Problem 12 - 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 solid like the one in Exercise 3 has \(a=4, b=2,\) and \(c=1\) . Make a quick sketch to check for yourself that the square of the distance between a typical point \((x, y, z)\) of the solid and the line \(L : x=4, y=0\) is \(r^{2}=(x-4)^{2}+y^{2} .\) Then find the moment of inertia and radius of gyration of the solid about \(L .\)

Step-by-Step Solution

Answer

The moment of inertia is 53.33, and the radius of gyration is approximately 2.58.

Step 1: Understand the Solid's Parameters

The solid in question has the dimensions given by \( a = 4 \), \( b = 2 \), and \( c = 1 \). This indicates that the solid can be visualized as a cuboid extending 4 units along the x-axis, 2 units along the y-axis, and 1 unit along the z-axis.

Step 2: Sketch and Determine Distance Formula

Sketch a 3D coordinate system and draw the solid. The line \(L : x = 4, y = 0\) is parallel to the z-axis and located at \((4, 0, z)\). The square of the distance from any point \((x, y, z)\) to the line \(L\) is calculated as \( r^2 = (x - 4)^2 + y^2 \).

Step 3: Set Up the Moment of Inertia Integral

For a solid with constant density \(\delta = 1\), the moment of inertia about line \(L\) is given by the integral \( I = \int \int \int_{V} r^2 \, dV \), where \(dV = dx\, dy\, dz\) is the volume element. Substitute \( r^2 = (x - 4)^2 + y^2 \) in the integrand.

Step 4: Define Limits of Integration

The volume of the solid is bound by \(0 \leq x \leq 4\), \(-1 \leq y \leq 1\), and \(-0.5 \leq z \leq 0.5\). These limits correspond to \(a\), \(b\), and \(c\) dimensions of the solid, centered at the origin in the y and z directions.

Step 5: Compute the Moment of Inertia

Evaluate the integral \( I = \int_{0}^{4} \int_{-1}^{1} \int_{-0.5}^{0.5} ((x - 4)^2 + y^2) \, dz \, dy \, dx \). This involves performing integration with respect to \(z\), \(y\), and finally \(x\).

Step 6: Calculate the Radius of Gyration

The radius of gyration \(k\) is defined as \(k = \sqrt{\frac{I}{M}}\), where \(M\) is the total mass of the solid given by \(\delta = 1\), hence \( M = \int \int \int_{V} 1\, dV \). Integrate over the volume to find \(M = 4 \times 2 \times 1 = 8\). To find \(k\), plug \(I\) and \(M\) into the formula.

Key Concepts

Radius of Gyration

When it comes to the moment of inertia of a solid object, one term you might come across is the "radius of gyration." This is a measure that helps understand how mass is distributed in the solid concerning an axis of rotation. Simply put, the radius of gyration is the distance from the axis at which the entire mass of the body could be concentrated without altering its rotational characteristics.
The formula for the radius of gyration, \( k \, is \, k = \sqrt{\frac{I}{M}} \, where \, I \, is the \, moment \, of \, inertia, \, and \, M \, is the \, mass \, of the \, solid. \)
For solids of constant density, calculating the radius of gyration gives insights into the object's stability while rotating. In exercises like the one at hand, the task involves not just understanding how the mass spreads over the volume but associating it with ease of rotation around a specified line.

Cuboid Solids

The discussed shape in the problem is a cuboid, which is a three-dimensional solid made up of six rectangular faces. In our example, it stretches over dimensions defined by lengths of 4 units, 2 units, and 1 unit along the x, y, and z axes, respectively.
A cuboid's geometry involves simple yet significant features. When dealing with calculations such as volume or moment of inertia, identifying these attributes becomes crucial. It holds particular importance in physical and engineering applications.
Each axis dimension dictates how the shape extends and is a fundamental factor when setting up limits for integration, volume calculation, or analyzing mass distribution, as seen in determining the moment of inertia and radius of gyration.

Integral Calculus

Integral calculus is integral, pun intended, to finding the moment of inertia for a continuous mass like a cuboid. When a body has constant density, as in our example, its mass is spread out evenly across its volume, making calculus the tool to sum an infinite number of infinitesimally small mass elements.
Using the integral \( I = \int \int \int_{V} r^2 \, dV \), you set up a triple integral to compute the moment of inertia. Each integral corresponds to one spatial dimension: x, y, or z. Through integration, the contribution of every small element, derived by \( dV = dx \, dy \, dz \), accumulated into the total rotational inertia with respect to a given line.
Integral calculus helps move from a simple sketch and setup of equations to a defined solution that provides insights into the dynamic behavior of the solid around specified lines of rotation.

Constant Density

Considering constant density adds simplicity to calculations as it assumes the distribution of mass is uniform throughout the object. In our exercise, the density \( \delta \) is given as 1, which streamlines the process since mass becomes equivalent to the volume of the solid.
This assumption transitions complex physical properties into manageable mathematical representations. It means the entire volume contributes equally to properties like the moment of inertia, without additional calculations to adjust for varying densities across the shape.
With constant density, the mass \( M \) of a cuboid is straightforward to determine by the product of its dimensions, i.e., \( M = 4 \times 2 \times 1 = 8 \). This provides a clear, calculative advantage, easing the integration process for kinetic and rotational studies.