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Chapter 10: Problem 45

A 500 -g wheel that has a moment of inertia of \(0.015 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is initially turning at \(30 \mathrm{rev} / \mathrm{s}\). It coasts uniformly to rest after 163 rev. How large is the torque that slowed it?

Short Answer

Expert verified
The torque that slowed the wheel is \(0.130 \, \text{N}\cdot\text{m}\).

Step by step solution

01

Convert Units

First, convert the initial angular speed from revolutions per second (rev/s) to radians per second (rad/s) since we need it in radians for calculations. We know that one revolution equals \(2\pi\) radians. Therefore, the initial angular velocity \(\omega_i\) is given by: \[ \omega_i = 30 \times 2\pi = 60\pi \, \text{rad/s} \]
02

Calculate Total Angular Displacement

Convert the angular displacement from revolutions to radians. Since there are \(2\pi\) radians in one revolution, the total angular displacement \(\theta\) is: \[ \theta = 163 \times 2\pi = 326\pi \, \text{radians} \]
03

Determine Final Angular Velocity

As the wheel comes to rest, the final angular velocity \(\omega_f\) is \(0\, \text{rad/s}\).
04

Use Kinematic Equation for Rotation

Apply the rotational kinematics equation to find the angular acceleration \(\alpha\). The equation is: \[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \] Plug in the known values: \[ 0 = (60\pi)^2 + 2\alpha(326\pi) \] Solve for \(\alpha\): \[ \alpha = -\frac{(60\pi)^2}{2 \times 326\pi} = -\frac{3600\pi}{652} \approx -8.668 \, \text{rad/s}^2 \]
05

Calculate Torque Using Newton's Second Law for Rotation

Use the formula for torque \(\tau\), which is based on angular acceleration: \[ \tau = I\alpha \] where \(I = 0.015\, \text{kg}\cdot\text{m}^2\) and \(\alpha \approx -8.668\, \text{rad/s}^2\). Find \(\tau\): \[ \tau = 0.015 \times -8.668 \approx -0.130 \, \text{N}\cdot\text{m} \]
06

Interpret the Solution

The negative sign in torque indicates that it acts opposite to the wheel's direction of motion to bring it to rest. Therefore, the magnitude of the torque slowing the wheel is \(0.130 \, \text{N}\cdot\text{m}\).

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Key Concepts

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

Rotational Kinematics
Rotational kinematics involves the study of objects in rotational motion. It is similar to linear kinematics but focused on rotating bodies. Just like in linear motion where we have distance, speed, and acceleration, in rotational motion we have angular displacement, angular velocity, and angular acceleration.

Key equations include those that relate angular displacement, velocity, and acceleration. These relationships are crucial for predicting the behavior of rotating objects over time. Understanding these principles allows us to analyze how torque and other forces affect rotational motion.
  • Angular Displacement: The angle through which a point or line has been rotated in a specified sense about a specified axis.
  • Angular Velocity: Defines how fast something is spinning around an axis.
  • Angular Acceleration: Describes how quickly angular velocity changes.
All these elements interplay to explain the rotating aspects of objects.
Angular Displacement
Angular displacement is the measure of the angle through which an object has rotated about a fixed point or axis. This concept is especially important in rotational dynamics because it helps determine other rotational properties like angular velocity and acceleration.

To convert revolutions into radians, a full revolution is equivalent to an angle of \(2\pi\) radians. In the exercise, converting 163 revolutions into radians gave us:
  • Total angular displacement \(\theta = 163 \times 2\pi = 326\pi\) radians.
Angular displacement can depict the entire path traced by a rotating object, allowing quicker determination of changes in motion.
Angular Velocity
Angular velocity denotes how fast an object rotates or revolves relative to another point. It is different from linear velocity as it describes rotation around a specific axis rather than linear movement from A to B.

In our context, converting angular velocity from revolutions per second to radians per second is essential since radian is the standard unit in rotational calculations:
  • Initial angular velocity \(\omega_i = 30 \times 2\pi = 60\pi\, \text{rad/s}\)
When solving rotational motion problems, angular velocity provides insight into how quickly the object reaches its rest state or another rotational speed. It is a critical component for understanding motion behavior in wheels, gears, and other rotating systems.
Moment of Inertia
Moment of inertia is a fundamental concept in rotational dynamics. It measures an object's resistance to changes in its rotation. You might think of it as the rotational equivalent of mass in linear motion.

Higher moment of inertia means an object is more difficult to start or stop rotating. For instance, our exercise used a wheel's moment of inertia \(I = 0.015\, \text{kg}\cdot\text{m}^2\). This value was critical in determining the torque required to alter the wheel's rotation:
  • Torque expression derived from \(\tau = I\alpha\)
Knowing the moment of inertia allows you to predict how much torque is necessary to achieve a desired angular acceleration, providing better control over mechanical devices involved in rotational dynamics.

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Most popular questions from this chapter

An airplane propeller has a mass of \(70 \mathrm{~kg}\) and a radius of gyration of \(75 \mathrm{~cm}\). Find its moment of inertia. How large a torque is needed to give it an angular acceleration of \(4.0 \mathrm{rev} / \mathrm{s}^{2}\) ? $$I=M k^{2}=(70 \mathrm{~kg})(0.75 \mathrm{~m})^{2}=39 \mathrm{~kg} \cdot \mathrm{m}^{2}$$ To be able to use \(\tau=I \alpha\), we must have \(\alpha\) in \(\mathrm{rad} / \mathrm{s}^{2}\) :

When \(100 \mathrm{~J}\) of work is done on a stationary flywheel (that is otherwise free to rotate in place), its angular speed increases from 60 rev/min to 180 rev/min. What is its moment of inertia?

A 0.75-hp motor acts for \(8.0\) s on an initially nonrotating wheel having a moment of inertia \(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Find the angular speed developed in the wheel, assuming no losses. Work done by motor in \(8.0 \mathrm{~s}=\mathrm{KE}\) of wheel after \(8.0 \mathrm{~s}\) $$\begin{array}{c} (\text { Power })(\text { Time })=\frac{1}{2} I \omega^{2} \\ (0.75 \mathrm{hp})(746 \mathrm{~W} / \mathrm{hp})(8.0 \mathrm{~s})=\frac{1}{2}\left(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\right) \omega^{2} \end{array}$$ from which \(\omega=67 \mathrm{rad} / \mathrm{s}\).

Compute the radius of gyration of a solid disk of diameter \(24 \mathrm{~cm}\) about an axis through its center of mass and perpendicular to its face.

A disk like the lower one in Fig. \(10-11\) has a moment of inertia \(I_{1}\) about the vertical axis shown. What will be the new moment of inertia if a tiny mass \(M\) is placed on it at a distance \(R\) from its center? The definition of moment of inertia tells us that, for the disk plus an added point mass \(M\), $$I=\sum_{\text {disk }} m_{i} r_{i}^{2}+M R^{2}$$ where the sum extends over all the point masses composing the original disk. With the value of that sum given as \(I_{1}\), the new moment of inertia is \(I=I_{1}+M R^{2}\).
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