/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 In the following exercises, cons... [FREE SOLUTION] | 91影视

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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 Find the moments of inertia \(I_{x}, I_{y}\), and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. Find the radii of gyration with respect to the \(x\) -axis, \(y\) -axis, and origin, respectively.\(R\) is the rectangular region with vertices \((0,1),(0,3),(3,3)\), and \((3,1) ; \rho(x, y)=x^{2} y\).

Short Answer

Expert verified
Moments: \(I_x = 180\), \(I_y = 194.4\), \(I_0 = 374.4\). Radii: \(k_x = \sqrt{5}\), \(k_y = \sqrt{5.4}\), \(k_0 = \sqrt{10.4}\).

Step by step solution

01

Define the Region and Density Function

The region of the lamina is defined by the rectangular area with vertices at \((0,1),(0,3),(3,3)\), and \((3,1)\). The given density function is \(\rho(x, y) = x^2 y\). The bounds for integration in the \(x\)-direction are from \(0\) to \(3\) and in the \(y\)-direction are from \(1\) to \(3\).
02

Calculate Moment of Inertia About the x-axis (I_x)

The moment of inertia about the \(x\)-axis is calculated using the integral \[ I_x = \int_{a}^{b} \int_{c}^{d} \rho(x, y) y^2 \, dy \, dx. \]Here, \(a = 0\), \(b = 3\), \(c = 1\), and \(d = 3\). Substitute \(\rho(x, y) = x^2 y\), giving:\[ I_x = \int_{0}^{3} \int_{1}^{3} (x^2 y) y^2 \, dy \, dx = \int_{0}^{3} x^2 \left( \int_{1}^{3} y^3 \, dy \right) \, dx. \]Evaluate the inner integral:\[ \int_{1}^{3} y^3 \, dy = \left[ \frac{y^4}{4} \right]_1^3 = \frac{81}{4} - \frac{1}{4} = 20. \]Then, evaluate the outer integral:\[ I_x = \int_{0}^{3} x^2 \cdot 20 \, dx = 20 \left[ \frac{x^3}{3} \right]_0^3 = 20 \cdot 9 = 180. \]
03

Calculate Moment of Inertia About the y-axis (I_y)

The moment of inertia about the \(y\)-axis is calculated using the integral \[ I_y = \int_{a}^{b} \int_{c}^{d} \rho(x, y) x^2 \, dy \, dx. \]Substitute the given values and evaluate the inner integral:\[ I_y = \int_{0}^{3} x^2 \left( \int_{1}^{3} x^2 y \, dy \right) \, dx = \int_{0}^{3} x^2 \left( x^2 \left[ \frac{y^2}{2} \right]_1^3 \right) \, dx. \]Evaluate:\[ \int_{1}^{3} y \, dy = \left[ \frac{y^2}{2} \right]_1^3 = \frac{9}{2} - \frac{1}{2} = 4. \]Now, replace and solve the outer integral:\[ I_y = \int_{0}^{3} x^4 \times 4 \, dx = 4 \left[ \frac{x^5}{5} \right]_0^3 = 4 \cdot \frac{243}{5} = 194.4. \]
04

Calculate Moment of Inertia About the Origin (I_0)

The moment of inertia about the origin is given by \[ I_0 = I_x + I_y. \]From previous calculations, \(I_x = 180\) and \(I_y = 194.4\), so \[ I_0 = 180 + 194.4 = 374.4. \]
05

Compute Radii of Gyration

The radius of gyration \(k_x\) with respect to the \(x\)-axis can be calculated using \[ k_x = \sqrt{\frac{I_x}{m}}, \]where \(m\) is the mass of the lamina:\[ m = \int_{0}^{3} \int_{1}^{3} x^2 y \, dy \, dx = \int_{0}^{3} x^2 \left( \frac{y^2}{2} \right)_1^3 \, dx = \int_{0}^{3} x^2 \cdot 4 \, dx = 4 \cdot 9 = 36. \]Now, calculate \(k_x\), \(k_y\), and \(k_0\):\[ k_x = \sqrt{\frac{180}{36}} = \sqrt{5}. \]The radius of gyration \(k_y\) is \[ k_y = \sqrt{\frac{I_y}{m}} = \sqrt{\frac{194.4}{36}} = \sqrt{5.4}. \]Finally, for \(k_0\):\[ k_0 = \sqrt{\frac{I_0}{m}} = \sqrt{\frac{374.4}{36}} = \sqrt{10.4}. \]

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Key Concepts

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

Density Function
Density function is a crucial concept in physics and mathematics when studying continuous mass distributions, such as a lamina. It describes how mass is distributed across an object. In this exercise, the density function is given as \(\rho(x, y) = x^2 y\).
This function takes two variables, \(x\) and \(y\), which represent the coordinates in the region \(R\). The density at any point \((x, y)\) is calculated by substituting these coordinates into the density function to get the mass per unit area at that point.

Understanding the density function is important because:
  • It allows us to compute the mass of the lamina by integrating over the defined region.
  • It plays a key role in calculating the moments of inertia, which describe how the mass is distributed relative to different axes.
So, the shape of the density function can remarkably change the physical properties of the lamina being studied."},{"concept_headline":"Lamina","text":"A lamina is a two-dimensional flat sheet of material. In mathematical problems, it often represents a region with a density distribution used to solve physical problems, such as finding centers of mass or inertia.

For this exercise, the lamina occupies a rectangular region \(R\) defined by vertices \((0,1), (0,3), (3,3),\) and \((3,1)\). The density function \(\rho(x, y) = x^2 y\) indicates that the lamina's mass varies at different parts of this rectangle.

The concept of a lamina is vital because:
  • It simplifies three-dimensional problems by considering only their two-dimensional aspects.
  • Describing physical properties over a defined area helps in calculating parameters such as mass and inertia.
Using an idealized lamina allows students to focus on understanding various physical phenomena without complex calculations involving three-dimensional bodies."},{"concept_headline":"Radii of Gyration","text":"The radii of gyration quantify how far the mass of an object is distributed from an axis or point about which it rotates. They help determine how much resistance an object will have to rotational motion.

In the solution, the radii of gyration are calculated with respect to the \(x\)-axis, \(y\)-axis, and the origin, using the moments of inertia and the total mass of the lamina:
  • \(k_x = \sqrt{\frac{I_x}{m}}\)
  • \(k_y = \sqrt{\frac{I_y}{m}}\)
  • \(k_0 = \sqrt{\frac{I_0}{m}}\)
Where \(I_x\), \(I_y\), and \(I_0\) are the moments of inertia about the \(x\)-axis, \(y\)-axis, and the origin, respectively, and \(m\) is the mass of the lamina.

Radii of gyration are important concepts because they:
  • Help in understanding how mass distribution affects the moment of inertia of a body.
  • Provide insights into the stability of rotational systems.
Basically, they offer a simple, visual measure of how mass is spread around a particular axis or point."},{"concept_headline":"Integration Bounds","text":"Integration bounds are the limits within which the integration is performed, specifying the area or volume over which a function is integrated. In problems concerning laminae, they define the region where the mass distribution is being analyzed.

In this exercise, the bounds for integration are:
  • \(0\) to \(3\) in the \(x\)-direction
  • \(1\) to \(3\) in the \(y\)-direction
These bounds correspond to the defined rectangular region \(R\) with vertices \((0,1), (0,3), (3,3),\) and \((3,1)\).

Understanding integration bounds is key because:
  • They ensure the calculations consider the correct region, integral for accurate results.
  • They help students visualize the limits of their integrals, making the abstract concept of integration more concrete.
The correct setting of integration bounds is crucial to obtaining the right physical parameters in any given mathematical physics problem."}]}]} 啶侧た啶傕 啶膏啶侧ぞ啶囙ぁ 啶膏す啶距く啶むぞ 啶忇啷嵿じ啶啶侧啶 啶⑧啶佮あ啷囙 啶啶膏啶傕啶 啶膏啶掂啶曕啶むた 啶曕啶掂た啶-19 啶侧ぞ啶囙さ 啶膏ぞ啶囙え 啶囙え 啶啶椸う啶距え 啶啶班ざ啷嵿え 啶曕ぞ啶夃啶熰ぐ 啶椸啶椸げ 啶呧え啷佮ざ啶距じ啶ㄠ啶 啶曕 啶ㄠ啶むた 啶啶むぐ啶む啶 啶夃う啷嵿く啷嬥 啶げ啷嵿じ 啶ぞ啶ㄠさ啶むぞ 啶曕ぞ 啶曕ぞ啶班啶 kwbgaeb.urs.edu nitrogen.io 啶囙げ啷囙啷嵿啷嵿ぐ啶苦 啶膏啶曕ぞ啶熰ぐ 啶膏啶曕啶 啶ぞ啶班啶 啶囙さ啷囙啶 啶ぐ啶ぞ啶む啶ぞ 啶ぞ啶膏啶熰ぐ啶距げ 啶膏啶膏啶ムぞ啶 啶︵ぞ啶 啶椸ぞ啶傕が啷 啶侧ぞ啶囙さ 啶曕啶粪啶む啶班啶 啶呧え啷佮じ啶傕ぇ啶距え 啶啶班啶班ぃ啶 啶嗋啶苦ぐ啷 啶ざ啷嵿啶苦ぎ 啶啶椸啶 啶掂啶灌啶曕げ crem.co meseleting.sjk 123 啶︵啶掂啶 啶班啶∴た啶 啶え啷 啶︵た啶栢ぞ啶 啶班啶 啶曕ぞ啶班啶 啶⑧啶佮あ啶ㄠぞ 啶膏啶ムぞ啶 啶嗋ざ啶 啶氞啶ㄠ 啶曕啶熰啶傕 啶掂啶 啶班じ啶膏た啶む啶氞啶溹啶溹啶ㄠた啷囙お啷嵿お 啶啶多啶掂ぐ 啶膏啶曕啶む 啶溹た啶 啶班た啶膏ぐ啷嵿啶班啶 啶涏さ啶 啶膏ぞ啶囙え啶呧お 啶啶椸う啶距え啶曕ぐ啷嵿い啶 啶椸啶 啶夃い啷嵿お啶距う vysok.si 啶掂ぞ啶囙 啶曕啶 啶むぐ啶距 啶ぐ啶苦啶 啶膏す啶距く啶 啶曕啶熰啶 ancialscience.dyn.mr 啶む啶溹 啶掂啶 啶ㄠ啶啶溹げ啷囙 啶溹啶栢た啶 啶侧ぞ啶げ啷嵿い啷嵿く 啶啶班啶椸啶班ぞ啶

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