Question

The moment of inertia of a continuous mass distribution with density \(\varrho\) is obtained by converting the sum of (10.25) into the volume integral \(\int \rho^{2} d m=\int \rho^{2} \varrho d V\). (Note the two forms of the Greek "rho": \(\rho=\) distance from \(z\) axis, \(\varrho=\) mass density.) Find the moment of inertia of a uniform circular cylinder of radius \(R\) and mass \(M\) for rotation about its axis. Explain why the products of inertia are zero.

Short answer

The moment of inertia is \( \frac{MR^2}{2} \); products of inertia are zero due to symmetry.

Step-by-step solution

Understand the Problem

We need to find the moment of inertia of a uniform circular cylinder of radius \( R \) and mass \( M \) about its axis of symmetry. This requires integrating the mass elements over the volume of the cylinder. Additionally, we'll explain why the products of inertia are zero for rotation about this axis.

Volume and Mass Density Relations

For a uniform cylinder with mass \( M \) and volume \( V \) (with height \( h \)), the mass density \( \varrho \) is given by \( \varrho = \frac{M}{V} = \frac{M}{\pi R^2 h} \). The volume differential element is \( dV = \rho \, d\rho \, d\theta \, dz \) in cylindrical coordinates.

Integrate to Find the Moment of Inertia

The moment of inertia \( I \) is calculated by the integral \( \int \rho^2 \varrho \, dV \), which becomes \[ I = \int_{0}^{h} \int_{0}^{2\pi} \int_{0}^{R} \rho^2 \left( \frac{M}{\pi R^2 h} \right) \rho \, d\rho \, d\theta \, dz. \] Simplifying the expression, we focus on integrating with limits for \( \rho, \theta, \) and \( z \).

Evaluate the Integral

First, integrate with respect to \( \rho \): \[ \int_{0}^{R} \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_{0}^{R} = \frac{R^4}{4}. \]Then integrate with respect to \( \theta \): \[ \int_{0}^{2\pi} d\theta = 2\pi. \]Finally, integrate with respect to \( z \): \[ \int_{0}^{h} dz = h. \]Combine these results to find the moment of inertia.

Combine Results

Substitute the integration results back into the equation for \( I \): \[ I = \left( \frac{M}{\pi R^2 h} \right) \left( \frac{R^4}{4} \right) (2\pi) (h) = \frac{MR^2}{2}. \] This is the moment of inertia of the cylinder about its axis.

Explain Why Products of Inertia Are Zero

For a cylinder rotating about its symmetry axis, any cross-products involving axes perpendicular to the axis of rotation are zero because the mass distribution is symmetric about the axis. The products of inertia, which represent the coupling between different axes, are zero due to this symmetry.

Key concepts

Cylindrical Coordinates

Understanding cylindrical coordinates is essential to solving problems that involve circular or cylindrical shapes. Cylindrical coordinates form a three-dimensional coordinate system that functions like an extension of polar coordinates. In this system, a point in space is represented using three values:

• \( \rho \), the radial distance from the axis of rotation, often represented as distance from the \( z \)-axis.
• \( \theta \), the angular coordinate which defines the direction around the cylinder, analogous to longitude on Earth.
• \( z \), the height above or below the reference plane.

Cylindrical coordinates are particularly useful in problems involving cylinders and circular paths because they align naturally with the symmetry of these shapes. This coordinate system simplifies the integration process by transforming three-dimensional volumes into relatively simple geometrical elements in terms of \( \rho \), \( \theta \), and \( z \). Breaking down the volume integral into these components allows for easier computation of quantities like the moment of inertia.

Mass Density

Mass density, denoted by \( \varrho \), measures how much mass is contained within a certain volume. It's a crucial concept when dealing with continuous mass distributions, such as the uniform cylinder in the exercise. For a uniform object, the mass density is constant throughout.

• For the cylinder, mass density \( \varrho \) is calculated as \( \varrho = \frac{M}{V} \), where \( M \) is the total mass and \( V \) is the total volume.
• The volume of a cylinder is given by \( V = \pi R^2 h \), where \( h \) is the height and \( R \) is the radius.

Thus, the mass density of a uniform cylinder is expressed as \( \varrho = \frac{M}{\pi R^2 h} \). Understanding mass density allows us to distribute the mass correctly across the volume in calculations, which is essential for finding physical quantities like the moment of inertia.

Volume Integral

The volume integral is a mathematical tool used to sum infinitesimal volume elements over a specified region. It is represented by \( \int dV \) and incorporates density (\( \varrho \)) and geometric parameters (\( \rho \), \( \theta \), \( z \)). For the cylinder, the volume element \( dV \) in cylindrical coordinates is given by \( dV = \rho \, d\rho \, d\theta \, dz \).When calculating the moment of inertia \( I \), the volume integral becomes \( \int \rho^2 \varrho \, dV \). This expression breaks into three integrals:

• \( \int_{0}^{R} \rho^3 \, d\rho \) integrates the radial volume component.
• \( \int_{0}^{2\pi} d\theta \) sums over the angular part, covering the entire circumference.
• \( \int_{0}^{h} dz \) evaluates along the height of the cylinder.

Performing these integrations step-by-step yields the final expression for the moment of inertia for the full volume. Volume integrals allow us to evaluate complex distributions by summing over tiny pieces.

Uniform Cylinder

A uniform cylinder is characterized by having the same mass density \( \varrho \) throughout its entire volume. This uniformity simplifies calculations, making analytical evaluations achievable. Since every cross-section along the height (\( h \)) is identical in a uniform cylinder:

• The density \( \varrho = \frac{M}{\pi R^2 h} \) remains constant.
• The moment of inertia calculation process becomes straightforward due to this uniform distribution.
• The symmetry ensures no cross-products of inertia due to any skewed mass, as such components are inherently canceled out.

Uniform cylinders are commonly used in mechanical and engineering problems because their symmetrical properties allow for easy modeling of rotation and inertia. The predictable nature aids in designing structures and systems that rely on rotational dynamics. For this reason, understanding and computing the moment of inertia for uniform cylinders is vital for applications ranging from mechanical design to theoretical physics.