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I Prove centre of mass of an arc is rotationally invariant

  1. Jul 13, 2016 #1
    Suppose the coordinates ##(\bar{x}, \bar{y})## of the centroid (or the centre of mass) of an arc is defined as follows

    Screen Shot 2016-07-13 at 3.53.40 pm.png

    ##\bar{x}=\frac{1}{L}\int x\,ds## and ##\bar{y}=\frac{1}{L}\int y\,ds##, where ##L## is the arc length.

    Could you prove that the centroid is invariant under a rotation of the coordinate axes?
     
  2. jcsd
  3. Jul 13, 2016 #2

    chiro

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    Hey Happiness.

    Have you tried doing a substitution that allows you to rotate everything?

    Hint - Try using polar co-ordinates.
     
  4. Jul 13, 2016 #3
    Hey chiro

    Yes I've considered doing that for a specific example, ##y=x^3-4x^2+x+12##, but it is very tedious and the integration we get is most likely not do-able. And even if we can perform the integration, we would only prove it for a specific case.

    I'm thinking using matrices may help, but I'm not sure how.
     
    Last edited: Jul 13, 2016
  5. Jul 13, 2016 #4

    chiro

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    If you want to use matrices then just expand the linear combination in terms of x as a function of r and theta.

    What you will find is that it's exactly the same as the integral.

    When you did the substitution - did you go to (r,theta) space and if so what did you get when doing so?

    Note that for 2D polar you have x = rcos(theta), y = rsin(theta) and the Jacobian is r.

    What you will have to show is that your integral as a function of theta is the same so this means your integral won't be a function of theta at all (i.e. it disappears).

    If you show this in general you have proved the invariance property.
     
  6. Jul 13, 2016 #5
    Is this correct?

    Under polar coordinates, the definitions of ##\bar{x}## and ##\bar{y}## become

    ##\bar{r}\cos\bar{\theta}=\frac{1}{L}\int_{\theta_1}^{\theta_2}r(\theta)\cos\theta\sqrt{r^2(\theta)+(\frac{dr(\theta)}{d\theta})^2}d\theta##

    ##\bar{r}\sin\bar{\theta}=\frac{1}{L}\int_{\theta_1}^{\theta_2}r(\theta)\sin\theta\sqrt{r^2(\theta)+(\frac{dr(\theta)}{d\theta})^2}d\theta##

    where ##\bar{r}## is the magnitude of the position vector ##\vec{r}## of the centroid and ##\bar{\theta}## is the angle the position vector ##\vec{r}## makes with the positive x-axis. (##\bar{r}## and ##\bar{\theta}## may not necessarily be the average values of ##r## and ##\theta##. Would they necessarily and sufficiently be so?) ##r^2(\theta)## means ##r^2## is a function of ##\theta##.

    Consider a rotation of the coordinate axes by an angle ##\alpha## clockwise. Then the point ##(r, \theta)\to(r, \theta+\alpha)## and the value ##r(\theta)\to r(\theta+\alpha)##. [##r(\theta+\alpha)## means for every ##\theta## in ##r(\theta)## we substitute ##\theta+\alpha##.] We have

    ##\bar{x}'=\frac{1}{L}\int_{\theta_1+\alpha}^{\theta_2+\alpha}r(\theta+\alpha)\cos(\theta+\alpha)\sqrt{r^2(\theta+\alpha)+(\frac{dr(\theta+\alpha)}{d\theta})^2}d\theta##

    where the ##\bar{x}'## is the x-coordinate of the centroid in the new rotated coordinate system.

    Using the substitution ##\beta=\theta+\alpha##, ##\beta_1=\theta_1+\alpha##, ##\beta_2=\theta_2+\alpha## and ##\frac{d\theta}{d\beta}=1##, we have

    ##\bar{r}'\cos\bar{\theta}'=\bar{x}'=\frac{1}{L}\int_{\beta_1}^{\beta_2}r(\beta)\cos(\beta)\sqrt{r^2(\beta)+(\frac{dr(\beta)}{d\beta})^2}d\beta##

    where the primed quantities are the corresponding quantities in the new rotated coordinate system.

    Recall that ##\bar{r}\cos\bar{\theta}=\frac{1}{L}\int_{\theta_1}^{\theta_2}r(\theta)\cos\theta\sqrt{r^2(\theta)+(\frac{dr(\theta)}{d\theta})^2}d\theta## is a function of ##\theta_1, \theta_2## and ##L##. Let's call it ##f=f(\theta_1, \theta_2, L)##. Then

    ##\bar{r}'\cos\bar{\theta}'=\bar{x}'=f(\beta_1, \beta_2, L)=f(\theta_1+\alpha, \theta_2+\alpha, L)##.

    But it remains to show why ##\bar{r}'=\bar{r}## and ##\cos\bar{\theta}'=\cos(\bar{\theta}+\alpha)##. That is, why

    ##f(\theta_1+\alpha, \theta_2+\alpha, L)=\bar{r}\cos(\bar{\theta}+\alpha)##.
     
    Last edited: Jul 13, 2016
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