The following notation is used in the problems: \(M=\) mass, \(\bar{x}, \bar{y}, \bar{z}=\) coordinates of center of mass (or centroid if the density is constant), \(I=\) moment of inertia (about axis stated), \(I_{x}, I_{y}, I_{z}=\) moments of inertia about \(x, y, z\) axes, \(I_{m}=\) moment of inertia (about axis stated) through the center of mass. Note: It is customary to give answers for \(I, I_{m}, I_{x},\) etc., as multiples of \(M\) (for example, \(I=\frac{1}{3} M l^{2}\) ). A thin rod \(10 \mathrm{ft}\) long has a density which varies uniformly from 4 to \(24 \mathrm{lb} / \mathrm{ft}\). Find (a) \(M\), (b) \(\bar{x}\), (c) \(I_{m}\) about an axis perpendicular to the rod, (d) \(I\) about an axis perpendicular to the rod passing through the heavy end.

The moment of inertia, often denoted as \(I\), is a measure of an object's resistance to changes in its rotation about a particular axis. It's crucial in understanding rotational dynamics. Just like mass measures an object's resistance to linear motion, moment of inertia measures resistance to rotational motion. The larger the moment of inertia, the harder it is to rotate the object.
For a thin rod, the moment of inertia about an axis perpendicular to the rod and passing through its center is calculated differently from an axis at one end. Using the center of mass simplifies integrals and calculations.
The moment of inertia when calculated around the center of mass often utilizes integrals involving the density function and distance from the center of mass. The formula used is:
\[ I_{m} = \int_{0}^{10} \rho(x) (x - \bar{x})^2 \, dx \]
Here, \(\rho(x)\) is the linear density function, and \(x - \bar{x}\) is the distance from the center of mass. Expansion and integration of this polynomial will yield the exact value of \(I_m\).

A linear density function, denoted as \(\rho(x)\), represents how mass is distributed along a one-dimensional object like a rod. For this exercise, the density changes uniformly from 4 lb/ft to 24 lb/ft over a length of 10 feet.
The density function can be described as:
\[ \rho(x) = 4 + 2x \]
where \(x\) is the distance from the lighter end. This means the farther from the light end, the denser the rod.
By integrating this density function over the length of the rod, one can find the total mass \(M\) of the rod. This integration is key to solving for mass and other properties related to mass distribution.

Integrating the density function helps find the total mass of an object when density is not uniform. The mass integration for this rod is given by:
\[ M = \int_{0}^{10} (4 + 2x) \, dx \]
This integral sums up all the tiny segments of mass along the rod, ultimately yielding the total mass.
By solving the integral, we get:
\[ M = [4x + x^2]_{0}^{10} = 4(10) + 10^2 = 40 + 100 = 140 \text{ lb} \]
Thus, the rod's total mass is 140 lb. This foundational step is critical for further calculations, such as finding the center of mass and moments of inertia.

The parallel axis theorem is a vital concept when calculating the moment of inertia about an axis that is parallel to one passing through the center of mass but not coinciding with it. The theorem states:
\[ I = I_m + M \bar{x}^2 \]
where:

• \(I\) is the moment of inertia about the new axis.
• \(I_m\) is the moment of inertia about the center of mass axis.
• \(M\) is the total mass.
• \(\bar{x}\) is the distance between the two axes.

This theorem allows one to easily switch from the center of mass axis to another parallel one. For instance, for the rod problem, moving from the center of mass to an axis at the heavy end involves:
\[ I = I_m + 140(6.19)^2 \]
By using this theorem, complex adjustments are simplified and calculations are more manageable.