Problem 22 - Advanced Engineering Mathematics

Question

In Problems, find the polar moment of inertia $$ I_{0}=\iint_{R} r^{2} \rho(r, \theta) d A=I_{x}+I_{y} $$ of the lamina that has the given shape and density. \(r=\theta, 0 \leq \theta \leq \pi, y=0 ;\) density at a point \(P\) proportional to the distance from the pole

Step-by-Step Solution

Answer

The polar moment of inertia is \( I_0 = \frac{k \pi^6}{30} \).

Step 1: Understand the Problem

The given problem asks us to find the polar moment of inertia of a lamina with density proportional to its distance from the pole. The lamina is defined by the region where \( r = \theta \) and \( 0 \leq \theta \leq \pi \), with \( y = 0 \).

Step 2: Express the Density Function

Given that the density is proportional to the distance from the pole, we express the density function as \( \rho(r, \theta) = k \cdot r \), where \( k \) is a constant of proportionality.

Step 3: Set Up the Integral for Polar Moment of Inertia

The polar moment of inertia is defined as \[ I_0 = \iint_R r^2 \rho(r, \theta) \, dA \]. Substituting our density function gives \[ I_0 = \iint_R r^3 k \, dA \].

Step 4: Convert to Polar Coordinates

In polar coordinates, \( dA = r \, dr \, d\theta \). For the region specified by \( r = \theta \) and \( 0 \leq \theta \leq \pi \), the limits of integration for \( r \) and \( \theta \) are \( 0 \leq r \leq \theta \) and \( 0 \leq \theta \leq \pi \).

Step 5: Integrate with Respect to r

First, integrate with respect to \( r \):\[ \int_0^\theta r^4 \, dr = \frac{r^5}{5} \bigg|_0^\theta = \frac{\theta^5}{5} \].

Step 6: Integrate with Respect to θ

Now, integrate with respect to \( \theta \):\[ k \int_0^\pi \frac{\theta^5}{5} \, d\theta = \frac{k}{5} \cdot \frac{\pi^6}{6} \].

Step 7: Simplify the Result

Simplify the expression to get the polar moment of inertia:\[ I_0 = \frac{k \pi^6}{30} \].

Key Concepts

Lamina Density

Understanding the concept of lamina density is essential when working with solid objects in physics. A lamina is a two-dimensional plate or sheet, having mass distributed over an area. In this problem, the density is not fixed but varies based on another factor — specifically, its distance from a reference point, known as the pole.

The density function for this problem is expressed as \( \rho(r, \theta) = k \cdot r \), where \( k \) is a constant, and \( r \) is the radial distance from the pole to the point \( (r, \theta) \) on the lamina. This means that the density of the material increases as you move further away from the pole.

Identifying how density varies over a lamina is crucial for calculating moments of inertia, which have many practical applications in real-world physics and engineering fields.

Integral Calculus

Integral calculus is the branch of mathematics that deals with finding the quantity where the basic method is to add up small parts over a defined region. In the context of this problem, it relates to finding the polar moment of inertia through integration.

The moment of inertia involves integrating the radial distance squared, multiplied by the density function across the region of interest. The integral to be computed is \[ I_0 = \iint_R r^3 k \, dA \] which simplifies when the proper limits and functions are applied.

Integration proceeds in steps: first with respect to \( r \), ensuring the function \( r^4 \) is calculated from \( 0 \) to the upper limit \( \theta \), resulting in \( \frac{\theta^5}{5} \). Then, integrate with respect to \( \theta \), evaluating from \( 0 \) to \( \pi \), which leads to the final solution.

• Integrate first with \( r \) within the range of \( 0 \) to \( \theta \).
• Second, carry out the integration with \( \theta \) from \( 0 \) to \( \pi \).

Mastering integration steps effectively is vital for solving problems across various mathematical and physical applications.

Polar Coordinates

Polar coordinates provide a way of describing a point's position in a plane using a radius and an angle. Unlike the Cartesian coordinate system, which uses \( x \) and \( y \) coordinates, polar coordinates offer a different perspective particularly suited to circular or symmetrical problems.

In this exercise, the lamina's boundary is set by \( r = \theta \) between \( 0 \leq \theta \leq \pi \). This condition defines a spiral-like shape where the radius increases linearly with the angle, creating a region to be evaluated.

• Radial distance \( r \) represents how far from the origin the point is located.
• Angle \( \theta \) denotes the counter-clockwise angle measured from the positive x-axis.

Polar coordinates are especially useful in scenarios involving circular or symmetric shapes, as they simplify expressions and calculations of areas, regions, and other properties essential in mathematical physics.

Mathematical Physics

Mathematical physics applies mathematical methods to solve problems in physics. It often involves complex calculations, such as those for moments of inertia, that require an understanding of both physical and mathematical principles.

The polar moment of inertia is a key concept in engineering and physics. It describes how a body resists twisting or rotational motion about a specific point or axis. This problem exemplifies how mathematical physics operates to provide insights into physical phenomena, using tools from calculus and geometry.

Calculating the polar moment of inertia involves understanding and applying

• how mass and density distribute over planes,
• the application of calculus for spatial calculations,
• and the functionality of different coordinate systems like polar coordinates.

These calculations serve critical roles such as determining the stability and performance of structures subjected to rotational forces. It helps students appreciate how mathematical ideals translate into tangible solutions in the real world.