Problem 16 - 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 first-quadrant region bounded by \(x=1, y=2-x,\) and the \(y\) -axis with respect to the \(y\) -axis.

Step-by-Step Solution

Answer

The radius of gyration is approximately 0.527 units.

Step 1: Identify the Boundaries and Set Up the Integral

The region covers the first quadrant bounded by the lines \(x=1\), \(y=2-x\), and the \(y\)-axis. The boundaries in terms of \(x\) are from \(x=0\) to \(x=1\), and in terms of \(y\), they are from \(y=0\) to \(y=2-x\). This forms a triangular region. We will use integration to find the mass \(m\) and moment of inertia \(I\) to determine the radius of gyration \(k\).

Step 2: Determine the Mass of the Plate

Assume unit density for simplicity. Compute the mass \(m\) using the integral \[ m = \int_{0}^{1} \int_{0}^{2-x} \, dy \, dx. \] Perform the integration over \(y\) first, then \(x\):\[ m = \int_{0}^{1} [y]_{y=0}^{y=2-x} \, dx = \int_{0}^{1} (2-x) \, dx. \] Solving gives \[ m = [2x - \frac{x^2}{2}]_{0}^{1} = 2(1) - \frac{1^2}{2} - (2(0) - \frac{0^2}{2}) = 2 - \frac{1}{2} = \frac{3}{2}. \]

Step 3: Calculate the Moment of Inertia about the y-Axis

To find the moment of inertia \(I\) about the \(y\)-axis, use:\[ I = \int_{0}^{1} \int_{0}^{2-x} x^2 \, dy \, dx. \] Again, integrate over \(y\) first:\[ I = \int_{0}^{1} [x^2 y]_{y=0}^{y=2-x} \, dx = \int_{0}^{1} x^2(2-x) \, dx. \] Expanding and integrating:\[ I = \int_{0}^{1} (2x^2 - x^3) \, dx = [\frac{2x^3}{3} - \frac{x^4}{4}]_{0}^{1} = (\frac{2(1)^3}{3} - \frac{1^4}{4}) - (0) = \frac{2}{3} - \frac{1}{4} = \frac{8}{12} - \frac{3}{12} = \frac{5}{12}. \]

Step 4: Compute the Radius of Gyration

The radius of gyration \(k\) is given by the formula\[ k = \sqrt{\frac{I}{m}}. \]Substitute \( I = \frac{5}{12} \) and \( m = \frac{3}{2} \):\[ k = \sqrt{\frac{\frac{5}{12}}{\frac{3}{2}}} = \sqrt{\frac{5}{12} \times \frac{2}{3}} = \sqrt{\frac{5}{18}}. \]Compute the numerical value to get\[ k \approx \sqrt{\frac{5}{18}} \approx 0.527. \]

Step 5: Conclusion

The radius of gyration of the plate with respect to the \(y\)-axis is approximately 0.527 units.

Key Concepts

Understanding Moment of Inertia

The moment of inertia is a crucial concept in physics and engineering that measures an object's resistance to rotation about an axis. It depends on the distribution of an object's mass relative to the axis. In mathematical terms, it's an integral that calculates the sum of all tinny mass elements each multiplied by the square of its distance from the axis. This is why the formula for moment of inertia can seem complex, because it takes into account these distances and masses:

• For a continuous mass distribution, like in our exercise, the moment of inertia, denoted as \( I \), is calculated using definite integrals.
• For example, in our solved exercise, we see how we apply \( I = \int x^2 \, dm \) for rotation about the \(y\)-axis. We express the mass element \( dm \) in terms of \( dx \) or \( dy \) as suitable for our boundaries.

These calculations require knowledge of integral calculus, as they involve integrating both with respect to volume or area and specific boundaries, depending on the shape and position of the object.

The Role of Integral Calculus

Integral calculus is a powerful part of mathematics that helps us find quantities like area, volume, and, in physics, things like center of mass and moment of inertia. It's all about summing tiny, infinitely small parts to find a whole.

• In the exercise, integral calculus was used to find the mass \( m \) and moment of inertia \( I \) of a region. We did this by setting up double integrals, which handle shapes defined in two dimensions.
• These steps break the problem into manageable pieces: integrating first with respect to \( y \), then with respect to \( x \), given the triangular region's boundaries.
• Ultimately, through integration, we calculated the radius of gyration, \( k \), a derived measure of distribution of mass relative to an axis, by dividing \( I \) by \( m \), and taking the square root.

Understanding the process of setting up and solving integrals is key to solving problems involving physical bodies and their properties.

Exploring Plane Regions

Plane regions, in the context of geometry and calculus, refer to areas on a flat surface defined by boundaries. They are integral to understanding shapes and configurations in mathematics and physics.

• In our specific problem, the plane region is bounded by the lines \( x = 1 \), \( y = 2-x \), and the \( y\)-axis. This forms a right triangle, crucial for our calculations.
• Understanding these boundaries is important for setting up the problem. They determine the limits of our integrals in finding \( m \) and \( I \).
• Recognizing the shape as a triangle simplifies the solution process. It lays the groundwork for using appropriate mathematical tools such as double integrals.

By identifying and analyzing plane regions, we can systematically solve physics and engineering problems, ensuring each step builds logically on the last.