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If the placement of a rope is defined by an even function $y (x) $ from $x=-a $ to $x=a $, what is the the center of mass of the rope?)

I imagined it would be the average value of the function over the interval, but that is not the case. Or if it is I am making an error in a larger problem I'm working on.

Can anyone explain more about the integral definition of center of mass? The Wikipedia page isn't very helpful.

2 Answers2

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You'll need line integrals: $$ y_c = \frac{\int y(x)\,d\ell}{\int d\ell} = \frac{\int y(x)\sqrt{(y'(x))^2+1}\,dx}{\int\sqrt{(y'(x))^2+1}\,dx}, $$ where $d\ell=\sqrt{dx^2+dy^2}$ is the little portion of length (and thus mass).

colt_browning
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If $\rho$ is the density mass and $S$ is the surface of the rope, the mass $M$ is $$ M=\iint_S\rho\,\mathrm dA $$ and the moments are $$ M_x=\iint_S\rho\,y\,\mathrm dA\qquad M_y=\iint_S\rho\,x\,\mathrm dA $$ Hence the center of mass coordinates are $$ \begin{align} \bar x&=\frac{M_y}{M}=\frac{\iint_S x\,\mathrm dA}{\iint_S\mathrm dA}\\ \bar y&=\frac{M_x}{M}=\frac{\iint_S y\,\mathrm dA}{\iint_S\mathrm dA} \end{align} $$ If the rope is between $f(x)$ and $g(x)$ and between $a$ and $b$ the center of mass coordinates are $$ \begin{align} \bar x&=\frac{M_y}{M}=\frac{\int_{a}^b \int_{g(x)}^{f(x)} x\,\mathrm dy\,\mathrm dx}{\int_{a}^b \int_{g(x)}^{f(x)} \mathrm dy\,\mathrm dx}=\frac{\int_{a}^b x\left[f(x)-g(x)\right]\mathrm dx}{\int_{a}^b \left[f(x)-g(x)\right]\mathrm dx}=\frac{1}{S}\int_{a}^b x\left[f(x)-g(x)\right]\mathrm dx\\ \\ \bar y&=\frac{M_x}{M}=\frac{\int_{a}^b \int_{g(x)}^{f(x)} y\,\mathrm dy\,\mathrm dx}{\int_{a}^b \int_{g(x)}^{f(x)} \mathrm dy\,\mathrm dx}=\frac{\int_{a}^b \frac{1}{2}\left[f^2(x)-g^2(x)\right]\mathrm dx}{\int_{a}^b \left[f(x)-g(x)\right]\mathrm dx}=\frac{1}{S}\int_{a}^b \frac{1}{2}\left[f^2(x)-g^2(x)\right]\mathrm dx \end{align} $$ where $S=\int_{a}^b \left[f(x)-g(x)\right]\mathrm dx$ is the area of the rope.

If the rope is the surface between $y=f(x)$ even, $y=0$, $x=-a$ and $x=a$ we have $$ \begin{align} \bar x&=\frac{M_y}{M}=\frac{\int_{-a}^a xf(x)\mathrm dx}{\int_{-a}^a f(x)\mathrm dx}=\frac{\int_{-a}^a xf(x)\mathrm dx}{2\int_{0}^a f(x)\mathrm dx}=0\\ \\ \bar y&=\frac{M_x}{M}=\frac{\int_{-a}^a \frac{1}{2}f^2(x)\mathrm dx}{\int_{-a}^a f(x)\mathrm dx}=\frac{1}{2}\frac{\int_{0}^a f^2(x)\mathrm dx}{\int_{0}^a f(x)\mathrm dx} \end{align} $$

alexjo
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  • Isn't rope a one-dimensional object (as implicitly assumed in the question)? If so, surface integrals become line integrals, and your formula for $\bar{y}$ becomes my answer. – colt_browning Sep 16 '17 at 07:44
  • Are you sure? Consider $f=x^2$... – alexjo Sep 16 '17 at 07:50
  • I don't understand how a particular example helps. Our understanding of the problem is different. – colt_browning Sep 16 '17 at 08:11
  • But if you want to see a direct connection: the width of the rope must be infinitesimal and constant; and if "lower border of the rope" is $g(x)$, then width at point $x$ is $w=(f(x)-g(x))\cos\arctan f'(x)=(f(x)-g(x))/\sqrt{1+(f'(x))^2}$ (when $(f-g)\to0$ and $(f'-g')\to0$), so $g(x)=f(x)-w\sqrt{1+(f'(x))^2}$. Substitute it to your answer and set $w\to0$, and you'll get mine. – colt_browning Sep 16 '17 at 08:25