/*! 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 10 Starting from rest, a disk rotat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In \(5.0 \mathrm{~s}\), it rotates \(25 \mathrm{rad}\). During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the \(5.0 \mathrm{~s}\) ? (d) With the angular acceleration unchanged, through what additional angle will the disk turn during the next \(5.0 \mathrm{~s}\) ?

Short Answer

Expert verified
(a) 2 rad/s² (b) 5 rad/s (c) 10 rad/s (d) 75 rad.

Step by step solution

01

Identify Known Variables

We know that the disk starts from rest, so the initial angular velocity \( \omega_0 = 0 \). The time \( t = 5.0 \ \text{s} \), and the disk rotates through an angle \( \theta = 25 \ \text{rad} \).
02

Calculate Angular Acceleration

Use the equation for angular displacement \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \) where \( \theta = 25 \ \text{rad} \), \( \omega_0 = 0 \), and \( t = 5 \ \text{s} \). Substitute these values:\[ 25 = 0 \cdot 5 + \frac{1}{2} \alpha (5)^2 \]Solving for \( \alpha \):\[ 25 = \frac{1}{2} \alpha \times 25 \]\[ \alpha = \frac{2 \times 25}{25} = 2 \ \text{rad/s}^2 \]
03

Calculate Average Angular Velocity

Use the formula for average angular velocity \( \overline{\omega} = \frac{\theta}{t} \), where \( \theta = 25 \ \text{rad} \) and \( t = 5 \ \text{s} \).\[ \overline{\omega} = \frac{25}{5} = 5 \ \text{rad/s} \]
04

Calculate Instantaneous Angular Velocity

Use the kinematic equation \( \omega = \omega_0 + \alpha t \), where \( \omega_0 = 0 \), \( \alpha = 2 \ \text{rad/s}^2 \) and \( t = 5 \ \text{s} \).\[ \omega = 0 + 2 \times 5 = 10 \ \text{rad/s} \]
05

Calculate Additional Angle Turned

For the next 5 seconds, use the equation for angular displacement with the initial angular velocity now being \( \omega_0 = 10 \ \text{rad/s} \). The time \( t \) for this next phase is again 5 seconds.\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 = 10 \times 5 + \frac{1}{2} \times 2 \times 5^2 \]\[ \theta = 50 + 25 = 75 \ \text{rad} \]

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
Angular acceleration is a measure of how rapidly the angular velocity of an object is changing. In simple terms, it's like how much faster or slower something is spinning per second.
Imagine starting up a merry-go-round. Initially, when you push it, it speeds up, and this speeding up is due to angular acceleration. The moment you stop pushing, it starts to slow down unless acted upon by more force.
For the problem at hand, the disk experiences angular acceleration because it's going from rest (not moving) to rotating over time. The calculation used here applies the formula for angular displacement, which is given by:- \[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]Since the initial angular velocity \(\omega_0 = 0\) (because it starts from rest), this simplifies to \(\theta = \frac{1}{2} \alpha t^2\). Solving for \(\alpha\) allows us to determine that the angular acceleration is 2 \(\text{rad/s}^2\). This means every second, the spinning speed of the disk increases by 2 radians per second.
Average Angular Velocity
Average angular velocity provides an overall picture of how fast something was rotating over a period of time. It's different from instantaneous angular velocity because it doesn't tell you about the speed at any specific moment, but rather gives the speed averaged over the entire time.
If you think about a car trip, the average speed is like saying how fast you were traveling for the whole trip, even if sometimes you were stopped or sometimes going faster.
In the situation described, the average angular velocity can be determined using:- \[ \overline{\omega} = \frac{\theta}{t} \]Here, \(\theta\) is the total angle the disk turned (25 radians) over time \(t\) of 5 seconds. Plugging in these values, the average angular velocity is 5 \(\text{rad/s}\). This tells us that on average, the disk turned 5 radians every second over the 5 seconds.
Instantaneous Angular Velocity
Instantaneous angular velocity is the speed of rotation at a particular moment. It’s like checking your car’s speedometer to see exactly how fast you’re going at that very instant.
While average angular velocity considers something like the whole trip of a disk, the instantaneous angular velocity tells us exactly how fast it was spinning at the end of the 5 seconds.
In this problem, to find the instantaneous angular velocity at the end of 5 seconds, we use the formula:- \[ \omega = \omega_0 + \alpha t \]Given that the initial velocity \(\omega_0\) is 0 and angular acceleration \(\alpha\) is 2 \(\text{rad/s}^2\), we substitute to get \(\omega = 2 \times 5 = 10 \text{rad/s}\). This means that right at the end of 5 seconds, the disk is spinning at 10 radians per second. This value might offer more detail than the average because it tells us exactly how the rotational speed builds to this point.

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 golf ball is launched at an angle of \(20^{\circ}\) to the horizontal, with a speed of \(60 \mathrm{~m} / \mathrm{s}\) and a rotation rate of \(90 \mathrm{rad} / \mathrm{s}\). Neglecting air drag, determine the number of revolutions the ball makes by the time it reaches maximum height.

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a \(30 \mathrm{~kg}\) box as shown in Fig. \(10-59 .\) The outer radius \(R\) of the device is \(0.50 \mathrm{~m}\), and the radius \(r\) of the hub is \(0.20 \mathrm{~m}\). When a constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(140 \mathrm{~N}\) is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude \(0.80\) \(\mathrm{m} / \mathrm{s}^{2}\). What is the rotational inertia of the device about its axis of rotation?

A uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

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?

Cheetahs running at top speed have been reported at an astounding \(114 \mathrm{~km} / \mathrm{h}\) (about \(71 \mathrm{mi} / \mathrm{h})\) by observers driving alongside the animals. Imagine trying to measure a cheetah's speed by keeping your vehicle abreast of the animal while also glancing at your speedometer, which is registering \(114 \mathrm{~km} / \mathrm{h}\). You keep the vehicle a constant \(8.0 \mathrm{~m}\) from the cheetah, but the noise of the vehicle causes the cheetah to continuously veer away from you along a circular path of radius \(92 \mathrm{~m}\). Thus, you travel along a circular path of radius \(100 \mathrm{~m} .\) (a) What is the angular speed of you and the cheetah around the circular paths? (b) What is the linear speed of the cheetah along its path? (If you did not account for the circular motion, you would conclude erroneously that the cheetah's speed is \(114 \mathrm{~km} / \mathrm{h}\), and that type of error was apparently made in the published reports)

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.