Problem 9 - Basic Technical Mathematics with Calculus

Question

Find the indicated moment of inertia or radius of gyration. Find the moment of inertia of a plate covering the first-quadrant region bounded by \(y^{2}=x, x=9,\) and the \(x\) -axis with respect to the \(x\) -axis.

Step-by-Step Solution

Answer

The moment of inertia is \( \frac{486}{5} \).

Step 1: Identify the Region

The given region is in the first quadrant of the coordinate system. We have the curve defined by the equation \( y^2 = x \), and it is bounded by \( x = 9 \) and the \( x \)-axis. The region is essentially the area under the curve \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 9 \).

Step 2: Set Up the Integral for Moment of Inertia

The formula for the moment of inertia \( I \) about the \( x \)-axis is given by \( I = \int y^2 \, dm \). For a continuous region, this is expressed as \( I = \int (y^2) y \, dx \), because \( y \) is the distance from the \( x \)-axis and \( y \, dx \) represents an infinitesimal mass element, assuming constant density which can be factored out later. Substituting \( y = \sqrt{x} \), the integral becomes \[ I = \int_{0}^{9} ((\sqrt{x})^2)(\sqrt{x}) \, dx = \int_{0}^{9} x^{3/2} \, dx. \]

Step 3: Solve the Integral

Compute the definite integral \( \int_{0}^{9} x^{3/2} \, dx \) as follows.1. Find the antiderivative of \( x^{3/2} \), which is \( \frac{2}{5}x^{5/2} \).2. Evaluate the antiderivative from 0 to 9:\[ \frac{2}{5}(9^{5/2}) - \frac{2}{5}(0^{5/2}) = \frac{2}{5}(243) = \frac{486}{5}. \]

Step 4: Conclusion

The moment of inertia \( I \) of the plate with respect to the \( x \)-axis is found to be \( \frac{486}{5} \). This value assumes a unit density; any specific density can be multiplied with this result if needed for another application.

Key Concepts

Understanding Integral Calculus in Moment of Inertia Calculations

Integral calculus plays a pivotal role in dealing with continuous quantities, such as areas or masses distributed over a plane. In our exercise, we use integral calculus to find the moment of inertia. The moment of inertia about the x-axis is found by integrating a function of mass, taking into account how far the mass is distributed from the axis. In simpler terms, it quantifies how much of the mass is spread out along a particular axis.

The specific integration formula used here is \( I = \int y^2 \, dm \), where each \( y^2 \) term represents how far a point is from the axis, squared. Each small piece of area or mass \( dm \) contributes to the total moment of inertia.

• First, identify the limits of integration, which define the range over which we sum up small contributions to the total inertia.
• The quantity \( y = \sqrt{x} \) is substituted, simplifying the integral to \( I = \int_{0}^{9} x^{3/2} \, dx \).
• Finally, calculating this integral provides the moment of inertia for the defined region.

Exploring the Radius of Gyration

The radius of gyration provides a way to visualize how mass is distributed relative to an axis of rotation. It is defined as the distance from the axis at which the entire area or mass could be concentrated, resulting in the same moment of inertia.

To calculate it, we use the relationship \( k = \sqrt{\frac{I}{A}} \), where \( I \) is the moment of inertia, and \( A \) is the area of the plane region. This essentially gives us a way to condense the geometry of an object's surface area into a single, understandable quantity.

• Here, one should first calculate the moment of inertia \( I \) which was found to be \( \frac{486}{5} \) in our exercise.
• The area of the shape under consideration must be determined, here it would be \( \int_{0}^{9} \sqrt{x} \, dx \).
• Finally, by solving \( k = \sqrt{\frac{I}{A}} \), we get the radius which provides a useful measure of how mass or area distribution affects rotational properties.

Applying Coordinate Geometry to Define Regions

Coordinate geometry is essential for effectively defining regions over which we can perform calculus operations. It helps in visualizing the defined boundaries and regions on a plane, facilitating the application of integral calculus.

In the exercise provided, coordinate geometry helps us define the specific area of interest clearly:

• The problem specifies the curve \( y^2 = x \), which allows us to understand the shape under consideration.
• Boundary definitions like \( x = 9 \) and the x-axis itself delimit the region over which the calculation occurs.
• By plotting the function, it becomes evident what area we're dealing with, highlighting the importance of a visual understanding of mathematical concepts.

Ultimately, such a graphical understanding aids in setting up the necessary integrals and ensures computations are correctly aligned with the defined physical region.