Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the moment of inertia and radius of gyration of a homogeneous ( \(\delta\) a constant) circular lamina of radius \(a\) with respect to a diameter.

Step-by-Step Solution

Answer

The moment of inertia is \( \frac{1}{4} \pi a^4 \delta \) and the radius of gyration is \( \frac{a}{2} \).

Step 1: Understand the Problem

The problem asks for the moment of inertia and radius of gyration of a circular lamina. A lamina is a thin flat plate, and as mentioned, it is homogeneous with uniform density. We need to calculate the moment of inertia with respect to one of its diameters.

Step 2: Use the Moment of Inertia Formula for Lamina

The formula for the moment of inertia of a circular lamina about its diameter is given by \( I = \frac{1}{4} m a^2 \), where \( m \) is the mass of the lamina, and \( a \) is the radius of the circle. For this lamina, because it is homogeneous, the mass \( m \) is determined by \( m = \pi a^2 \delta \), where \( \delta \) is the density.

Step 3: Calculate Moment of Inertia

Substitute \( m = \pi a^2 \delta \) into the inertia formula \( I = \frac{1}{4} m a^2 \): \[ I = \frac{1}{4} (\pi a^2 \delta) a^2 = \frac{1}{4} \pi a^4 \delta \]. This is the moment of inertia of the lamina with respect to the diameter.

Step 4: Understand Radius of Gyration

The radius of gyration \( k \) relates to the moment of inertia and mass by the formula \( I = m k^2 \). We rearrange this formula to find \( k \), the radius of gyration.

Step 5: Calculate Radius of Gyration

Using \( I = \frac{1}{4} \pi a^4 \delta \) and \( m = \pi a^2 \delta \), substitute into the radius of gyration formula \( k^2 = \frac{I}{m} \): \[ k^2 = \frac{\frac{1}{4} \pi a^4 \delta}{\pi a^2 \delta} = \frac{1}{4} a^2 \]. Thus, \( k = \frac{1}{2} a \). The radius of gyration about the diameter is \( \frac{a}{2} \).

Key Concepts

Radius of Gyration

The radius of gyration, denoted as \( k \), is a key concept in understanding how mass is distributed in a rigid body. In simple terms, it measures the distribution of the area or mass around an axis of rotation. It is the imaginary point where the entire mass of the body can be thought to be concentrated to give the same moment of inertia as the actual mass.

• The formula for the radius of gyration is given by \( k = \sqrt{\frac{I}{m}} \), where \( I \) is the moment of inertia and \( m \) is the mass of the body.
• In the context of a homogeneous circular lamina, the radius of gyration about a diameter was found to be \( \frac{a}{2} \), where \( a \) is the radius of the circle.

Therefore, the radius of gyration helps in simplifying complex shapes into easier to understand points for theoretical calculations. This is crucial when dealing with rotational dynamics.

Circular Lamina

A circular lamina is essentially a thin, flat, disk-shaped object. It is often modeled in physics and engineering as having no thickness, which helps simplify calculations regarding its behavior under forces, especially rotational forces.

• The lamina is crucial in understanding the distribution of area or mass along its radius, which contributes to its moment of inertia calculations.
• Its geometry is simple yet fundamental in physics to explain complex concepts like torque and angular momentum.

The study of such lamina focuses on their moment of inertia, especially when calculating their properties relative to an axis, like a diameter, as seen in our exercise.

Homogeneous Density

In the context of our circular lamina, homogeneous density refers to the uniform distribution of mass throughout the object. This means that the density, denoted as \( \delta \), remains constant across the entire lamina.

• This uniformity simplifies calculations as it means mass \( m \) can be calculated by \( m = \pi a^2 \delta \).
• When calculating the moment of inertia for uniform objects like our lamina, homogeneous density means that each point on the circle contributes equally to the described axis.

This characteristic is critical as it allows us to derive simplified, formulaic solutions for physical properties, facilitating practical applications in engineering and physics.

Diameter

The diameter of a circle is an essential dimension in the context of our problem. It represents the longest straight line that can be drawn through the center of the circle, touching two points on its boundary.

• In calculations of moment of inertia, using diameter as a reference axis is common because it bisects the object symmetrically.
• For a circular lamina, analyzing by diameter helps in distributing mass equally across the axis during rotational motion calculations.

In our exercise, the diameter was used as the axis around which the moment of inertia and the radius of gyration were computed, providing insights into how mass is distributed along the circle.