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Chapter 10: Problem 74

A machine part is made from a uniform solid disk of radius \(R\) and mass \(M\). A hole of radius \(R / 2\) is drilled into the disk, with the center of the hole at a distance \(R / 2\) from the center of the disk (the diameter of the hole spans from the center of the disk to its outer edge). What is the moment of inertia of this machine part about the center of the disk in terms of \(R\) and \(M ?\)

Short Answer

Expert verified
Answer: The moment of inertia of the machine part about the center of the disk is $\frac{3}{8}MR^2$.

Step by step solution

01

Recall the moment of inertia formula for a solid disk

The moment of inertia \(I\) for a solid disk of mass \(M\) and radius \(R\) about its center is given by the formula \(I = \frac{1}{2}MR^2\).
02

Calculate the mass of the entire solid disk

The mass of the entire solid disk \(M_1\) is given. So, \(M_1 = M\).
03

Calculate the mass of the hole

The hole is also a uniform solid disk of radius \(\frac{R}{2}\). Since the mass is uniformly distributed, we can calculate the mass of the hole \(M_2\) in terms of \(M\) and the ratio of their areas: $$ M_2 = M \times \frac{\pi(\frac{R}{2})^2}{\pi R^2} = M \times \frac{R^2 / 4}{R^2} = \frac{1}{4}M. $$
04

Find the mass of the remaining part

To find the mass of the remaining part after the hole is drilled, we need to subtract the mass of the hole from the mass of the entire disk: $$ M' = M - M_2 = M - \frac{1}{4}M = \frac{3}{4}M. $$
05

Calculate the moment of inertia for the remaining part

Now that we know the mass of the remaining part, we can find its moment of inertia using the formula from Step 1: $$ I = \frac{1}{2}(M')R^2 = \frac{1}{2}\left(\frac{3}{4}M\right)R^2 = \frac{3}{8}MR^2. $$So, the moment of inertia of the machine part about the center of the disk is \(\frac{3}{8}MR^2\).

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Key Concepts

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

Understanding Physics and Moment of Inertia
Physics is the natural science that studies matter, its motion and behavior through space and time, and the related entities of energy and force. One of the fundamental concepts in the physics of rotational motion is the moment of inertia, which is essentially the rotational equivalent of mass for linear motion.

Moment of inertia represents the distribution of mass within a body and how that mass is located relative to a given rotational axis. The farther the mass is from the axis of rotation, the higher the moment of inertia, making it harder to change the rotational speed of the object. It plays a vital role in understanding how objects will behave when subjected to torques and angular accelerations.
Mechanical Engineering and Design
In mechanical engineering, the moment of inertia isn't just a theoretical concept鈥攊t鈥檚 a crucial factor in the design and analysis of mechanical components. Engineers need to calculate the moment of inertia to ensure that machine parts, like gears or flywheels, can withstand the stresses of operation without failure.

For example, in our exercise, the moment of inertia of a modified solid disk (with a hole drilled in it) is sought to ensure stability and balance in the machine鈥檚 design. Engineers often employ computational tools or apply mathematical formulas to obtain accurate measurements of these physical properties to design efficient and safe mechanical systems.
Rotational Motion Explained
Rotational motion is the circular movement of an object around a central point or axis. This type of motion is characterized by parameters such as angular velocity, angular acceleration, torque, and, critically, the moment of inertia. When an external force is applied, the object's moment of inertia dictates how easily it can be set into rotational motion or be stopped. Greater moments of inertia mean that more torque or a longer duration of force application is required to achieve a certain angular acceleration.

Understanding the interplay between force and the moment of inertia is essential when analyzing systems in rotational motion, much like the solid disk in our exercise, which is expected to rotate around its central axis.
Solid Disk Inertia Calculation
The moment of inertia for a solid disk provides a practical example of how matter distribution affects rotational properties. In the exercise, the original solid disk has a known moment of inertia formula, which changes when a segment of the disk is removed. Knowing how to calculate the new moment of inertia by subtracting the inertia of the removed section from the original disk's inertia is essential for tasks in both physics and engineering.

In the context of drilling a hole offset from the center of a disk, you must calculate the mass and corresponding moment of inertia for the missing segment. The final moment of inertia is vital for analyzing the dynamic response of the modified disk when integrated into a mechanical system. The precise calculation ensures the machine part operates as intended without unexpected stresses or imbalances, highlighting the crossroad where theoretical physics meets real-world engineering applications.

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Most popular questions from this chapter

Using the conservation of mechanical energy, calculate the final speed and the acceleration of a cylindrical object of mass \(M\) and radius \(R\) after it rolls a distance \(s\) without slipping along an inclined plane of angle \(\theta\) with respect to the horizontal

A thin uniform rod (length \(=1.00 \mathrm{~m},\) mass \(=2.00 \mathrm{~kg})\) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is \(\frac{1}{3} m L^{2} .\) The rod is released when it is \(60.0^{\circ}\) below the horizontal. What is the angular acceleration of the rod at the instant it is released?

A force, \(\vec{F}=(2 \hat{x}+3 \hat{y}) \mathrm{N},\) is applied to an object at a point whose position vector with respect to the pivot point is \(\vec{r}=(4 \hat{x}+4 \hat{y}+4 \hat{z}) \mathrm{m} .\) Calculate the torque created by the force about that pivot point.

You are unwinding a large spool of cable. As you pull on the cable with a constant tension, what happens to the angular acceleration and angular velocity of the spool, assuming that the radius at which you are extracting the cable remains constant and there is no friction force? a) Both increase as the spool unwinds. b) Both decrease as the spool unwinds. c) Angular acceleration increases, and angular velocity decreases. d) Angular acceleration decreases, and angular velocity increases. e) It is impossible to tell.

Why does a figure skater pull in her arms while increasing her angular velocity in a tight spin?
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