/*! 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 52 A disk rotates with constant ang... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 52

A disk rotates with constant angular acceleration. The initial angular speed of the disk is \(2 \pi\) rad/s. After the disk rotates through \(10 \pi\) radians, the angular speed is \(7 \pi \mathrm{rad} / \mathrm{s} .\) (a) What is the magnitude of the angular acceleration? (b) How much time did it take for the disk to rotate through \(10 \pi\) radians? (c) What is the tangential acceleration of a point located at a distance of \(5.0 \mathrm{cm}\) from the center of the disk?

Short Answer

Expert verified
a) The magnitude of the angular acceleration is \(\frac{9}{4}蟺 \,\text{rad/s}^2\). b) The time it took for the disk to rotate through \(10蟺\) radians is \(\frac{20}{9}\,\text{s}\). c) The tangential acceleration of a point located 5.0 cm from the center of the disk is \(0.1125蟺 \,\text{m/s}^2\).

Step by step solution

01

Write down the given information

Write down the initial angular speed (\(蠅_0\)), the final angular speed (\(蠅\)), and the change in angle (\(胃-胃_0\)): \(蠅_0 = 2蟺 \,\text{rad/s}\) \(蠅 = 7蟺 \,\text{rad/s}\) \(胃 - 胃_0 = 10蟺 \,\text{rad}\)
02

Determine the angular acceleration (a)

Use equation 3 to solve for the angular acceleration (\(伪\)): \(蠅^2 = 蠅_0^2 + 2伪(胃 - 胃_0)\) Solve for \(伪\): \(伪 = \frac{蠅^2 - 蠅_0^2}{2(胃 - 胃_0)}\) Plug in the given values: \(伪 = \frac{(7蟺)^2 - (2蟺)^2}{2(10蟺)}\) \(伪 = \frac{45蟺^2}{20蟺}\) \(伪 = \frac{9}{4}蟺 \,\text{rad/s}^2\) The magnitude of the angular acceleration is \(\frac{9}{4}蟺 \,\text{rad/s}^2\).
03

Determine the time it took (b)

Use equation 2 to solve for the time it took for the disk to rotate through \(10蟺\) radians (\(t\)): \(蠅 = 蠅_0 + 伪t\) Solve for \(t\): \(t = \frac{蠅 - 蠅_0}{伪}\) Plug in the given values: \(t = \frac{(7蟺) - (2蟺)}{\frac{9}{4}蟺}\) \(t = \frac{5蟺}{\frac{9}{4}蟺}\) \(t = \frac{20}{9}\,\text{s}\) The time it took for the disk to rotate through \(10蟺\) radians is \(\frac{20}{9}\,\text{s}\).
04

Determine the tangential acceleration (c)

The tangential acceleration (\(a_t\)) of a point located at a distance \(r\) from the center of the disk is related to the angular acceleration by the following equation: \(a_t = 伪r\) We are given that the distance \(r\) is 5.0 cm, which we must convert to meters: \(r = 5.0 \,\text{cm} = 0.050 \,\text{m}\) Plug in the values for \(伪\) and \(r\): \(a_t = \frac{9}{4}蟺 \,\text{rad/s}^2 \cdot 0.050 \,\text{m}\) \(a_t = \frac{9}{4} \cdot(0.050) \cdot 蟺 \,\text{m/s}^2\) \(a_t = 0.1125蟺 \,\text{m/s}^2\) The tangential acceleration of a point located 5.0 cm from the center of the disk is \(0.1125蟺 \,\text{m/s}^2\).

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

A conical pendulum consists of a bob (mass \(m\) ) attached to a string (length \(L\) ) swinging in a horizontal circle (Fig. 5.11 ). As the string moves, it sweeps out the area of a cone. The angle that the string makes with the vertical is \(\phi\) (a) What is the tension in the string? (b) What is the period of the pendulum?
A person of mass \(M\) stands on a bathroom scale inside a Ferris wheel compartment. The Ferris wheel has radius \(R\) and angular velocity \(\omega\). What is the apparent weight of the person (a) at the top and (b) at the bottom?
What's the fastest way to make a U-turn at constant speed? Suppose that you need to make a \(180^{\circ}\) turn on a circular path. The minimum radius (due to the car's steering system) is \(5.0 \mathrm{m},\) while the maximum (due to the width of the road) is \(20.0 \mathrm{m}\). Your acceleration must never exceed \(3.0 \mathrm{m} / \mathrm{s}^{2}\) or else you will skid. Should you use the smallest possible radius, so the distance is small, or the largest, so you can go faster without skidding, or something in between? What is the minimum possible time for this U-turn?
A curve in a stretch of highway has radius \(R .\) The road is unbanked. The coefficient of static friction between the tires and road is \(\mu_{s} .\) (a) What is the fastest speed that a car can safely travel around the curve? (b) Explain what happens when a car enters the curve at a speed greater than the maximum safe speed. Illustrate with an FBD.
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

Solid State Physics

Read Explanation

Translational Dynamics

Read Explanation

Fluids

Read Explanation

Radiation

Read Explanation

Classical Mechanics

Read Explanation

Medical 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.