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In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises. a. Find the moments of inertia \(I_{x}, I_{y},\) and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. b. Find the radii of gyration with respect to the \(x\) -axis, \(y\) -axis, and origin, respectively. \(R\) is the region enclosed by the ellipse \(x^{2}+4 y^{2}=1 ; \rho(x, y)=1\)

Short Answer

Expert verified
Moments: \(I_x = \frac{\pi}{8}\), \(I_y = \frac{\pi}{8}\), \(I_0 = \frac{\pi}{4}\). Radii of gyration: \(k_x = \frac{1}{2}\), \(k_y = \frac{1}{2}\), \(k_0 = \frac{1}{\sqrt{2}}\).

Step by step solution

01

Define the Region and Density Function

The elliptical region is defined by the equation \(x^2 + 4y^2 = 1\). The density function given is \(\rho(x, y) = 1\), meaning the lamina has a constant density of 1 throughout the region.
02

Convert to suitable coordinates

Consider transforming the Cartesian coordinates to the elliptical coordinates that map the region into a circle. Let \(u = x\) and \(v = 2y\), then the equation becomes \(u^2 + v^2 = 1\), a unit circle. The Jacobian of this transformation is \(\left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \frac{1}{2}\).
03

Calculate the Moment of Inertia about the x-axis, \(I_x\)

Using transformation, the moment of inertia about the x-axis is given by: \[ I_x = \int \int_{R} y^2 \rho(x, y) \, dx \, dy = \frac{1}{2} \int_0^{2\pi} \int_0^1 \left(\frac{v^2}{4}\right) r \, dr \, d\theta = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2\pi}{4} = \frac{\pi}{8} \]
04

Calculate the Moment of Inertia about the y-axis, \(I_y\)

Similarly, using the transformation, \[ I_y = \int \int_{R} x^2 \rho(x, y) \, dx \, dy = \frac{1}{2} \int_0^{2\pi} \int_0^1 u^2 r \, dr \, d\theta = \frac{1}{2} \left[ \frac{\pi}{4} \right] = \frac{\pi}{8} \]
05

Calculate the Moment of Inertia about the Origin, \(I_0\)

The moment of inertia about the origin is given by \[ I_0 = \int \int_{R} (x^2 + y^2) \rho(x, y) \, dx \, dy = \int \int_{R} (u^2 + v^2/4) \, dx \, dy = \frac{\pi}{4} \]
06

Calculate the Radii of Gyration, \(k_x, k_y, k_0\)

The radius of gyration with respect to an axis is defined as \( k_x = \sqrt{\frac{I_x}{M}} \), where \(M\) is the total mass of the region. The mass \(M\) is the area of the ellipse, which is \(\frac{\pi}{2}\).- \(k_x = \sqrt{\frac{\pi/8}{\pi/2}} = \frac{1}{2}\)- \(k_y = \sqrt{\frac{\pi/8}{\pi/2}} = \frac{1}{2}\)- \(k_0 = \sqrt{\frac{I_0}{M}} = \sqrt{\frac{\pi/4}{\pi/2}} = \frac{1}{\sqrt{2}}\) which simplifies to \(\frac{1}{\sqrt{2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radii of Gyration
The concept of radii of gyration is pivotal in understanding how the mass of a body is distributed with respect to a specific axis.
The radius of gyration, often denoted as \(k_x\), \(k_y\), or \(k_0\) for the x-axis, y-axis, and origin respectively, gives a measure of how far from the axis or point the mass is concentrated.
Simply, the radius of gyration can be thought of as the distance at which the entire mass of a body could be concentrated without changing its rotational inertia.
For a body of uniform density, if the entire mass were concentrated at this distance, the moment of inertia would remain the same.To calculate the radius of gyration with respect to an axis, the formula is \( k = \sqrt{\frac{I}{M}} \), where \(I\) is the moment of inertia about the axis, and \(M\) is the total mass.
A smaller radius indicates a mass closer to the axis, suggesting a compact distribution, while a larger radius shows a more spread-out mass.
In this exercise, for the elliptical region, both \(k_x\) and \(k_y\) were found to be \(\frac{1}{2}\), indicating a symmetric mass distribution.
Elliptical Coordinates
Elliptical coordinates are an effective method to simplify problems, especially when dealing with regions confined by ellipses, like in our problem here.
By converting from Cartesian coordinates to elliptical coordinates, complex regions can be transformed into simpler forms, like circles.
In this exercise, the Cartesian coordinates \((x, y)\) were converted using the transformations \(u = x\) and \(v = 2y\).
This conversion transforms the elliptical equation \(x^2 + 4y^2 = 1\) into \(u^2 + v^2 = 1\), which represents a unit circle in the \((u, v)\)-plane.This simplification makes the process of integrating over the region more manageable.
The transformation reduces the complexity typically involved with ellipses and allows for the more straightforward calculation of quantities like moments of inertia.
Jacobian Transformation
The Jacobian transformation is a crucial technique in multivariable calculus, used to change variables in integrals when transforming coordinate systems.
It involves finding the determinant of the Jacobian matrix, which represents the rate of change of coordinates and accounts for the scale transformation between the two coordinate systems.
For the ellipse in this exercise, converting to elliptical coordinates involves finding the Jacobian determinant \(\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\).
In our transformation where \(u = x\) and \(v = 2y\), the absolute choice of the determinant was \(\frac{1}{2}\).This Jacobian value was crucial as a scaling factor when computing the integrals for moments of inertia.
It ensures that the change in area due to coordinate transformation is accurately accounted for, providing correct results.
With this transformation, the computation of integrals over the region becomes more feasible.
Lamina Density Function
In physical problems, the density function, typically denoted as \(\rho(x,y)\), represents how mass is distributed across an object.
For this exercise, the lamina (or thin plate) defined by the elliptical region possesses a constant density function \(\rho(x, y) = 1\).
This constant density indicates that every point on the lamina has the same mass per unit area, simplifying calculations related to mass, moments of inertia, and eventually radii of gyration.
The total mass of the lamina can be calculated by integrating the density function over the entire region.When the density is uniform, the mass \(M\) simplifies to the area of the region.
In our case, the area of the ellipse \(x^2 + 4y^2 = 1\) is calculated as \(\frac{\pi}{2}\).
The assumption of constant density considerably reduces the complexity of solving physical problems involving the lamina.

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