Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

The sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) has constant area density \(k\). Find each moment of inertia. (a) About a diameter (b) About a tangent line (assume the Parallel Axis Theorem from Problem 28 of Section 13.5).

Step-by-Step Solution

Answer

(a) \(\frac{2}{3}Ma^2\), (b) \(\frac{5}{3}Ma^2\)

Step 1: Moment of Inertia Mass Element Consideration

The sphere's surface is described by the equation \(x^2 + y^2 + z^2 = a^2\). With a constant area density \(k\), the mass element on the sphere's surface is \(dm = k \, dA\). Here, \(dA\) is the differential area element on the surface of the sphere.

Step 2: Calculate Total Mass of the Sphere

To find the total mass \(M\) of the sphere, integrate the mass element over its surface: \(M = \int_S dm = k \int_S dA = k \times 4\pi a^2\). Thus, \(M = 4\pi a^2 k\).

Step 3: Moment of Inertia About a Diameter

The moment of inertia \(I_d\) about a diameter is given by the integral \(I_d = \int_S r^2 \, dm\), where \(r\) is the perpendicular distance from the axis of rotation. For a sphere about a diameter: \(I_d = \frac{2}{3} Ma^2\) by using symmetry properties and known formula for a solid sphere.

Step 4: Applying Parallel Axis Theorem

For the moment of inertia about a tangent line, use the Parallel Axis Theorem, which states that \(I_t = I_d + Md^2\), where \(d\) is the distance from the center of mass to the new axis. Here, \(d = a\), the radius of the sphere.

Step 5: Moment of Inertia About a Tangent Line

Using the parallel axis theorem, the moment of inertia about a tangent line is: \[ I_t = I_d + M \cdot a^2 = \frac{2}{3} M a^2 + M a^2 = \frac{5}{3} M a^2. \]

Key Concepts

Area Density

When dealing with continuous objects like a sphere, understanding how mass is distributed over an area is crucial. Area density, often denoted by the symbol \(k\), represents how much mass is contained within a unit area on a surface. For a sphere with a constant area density, this means that every small patch on its surface has the same proportion of mass.

• The area density \(k\) is given in units of mass per unit area (e.g., kg/m²).
• It allows us to express the mass element of the sphere's surface as \(dm = k \, dA\), where \(dA\) is a differential area element.
• This is pivotal in determining quantities like total mass and moment of inertia that depend on how mass is spread over an object.

Parallel Axis Theorem

The Parallel Axis Theorem is a powerful tool in rotational dynamics, especially when calculating the moment of inertia of an object about an axis that is parallel to an axis through its center of mass. The theorem states:

\[I_t = I_d + Md^2\]

Here, \(I_t\) is the moment of inertia about the tangent line (or new axis), \(I_d\) is the moment of inertia about the axis through the center of mass, \(M\) is the total mass of the sphere, and \(d\) is the distance between the two axes.

• This theorem simplifies computations where direct calculation would be complex, such as finding the inertia about a tangent line on a sphere.
• It allows us to leverage known values (e.g., \(I_d = \frac{2}{3} Ma^2\) for a sphere) to find new inertial characteristics.
• In practical applications, this helps engineers predict how different parts rotate, which is crucial in design processes.

Sphere

A sphere is a perfectly symmetrical three-dimensional object. It can be defined mathematically by the equation \(x^2 + y^2 + z^2 = a^2\), where \(a\) is the radius of the sphere. Its properties make it an essential topic in geometry and physics.

• The sphere's surface area is \(4\pi a^2\) and the volume is \(\frac{4}{3}\pi a^3\).
• It is uniquely smooth and continuous, with its entire surface equidistant from the center point. This symmetry is beneficial for calculations.
• Inertia-related calculations are simpler because of this symmetry, with known results like \(I_d = \frac{2}{3} Ma^2\) for a sphere about a diameter aiding quick solutions.

Understanding these properties allows for more straightforward calculations both in theoretical and applied mechanics.

Integral Calculus

Integral calculus plays a crucial role in calculating physical properties over a continuous domain. In the context of a sphere, it is used to find the total mass and the moment of inertia.

• To find the total mass \(M\) of the sphere, we integrate the mass element over its surface: \(M = \int_S dm = k \int_S dA = k \times 4\pi a^2\).
• For the moment of inertia about a diameter, \(I_d = \int_S r^2 \, dm\), where \(r\) is the perpendicular distance from the axis of rotation.
• These integrals involve breaking down the continuous surface into small elements and summing, a concept central to integral calculus.

Integral calculus thus provides a method to describe and compute physical quantities for objects with continuous mass distribution.