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I have a problem which simply states:

Consider a circle (lamina) of radius 1 with centre (0,0) where the left half is twice as heavy as the right. Find its centre of mass. Extend your solution to consider the left half being $n$ times as heavy as the right.

So I know that the formulae for the coordinate of centre of mass (for uniform density are):

$$x = \frac{\int_{A} x\rho dA}{\int_{A} \rho dA}$$ $$y = \frac{\int_{A} y\rho dA}{\int_{A} \rho dA}$$

I'm quite stuck but the only thing I can think of is to convert to polar coordinates to compute the actual integration but as for setting up the problem, I'm not really sure. Should I compute each half separately and just compute the weighted average? Or just compute for one set of coordinates, and then multiply the coordinate by 2 (or $n$) and find an average from there?

EDIT: so I've gone ahead and used polar coordinates to find the centre of mass for each half, here they are:

Using $\frac{\pi}{2} < \theta < \frac{3\pi}{2}$ and $0 < r < 1$

right half:

$$ x = \frac{1}{\pi2\rho}$$$$y = 0$$

and left half I could easily derive by looking at the above and dividing by the factor of two and reversing the sign:

$$x = \frac{-1}{\pi\rho}$$$$y = 0$$

What can I do now?

I'm still stuck

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  • $\begingroup$ Is this a circular ring or a circular lamina? $\endgroup$ – David Quinn Aug 20 '15 at 5:48
  • $\begingroup$ @DavidQuinn it's a lamina. I'll add that to the question $\endgroup$ – PythonNewb Aug 20 '15 at 5:49
  • $\begingroup$ Can you use the standard formula for the centroid of a semicircular lamina, or do you need to derive the final result from first principles? $\endgroup$ – David Quinn Aug 20 '15 at 5:51
  • $\begingroup$ @DavidQuinn it should be derived using calculus. I will have a look at the formula mentioned in my own time though so I know multiple ways to solve the problem. $\endgroup$ – PythonNewb Aug 20 '15 at 5:53
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Using calculus, you can derive the following formula for the distance, along the line of symmetry, from the centre of the semicircle to the centroid: $$\bar{x}=\frac{4r}{3\pi}=\frac{4}{3\pi}$$

Assuming this result for the moment, let the area of the semicircles be $A$, the mass of the lighter one be $A\rho$, and the heavier one (on the left) be $nA\rho$.

Applying Varignon's Principle, we have $$A\rho\bar{x}-An\rho\bar{x}=A(n+1)\rho\bar{X},$$ where $\bar{X}$ is the combined centroid distance from the centre.

so this gives the general result $$\bar{X}=\frac{4(1-n)}{3\pi(n+1)}$$

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  • $\begingroup$ How did you derive $x = \frac{4}{3\pi}$? Did you use the formula in my question or from some other method? $\endgroup$ – PythonNewb Aug 21 '15 at 2:17
  • $\begingroup$ It's a standard result, en.m.wikipedia.org/wiki/List_of_centroids and the proof if fairly standard as well. $\endgroup$ – David Quinn Aug 21 '15 at 8:07

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