Question

Moment of Inertia In Exercises 37 and \(38,\) find \(I_{z}\) for the given lamina with a uniform density of 1. Use a computer algebra system to verify your results. $$ x^{2}+y^{2}=a^{2}, \quad 0 \leq z \leq h $$

Short answer

The moment of inertia \(I_z\) for the given lamina is \(\frac{1}{2} \pi a^{4} h\).

Step-by-step solution

Change to Polar Coordinates

We can simplify the integral by changing from Cartesian to polar coordinates. In polar coordinates the equation of the lamina becomes \(r^{2} = a^{2}\), which is a circle of radius \(a\). Now, \(x^{2}+y^{2}\) can be represented as \(r^2\) and \(dm\) as \(\rho\ dr\ d\theta\ dz\), where \(\rho\) is the density and given as 1 in this case.

Setup the Integral for the Moment of Inertia

The moment of inertia \(I_z\) is the integral of \(r^{2} dm\) over the whole lamina. In this case we can set up the integral as \(I_z = \int_0^h \int_0^{2\pi} \int_0^a r^{3} dr d\theta dz\).

Evaluate the Integral

Evaluate the integral from inside out. First integrate with respect to \(r\), then \(\theta\), and finally \(z\). After evaluating the integrals successively you will find the moment of inertia \(I_z = \frac{1}{2} \pi a^{4} h\).

Key concepts

Polar Coordinates

When studying the moment of inertia for various shapes, we often encounter complex integrals that are difficult to solve using Cartesian coordinates (x, y). That's where polar coordinates come into play. They provide a simpler alternative for dealing with circular symmetry.

In polar coordinates, any point in the plane is represented by two values: the radius r, which is the distance from the origin to the point, and the angle θ (theta), which is measured from the positive x-axis to the line segment connecting the origin to the point. With this system, circular and radial shapes become much easier to describe.

For example, the equation of a circle with radius a centered at the origin in Cartesian coordinates, x^2 + y^2 = a^2, becomes simply r = a in polar coordinates. This representation removes the two-dimensional constraint and allows us to focus on the radial component for calculations, greatly simplifying the process of finding integrals like the moment of inertia.

Integral Calculus

Integral calculus is fundamental in physics and engineering, particularly when determining various properties of shapes and objects, such as the moment of inertia. In the context of the exercise, we use integral calculus to sum up infinitesimal contributions of mass elements to the moment of inertia.

The moment of inertia, I_z, involves integrating the square of the distance from each mass element to the axis of rotation. In calculus, this is expressed as a definite integral, where each part of the lamina is considered in the integration process. When dealing with objects of uniform density, as in this exercise, the integration becomes more manageable as ρ remains constant.

Using polar coordinates simplifies the integral setup for the moment of inertia, as it allows for a one-dimensional radial integral along with the integration over the angle θ and the height z. Thus, integral calculus helps us to solve for I_z by systematically integrating over the dimensions of the lamina.

Lamina Density

In physics, lamina density is a measure of mass per unit area for a thin, flat layer or a 'lamina.' The concept is vital when evaluating the mass distribution of an object and affects calculations such as center of mass and the moment of inertia. In our exercise, the lamina is assumed to have a uniform density of 1, which implies that the mass is spread evenly across the surface of the lamina.

This uniform density simplifies calculations because we don't need to account for varying mass distributions across the lamina. The mass element, dm, for a small segment of the lamina is described as a product of the density, ρ, and the infinitesimal area element, which, in polar coordinates, is ρ (r dr dθ).

Thus, by integrating these infinitesimally small mass elements over the entire volume of the lamina, we get the total mass, and by integrating r² dm across the lamina, which involves the square of the distance from the axis, we find the moment of inertia. It's fascinating how these concepts interconnect to allow for the precise calculation of such an important physical property.