1.4: Plane Curves

Plane Curves Expressed in $x-y$ coordinates

Figure I.7 shows how an elemental length $\delta s$ is related to the corresponding increments in $x$ and $y$:

$$\delta s = \sqrt{\delta x^{2} + \delta y^{2}} = \sqrt{1+\left(\dfrac{dy}{dx}\right)^{2}} \delta x = \sqrt{ \left(\dfrac{dx}{dy}\right)^{2} + 1} \, dy \label{eq:1.4.1}$$

Consider a wire of mass per unit length (linear density) $\lambda$ bent into the shape $y = y(x)$ between $x = a$ and $x=b$. The mass of an element $ds$ is $\lambda \delta s$, so the total mass is

$$\int \lambda \,ds = \int_a^b \lambda \sqrt{1 + \left(\dfrac{dy}{dx}\right)^{2}} \, dx \label{eq:1.4.2}$$

The first moments of mass about the $y$ - and $x$ -axes are respectively

$$\int_a^b \lambda x \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} \, dx \label{eq:1.4.3A}$$

and

$$\int_a^b \lambda y \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} \, dx \label{eq:1.4.3B}$$

If the wire is uniform and $\lambda$ is therefore not a function of $x$ or $y$, $\lambda$ can come outside the integral signs in Equations $\ref{eq:1.4.2}$ - $\ref{eq:1.4.3B}$, and we hence obtain

$$\overline{x} = \dfrac{\displaystyle \int_a^b x \sqrt{ 1+\left( \dfrac{dy}{dx} \right)^2} dx } { \displaystyle \int_a^b \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} dx} \label{eq:1.4.4A}$$

and

$$\overline{y} = \dfrac{\displaystyle \int_a^b y \sqrt{ 1+\left( \dfrac{dy}{dx} \right)^2} dx } { \displaystyle \int_a^b \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} dx} \label{eq:1.4.4B}$$

the denominator in each of these expressions merely being the total length of the wire.

Example $\PageIndex{1}$

Consider a uniform wire bent into the shape of the semicircle $x^{2} + y^{2} = a^{2}$, $x >0$.

First, it might be noted that one would expect $\overline{x} > 0.4244a$ (the value for a plane semicircular lamina).

The length (i.e. the denominators in Equations $\ref{eq:1.4.4A}$ and $\ref{eq:1.4.4B}$) is just $\pi a$. Since there are, between $x$ and $x + \delta x$, two elemental lengths to account for, one above and one below the $x$ axis, the numerator of Equation $\ref{eq:1.4.4A}$ must be

$$2 \int_0^a x \sqrt{1+ \left(\dfrac{dy}{dx}\right)^{2}} dx \nonumber$$

In this case

$$y = \sqrt{a^2 - x^2} \nonumber$$

and

$$\dfrac{dy}{dx} = \dfrac{-x}{ \sqrt{a^2 - x^2} } \nonumber$$

The first moment of length of the entire semicircle is

$$\overline{x} = 2 \int_0^a x \sqrt{ 1 + \dfrac{x^2}{a^2-x^2} } dx = 2a \int_0^a \dfrac{x\,dx}{ \sqrt{ a^2-x^2 }} \nonumber$$

From this point the student is left to his or her own devices to solve this integral and derive $\overline{x} = \dfrac{2a}{\pi} = 0.6366a$.

Plane Curves Expressed in Polar Coordinates

Figure I.8 shows how an elemental length $\delta s$ is related to the corresponding increments in $r$ and $\theta$:

$$\delta s = \sqrt{( \delta r)^{2} + (r \delta \theta )^{2}} = \sqrt{\left( \dfrac{dr}{d \theta }\right)^{2} + r^{2}} \, \delta \theta = \sqrt{1+\left(r \dfrac{d \theta }{dr}\right)^{2}} \,\delta r. \label{eq:1.4.5}$$

The mass of the curve (between $\theta = a$ and $\theta = b$ ) is

$$\int_ \alpha ^ \beta \lambda \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta. \nonumber$$

The first moments about the $y$ - and $x$ -axes are (recalling that $x = r \cos \theta$ and $y = r \sin \theta$ )

$$\int_ \alpha ^ \beta \lambda r \cos \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta \nonumber$$

and

$$\int_ \alpha ^ \beta \lambda r \sin \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta. \nonumber$$

If $\lambda$ is not a function of $r$ or $\theta$ , we obtain

$$\overline{x} = \dfrac{1}{L} \int_ \alpha ^ \beta r \cos \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta \label{eq:1.4.6A}$$

and

$$\overline{y} = \dfrac{1}{L} \int_ \alpha ^ \beta r \sin \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} d \theta \label{eq:1.4.6B}$$

where $L$ is the length of the wire.

Example $\PageIndex{2}$

Again consider the uniform wire of Figure I.8 bent into the shape of a semicircle. The equation in polar coordinates is simply $r = a$, and the integration limits are $\theta = \dfrac{- \pi}{2}$ to $\theta = \dfrac{+ \pi}{2}$ and the length is $\pi a$.

Thus

$$\overline{x} = \dfrac{1}{ \pi a} \int_{- \pi/2} ^{+ \pi/2} acos \theta [ 0 - a^{2}]^ \frac{1}{2} d \theta = \dfrac{2a}{ \pi } . \nonumber$$

The reader should now find the position of the center of mass of a wire bent into the arc of a circle of angle $2 \alpha$. The expression obtained should go to $\dfrac{2 a }{\pi}$ as $\alpha$ goes to $\dfrac{\pi}{2}$, and to $a$ as $\alpha$ goes to zero.