Problem 37 - Thomas Calculus

Question

Moment of inertia of wire hoop A circular wire hoop of constant density \(\delta\) lies along the circle \(x^{2}+y^{2}=a^{2}\) in the \(x y\) -plane. Find the hoop's moment of inertia about the \(z\) -axis.

Step-by-Step Solution

Answer

The moment of inertia is \( I = 2\pi a^3 \delta \).

Step 1: Identify the Basics

The moment of inertia of the wire hoop around the \(z\)-axis involves calculating how mass elements are distributed at varying distances from the axis. For a circular hoop, all mass elements are at the same radial distance, \(a\), from the \(z\)-axis.

Step 2: Define the Mass Element

Consider a small mass element, \(dM\), of the wire hoop. Since the hoop is uniform, this mass element can be defined using the linear mass density \(\delta\) and an infinitesimal arc length \(ds\):\[ dM = \delta \cdot ds \]where \(ds\) is an infinitesimal part of the circumference of the circle.

Step 3: Express Arc Length Element

For a circle defined by \(x^2 + y^2 = a^2\), the arc length element \(ds\) can be expressed in terms of the angle \(\theta\):\[ ds = a \, d\theta \]

Step 4: Calculate the Moment of Inertia

The moment of inertia \(I\) around the \(z\)-axis is given by the integral:\[ I = \int r^2 \, dM \]Replace \(dM\) with \(\delta \, ds\):\[ I = \int a^2 \, (\delta \, ds) \]Substitute \(ds = a \, d\theta\):\[ I = \delta \, a^2 \, \int_{0}^{2\pi} a \, d\theta \]Thus, the final integral becomes:\[ I = \delta \, a^3 \, \int_{0}^{2\pi} d\theta = \delta \, a^3 \, [\theta]_{0}^{2\pi} \]\[ I = \delta \, a^3 \, (2\pi) \]

Step 5: Express the Final Result

The moment of inertia of the wire hoop about the \(z\)-axis simplifies to:\[ I = 2\pi a^3 \delta \]This formula confirms that the moment of inertia is dependent on the radius of the circle and the linear mass density.

Key Concepts

Circular Wire Hoop

A circular wire hoop is a fascinating object in physics. Picture it as a thin loop of wire shaped like a perfect circle. It's flat, lying in a plane, with all points equally spaced from the center.

The hoop we are considering lies in the xy-plane, and its shape can be mathematically described by the equation of a circle:

• \(x^2 + y^2 = a^2\).

Here, \(a\) represents the radius of the hoop. When discussing a circular wire hoop, it’s important to recognize that every point on this hoop is equidistant from the center, all lying along the circumference of the circle. This feature significantly simplifies the computation of moments of inertia, as the uniform layout ensures symmetry around the center.

Mass Distribution

Mass distribution in a circular wire hoop is fairly straightforward. Since the hoop is symmetrical, all points on it are uniformly distributed along the circle.

For a uniform hoop, this means:

• Every tiny segment of the hoop has the same mass per unit length.
• This uniformity leads us to characterize this distribution using a constant known as linear mass density, denoted as \(\delta\).

Knowing the density helps in forming expressions for small mass elements necessary for computations like moment of inertia. As you perceive this hoop in the xy-plane, think of each tiny section of the wire, its contribution always aligned with \(\delta\), giving consistency to mass distribution.

Integral Calculation

The integral calculation is central to determining the moment of inertia of our hoop about the z-axis. Integrals help us sum up infinitely small contributions to this moment of inertia around the axis.

Here's how we break it down for the hoop:

• We take every tiny segment (best described using calculus as \(dM = \delta ds\)).
• The contribution to inertia by each element is determined by its mass and its distance from the axis, in this case, expressed as \(a^2 dM\).
• We then integrate this around the entire hoop, from 0 to \(2\pi\) radians, covering the full circle.

Thus, the integral encapsulates the comprehensive effect of every infinitesimal element of the hoop's mass, calculated as \(I = \delta a^3 \int_0^{2\pi} d\theta\). This process is at the heart of figuring out the moment of inertia.

Linear Mass Density

Linear mass density is fundamental when dealing with elongated, thin objects like wire hoops. It’s essentially a measure of how much mass is distributed along a unit length of the object, expressed as \(\delta\).

For a hoop:

• Since it is uniform, \(\delta\) remains constant across its entire length.
• This uniform density allows us to simplify expressions related to mass, such as \(dM = \delta ds\).

Why is \(\delta\) crucial? Because it allows one to consider just the geometry (like the radius \(a\) and arc length \(ds\)) when performing calculations for total mass or moment of inertia.

The consistent value of \(\delta\) forms the backbone of calculating the moment of inertia, by ensuring each tiny segment of the hoop contributes equally proportionately to the overall inertia.