Question

Moment of Inertia In Exercises 35 and \(36,\) use the following formulas for the moments of inertia about the coordinate axes of a surface lamina of density \(\rho .\) \(I_{x}=\int_{S} \int\left(y^{2}+z^{2}\right) \rho(x, y, z) d S\) \(I_{y}=\int_{S} \int\left(x^{2}+z^{2}\right) \rho(x, y, z) d S\) \(I_{z}=\int_{S} \int\left(x^{2}+y^{2}\right) \rho(x, y, z) d S\) Verify that the moment of inertia of a spherical shell of uniform density about its diameter is \(\frac{2}{3} m a^{2},\) where \(m\) is the mass and \(a\) is the radius.

Short answer

The moment of inertia of a spherical shell of uniform density about its diameter is \(\frac{2}{3} m a^{2}\) where \(m\) is the mass and \(a\) is the radius. It verifies the claim made in the exercise.

Step-by-step solution

Understand the Sphere's Coordinate System

Consider a spherical coordinate system where the shell is centered at the origin. In this system, any element \(dS\) of mass \(dm=\rho dS\) is located at coordinates \((r, \theta, \phi)\) where r is the radius, \(\theta\) is the angle with the x-axis and \(\phi\) is the angle from the z-axis. Here \(dS = r^2 \sin \phi \, d\phi \, d\theta\). Also, remember that we need to use the moment of inertia formula \(I_{z}=\int_{S} \int\left(x^{2}+y^{2}\right) \rho(x, y, z) d S\) where \(\rho\) is the density which is constant for a uniform spherical shell.

Converting to Spherical Coordinates

In spherical coordinates, \(x^{2}+y^{2}=r^{2}\sin^{2}\phi\). Substituting this into the formula for \(I_{z}\) gives \(I_{z}=\rho\int_{S}\int r^{2}\sin^{2}\phi \, d S\).

Solving the Integral

Substituting the value of \(dS\) and integrating over the spherical shell, we get \(I_{z}=\rho\int_{0}^{2\pi}\int_{0}^{\pi} r^{4}\sin^{3}\phi \, d\phi \, d\theta\). The mass m of the sphere can be written as \(m=\rho \int_{0}^{2\pi}\int_{0}^{\pi} r^{2}\sin\phi \, d\phi \, d\theta\), so we can express the moment of inertia as \(I_{z}=\frac{2}{3}\int_{0}^{2\pi}\int_{0}^{\pi} r^{2}\sin\phi\, d\phi\,d\theta =\frac{2}{3}ma^{2}\) where \(a=r\) is the radius of the shell.

Key concepts

Spherical Coordinates

In the realm of geometry and calculus, spherical coordinates are akin to a beacon in a three-dimensional space, guiding us to the position of a point using a trio of numbers. These three coordinates are usually denoted as \(r, \theta, \phi\). Here, \(r\) represents the distance from the origin to the point, \(\theta\) is the angle in the xy-plane from the x-axis, and \(\phi\) is the angle from the positive z-axis to the point.

Imagine you're holding a globe. The 'r' is like the straight line from the center of the Earth to any point on the surface. The angle \(\theta\) can be seen as the longitude, defining the rotation around the globe, while \(\phi\) resembles the latitude, only it measures the angle from the North Pole instead of the equator. In the context of our exercise, these coordinates simplify the process of calculating the moment of inertia for objects like spherical shells by focusing our calculations to the interior structure of the sphere.

A Practical Illustration

In the study case of a spherical shell, we're examining a hollow ball. By employing spherical coordinates, we can understand how to calculate the properties of this shell by peering into its 'invisible' axes, as the shell is not flat like the lamina nor linear like a line or rod. The elegance of this system shines when we integrate over a spherical surface, since it allows us to account for the curvature of the sphere with ease.

Surface Lamina Density

When it comes to objects that have only two dimensions in terms of their mass distribution, like a sheet of paper or the shell of a hollow sphere, we describe this characteristic with surface lamina density, often symbolized by \(\rho\). This density measures the mass per area and is critical when deducing the moment of inertia for such objects. For a uniform spherical shell, this density is constant, which implies that any same-sized patch on the sphere's surface has the same mass.

The concept of surface lamina density transfigures a potentially confusing array of mass distributions into a single value that enables easier calculations. It allows us to use an integral to sum up these tiny patches of mass into a total mass, crucial to finding the moment of inertia, a measure of how a body resists rotational changes about an axis.

Fitting The Pieces Together

For our spherical shell example, since the surface lamina density \(\rho\) is constant, we can pull it out of the integral when calculating the moment of inertia. This makes solving the integral simpler and illustrates the harmony of using surface lamina density in unison with spherical coordinates while applying integral calculus.

Integral Calculus

The branch of mathematics known as integral calculus is a powerful tool for computing areas, volumes, and in cases such as ours, the moment of inertia. It allows us to sum an infinite number of infinitesimally small quantities to accumulate into a finite value. Imagine you're painting a masterpiece on a canvas, where each brush stroke is so tiny it's barely visible. Integral calculus helps to view the entire picture that emerges from these numerous strokes combined.

In our current exercise scenario, integral calculus is used to integrate the density function over the sphere's surface, taking into account every tiny element of area \(dS\) to find the overall moment of inertia. The formulae provided are like recipes telling us how to mix together these elements with the density to get our final result.

Connecting The Dots

By treating the surface of the sphere as a collection of infinitesimal elements \(dS\), and with the knowledge of spherical coordinates and surface lamina density, integral calculus empowers us to sum up these elements to derive the moment of inertia of the spherical shell. The integral transforms the abstract concept of a rotating body's inertia into a concrete numerical value that can be used for engineering and physics calculations.