/*! 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 4 A flywheel rotates with a unifor... [FREE SOLUTION] | 91Ó°ÊÓ

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A flywheel rotates with a uniform angular acceleration. Its angular velocity increases from \(20 \pi \operatorname{rads}^{-1}\) to \(40 \pi \mathrm{rads}^{-1}\) in \(10 \mathrm{~s}\). How many rotations did it make in this period? (a) 80 (b) 100 (c) 120 (d) 150

Short Answer

Expert verified
150 rotations; option (d) is correct.

Step by step solution

01

Identify the Given Values

From the problem, we know the initial angular velocity \( \omega_i = 20\pi \ \mathrm{rad/s} \), the final angular velocity \( \omega_f = 40\pi \ \mathrm{rad/s} \), and the time period \( t = 10\, \mathrm{s} \).
02

Calculate Angular Acceleration

Angular acceleration \( \alpha \) can be found using the formula \( \alpha = \frac{\omega_f - \omega_i}{t} \). Substituting the given values, \( \alpha = \frac{40\pi - 20\pi}{10} = 2\pi \ \mathrm{rad/s^2} \).
03

Determine Total Angular Displacement

The angular displacement \( \theta \) can be calculated using the formula \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \). Substituting the known values, \( \theta = 20\pi \times 10 + \frac{1}{2} \times 2\pi \times 10^2 \). Simplifying gives \( \theta = 200\pi + 100\pi = 300\pi \ \mathrm{radians} \).
04

Convert Angular Displacement to Rotations

Since there are \( 2\pi \) radians in one rotation, the number of rotations is given by \( \frac{\theta}{2\pi} = \frac{300\pi}{2\pi} \). Simplifying gives 150 rotations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves around another point. It is expressed in radians per second (rad/s). In the context of our problem, we have two key values: an initial angular velocity of \(20\pi \ \mathrm{rad/s}\) and a final angular velocity of \(40\pi \ \mathrm{rad/s}\).
The change in angular velocity indicates that our system experiences acceleration. This is because the final speed is higher than the initial, showing an increase in the rotation rate.Understanding angular velocity helps us describe the motion of objects that spin or rotate. It is a crucial aspect of rotational dynamics, alongside angular displacement and angular acceleration.
In practical terms, angular velocity is analogous to velocity in linear motion, providing the speed and direction of an object's rotation. Just as linear velocity describes how quickly an object moves from one point to another, angular velocity describes how quickly an object rotates around a fixed axis.
Rotational Dynamics
Rotational dynamics deals with the motion of objects that rotate. It considers forces and torques that induce rotation, helping us understand how objects behave when they spin. In our exercise, rotational dynamics is at play, as we examine a flywheel accelerating under constant angular acceleration.
Rotational dynamics is governed by similar principles as linear dynamics, but adapted for rotation:
  • Angular Velocity: how fast the object rotates.
  • Angular Acceleration: how the rotational speed changes over time.
  • Torque: the rotational equivalent of force, which influences angular acceleration.
To analyze the flywheel's motion, we apply Newton's second law of rotation, which stated for angular quantities is \( I\alpha = \tau \), where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration. Although torque (\(\tau\)) isn't directly mentioned in our problem, understanding that acceleration results from applied torque enhances comprehension of rotational dynamics overall.The step-by-step solution provided gives insight into rotational dynamics by calculating angular acceleration based on changes in angular velocity and then applying that information to find the angular displacement.
Angular Displacement
Angular displacement refers to the angle through which an object rotates about a specific axis over time. It's measured in radians and conveys how much "turn" has happened.
In our example, the objective was to find how many rotations the flywheel completed as its angular velocity increased. We used the formula for angular displacement: \(\theta = \omega_i t + \frac{1}{2} \alpha t^2\). In this case, \(\theta\) came out to be 300\(\pi\) radians.The unit radian measures angles based on the radius length wrapped exactly around a circle. Since one full rotation equates to \(2\pi\) radians, to convert angular displacement into rotations, we divide \(\theta\) by \(2\pi\). In our solution, simplifying \(\frac{300\pi}{2\pi}\) gave 150 rotations.
This aspect of rotational dynamics helps relate angular changes to the physical, tangible rotations we observe, converting mathematical insights into practical, measurable outcomes. Angular displacement not only tells us the extent of rotation but aids in linking back to real-world quantities like revolutions or cycles.

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Most popular questions from this chapter

A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with a constant angular velocity @. Two objects each of mass \(M\) are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity (a) \(\frac{\omega(M-2 m)}{M+2 m}\) (b) \(\frac{\omega M}{M+2 m}\) (c) \(\frac{\omega M}{M+m}\) (d) \(\frac{\omega(M+2 M)}{M}\)

The spin driver of a washing machine revolving at 15 rps slow down to 5 rps, while making 50 revolutions. Angular acceleration of the driver is (a) \(-4 \pi \mathrm{rads}^{-2}\) (b) \(-4 \pi \mathrm{rads}^{-2}\) (c) \(8 \pi \mathrm{rads}^{-2}\) (d) \(-8 \pi \mathrm{rads}^{-2}\)

Two dises of the same material and thickness have radii \(0.2 \mathrm{~m}\) and \(0.6 \mathrm{~m}\). Their moments of inertia about the axes will be in the ratio of [BVP Engs, 2005] (a) \(1: 81\) (b) \(1: 37\) (c) \(1: 9\) (d) \(1: 3\)

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Moment of inertia of a solid cylinder of length \(L\) and diameter \(D\) about an axis passing through its centre of gravity and perpendicular to its geometric axis is (a) \(M\left(\frac{D^{2}}{4}+\frac{L^{2}}{12}\right)\) (b) \(M\left(\frac{L^{2}}{16}+\frac{D^{2}}{8}\right)\) (c) \(M\left(\frac{D^{2}}{4}+\frac{L^{2}}{6}\right)\) (d) \(M\left(\frac{L^{2}}{12}+\frac{D^{2}}{16}\right)\)

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