/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 A \(5.0-\mathrm{kg}\) wheel with... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 37

A \(5.0-\mathrm{kg}\) wheel with a radius of gyration of \(20 \mathrm{~cm}\) is to be given an angular frequency of \(10 \mathrm{rev} / \mathrm{s}\) in 25 revolutions from rest. Find the constant unbalanced torque required.

Short Answer

Expert verified
The constant unbalanced torque required is approximately \(2.512\,\text{N}\cdot\text{m}\).

Step by step solution

01

Understand the Problem

We are given a wheel with a mass of \(5.0\,\text{kg}\) and a radius of gyration \(k = 20\,\text{cm} = 0.2\,\text{m}\). It needs to reach an angular frequency of \( \omega = 10\,\text{rev/s} = 10 \times 2\pi \text{ rad/s} = 20\pi \text{ rad/s} \) after making 25 revolutions. We need to find the constant torque required to achieve this from rest.
02

Calculate Moment of Inertia

The moment of inertia \(I\) of the wheel can be calculated using its mass \(m\) and the radius of gyration \(k\) with the formula: \[ I = m k^2 \]Plugging in the given values:\[ I = 5.0 \times (0.2)^2 = 0.2\,\text{kg}\cdot\text{m}^2 \]
03

Use Kinematic Equation for Angular Motion

Use the angular kinematics equation which relates initial angular velocity \(\omega_0 = 0\), final angular velocity \(\omega = 20\pi \text{ rad/s}\), number of revolutions (converted to radians), and angular acceleration \(\alpha\):\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]Convert 25 revolutions to radians:\[ \theta = 25 \times 2\pi = 50\pi \text{ rad} \]Since \(\omega = \omega_0 + \alpha t\) and \(\omega_0 = 0\), we have:\[ \omega = \alpha t \]
04

Solve for Angular Acceleration

Combining the equations from Step 3, we solve for \(\alpha\):\[ 50\pi = \frac{1}{2} \alpha t^2 \]Since \( \omega = \alpha t \Rightarrow t = \frac{\omega}{\alpha} \), substitute \(t\) into the \(\theta\) equation:\[ 50\pi = \frac{1}{2} \alpha \left( \frac{20\pi}{\alpha} \right)^2 \]Simplifying gives:\[ 50\pi = \frac{1}{2} \alpha \left( \frac{400\pi^2}{\alpha^2} \right) \]\[ 50\pi = 200\pi^2/\alpha \]Solve for \(\alpha\):\[ \alpha = \frac{200\pi}{50} = 4\pi \text{ rad/s}^2 \]
05

Calculate Torque

Now calculate the torque using the relationship of torque \(\tau\), moment of inertia \(I\), and angular acceleration \(\alpha\):\[ \tau = I \alpha \]Substitute \(I = 0.2\,\text{kg}\cdot\text{m}^2\) and \(\alpha = 4\pi\,\text{rad/s}^2\):\[ \tau = 0.2 \times 4\pi = 0.8\pi \,\text{N}\cdot\text{m} \]
06

Final Calculation of Torque

We can calculate the numerical value of the torque by approximating \(\pi\):\[ \tau \approx 0.8 \times 3.14 = 2.512 \,\text{N}\cdot\text{m} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Moment of Inertia
Understanding the moment of inertia is crucial to solving problems involving angular motion. It can be thought of as the rotational equivalent of mass in linear motion. The moment of inertia measures how difficult it is to change the rotational speed of an object about an axis. To calculate it, we use the formula: \( I = m k^2 \), where:
  • \( I \) is the moment of inertia,
  • \( m \) is the mass of the object, and
  • \( k \) is the radius of gyration.
In our exercise, the moment of inertia for the wheel is calculated using its mass \(5.0\,\text{kg}\) and its radius of gyration \(k = 0.2\, \text{m}\). After substituting these values into the formula, we find:\[ I = 5.0 \times (0.2)^2 = 0.2\,\text{kg} \cdot \text{m}^2 \]This value helps us to understand how the wheel will respond to the applied torque, making it a critical part of calculating rotational acceleration.
Angular Acceleration
Angular acceleration is a key component when analyzing rotating objects. It indicates how quickly an object's rotational speed is changing with time. It's analogous to linear acceleration in straight-line motion and is represented by \( \alpha \).Angular acceleration can be derived using the angular kinematics equation:\[ \omega = \omega_0 + \alpha t \]where:
  • \( \omega \) is the final angular velocity,
  • \( \omega_0 \) is the initial angular velocity (generally 0 if starting from rest), and
  • \( t \) is the time taken for the change.
In the given problem, the wheel is accelerated from rest to \( \omega = 20\pi \text{ rad/s} \). We equate the angular displacement equation \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \) and use the provided revolutions (converted to radians) to solve for \( \alpha \). After doing the math, we discover that:\[ \alpha = 4\pi \text{ rad/s}^2 \]This result illuminates how rapidly the wheel's rotational velocity increases as it spins.
Torque Calculation
Torque is the twisting force that causes an object to rotate. Calculating torque is essential in understanding the dynamics of rotational systems and is directly linked to angular acceleration and moment of inertia.The relationship between torque \( \tau \), moment of inertia \( I \), and angular acceleration \( \alpha \) is expressed by the formula:\[ \tau = I \alpha \]This formula shows that torque depends on both the distribution of mass in the object (moment of inertia) and how quickly it is spinning up (angular acceleration).In the exercise, using the previously determined moment of inertia \( I = 0.2\,\text{kg} \cdot \text{m}^2 \) and the angular acceleration \( \alpha = 4\pi \,\text{rad/s}^2 \), we apply the formula:\[ \tau = 0.2 \times 4\pi = 0.8\pi \,\text{N} \cdot \text{m} \]Approximating \( \pi \) as 3.14, we calculate the torque to be approximately:\[ \tau \approx 2.512\,\text{N} \cdot \text{m} \]This torque value signifies the constant rotational force required to achieve the desired angular motion for the wheel, emphasizing the interplay of all discussed elements.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(0.75-\) hp motor acts for \(8.0 \mathrm{~s}\) 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 \(8.0 \mathrm{~s}\) $$ \begin{aligned} (\text { Power) }(\text { Time })&=\frac{1}{2} I \omega^{2} \\ (0.75 \mathrm{hp})(746 \mathrm{~W} / \mathrm{hp})(8.0 \mathrm{~s}) &=\frac{1}{2}\left(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\right) \omega^{2} \end{aligned} $$ from which \(\omega=67 \mathrm{rad} / \mathrm{s}\).

A large roller in the form of a uniform cylinder is pulled by a tractor to compact earth; it has a 1.80-m diameter and weighs \(10 \mathrm{kN}\). If frictional losses can be ignored, what average horsepower must the tractor provide to accelerate the cylinder from rest to a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) in a horizontal distance of \(3.0 \mathrm{~m}\) ? The power required is equal to the work done by the tractor divided by the time it takes. The tractor does the following work: $$ \text { Work }=(\Delta \mathrm{KE})_{r}+(\Delta \mathrm{KE})_{t}=\frac{1}{2} I \omega_{f}^{2}+\frac{1}{2} m v_{f}^{2} $$ We have \(v_{f}=4.0 \mathrm{~m} / \mathrm{s}, \omega_{f}=v_{f} / r=4.44 \mathrm{rad} / \mathrm{s}\), and \(m=10000 / 9.81=1019 \mathrm{~kg} .\) The moment of inertia of the cylinder is $$ I=\frac{1}{2} m r^{2}=\frac{1}{2}(1019 \mathrm{~kg})(0.90 \mathrm{~m})^{2}=413 \mathrm{~kg} \cdot \mathrm{m}^{2} $$ Substituting these values, the work required tums out to be \(12.23 \mathrm{~kJ}\). We still need the time taken to do this work. Because the roller went \(3.0 \mathrm{~m}\) with an average velocity \(v_{\mathrm{av}}=\frac{1}{2}(4+0)=2.0 \mathrm{~m} / \mathrm{s}\) $$ t=\frac{s}{v_{a v}}=\frac{3.0 \mathrm{~m}}{2.0 \mathrm{~m} / \mathrm{s}}=1.5 \mathrm{~s} $$ Then \(\quad\) Power \(=\frac{\text { Work }}{\text { Time }}=\frac{12230 \mathrm{~J}}{1.5 \mathrm{~s}}=(8150 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=11 \mathrm{hp}\)

A flywheel (i.e., a massive disk capable of rotating about its central axis) has a moment of inertia of \(3.8 \mathrm{~kg} \cdot \mathrm{m}^{2}\). What constant torque is required to increase the wheel's frequency from \(2.0 \mathrm{rev} / \mathrm{s}\) to \(5.0 \mathrm{rev} / \mathrm{s}\) in \(6.0\) revolutions? Neglect friction. Given $$ \begin{array}{lll} \theta=12 \pi \mathrm{rad} & \omega_{i}=4.0 \pi \mathrm{rad} / \mathrm{s} & \text { and } & \omega_{f}=10 \pi \mathrm{rad} / \mathrm{s} \end{array} $$ we can write Work done on wheel = Change in \(\mathrm{KE}_{r}\) of wheel $$ \begin{aligned} \tau \theta &=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} I \omega_{i}^{2} \\ (\tau)(12 \pi \mathrm{rad}) &=\frac{1}{2}\left(3.8 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left[\left(100 \pi^{2}-16 \pi^{2}\right)(\mathrm{rad} / \mathrm{s})^{2}\right] \end{aligned} $$ which leads to \(\tau=42 \mathrm{~N} \cdot \mathrm{m}\). Notice in all of these problems that radians and seconds must be used.

A 90 -kg person stands at the edge of a stationary children's merry-go-round (essentially a disk) at a distance of \(5.0 \mathrm{~m}\) from its center. The person starts to walk around the perimeter of the disk at a speed of \(0.80 \mathrm{~m} / \mathrm{s}\) relative to the ground. What rotation rate does this motion impart to the disk if \(I_{\text {disk }}=20000 \mathrm{~kg} \cdot \mathrm{m}^{2} ?\left[\right.\) Hint \(:\) For the person, \(\left.I=m r^{2} .\right]\)

A cord \(3.0 \mathrm{~m}\) long is wrapped around the axle of a wheel. The cord is pulled with a constant force of \(40 \mathrm{~N}\), and the wheel revolves as a result. When the cord leaves the axle, the wheel is rotating at \(2.0 \mathrm{rev} / \mathrm{s}\). Determine the moment of inertia of the wheel and axle. Neglect friction. [Hint: The easiest solution is obtained via the energy method.]
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Electric Charge Field and Potential

Read Explanation

Thermodynamics

Read Explanation

Mechanics and Materials

Read Explanation

Medical Physics

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.