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 \(2\pi \delta a^3\).

Step 1: Understand Moment of Inertia

The moment of inertia (MI) of an object about an axis is a measure of its resistance to rotational motion around that axis. For a continuous mass distribution, the MI is given by the integral \(I = \int r^2 \, dm\), where \(r\) is the distance from the axis of rotation and \(dm\) is an infinitesimal element of mass.

Step 2: Define Mass Element

The hoop is a ring with radius \(a\), lying in the plane \(x^2 + y^2 = a^2\). The mass element \(dm\) can be expressed as \(\delta \, ds\), where \(ds\) is the differential arc length around the circle. Since \(s = a\theta\) for angle \(\theta\), \(ds = a \, d\theta\).

Step 3: Express dm and r

With \(ds = a \, d\theta\), the mass element is \(dm = \delta a \, d\theta\). The distance \(r\) from a point on the hoop to the z-axis (center of the circle) is \(a\), the radius of the circle.

Step 4: Set Up the Integral

The moment of inertia is thus \(I = \int r^2 \, dm = \int (a^2) \, (\delta a \, d\theta) = \delta a^3 \int \, d\theta\). We integrate \(\theta\) from 0 to \(2\pi\), covering the entire hoop.

Step 5: Solve the Integral

Calculate \(I = \delta a^3 \int_{0}^{2\pi} \, d\theta = \delta a^3 [\theta]_{0}^{2\pi} = \delta a^3 (2\pi) = 2\pi \delta a^3\).

Step 6: Conclusion

Thus, the moment of inertia of the circular wire hoop about the z-axis is \(I = 2\pi \delta a^3\).

Key Concepts

Circular Wire Hoop

A circular wire hoop is essentially a ring-shaped object with consistent thickness and material. Imagine taking a thin piece of wire and bending it into a perfect circle. This structure is then arranged with its center at the origin in the 2D plane.
This means every point on the hoop is equidistant from the center. If you picture it lying flat on the ground, like the rim of a glass, it forms what mathematicians describe as a circle.
In our problem, this circle is defined by the equation \(x^2 + y^2 = a^2\). This tells us that the radius of the hoop is \(a\). The wire itself has a uniform density \(\delta\), meaning the mass is evenly distributed along the length of the hoop.

Mass Distribution

Understanding mass distribution is crucial when dealing with physical objects' motion. For the circular wire hoop, mass distribution refers to how the mass is spread entirely around its circumference.
Because the hoop is a perfect circle with uniform density \(\delta\), the mass distribution is even. Every tiny piece, or differential element \(dm\), that makes up the hoop contributes equally to the total mass.
If you were to slice the hoop into tiny segments, each would have the same mass according to its length \(ds\), which is essential for calculating the moment of inertia.
This leads us to express \(dm\) as \(\delta \, ds\), accommodating the uniform distribution through the arc length \(ds = a \, d\theta\), where \(d\theta\) is the angular component.

Rotational Motion

Rotational motion is the motion of an object about an axis. For our hoop, when it rotates about the \(z\)-axis, it spins like a wheel or a hula hoop around your waist.
The concept of moment of inertia (\