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Chapter 5: Problem 330

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, \(\quad y\) -axis, and origin, respectively. $$ R=\left\\{(x, y) \mid 9 x^{2}+y^{2} \leq 1, x \geq 0, y \geq 0\right\\} ; \rho(x, y)=\sqrt{9 x^{2}+y^{2}} $$

Short Answer

Expert verified
Compute moments \(I_x\), \(I_y\), \(I_0\) using region R and density \(\rho\). Calculate radii using \(\overline{k} = \sqrt{I/m}\).

Step by step solution

01

Define the Region R

The region \( R \) is the first quadrant of the ellipse obtained from the inequality \( 9x^2 + y^2 \leq 1 \). This can be rewritten as \( \frac{x^2}{\frac{1}{9}} + \frac{y^2}{1} \leq 1 \), indicating it is an ellipse with semi-major axis 1 and semi-minor axis \( \frac{1}{3} \). The lamina occupies the first quadrant of this ellipse.
02

Express \( \rho(x,y) \)

The density function provided is \( \rho(x, y) = \sqrt{9x^2 + y^2} \). This represents the distance of any point \((x, y)\) in the ellipse from the origin.
03

Calculate the Moment of Inertia \(I_x\)

The moment of inertia with respect to the \(x\)-axis is calculated using the integral: \[I_x = \int \int_R y^2 \rho(x, y) \, dA\]Where:- \(dA\) is the differential area element of the region, which in polar coordinates for the ellipse can be taken as \( dA = r \cdot \frac{1}{9} \cdot dr \cdot d\theta\).
04

Calculate the Moment of Inertia \(I_y\)

Similarly, the moment of inertia about the \(y\)-axis is:\[I_y = \int \int_R x^2 \rho(x, y) \, dA\]Converting to polar coordinates, we use the expressions for \( x = \frac{r}{3} \cos\theta \) and \( y = r \sin\theta \).
05

Calculate the Moment of Inertia \(I_0\)

For the moment of inertia about the origin, the formula is:\[I_0 = \int \int_R (x^2 + y^2) \rho(x, y) \, dA\]This integral covers both axes using the density function. We perform the integrals using suitable limits for \(r\) and \(\theta\) for the first quadrant.
06

Convert to Polar Coordinates and Evaluate

Convert using \(x = \frac{1}{3}r\cos\theta\), \(y = r\sin\theta\) and \(dA = r\,dr\,d\theta\). The limits for \(r\) go from 0 to 1 and for \(\theta\) from 0 to \(\frac{\pi}{2}\). Evaluate each integral.
07

Compute Radii of Gyration

The radius of gyration about an axis \(a\) is given by:\[\overline{k_a} = \sqrt{\frac{I_a}{m}},\]where \(m\) is the mass of the lamina. Compute \(\overline{k_x}, \overline{k_y},\) and \(\overline{k_0}\) using respective moments and total mass obtained from the density function.

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Key Concepts

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

Lamina Density
When discussing the lamina density, we refer to the density function, often denoted by \( \rho(x, y) \). In simple terms, this function provides us with information about how tightly mass is packed at a particular point \((x, y)\) within the region of the lamina. For this particular problem, the density function is given as \( \rho(x, y) = \sqrt{9x^2 + y^2} \).
To understand it intuitively:
  • The value of \( \rho(x, y) \) increases as one moves further from the origin because it represents the distance from the origin.
  • This function ensures that points further from the center have a higher density, which influences how the mass is distributed throughout the lamina.
Overall, the density function is essential for calculating other properties of the lamina such as its mass and moments of inertia. Since the density varies across the lamina, it impacts how the material properties are calculated.
Radii of Gyration
Radii of gyration offer a measure of how the mass of the lamina is distributed relative to a specified axis. It defines a hypothetical radius where the entire mass can be assumed to be concentrated while keeping the moment of inertia unchanged. Mathematically, the radius of gyration \( \overline{k_a} \) about an axis \( a \) is expressed as follows:\[ \overline{k_a} = \sqrt{\frac{I_a}{m}} \]Where:
  • \( I_a \) is the moment of inertia about axis \( a \).
  • \( m \) is the total mass of the lamina.
To find the radii of gyration \( \overline{k_x}, \overline{k_y}, \) and \( \overline{k_0} \):
  • Calculate the moments of inertia \( I_x, I_y, \) and \( I_0 \) using the given density function \( \rho(x, y) \).
  • Compute the lamina's total mass from the density \( \rho(x, y) \).
  • Apply the formula for each axis to obtain the respective radii of gyration.
The radii of gyration provide a useful way to conceptualize how the mass is distributed around an axis.
Ellipse in Polar Coordinates
Understanding the ellipse in polar coordinates helps in simplifying integrals for calculating moments of inertia and mass. The given region is defined by the inequality \( 9x^2 + y^2 \leq 1 \) and is identified as an ellipse. By converting Cartesian coordinates to polar coordinates, we make the integration over the ellipse much more manageable.
In polar coordinates, points on the ellipse are expressed using \((r, \theta)\):
  • The equations for conversion are \( x = \frac{1}{3}r\cos\theta \) and \( y = r\sin\theta \).
  • The differential area element becomes \( dA = r\,dr\,d\theta \).
  • The limits for \( r \) are from 0 to 1, and for \( \theta \), it ranges from 0 to \( \frac{\pi}{2} \) since we're considering only the first quadrant.
Using polar coordinates ensures the calculation of integrals over the elliptical region is straightforward, simplifying the process of finding mass and inertia of the lamina.
Mass Calculation of Lamina
To calculate the mass of the lamina, we integrate the density function \( \rho(x, y) \) over the specified region \( R \).
Here's a step-by-step overview:
  • Firstly, recognize that the mass \( m \) of the lamina can be found using the formula:
\[ m = \int \int_R \rho(x, y) \, dA \]
  • Convert this double integral into polar coordinates, resulting in:
\[ m = \int_0^{\frac{\pi}{2}} \int_0^1 \sqrt{9(\frac{r}{3}\cos\theta)^2 + (r\sin\theta)^2} \cdot r \, dr \, d\theta \]
  • Simplify the integrand, realizing that \( 9 x^2 + y^2 = r^2 \), turning the integrand into just \( r^2 \).
  • Calculate the integral over \( r \) from 0 to 1 and \( \theta \) from 0 to \( \frac{\pi}{2} \).
  • The resulting integral provides the total mass of the lamina.
This mass is critical for other calculations, such as determining the radii of gyration and moments of inertia, making it a foundational step in the analysis.

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Most popular questions from this chapter

Let \(Q\) be the solid unit cube. Find the mass of the solid if its density \(\rho\) is equal to the square of the distance of an arbitrary point of \(Q\) to the \(x y\) -plane.

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