Problem 13 - Basic Technical Mathematics with Calculus

Question

Find the indicated moment of inertia or radius of gyration. Find the radius of gyration of a plate covering the region bounded by \(y=x^{2}, x=3,\) and the \(x\) -axis with respect to the \(x\) -axis.

Step-by-Step Solution

Answer

The radius of gyration is approximately 2.323.

Step 1: Understand the Problem

We need to find the radius of gyration, denoted as \( k_x \), with respect to the \( x \)-axis for a given region. The region is bounded by the parabola \( y = x^2 \), the line \( x = 3 \), and the \( x \)-axis (\( y = 0 \)).

Step 2: Define the Moment of Inertia Formula

The moment of inertia \( I_x \) about the \( x \)-axis for an area \( A \) is given by:\[ I_x = \int (y^2) \, dA \]For our region, the differential area element \( dA \) can be considered as the strip \( y \, dx \), hence \( dA = y \, dx = x^2 \, dx \).

Step 3: Set Up Integral for Moment of Inertia

Substitute \( dA = x^2 \, dx \) into the formula:\[ I_x = \int_0^3 (x^2) \, (x^2) \, dx = \int_0^3 x^4 \, dx \]

Step 4: Compute the Integral for Moment of Inertia

Calculate the integral:\[ I_x = \int_0^3 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^3 = \frac{243}{5} \]

Step 5: Find the Area of the Region

The area \( A \) under the curve \( y = x^2 \) from \( x = 0 \) to \( x = 3 \) is given by:\[ A = \int_0^3 x^2 \, dx \]Calculate the integral:\[ A = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} = 9 \]

Step 6: Apply Formula for Radius of Gyration

The radius of gyration \( k_x \) with respect to the \( x \)-axis is:\[ k_x = \sqrt{\frac{I_x}{A}} = \sqrt{\frac{\frac{243}{5}}{9}} = \sqrt{\frac{243}{45}} = \sqrt{\frac{27}{5}} \]

Step 7: Simplify the Radius of Gyration Expression

Calculate the numerical value:\[ k_x = \sqrt{5.4} \approx 2.323 \]

Key Concepts

Moment of Inertia

Moment of inertia is a fundamental concept in physics and engineering that describes how mass is distributed in relation to an axis of rotation. It is crucial in problems involving rotational dynamics. To visualize this, imagine a seesaw, where the distribution of weight affects how easily it can turn. The formula for the moment of inertia about the x-axis, in terms of area A, is given by:

• \( I_x = \int y^2 \, dA \).

This formula implies that every little piece of area contributes to the total moment, depending on its distance from the axis of rotation. For our problem, where the region is defined by a parabola and straight borders, we derive \( I_x \) by integrating over the entire region, taking advantage of symmetry and the known function that describes the curve.

Parabola

A parabola is a symmetric curve defined as the set of all points that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix. In algebraic terms, a simple parabola aligned with the x-axis is represented by the equation:

• \( y = ax^2 \).

In our exercise, the parabola is given by \( y = x^2 \). This particular parabolic equation represents a curve opening upwards, with the x-axis as the line of symmetry. The primary interest is the region beneath this curve from \( x = 0 \) to \( x = 3 \). Understanding the nature of the parabola is essential for determining the limits and setup of the integrals used in calculus operations for computing both area and moment of inertia.

Integral Calculus

Integral calculus deals with the calculation of the area, volume, and other quantities under a curve that is defined mathematically. In our context, we are focused on finding both the moment of inertia and the total area bounded by the curve (parabola), the line, and the axis. Integrals help in summing infinitely many small quantities:

• The integral for the moment of inertia in our problem is \( I_x = \int_0^3 x^4 \, dx \), which simplifies to \( \frac{243}{5} \). These calculations provide not just a numerical value, but a deeper understanding of the distribution of the ‘mass’ of the surface in question.
• Separately, you also have the integral for the area \( A = \int_0^3 x^2 \, dx \), which results in the value \( 9 \).

Each integral handles a different aspect of object properties important for mechanical design and analysis.

Area Under a Curve

The area under a curve is a vital concept in calculus, representing a fundamental way to quantify the space that a function encloses with respect to an axis over a certain interval. It's essentially the sum of the y-values (function values) over each small slice of the x-interval. For the parabola \( y = x^2 \) from \( x = 0 \) to \( x = 3 \), we calculate the area as follows:

• Defined mathematically by the integral, \( A = \int_0^3 x^2 \, dx \), meaning we are summing up all the vertical slices from the x-axis up to the curve.
• The calculated area is \( 9 \). This operation involves breaking the area into infinitely small rectangles, summing them, then solving for the definite integral.

Knowing how to find the area under a curve is essential not just in mathematics, but also in fields like physics and engineering, where understanding the behavior of different functions over intervals can influence design and functional constraints.