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

Short Answer

Expert verified
The moment of inertia is given by the sum of the moments of inertia of the inner disk and the concentric ring. Compute each component and add them.

Step by step solution

01

Understand the Components

The compound disk consists of two parts: a solid inner disk and a concentric ring. The inner disk has a radius of 50.0 cm and an area density of 3.00 g/cm². The ring has an inner radius of 50.0 cm, an outer radius of 70.0 cm, and an area density of 2.00 g/cm².
02

Calculate Moment of Inertia of Inner Disk

The moment of inertia for a solid disk about its central axis is given by the formula \(I = \frac{1}{2} m r^2\). First, calculate the mass of the inner disk using its area \( A = \pi r^2 = \pi (50.0)^2 \text{ cm}^2\) and area density: \(m = \text{Area density} \times A = 3.00 \times \pi (50.0)^2 \text{ g}.\) The moment of inertia \( I_1 \) is then \( I_1 = \frac{1}{2} (3.00 \times \pi (50.0)^2) (50.0)^2 \text{ g} \cdot \text{cm}^2.\)
03

Calculate Moment of Inertia of the Ring

The moment of inertia for a ring about its central axis is given by \(I = \frac{1}{2} m (r_{outer}^2 + r_{inner}^2)\). Calculate the mass of the ring using its area \( A = \pi (r_{outer}^2 - r_{inner}^2) = \pi ((70.0)^2 - (50.0)^2) \text{ cm}^2\) and area density: \(m = 2.00 \times \pi ((70.0)^2 - (50.0)^2) \text{ g}.\) The moment of inertia \( I_2 \) is \( I_2 = \frac{1}{2} (2.00 \times \pi ((70.0)^2 - (50.0)^2)) (70.0^2 + 50.0^2) \text{ g} \cdot \text{cm}^2.\)
04

Add Moments of Inertia

The total moment of inertia about the central axis is the sum of the moments of inertia of the inner disk and the ring: \( I = I_1 + I_2 \). Substitute the calculated values from Step 2 and Step 3 to find the total moment of inertia.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

compound disk
A compound disk is an interesting object made up of more than one component, designed to work together as a single unit. In our exercise, the compound disk consists of two main parts: a solid central disk and an outer concentric ring.

Think of it like a cake with layers; the inner part is one layer, and the surrounding part is another layer. This kind of setup is often found in real-world applications like flywheels, CDs, or certain types of machinery.
  • The solid inner disk acts as the core structure.
  • The concentric ring adds extra material and design to the outer part, enhancing its properties.
Understanding the compound disk is essential because calculating its properties, like moment of inertia, requires analyzing each part individually and then combining the results.
area density
Area density is a crucial concept when dealing with objects like compound disks. It refers to the mass of an object per unit area. In simple terms, it tells us how much material is packed into a given space on a surface.
  • In this exercise, the inner disk has an area density of 3.00 g/cm², while the concentric ring has an area density of 2.00 g/cm².
  • This difference in area densities means that material distribution in the two sections of the compound disk varies.
The area density allows us to calculate the mass of each section when given the area. This is important as the mass directly influences other calculations, such as finding the moment of inertia. Calculations start with determining the area first and then follow with mass calculations using area density.
solid disk
The solid disk is a fundamental shape in physics known for its straightforward properties. In our exercise, it forms the central part of the compound disk.

To compute the moment of inertia of a solid disk about its central axis, we use the formula: \[I = \frac{1}{2} m r^2\]Here, \(m\) is the mass of the disk, and \(r\) is its radius. The moment of inertia is a measure of an object's resistance to change in its rotational motion. Think of it as rotational mass—a heavier or larger object will naturally resist rotation more than a lighter or smaller one.
  • For our solid disk, the radius is 50.0 cm, and it has an area density that helps in calculating its mass.
  • The calculated mass is then used to find the moment of inertia, which is an essential step in solving for the moment of inertia of the entire compound disk.
These steps highlight the importance of understanding each element to find the characteristics of more complex systems.
concentric ring
A concentric ring is like a donut-shaped object that circles another shape, such as the inner disk in our compound disk. In our exercise, the ring surrounds the solid disk and adds to the overall weight and rotational characteristics of the compound disk.

Calculating the moment of inertia for the concentric ring differs slightly from that of a solid disk. The formula used is:\[I = \frac{1}{2} m (r_{outer}^2 + r_{inner}^2)\]In this equation:
  • \(m\) represents the mass of the ring.
  • \(r_{outer}\) and \(r_{inner}\) are the outer and inner radii of the ring, respectively.
The concentric ring's mass is found with the help of its area and area density. Subtracting the inner area from the outer area gives the actual area of the ring, which is then multiplied by the area density to find its mass.
This step is imperative to determining how the ring affects the total moment of inertia of the compound disk. By combining the ring's properties with those of the solid disk, we achieve a thorough understanding of the object's rotational dynamics.

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Most popular questions from this chapter

A classic 1957 Chevrolet Corvette of mass 1240 \(\mathrm{kg}\) starts from rest and speeds up with a constant tangential acceleration of 3.00 \(\mathrm{m} / \mathrm{s}^{2}\) on a circular test track of radius 60.0 \(\mathrm{m}\) . Treat the car as a particle. (a) What is its angular acceleration? (b) What is its angular speed 6.00 s after it starts? (c) What is its radial acceleration at this time? (d) Sketch a view from above showing the circular track, the car, the velocity vector, and the acceleration component vectors 6.00 s after the car starts. (e) What are the magnitudes of the total acceleration and net force for the car at this time? (f) What angle do the total acceleration and net force make with the car's velocity at this time?

In a charming 19 thcentury hotel, an old-style elevator is connected to a counter- weight by a cable that passes over a rotaing disk 2.50 \(\mathrm{m}\) in diameter (Fig. 9.28 ). The elevator is raised and lowered by turning the disk, and the cable does not slip on the rim of the disk but turns with it. (a) At bow many rpm must the disk turn to raise the elevator at 25.0 \(\mathrm{cm} / \mathrm{s}\) ? (b) To start the elevator moving, it must be accelerated at \(\frac{1}{8} \mathrm{g}\) . What must be the angular acceleration of the disk, in rad/s'? (c) Through what angle (in radi- ans and degrees) has the disk turned when it has raised the elevator 3.25 \(\mathrm{m}\) between floors?

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

A thin, uniform rod is bent into a square of side length \(a\) . If the total mass is \(M\) , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

Calculate the moment of inertia of a uniform solid cone about an axis through its center (Fig. 9.40). The cone has mass \(M\) and altitude \(h\) . The radius of its circular base is \(R\) . figure can't copy

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