Question

Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

Short answer

The moment of inertia of a uniform solid sphere about a diameter is \(\frac{2}{5} M R^2\).

Step-by-step solution

Understanding Moment of Inertia Definition

The moment of inertia, \(I\), for a continuous body is defined by the integral \(I = \int r^2 \, dm\), where \(r\) is the distance from the axis of rotation, and \(dm\) is a mass element of the body.

Expressing Mass Element in Terms of Density

For a uniform solid sphere, use density, \(\rho\), where \(\rho = \frac{M}{\frac{4}{3}\pi R^3}\). A small mass element, \(dm\), is expressed in terms of volume element \(dV\) as \(dm = \rho \, dV\).

Setting Up the Integral

Substitute the given volume element \(dV = r^2 \, dr \, \sin \theta \, d\theta \, d\phi\) into the equation \(dm = \rho \, dV\). The effective distance from the axis, being \(z\) axis, is \(r \sin \theta\).

Writing the Integral in Spherical Coordinates

Substitute \(dm\) and the expression for \(r'\) (effective \(r\)) into the inertia integral: \(I = \int_0^R \int_0^{\pi} \int_0^{2\pi} \rho \, (r \sin \theta)^2 \, r^2 \, dr \, \sin \theta \, d\theta \, d\phi\).

Simplifying the Triple Integral

In the integral, separate the components that depend on \(r\), \(\theta\), and \(\phi\). The integral becomes \(I = \rho \int_0^R r^4 \, dr \int_0^{\pi} \sin^3 \theta \, d\theta \int_0^{2\pi} d\phi\).

Solving Each Integral Separately

Calculate each integral:1. \(\int_0^R r^4 \, dr = \frac{R^5}{5}\)2. \(\int_0^{2\pi} d\phi = 2\pi\)3. \(\int_0^{\pi} \sin^3 \theta \, d\theta = \frac{4}{3}\) (using the identity and integration by parts, or look-up).

Combining Results and Final Calculation

Combine the results: \(I = \rho \left( \frac{R^5}{5} \right) \left( \frac{4}{3} \right) \left( 2\pi \right)\). Substitute \(\rho = \frac{M}{\frac{4}{3}\pi R^3}\) to get \(I = \frac{2}{5} MR^2\).

Key concepts

Spherical Coordinates

Spherical coordinates offer a powerful alternative to Cartesian coordinates, especially when dealing with problems involving spheres or circular symmetry. Instead of using the familiar \(x\), \(y\), and \(z\) coordinates, spherical coordinates use \(r\), \(\theta\), and \(\phi\):

• \(r\) is the radius from the origin to the point in space.
• \(\theta\) is the polar angle measured from the positive \(z\)-axis.
• \(\phi\) is the azimuthal angle in the \(xy\)-plane from the positive \(x\)-axis.

When calculating volumes or integrating over a sphere, spherical coordinates greatly simplify the math as they naturally align with the symmetry of the sphere. This coordinate system captures the geometry of spherical objects neatly, making it easier to calculate properties like volume or surface area. In this exercise, spherical coordinates are instrumental in calculating the moment of inertia for a sphere.

Solid Sphere

A solid sphere is a three-dimensional object where every point on the outer surface is at the same distance, known as the radius \(R\), from a fixed central point. The symmetry of a solid sphere makes it ideal to study in spherical coordinates.

Key characteristics include:

• Uniform mass distribution if the sphere is solid and homogeneous.
• The volume of the sphere is calculated using the formula \(V = \frac{4}{3}\pi R^3\), where \(R\) is the radius of the sphere.
• All diameters in the sphere pass through the center point.

For physics problems, such as calculating the moment of inertia, the solid sphere has relevant properties that can be represented in simpler mathematical forms when spherical coordinates are applied.

Density of a Sphere

Density \(\rho\) is an essential concept in understanding how mass is distributed in an object like a sphere. For a uniform solid sphere, density can be considered constant throughout the sphere.

The density is calculated by dividing the total mass \(M\) by the total volume \(V\) of the sphere:

• \(\rho = \frac{M}{\frac{4}{3}\pi R^3}\)

By knowing the density, we can express any small mass element \(dm\) within the sphere as the product of density and a small volume element \(dV\), making the integration in spherical coordinates more feasible. It simplifies the calculation of the moment of inertia and other dynamical properties by ensuring that the mass element calculations are consistent with the continuous uniform distribution of mass within the sphere.

Mass Element Integration

Mass element integration involves integrating over all the small pieces of mass \(dm\) that make up an object to calculate properties such as moment of inertia. In the case of a sphere, the element of mass \(dm\) is given by \(dm = \rho \, dV\), where \(dV\) is the infinitesimal volume element.

To find the moment of inertia \(I\) of a sphere, we use the integral:

• \(I = \int r^2 \, dm = \int r^2 \, \rho \, dV\)
• In spherical coordinates, \(dV = r^2 \, dr \, \sin \theta \, d\theta \, d\phi\)

This integration considers all contributions to the moment of inertia by adding up all the small mass elements and their respective distances squared from the axis of rotation. Each component is integrated over its specific limits, accounting for the radius, angle along the sphere's surface, and rotation angle, ultimately giving a clear result that suits the problem's symmetrical nature.