/*! 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 34 A grinding wheel is in the form ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A grinding wheel is in the form of a uniform solid disk of radius 7.00 \(\mathrm{cm}\) and mass 2.00 \(\mathrm{kg}\) . It starts from rest and accelerates uniformly under the action of the constant torque of 0.600 \(\mathrm{N} \cdot \mathrm{m}\) that the motor exerts on the wheel. (a) How long does the wheel take to reach its final operating speed of 1200 \(\mathrm{rev} / \mathrm{min}\) ? (b) Through how many revolutions does it turn while accelerating?

Short Answer

Expert verified
The wheel takes approximately 1.02 seconds to reach its final operating speed of 1200 rev/min, and it turns through approximately 10.13 revolutions while accelerating.

Step by step solution

01

Convert final operating speed to radians per second

First, we need to convert the final operating speed from revolutions per minute (rev/min) to radians per second (rad/s). The conversion is given by the formula: \[\text{rad/s} = \text{rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}\]. Therefore, \[1200 \frac{\text{rev}}{\text{min}} = 1200 \times \frac{2\pi}{60} = 40\pi \text{ rad/s}\].
02

Compute the moment of inertia of the disk

As we are dealing with a solid disk, we use the formula for the moment of inertia of a disk rotating about its central axis: \[I = \frac{1}{2} m r^2\], where \(m\) is the mass of the disk and \(r\) is its radius. Substituting the given values, \[I = \frac{1}{2} \times 2.00 \text{ kg} \times (0.07 \text{ m})^2 = 0.0049 \text{ kg m}^2\].
03

Calculate angular acceleration

Torque (\(\tau\)) is related to angular acceleration (\(\alpha\)) and the moment of inertia (\(I\)) by the formula \[\tau = I\alpha\]. Solving for \(\alpha\) gives us \[\alpha = \frac{\tau}{I}\]. Substituting the given torque and the calculated moment of inertia, \[\alpha = \frac{0.600 \text{ N} \cdot \text{m}}{0.0049 \text{ kg} \cdot \text{m}^2} = 122.45 \text{ rad/s}^2\].
04

Calculate the time to reach final speed

Angular acceleration is also defined by the change in angular velocity (\(\omega\)) over time (\(t\)): \[\alpha = \frac{\omega_f - \omega_i}{t}\]. Since the wheel starts from rest, \(\omega_i = 0\), so we have \[t = \frac{\omega_f}{\alpha}\]. Using the final angular velocity and angular acceleration from previous steps, \[t = \frac{40\pi \text{ rad/s}}{122.45 \text{ rad/s}^2} \approx 1.02 \text{ s}\].
05

Calculate the number of revolutions while accelerating

The number of revolutions (N) can be found using the equation of motion for constant angular acceleration: \[\theta = \omega_i t + \frac{1}{2} \alpha t^2\]. \(\theta\) is the angular displacement in radians, which can be converted to revolutions by dividing by \(2\pi\). Using \(\omega_i = 0\), the equation simplifies to \[\theta = \frac{1}{2} \alpha t^2\]. Substituting \(\alpha\) and \(t\), we find the total angular displacement in radians and then convert to revolutions: \[\theta = \frac{1}{2} \times 122.45 \text{ rad/s}^2 \times (1.02 \text{ s})^2\], \[\theta = 63.66 \text{ rad}\], \[N = \frac{63.66 \text{ rad}}{2\pi} \approx 10.13 \text{ revolutions}\].

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 91Ó°ÊÓ!

Key Concepts

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

Angular Acceleration
In the realm of rotational motion physics, angular acceleration plays a fundamental role. It describes how fast an object's rotational speed changes with time. Similar to linear acceleration in translational motion, angular acceleration (denoted as \( \alpha \)) is the rate at which angular velocity changes. In mathematical terms, angular acceleration is given by the formula \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) represents the change in angular velocity, and \( \Delta t \) signifies the time interval over which this change occurs.

When a constant torque is applied to an object, as in the case of a grinding wheel being powered by a motor, the object experiences a constant angular acceleration. This relationship can be expressed by the equation \( \tau = I\alpha \), where \( \tau \) is the torque, and \( I \) is the moment of inertia. From this equation, we can deduce that a larger torque or a smaller moment of inertia would result in a greater angular acceleration, causing the object to reach its final angular velocity quicker.

Understanding angular acceleration is crucial in various practical applications, such as in the design of mechanical systems where rotational parts need to reach certain speeds within specific time frames.
Moment of Inertia
The concept of moment of inertia, symbolized as \( I \), is pivotal in understanding rotational dynamics. It represents how the mass of an object is distributed with respect to an axis of rotation and plays a similar role to mass in linear motion. Specifically, the moment of inertia quantifies an object's resistance to changes in its rotational motion.

The formula for the moment of inertia varies depending on the object's shape and the axis about which it is rotating. For a solid disk like a grinding wheel, the moment of inertia is calculated using \( I = \frac{1}{2} m r^2 \), where \( m \) is mass and \( r \) is the radius. A larger radius or mass means a greater moment of inertia, indicating that more torque is required to achieve the same angular acceleration.

By calculating the moment of inertia, engineers and physicists can predict how difficult it will be to set an object in motion or to bring it to a stop. When dealing with any rotating system, comprehending its moment of inertia is a crucial factor in its design and operation.
Uniform Circular Motion
Uniform circular motion describes the motion of an object that travels in a circular path at a constant speed. Although the speed is constant, the direction is continuously changing, resulting in the object undergoing centripetal acceleration directed towards the center of the circle. This type of motion is best visualized by imagining a point on the edge of a disk, like our grinding wheel, as it spins at a steady rate.

When examining the wheel's performance under the influence of constant torque, one key observable phenomenon is that the wheel eventually reaches a final operating speed. At this point, it exhibits uniform circular motion since the angular velocity \( \omega \) remains constant. This operational state is significant for machines and applications that rely on a consistent rotational output.

Understanding uniform circular motion is not only fundamental in physics but also essential in numerous practical contexts such as the design of wheels, gears, and other components of machinery which are expected to deliver a consistent performance during their operation.

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 car accelerates uniformly from rest and reaches a speed of 22.0 \(\mathrm{m} / \mathrm{s}\) in 9.00 \(\mathrm{s}\) . If the diameter of a tire is \(58.0 \mathrm{cm},\) find \((\mathrm{a})\) the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

(a) What is the rotational kinetic energy of the Earth about its spin axis? Model the Earth as a uniform sphere and use data from the endpapers. (b) The rotational kinetic energy of the Earth is decreasing steadily because of tidal friction. Find the change in one day, assuming that the rotational period decreases by 10.0\(\mu\) s each year.

The density of the Earth, at any distance \(r\) from its center, is approximately $$ \rho=[14.2-11.6(r / R)] \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} $$ where \(R\) is the radius of the Earth. Show that this density leads to a moment of inertia \(I=0.330 M R^{2}\) about an axis through the center, where \(M\) is the mass of the Earth.

(a) Determine the acceleration of the center of mass of a uniform solid disk rolling down an incline making angle \(\theta\) with the horizontal. Compare this acceleration with that of a uniform hoop. (b) What is the minimum coefficient of friction required to maintain pure rolling motion for the disk?

During a certain period of time, the angular position of a swinging door is described by \(\theta=5.00+10.0 t+2.00 t^{2}\) , where \(\theta\) is in radians and \(t\) is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at \(t=0\) (b) at \(t=3.00 \mathrm{s}\) .

See all solutions

Recommended explanations on Physics Textbooks

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.