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1.4: Curvas Planas

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    131057
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    Curvas Planas Expresadas en\( x-y\) Coordenadas

    alt

    La Figura I.7 muestra cómo una longitud elemental\( \delta s \) is related to the corresponding increments in \( x \) and \( y \):

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

    Considere un alambre de masa por unidad de longitud (densidad lineal)\( \lambda \) bent into the shape \( y = y(x) \) between \( x = a \) and \( x=b \). The mass of an element \( ds \) is \( \lambda \delta s \) , por lo que la masa total es

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

    Los primeros momentos de misa sobre la\( y \) - and \( x \) -axes are respectively

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

    y

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

    Si el cable es uniforme y\( \lambda \) 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

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

    y

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

    siendo el denominador en cada una de estas expresiones simplemente la longitud total del cable.

    Ejemplo\(\PageIndex{1}\)

    Considera un alambre uniforme doblado en la forma del semicírculo\( x^{2} + y^{2} = a^{2} \), \( x >0 \).

    En primer lugar, cabe señalar que uno esperaría\( \overline{x} > 0.4244a \) (el valor para una lámina semicircular plana).

    La longitud (es decir, los denominadores en Ecuaciones\( \ref{eq:1.4.4A}\) 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

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

    En este caso

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

    y

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

    El primer momento de longitud de todo el semicírculo es

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

    A partir de este punto el alumno se deja a sus propios dispositivos para resolver esta integral y derivar\( \overline{x} = \dfrac{2a}{\pi} = 0.6366a \).

    Curvas Planas Expresadas en Coordenadas Polares


    alt

    La Figura I.8 muestra cómo una longitud elemental\( \delta s \) is related to the corresponding increments in \( r \) and \( \theta \) :

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

    La masa de la curva (entre\( \theta = a \) y\( \theta = b \) ) es

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

    Los primeros momentos sobre el\( y \) - and \( x \) -axes are (recalling that \( x = r \cos \theta \) and \( y = r \sin \theta \) )

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

    y

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

    Si no\( \lambda \) es una función de\( r \) or \( \theta \) , obtenemos

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

    y

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

    donde e\(L\) es la longitud del cable.

    Ejemplo\(\PageIndex{2}\)

    Consideremos nuevamente el alambre uniforme de la Figura I.8 doblado en forma de semicírculo. La ecuación en coordenadas polares es simple\( r = a \), and the integration limits are \( \theta = \dfrac{- \pi}{2} \) to \( \theta = \dfrac{+ \pi}{2} \) y la longitud es\( \pi a \) .

    Así

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

    El lector debería ahora encontrar la posición del centro de masa de un alambre doblado en el arco de un círculo de ángulo a\( 2 \alpha \). The expression obtained should go to \( \dfrac{2 a }{\pi} \) medida que\( \alpha \) va a\( \dfrac{\pi}{2} \), and to \( a \) as \( \alpha \) va a cero.


    This page titled 1.4: Curvas Planas is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.