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Chapter 5: Problem 322

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 triangular region with vertices \((0,0),(1,1), \quad\) and \((0,5) ; \rho(x, y)=x+y .\)

Short Answer

Expert verified
Calculate \( I_x \), \( I_y \), and \( I_0 \) using defined integrals and solve for the radii of gyration using mass integration.

Step by step solution

01

Define Moments of Inertia Formulas

The moments of inertia for a lamina with density function \( \rho(x,y) \) are given by the following formulas:- Moment of inertia about the x-axis: \[ I_x = \int \int_R y^2 \rho(x,y) \ dA \]- Moment of inertia about the y-axis: \[ I_y = \int \int_R x^2 \rho(x,y) \ dA \]- Moment of inertia about the origin: \[ I_0 = \int \int_R (x^2 + y^2) \rho(x,y) \ dA \]
02

Define the Region of Integration

The region \( R \) is a triangular region with vertices at \((0,0)\), \((1,1)\), and \((0,5)\). By plotting these vertices and drawing lines between them, we notice that the region is bounded by the line \( y = x \) from \( (0,0) \) to \( (1,1) \), and by the line \( x = 0 \) from \( (0,0) \) to \( (0,5) \). Therefore, the region of integration can be described as \( 0 \leq x \leq 1 \) and \( x \leq y \leq 5 \).
03

Find Ix 鈥� Moment of Inertia about x-axis

Using the formula for \( I_x \), substitute \( \rho(x, y) = x+y \):\[ I_x = \int_0^1 \int_x^5 y^2 (x+y) \ dy \, dx \]Integrate with respect to \( y \):\[ = \int_0^1 \left[ \frac{y^3}{3}x + \frac{y^4}{4} \right]_x^5 \, dx \]Substitute the limits and then integrate with respect to \( x \).
04

Find Iy 鈥� Moment of Inertia about y-axis

Using the formula for \( I_y \), substitute \( \rho(x, y) = x+y \):\[ I_y = \int_0^1 \int_x^5 x^2 (x+y) \ dy \, dx \]Integrate with respect to \( y \):\[ = \int_0^1 \left[ x^2y + \frac{x^2y^2}{2} \right]_x^5 \, dx \]Substitute the limits and then integrate with respect to \( x \).
05

Find I0 鈥� Moment of Inertia about origin

Using the formula for \( I_0 \), substitute \( \rho(x, y) = x+y \):\[ I_0 = \int_0^1 \int_x^5 (x^2 + y^2)(x+y) \ dy \, dx \]Expand and integrate with respect to \( y \):\[ = \int_0^1 \left[ \text{expanded terms involving powers of } y \right]_x^5 \, dx \]Substitute the limits and integrate the resulting expression with respect to \( x \).
06

Define Radii of Gyration

The radius of gyration \( k_x \) with respect to the x-axis is given by \( k_x = \sqrt{\frac{I_x}{m}} \), \( k_y \) with respect to the y-axis by \( k_y = \sqrt{\frac{I_y}{m}} \), and \( k_0 \) with respect to the origin by \( k_0 = \sqrt{\frac{I_0}{m}} \), where \( m \) is the mass of the lamina.
07

Calculate Mass of the Lamina

The mass \( m \) of the lamina is calculated by integrating the density function over the region \( R \):\[ m = \int_0^1 \int_x^5 (x+y) \ dy \, dx \]Integrate with respect to \( y \) first, and then \( x \).
08

Calculate Radii of Gyration

Substitute the calculated moments \( I_x \), \( I_y \), and \( I_0 \), as well as the mass \( m \), into the formulas for \( k_x \), \( k_y \), and \( k_0 \) respectively. Calculate each to find the radii of gyration.

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Key Concepts

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

Lamina
A lamina can be visualized as a thin, flat, two-dimensional object, like a sheet of paper. It has a definite shape and occupies a certain area in space. However, unlike a typical geometric shape studied in pure mathematics, a lamina in physics also possesses a property known as density. This concept is important when calculating physical properties such as the mass or the moments of inertia of the object. The lamina in context here is defined to occupy the region enclosed by the vertices
  • (0,0)
  • (1,1)
  • (0,5)
This forms a triangular region, which is essential when setting up the integrals necessary to compute its physical characteristics.
Density Function
The density function, denoted as \( \rho(x, y) \), describes how mass is distributed over the lamina. For our specific problem, this function is given by \( \rho(x, y) = x + y \). This means that the density of the lamina varies depending on its position within the triangular region.
Understanding how to work with the density function is crucial. It involves setting up a double integral over the region of the lamina, where the function \( \rho(x, y) \) is multiplied by the infinitesimal area \( dA \).
  • Mass calculations: The total mass of the lamina is given by the integral of the density function over the area.
  • Impact on inertia: The density affects the moments of inertia, influencing how the mass is spread across the x-axis, y-axis, or origin.
Radii of Gyration
Radii of gyration are important concepts used to describe how the mass is spread out with respect to an axis. It gives an indication of the distribution of the lamina's mass around an axis and can be defined for the x-axis, y-axis, and the origin.
The formula involves the moments of inertia \( I \) and the total mass \( m \) of the lamina. For example, the radius of gyration with respect to the x-axis is given by:\[ k_x = \sqrt{\frac{I_x}{m}} \]The radius of gyration essentially helps in finding how concentrated or spread out the object鈥檚 mass is from the axis.
  • A smaller radius indicates that the mass is closer to the axis.
  • A larger radius signifies the mass is more spread out.
Understanding radii of gyration is crucial for engineering and physics applications, particularly in mechanical design and structural analysis.
Mass of the Lamina
Calculating the mass of the lamina is a fundamental task to understand its physical behavior. The mass is derived using the density function and the region R over which the lamina is spread. Here, the formula to find the mass \( m \) is: \[ m = \int \int_R \rho(x, y) \, dA \]In practical terms, this represents integrating the density function \( \rho(x, y) = x + y \) over the specified region. The specific region limits, defined by the coordinates (0,0), (1,1), and (0,5), play a key role in setting these integrals.
  • First, integrate with respect to \( y \), ensuring the right bounds.
  • Then, integrate the resulting expression with respect to \( x \).
This process gives us the overall mass, which is a vital constant in formulas for moments of inertia and radii of gyration.

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Most popular questions from this chapter

The transformation \(T_{\theta} : \mathrm{R}^{2} \rightarrow \mathrm{R}^{2}, T_{\theta}(u, v)=(x, y)\) where \(x=u \cos \theta-v \sin \theta, \quad y=u \sin \theta+v \cos \theta,\) is called a rotation of angle \(\theta .\) Show that the inverse transformation of \(T_{\theta}\) satisfies \(T_{\theta}^{-1}=T_{-\theta},\) where \(T_{-\theta}\) is the rotation of angle \(-\theta\)

Let \(Q\) be a solid of constant density \(k,\) where \(k>0,\) that is located in the first octant, inside the circular cone \(x^{2}+y^{2}=9(z-1)^{2},\) and above the plane \(z=0\) . Show that the moment \(M_{x y}\) about the \(x y\) -plane is the same as the moment \(M_{y z}\) about the \(x z\) -plane.

Convert the integral $$ \int_{0}^{4} \int_{0}^{\sqrt{16-x^{2}}} \int_{-\sqrt{16-x^{2}-y^{2}}}^{\sqrt{16-x^{2}-y^{2}}}\left(x^{2}+y^{2}+z^{2}\right)^{2} d z d y d x $$ into an integral in spherical coordinates.

Set up the integral that gives the volume of the solid \(E\) bounded by \(y^{2}=x^{2}+z^{2}\) and \(y=a^{2},\) where \(a > 0\) .

The following problems examine Mount Holly in the state of Michigan. Mount Holly is a landfill that was converted into a ski resort. The shape of Mount Holly can be approximated by a right circular cone of height 1100 ft and radius 6000 \(\mathrm{ft.}\) In reality, it is very likely that the trash at the bottom of Mount Holly has become more compacted with all the weight of the above trash. Consider a density function with respect to height: the density at the top of the mountain is still density 400 \(\mathrm{lb} / \mathrm{ft}^{3}\) and the density increases. Every 100 feet deeper, the density doubles. What is the total weight of Mount Holly?
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