Centre of mass of a semicircle using polar coordinates

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  • #1
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I am labelling this as undergraduate because I got it from an undergraduate physics book (Tipler and Mosca).

The uniform semicircle has radius R and mass M. I am getting the wrong answer but I can't see where I am going wrong. Any help would be appreciated.

My solution:

The centre of mass satisfies ##M\mathbf{r_{cm}}=\int \mathbf{r} dM##. In order to use polar coordinates, we will consider an element of area ##dA## and sum over all elements in the semicircle.

We have ##dA=\frac{1}{2}R^2 d\theta## and since the semicircle is uniform we have ##\frac{M}{A}=\frac{dM}{dA}## where ##A## is the area of the semicircle, ie ##dM=\frac{2M}{\pi R^2} dA##. Also, the position of this area along the semicircular arc we are sweeping out is ##R\cos \theta \mathbf{i}+R\sin \theta \mathbf{j}##. Substituting this into the integral we get

##M\mathbf{r_{cm}}=\int (R\cos \theta \mathbf{i}+R\sin \theta \mathbf{j}) (\frac{2M}{\pi R^2}(\frac{1}{2}R^2 d\theta))##

##=\frac{MR}{\pi} \int \cos \theta \mathbf{i}+\sin \theta \mathbf{j} d\theta##

and I simplified this to get ##M\mathbf{r_{cm}} = \frac{2RM}{\pi} \mathbf{j}##.

Where have I gone wrong?
 

Answers and Replies

  • #2
pasmith
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The area element in polar coordinates is [itex]r\,dr\,d\theta[/itex]. The position of the area element is [itex](r\cos\theta, r\sin\theta)[/itex].

Thus the integral is [tex]
\frac{2M}{\pi R^2} \int_0^\pi \int_0^R r(r\cos\theta, r\sin\theta)\,dr\,d\theta
= \frac{2M}{\pi R^2} \int_0^R r^2\,dr \int_0^\pi (\cos\theta, \sin\theta)\,d\theta.[/tex]

The mass is not concentrated at [itex]r = R[/itex].
 
  • #3
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The area element in polar coordinates is [itex]r\,dr\,d\theta[/itex]. The position of the area element is [itex](r\cos\theta, r\sin\theta)[/itex].

Thus the integral is [tex]
\frac{2M}{\pi R^2} \int_0^\pi \int_0^R r(r\cos\theta, r\sin\theta)\,dr\,d\theta
= \frac{2M}{\pi R^2} \int_0^R r^2\,dr \int_0^\pi (\cos\theta, \sin\theta)\,d\theta.[/tex]

The mass is not concentrated at [itex]r = R[/itex].
Sorry I don't understand the notation you used. Maybe I was not using polar coordinates (in which case I should change the title).

Could you please write the solution in terms of my notation?
 
  • #4
pasmith
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Sorry I don't understand the notation you used. Maybe I was not using polar coordinates (in which case I should change the title).

Could you please write the solution in terms of my notation?
[tex]
\frac{2M}{\pi R^2} \int_0^\pi \int_0^R r(r\cos\theta \mathbf{i} + r\sin\theta \mathbf{j})\,dr\,d\theta
= \frac{2M}{\pi R^2} \int_0^R r^2\,dr \int_0^\pi (\cos\theta\mathbf{i} + \sin\theta\mathbf{j})\,d\theta.[/tex]

The point is that you are effectively assuming that the mass is concentrated at [itex]r = R[/itex], but in fact it is uniformly distributed from [itex]r = 0[/itex] to [itex]r = R[/itex]. Hence you must do a double integral.
 
  • #5
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[tex]
\frac{2M}{\pi R^2} \int_0^\pi \int_0^R r(r\cos\theta \mathbf{i} + r\sin\theta \mathbf{j})\,dr\,d\theta
= \frac{2M}{\pi R^2} \int_0^R r^2\,dr \int_0^\pi (\cos\theta\mathbf{i} + \sin\theta\mathbf{j})\,d\theta.[/tex]

The point is that you are effectively assuming that the mass is concentrated at [itex]r = R[/itex], but in fact it is uniformly distributed from [itex]r = 0[/itex] to [itex]r = R[/itex]. Hence you must do a double integral.
Could you please post the full solution?

I have read a bit about double integrals using Kleppner and Kolenkow but I still don't see how it applies to the semicircle. Every time I try to evaluate it I end up with the centre of mass of each sector being at R/2 along its radius and this leads to a wrong answer again.
 
  • #6
pasmith
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I think you need to read up on surface and volume integrals and how to calculate them (as for example here) before proceeding further. Without that knowledge you will not be able to correctly answer your question.
 
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