Problem 35 - University Physics with Modern Physics

Question

A compound disk of outside diameter 140.0 cm is made up of a uniform solid disk of radius 50.0 cm and area density 3.00 g/cm\(^2\) surrounded by a concentric ring of inner radius 50.0 cm, outer radius 70.0 cm, and area density 2.00 g/cm\(^2\). Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

Step-by-Step Solution

Answer

The moment of inertia of the compound disk is the sum of the inner disk and ring's moments of inertia.

Step 1: Understanding the Disk Structure

The compound disk consists of a solid inner disk and an outer ring. The inner disk has a radius of 50.0 cm, and the outer ring has inner and outer radii of 50.0 cm and 70.0 cm, respectively. Each part has a different area density.

Step 2: Calculate the Moment of Inertia of the Inner Disk

The moment of inertia (I) of a solid disk is given by the formula: \( I = \frac{1}{2} m r^2 \). To find the mass (m), use the area (A) and area density (\( \sigma \)): \( m = \sigma \cdot A \). Here, \( A = \pi r^2 \) and \( \sigma = 3.00 \, \text{g/cm}^2 \). With \( r = 50.0 \, \text{cm} \), find \( m \) and substitute in the formula to find \( I \) of the inner disk.

Step 3: Calculate the Moment of Inertia of the Outer Ring

For the ring, the moment of inertia is \( I = \frac{1}{2} m (r_2^2 + r_1^2) \), where \( r_2 = 70.0 \, \text{cm} \) and \( r_1 = 50.0 \, \text{cm} \). First, calculate the area \( A = \pi (r_2^2 - r_1^2) \) and then the mass with \( \sigma = 2.00 \, \text{g/cm}^2 \). Use these in the moment of inertia formula for the ring.

Step 4: Sum the Moments of Inertia

The total moment of inertia for the compound disk is the sum of the moments of inertia for the inner disk and the outer ring. Add the results from Step 2 and Step 3 to find the total moment of inertia.

Key Concepts

Compound Disk

A compound disk is a system that consists of more than one component put together, where each part contributes to the overall properties of the system. In the context of the problem, the compound disk consists of two main components:

• The inner solid disk
• The outer ring

These two parts combine to give the compound disk its structural and mass properties. The moment of inertia, which tells us how hard it is to rotate the disk about a certain axis, depends on both of these components. Understanding each part of the disk structure is crucial for accurately determining the overall moment of inertia. Each component's contribution is calculated separately and then added together due to its respective distribution of mass and geometry.

Area Density

Area density is an essential concept when calculating the moment of inertia for objects that have complex shapes like the compound disk. It measures how much mass is distributed over a specific area; therefore, it is given in units like grams per square centimeter (g/cm²).

For the inner solid disk, the area density is 3.00 g/cm², while the outer ring has an area density of 2.00 g/cm². These values are used to determine the mass of each component. Once you know the area covered by each part of the disk, the area density allows you to find the corresponding mass, which then plays a role in calculating rotational inertia.

Mathematically, to find mass (\( m \)), you use the equation:\[ m = \sigma \times A \]where \( \sigma \) is the area density and \( A \) is the area of the particular disk part. This calculated mass is then substituted into the formula for the moment of inertia.

Solid Disk

The solid disk in this problem forms the inner part of the compound disk. This disk has a radius of 50.0 cm and an area density of 3.00 g/cm². The concept of a solid disk is essential because it dictates how to calculate its moment of inertia, which is a measure of its resistance to rotational acceleration.

To compute the moment of inertia (\( I \)) of a solid disk, the formula used is:\[ I = \frac{1}{2} m r^2 \]where \( m \) is the mass of the disk and \( r \) is its radius.

The first step involves calculating the disk's area using the formula:\[ A = \pi r^2 \]With this area and the given area density, the mass \( m \) is calculated, which is then used in the inertia formula above. Doing this provides the first part of the total rotational inertia of the whole compound disk.

Outer Ring

The outer ring surrounds the solid disk in the compound disk setup. It spans from an inner radius of 50.0 cm to an outer radius of 70.0 cm, having its own distinct area density of 2.00 g/cm². Since this is a ring, its moment of inertia involves a different formula compared to a solid disk due to its geometry.

For a ring, the moment of inertia (\( I \)) is given by:\[ I = \frac{1}{2} m (r_2^2 + r_1^2) \]where \( r_2 \) and \( r_1 \) are the outer and inner radii, respectively.

First, calculate the area of the outer ring using:\[ A = \pi (r_2^2 - r_1^2) \]With this area, the mass \( m \) can be found using the area density. Substituting the mass into the inertia formula allows for determining the ring's contribution to the total moment of inertia. Once you have both the inner disk and outer ring's moments, add them to find the entire system's rotational inertia.