/*! 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 9 A drum rotates around its centra... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 9

A drum rotates around its central axis at an angular velocity of \(12.60 \mathrm{rad} / \mathrm{s}\). If the drum then slows at a constant rate of \(4.20\) \(\mathrm{rad} / \mathrm{s}^{2}\), (a) how much time does it take and (b) through what angle does it rotate in coming to rest?

Short Answer

Expert verified
(a) 3 seconds, (b) 18.9 radians.

Step by step solution

01

Determine Time to Rest

First, we need to determine the time it takes for the drum to come to rest. Using the equation for angular velocity, \( \omega = \omega_0 + \alpha t \), where \( \omega_0 = 12.60 \ \mathrm{rad/s} \) is the initial angular velocity, \( \omega = 0 \ \mathrm{rad/s} \) is the final angular velocity (when the drum comes to rest), and \( \alpha = -4.20 \ \mathrm{rad/s}^2 \) is the angular deceleration. Rearranging the equation, we get \( t = \frac{\omega - \omega_0}{\alpha} = \frac{0 - 12.60}{-4.20} = 3 \ \mathrm{seconds} \).
02

Calculate Angle Rotated

Next, to find the angle through which the drum rotates while coming to rest, we use the angular displacement formula: \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \). Substituting the known values, \( \theta = 12.60 \times 3 + \frac{1}{2} \times (-4.20) \times (3)^2 \). Calculating each part, we have \( \theta = 37.8 - 18.9 = 18.9 \ \mathrm{radians} \).

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.

Understanding Angular Velocity
Angular velocity is a measure of how quickly an object rotates around an axis. It's akin to how "speed" works for linear motion but for rotation. It's measured in radians per second (\(\mathrm{rad/s}\)). This value gives you the angular change over time. For example, if a drum has an angular velocity of \(12.6 \, \mathrm{rad/s}\), it means it rotates through an angular distance of \(12.6\) radians every second.

Key points to remember about angular velocity include:
  • It can be positive or negative, depending on the direction of rotation.
  • An increase in angular velocity over time indicates angular acceleration.
  • A decrease, as in our example, indicates angular deceleration.
A practical example is a spinning drum, which initially moves at \(12.6 \, \mathrm{rad/s}\). Suppose no force acts upon it, it continues spinning indefinitely at the same velocity due to inertia.
The Role of Angular Deceleration
Angular deceleration refers to the rate at which an object's angular velocity decreases. It is essentially the opposite of angular acceleration, as it slows down the rotational speed. The unit for angular deceleration is also \(\mathrm{rad/s^2}\). In our context, the drum experiences an angular deceleration of \(-4.2 \, \mathrm{rad/s^2}\), meaning its speed reduces by \(4.2\) radians per second squared.

Let's break it down:
  • Angular deceleration is negative acceleration.
  • It tells you how quickly the object is slowing down in angular motion.
  • It helps in calculating the time required for a rotating object to come to rest.
To determine the time it takes for the drum to stop, use the formula \( \omega = \omega_0 + \alpha t \). Rearranging gives us \( t = \frac{\omega - \omega_0}{\alpha} \), solving to \(3\) seconds in this instance.
Exploring Angular Displacement
Angular displacement describes the change in the position of an object as it rotates around an axis, measured in radians. It's a core concept in rotational motion, akin to distance in linear motion. For the drum in our example, we calculated the angular displacement as it comes to rest.

Consider these aspects of angular displacement:
  • It is the total angle through which an object rotates during a given time period.
  • The formula used is \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), which gives a complete picture of both initial velocity and acceleration's role.
  • In our example, the drum rotated through \(18.9\) radians as it came to a halt.
This concept helps us understand not just the extent of rotation, but also how factors like initial velocity and deceleration contribute to the overall movement.

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 angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds \(A t\) \(t=0\), what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s}\) ? (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\) (e) Is its angular acceleration constant?

(a) Show that the rotational inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis is equal to the rotational inertia of a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\) about its central axis. (b) Show that the rotational inertia \(I\) of any given body of mass \(M\) about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass \(M\) and a radius \(k\) given by $$ k=\sqrt{\frac{I}{M}} $$ The radius \(k\) of the equivalent hoop is called the radius of gyration of the given body.

At \(t=0\), a flywheel has an angular velocity of \(4.7 \mathrm{rad} / \mathrm{s}, \mathrm{a}\) constant angular acceleration of \(-0.25 \mathrm{rad} / \mathrm{s}^{2}\), and a reference line at \(\theta_{0}=0 .\) (a) Through what maximum angle \(\theta_{\max }\) will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at \(\theta=\frac{1}{2} \theta_{\max } ?\) At what (d) negative time and (e) positive time will the reference line be at \(\theta=10.5 \mathrm{rad}\) ? (f) Graph \(\theta\) versus \(t\), and indicate your answers.

A disk rotates at constant angular acceleration, from angular position \(\theta_{1}=10.0\) rad to angular position \(\theta_{2}=70.0\) rad in \(6.00 \mathrm{~s}\). Its angular velocity at \(\theta_{2}\) is \(15.0 \mathrm{rad} / \mathrm{s}\). (a) What was its angular velocity at \(\theta_{1} ?\) (b) What is the angular acceleration? (c) At what angular position was the disk initially at rest? (d) Graph \(\theta\) versus time \(t\) and angular speed \(\omega\) versus \(t\) for the disk, from the beginning of the motion (let \(t=0\) then \()\).

The angular acceleration of a wheel is \(\alpha=6.0 t^{4}-4.0 t^{2}\), with \(\alpha\) in radians per second-squared and \(t\) in seconds. At time \(t=0\), the wheel has an angular velocity of \(+2.0 \mathrm{rad} / \mathrm{s}\) and an angular position of \(+1.0\) rad. Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s).
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Energy Physics

Read Explanation

Medical Physics

Read Explanation

Famous Physicists

Read Explanation

Physical Quantities And Units

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.