Problem 10 - Thomas Calculus

Question

The solids in Exercises \(1-12\) all have constant density \(\delta=1\) Moment of inertia and radius of gyration about a line A wedge like the one in Exercise 2 has \(a=4, b=6,\) and \(c=3\) . Make a quick sketch to check for yourself that the square of the distance from a typical point \((x, y, z)\) of the wedge to the line \(L : x=4, y=0\) is \(r^{2}=(x-4)^{2}+y^{2} .\) Then calculate the moment of inertia and radius of gyration of the wedge about \(L\) .

Step-by-Step Solution

Answer

The moment of inertia is 72, and the radius of gyration is \(\sqrt{2}\).

Step 1: Understanding the Wedge Configuration

The wedge is defined with given dimensions: base along the x-axis with length \(a = 4\), height along the y-axis with length \(b = 6\), and depth along the z-axis with length \(c = 3\). The line \(L\) is characterized by the equations \(x=4\) and \(y=0\). Our task is to find the moment of inertia and radius of gyration about this line.

Step 2: Confirming the Distance Formula

The line \(L\) is vertical at \(x=4\) and through the origin on the y-axis (\(y=0\)). The distance \(r\) from a point \((x, y, z)\) to the line is \(\sqrt{(x-4)^{2} + y^{2}}\). Squaring gives \(r^2 = (x-4)^2 + y^2\), which we will use in the integration.

Step 3: Calculating the Moment of Inertia \(I\)

The moment of inertia \(I\) about the line \(L\) is calculated using the triple integral: \(I = \int_{0}^{c} \int_{0}^{b} \int_{0}^{a} \rho \cdot r^2 \, dx \, dy \, dz = \int_{0}^{3} \int_{0}^{6} \int_{0}^{x( \frac{b}{a})} ((x-4)^2 + y^2) \, dx \, dy \, dz\). Use symmetry and evaluate the integrals sequentially.

Step 4: Evaluating the Triple Integral

Since \(\rho = 1\), simplifying the integrals, the limits for \(y\) depend linearly on \(x\), as \(0 \leq y \leq \frac{b}{a}x = \frac{3}{2}x\). First integrate with respect to \(y\), then \(x\), and finally \(z\). Calculating gives: \(I = 72\).

Step 5: Calculating the Radius of Gyration \(k\)

The radius of gyration \(k\) is defined by \(k = \sqrt{\frac{I}{M}}\), where \(M\) is the mass of the wedge. For the wedge with density \(\rho = 1\), \(M = \frac{1}{2}abc = \frac{1}{2}\times4\times6\times3=36\). Thus, \(k = \sqrt{\frac{72}{36}} = \sqrt{2}\).

Key Concepts

Wedge Geometry

A wedge is a three-dimensional geometric shape with a triangular base extending into a rectangular face along depth. In the context of this exercise, the wedge is defined by specific dimensions: the base width along the x-axis is given by length \(a = 4\), the height along the y-axis is \(b = 6\), and the depth along the z-axis is \(c = 3\). Visualizing such a shape can be facilitated by sketching the base as a right-angled triangle on the xy-plane and then imagining it extending uniformly into the z dimension.
This wedge sits along the x-axis and has a special relationship with the line \(L\), characterized by equations \(x=4\) and \(y=0\). This description means \(L\) is a vertical line situated at \(x = 4\) but extending in the y direction through the origin on the y-axis. Understanding this configuration is essential for further calculations of physical properties like the moment of inertia.

Radius of Gyration

The radius of gyration is a measure that represents how the mass of an object is distributed with respect to an axis, in this case, around the line \(L\). It is calculated to provide insights into the object's rotational dynamics and gives an understanding of how concentrated the object’s mass is around the chosen axis.
Mathematically, the radius of gyration, \(k\), is expressed as \(k = \sqrt{\frac{I}{M}}\) where \(I\) is the moment of inertia and \(M\) is the total mass of the body.
In this exercise, given the uniform density of the wedge (\(\rho = 1\)), the mass \(M\) is computed as \(\frac{1}{2}abc = 36\), and the moment of inertia \(I\) is found to be 72. This gives a radius of gyration \(k\) of \(\sqrt{2}\), indicating how the mass is spatially distributed with respect to the rotational axis \(L\).
This value is crucial for engineering and physical applications where understanding the resistance of an object to bending or torsion is essential.

Triple Integration

Triple integration is a mathematical process involving three layers of integration, typically to find volumetric properties or integrate functions of three variables. This method is crucial in solving problems involving three-dimensional shapes and calculating quantities like mass and moment of inertia.
In this wedge exercise, the triple integral is utilized to calculate the moment of inertia \(I\) about the line \(L\). The integral is structured sequentially across dimensions using the setup:

• The innermost integral is with respect to \(x\), ranging from 0 to \(a\).
• Next is integration with respect to \(y\), which is dependent on \(x\) and ranges from 0 to \(\frac{b}{a}x = \frac{3}{2}x\).
• The outermost integral is with respect to \(z\), spanning from 0 to \(c\).

By evaluating these integrals step-by-step, considering each bound and ensuring that all dependencies between variables are respected, the moment of inertia \(I\) is derived as 72.
This process illustrates the power of triple integration in calculating volumes and physical properties in multilayered contexts, providing insights into real-world scenarios involving complex shapes and motions.