/*! 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 14 An electric motor rotating a wor... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electric motor rotating a workshop grinding wheel at a rate of \(1.00 \times 10^{2} \mathrm{rev} / \mathrm{min}\) is switched off. Assume the wheel has a constant negative angular acceleration of magnitude \(2.00 \mathrm{rad} / \mathrm{s}^{2}\). (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in part (a)?

Short Answer

Expert verified
Use the step 2 and step 3 calculations to find the time taken for the grinding wheel to stop and the total angle turned by the wheel during this time respectively.

Step by step solution

01

- Conversion to Standard Units

Convert the given angular speed from revolutions per minute to radian per second. Given: \(1.00 \times 10^{2} \, rev/min\). We know, \(1 \, rev = 2\pi \, rad\) and \(1 \, min = 60 \, sec\). Therefore, initial angular speed, \(\omega_{i}\) = \(1.00 \times 10^{2} \times\frac{2\pi}{60}\) rad/sec.
02

- Calculate Time to Stop

Apply the kinematic equation for angular motion \(\omega_{f} = \omega_{i} + \alpha*t\) where \(\omega_{f}\) is the final angular velocity (which is zero, since the wheel stops), \(\omega_{i}\) is the initial angular velocity (calculated in step 1), \(\alpha\) is the angular acceleration and \(t\) is the time. Using the known values, \(t = \frac{\omega_{f} - \omega_{i}}{\alpha}\). This gives the time taken for the grinding wheel to stop.
03

- Calculate Radians Turned

Next, apply the equation \(\theta = \omega_{i} \times t + \frac{1}{2} \times \alpha \times t^2\) to find the total angle turned by the wheel during the time calculated in step 2. Here, \(\theta\) is the total angle, \(\omega_{i}\) is the initial angular velocity, \(t\) is the time and \(\alpha\) is the angular acceleration.

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.

Understanding Angular Velocity
Angular velocity is a fundamental concept in rotational motion which describes how fast an object rotates around a specific point or axis. It's analogous to linear velocity but for circular motion. The unit of angular velocity is radians per second (rad/s).

To compute angular velocity, a common formula is used: \( \omega = \frac{\Delta \theta}{\Delta t} \), where \( \Delta \theta \) is the change in angular position, and \( \Delta t \) is the change in time. This tells us the rate of rotation. For example, if an object completes one full revolution, it covers \( 2\pi \) radians. If it accomplishes this in one second, its angular velocity would be \( 2\pi \) rad/s.

In the exercise, the initial angular velocity is provided in revolutions per minute. To convert this to radians per second:
  • First, note that one revolution equals \( 2\pi \) radians.
  • Second, there are 60 seconds in a minute.
Consequently, converting \( 1.00 \times 10^{2} \) revolutions per minute into radians per second involves multiplying by \( \frac{2\pi}{60} \). This conversion aligns angular measurements with our standards for calculations concerning angular motion.
Exploring Angular Acceleration
Angular acceleration measures the rate at which angular velocity changes. It's particularly important when an object is either speeding up or slowing down as it rotates. The unit of angular acceleration is radians per second squared (rad/s²).

Use the formula \( \alpha = \frac{\Delta \omega}{\Delta t} \) to calculate it, where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the time taken for this change. In situations where an object decelerates, the angular acceleration takes a negative value, indicating a reduction in speed.

For our specific exercise, we deal with negative angular acceleration since the wheel slows down and eventually stops. With an initial angular velocity derived from our conversion and an angular acceleration given, this allows the use of kinematic equations to determine time and displacement in angular terms. By integrating these, we can calculate the total angle the wheel turned as it decelerated and came to a halt.
Solving with Kinematic Equations for Rotational Motion
Just like in linear motion, we use kinematic equations to solve problems in angular motion. These equations are adaptations that account for rotations. They help to determine various parameters like angular displacement, velocity, and acceleration.

One useful kinematic equation for angular motion is \( \omega_{f} = \omega_{i} + \alpha \times t \), where:
  • \( \omega_{f} \) is the final angular velocity.
  • \( \omega_{i} \) is the initial angular velocity.
  • \( \alpha \) is the angular acceleration.
For an object coming to a stop, \( \omega_{f} = 0 \). Rearrange the equation to solve for time \( t \) when given \( \omega_{i} \) and \( \alpha \).

To find out how much the wheel rotated before coming to rest, another kinematic equation is \( \theta = \omega_{i} \times t + \frac{1}{2} \times \alpha \times t^2 \). Here, \( \theta \) represents the angular displacement.

These equations not only provide a framework for solving rotational dynamics problems but also allow insights into understanding the motion characteristics of rotating systems, much like we do for linear kinematics.

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

radial and tangential components) just before it is released.One method of pitching a softball is called the "windmill" delivery method, in which the pitcher's arm rotates through approximately \(360^{\circ}\) in a vertical plane before the 198-gram ball is released at the lowest point of the circular motion. An experienced pitcher can throw a ball with a speed of \(98.0 \mathrm{mi} / \mathrm{h}\). Assume the angular acceleration is uniform throughout the pitching motion and take the distance between the softball and the shoulder joint to be \(74.2 \mathrm{~cm}\). (a) Determine the angular speed of the arm in rev/s at the instant of release. (b) Find the value of the angular acceleration in \(\mathrm{rev} / \mathrm{s}^{2}\) and the radial and tangential acceleration of the ball just before it is released. (c) Determine the force exerted on the ball by the pitcher's hand (both

A car rounds a banked curve where the radius of curvature of the road is \(R\), the banking angle is \(\theta\), and the coefficient of static friction is \(\mu\). (a) Determine the range of speeds the car can have without slipping up or down the road. (b) What is the range of speeds possible if \(R=100 \mathrm{~m}, \theta=10^{\circ}\), and \(\mu=0.10\) (slippery conditions)?

A \(40.0\)-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of \(18.0 \mathrm{~m}\). (a) What is the centripetal acceleration of the child? (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? (c) What force does the seat exert on the child at the highest point of the ride? (d) What force does the seat exert on the child when the child is halfway between the top and bottom?

(a) One of the moons of Jupiter, named Io, has an orbital radius of \(4.22 \times 10^{8} \mathrm{~m}\) and a period of \(1.77\) days. Assuming the orbit is circular, calculate the mass of Jupiter. (b) The largest moon of Jupiter, named Ganymede, has an orbital radius of \(1.07 \times 10^{9} \mathrm{~m}\) and a period of \(7.16\) days. Calculate the mass of Jupiter from this data. (c) Are your results to parts (a) and (b) consistent? Explain.

A \(50.0\)-kg child stands at the rim of a merry-go-round of radius \(2.00 \mathrm{~m}\), rotating with an angular speed of \(3.00 \mathrm{rad} / \mathrm{s}\). (a) What is the child's centripetal acceleration? (b) What is the minimum force between her feet and the floor of the carousel that is required to keep her in the circular path? (c) What minimum coefficient of static friction is required? Is the answer you found reasonable? In other words, is she likely to stay on the merry-go-round?

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.