Calculating the Center of Mass of a Uniform Thin Hoop Using Symmetry

[String theory guy]

Homework Statement

Is there a way of solving this problem without without using polar coordinate's.

Relevant Equations

CM formula

in

 

[haruspex]

An easy way is to pair up diametrically opposite identical segments and observe that the mass centre of each pair is at the origin.

 

[PeroK]

By rotational symmetry, the CM must be at the centre.

 

[String theory guy]

Sorry I was not clear. I would like to understand how to solve this problem without using polar coordinate's.

 

[kuruman]

Start with the definition of the x-position of the CM relative to the center, $$X_{\text{cm}}=\frac{1}{M}\int x~dm=\frac{1}{M}\int x~\lambda ds$$ where $ds$ is an arc element $ds=\sqrt{dx^2+dy^2}.$
Now $$x^2+y^2=R^2\implies xdx+ydy=0\implies dy=-\frac{xdx}{y}.$$Therefore $$ds=\sqrt{dx^2+\frac{x^2 dx^2}{y^2}}=dx\sqrt{\frac{x^2+y^2}{y^2}}=R\frac{dx}{|y|}.$$Thus if we integrate over a semicircle from the 3 o'clock position to the 9 o'clock position, $$X_{\text{cm}}=\frac{\lambda R}{M}\int_{R}^{-R}\frac{x dx}{|y|}=\frac{\lambda R}{M}\int_{R}^{-R}\frac{x dx}{\sqrt{R^2-x^2}}.$$At this point you can

1. Observe that you have an odd function integrated over symmetric limits which means that it vanishes. This is the symmetry argument that people have mentioned.
2. Actually do the integral and verify that it vanishes. Let $u^2=R^2-x^2$, etc. etc.

Obviously, the integral that completes the circle from the 9 o'clock position to the 3 o'clock position also vanishes and so do the integrals for $Y_{\text{cm}}$.

Is this what you were looking for?

 

[hutchphd]

Sorry I was not clear. I would like to understand how to solve this problem without using polar coordinate's.

None of the replies mention polar coordinates.

 

[jbriggs444]

This is the symmetry argument that people have mentioned.

Well, it is at least a symmetry argument that people have mentioned. I see three approaches via symmetry. There may be more.

@haruspex uses a symmetric pairing argument show that the COM is the average of a bunch of zero vectors.

One can do as you suggest and use bilateral symmetry in one dimension to get the x coordinate of the COM and then in the orthogonal dimension to get the y coordinate of the COM, both of which must be zero.

Or one can do as @PeroK seems to suggest and simply observe that by rotational symmetry, there is no preferred direction. So the COM, if any, can only be at the center. That one fits my personal sense of mathematical aesthetics best.

 

[String theory guy]

Start with the definition of the x-position of the CM relative to the center, $$X_{\text{cm}}=\frac{1}{M}\int x~dm=\frac{1}{M}\int x~\lambda ds$$ where $ds$ is an arc element $ds=\sqrt{dx^2+dy^2}.$
Now $$x^2+y^2=R^2\implies xdx+ydy=0\implies dy=-\frac{xdx}{y}.$$Therefore $$ds=\sqrt{dx^2+\frac{x^2 dx^2}{y^2}}=dx\sqrt{\frac{x^2+y^2}{y^2}}=R\frac{dx}{|y|}.$$Thus if we integrate over a semicircle from the 3 o'clock position to the 9 o'clock position, $$X_{\text{cm}}=\frac{\lambda R}{M}\int_{R}^{-R}\frac{x dx}{|y|}=\frac{\lambda R}{M}\int_{R}^{-R}\frac{x dx}{\sqrt{R^2-x^2}}.$$At this point you can

1. Observe that you have an odd function integrated over symmetric limits which means that it vanishes. This is the symmetry argument that people have mentioned.
2. Actually do the integral and verify that it vanishes. Let $u^2=R^2-x^2$, etc. etc.

Obviously, the integral that completes the circle from the 9 o'clock position to the 3 o'clock position also vanishes and so do the integrals for $Y_{\text{cm}}$.

Is this what you were looking for?

Yep, Thank You.