}) \Rightarrow (0, 80.409^\circ)[/tex]

    But since our region is symmetrical about the positive y-axis, how could this angle (or even the radius) by right? Though this is precisely the answer the solution guide that accompanied my text has and the calculations confirm it, how can this be the lamina's center of mass? There's no way you could balance this lamina horizontally at this point. So now I wonder how reliable this method is.

    Am I setting the first example up properly? Could my answer to the first example be correct? If so, could someone please explain the apparent flaw in the calculations/answer to the second example? I would think that the radial coordinate of the centroid for the second example would be farther "up" the positive y-axis and that the angular coordinate would be [tex]90^\circ[/tex], or at least [tex]0^\circ[/tex]. What's going on here? If everything is correct here, then how can one trust this method?

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    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Jun 13, 2008 #2


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    ?? Are you sure of these formulas? [itex]r cos(\theta)[/itex] and [itex]r sin(\theta)[/itex] are the x and y coordinates (respectively) of the point with polar coordinates r and [itex]\theta[/itex]. Notice that both of these formulas give the same units] as r- length units.

    Last edited by a moderator: Apr 23, 2017
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