/*! 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 49 A turntable reaches an angular s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 49

A turntable reaches an angular speed of \(33.3 \mathrm{rpm}\) in $2.0 \mathrm{s}$ starting from rest. (a) Assuming the angular acceleration is constant, what is its magnitude? (b) How many revolutions does the turntable make during this time interval?

Short Answer

Expert verified
Answer: The magnitude of the angular acceleration is 1.74 rad/s虏, and the turntable makes about 0.554 revolutions during the 2.0-second time interval.

Step by step solution

01

Convert to radians

First, we should convert the given angular velocity from rpm (revolutions per minute) to radians per second. $$ \omega_f = 33.3 \frac{rev}{min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \ min}{60 \ s} \approx 3.48 \; \frac{rad}{s} $$
02

Determine the angular acceleration

Now, we will use the linear acceleration formula: $$ \omega_f = \omega_i + \alpha t $$ Since the turntable starts from rest, the initial angular velocity (\(\omega_i\)) is 0. Given the final angular velocity \(\omega_f = 3.48 \; \frac{rad}{s}\) and the time interval \(t = 2.0 \ s\), we can solve for the angular acceleration \(\alpha\): $$ \alpha = \frac{\omega_f - \omega_i}{t} = \frac{3.48 \; \frac{rad}{s}}{2.0 \, s} = 1.74 \; \frac{rad}{s^2} $$ So, the magnitude of the angular acceleration is 1.74 rad/s虏.
03

Calculate the angular displacement

To find the number of revolutions made by the turntable during the 2.0-second time interval, we need to calculate the angular displacement \(\theta\). We can use the following formula for angular displacement: $$ \theta = \omega_i t + \frac{1}{2} \alpha t^2 $$ Since \(\omega_i = 0\), the equation simplifies to: $$ \theta = \frac{1}{2} \alpha t^2 $$ Plugging in the values for \(\alpha\) and \(t\), we have: $$ \theta = \frac{1}{2} (1.74 \; \frac{rad}{s^2})(2.0 \, s)^2 \approx 3.48 \, rad $$
04

Determine the number of revolutions

Finally, we need to convert the angular displacement in radians to the number of revolutions. We can do this by dividing the angular displacement by \(2\pi\) since there are \(2 \pi\) radians in one revolution: $$ \text{Number of revolutions} = \frac{3.48 \, rad}{2 \pi \, rad/rev} \approx 0.554 \, rev $$ So, the turntable rotates approximately 0.554 revolutions during the 2.0-second time interval. In conclusion, the magnitude of the angular acceleration is 1.74 rad/s虏, and the turntable makes about 0.554 revolutions during the 2.0-second time interval.

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!

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

The time to sunset can be estimated by holding out your arm, holding your fingers horizontally in front of your eyes, and counting the number of fingers that fit between the horizon and the setting Sun. (a) What is the angular speed, in radians per second, of the Sun's apparent circular motion around the Earth? (b) Estimate the angle subtended by one finger held at arm's length. (c) How long in minutes does it take the Sun to "move" through this same angle?
During normal operation, a computer's hard disk spins at 7200 rpm. If it takes the hard disk 4.0 s to reach this angular velocity starting from rest, what is the average angular acceleration of the hard disk in $\mathrm{rad} / \mathrm{s}^{2} ?$
In Chapter 19 we will see that a charged particle can undergo uniform circular motion when acted on by a magnetic force and no other forces. (a) For that to be true, what must be the angle between the magnetic force and the particle's velocity? (b) The magnitude of the magnetic force on a charged particle is proportional to the particle's speed, \(F=k v .\) Show that two identical charged particles moving in circles at different speeds in the same magnetic field must have the same period. (c) Show that the radius of the particle's circular path is proportional to the speed.
A carnival swing is fixed on the end of an 8.0 -m-long beam. If the swing and beam sweep through an angle of \(120^{\circ},\) what is the distance through which the riders move?
A car drives around a curve with radius \(410 \mathrm{m}\) at a speed of $32 \mathrm{m} / \mathrm{s} .\( The road is banked at \)5.0^{\circ} .$ The mass of the car is \(1400 \mathrm{kg}\). (a) What is the frictional force on the car? (b) At what speed could you drive around this curve so that the force of friction is zero?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Radiation

Read Explanation

Energy Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Wave Optics

Read Explanation

Modern Physics

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.