/*! 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 17 At \(t=0\), a flywheel has an an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
Max angle 44.18 rad; first time 4.5 s; second time 35.0 s; neg. time -1.5 s; pos. time 65.0 s.

Step by step solution

01

Determine Maximum Angle (θ_max)

The reference line will reach its maximum angle when the angular velocity becomes zero, i.e., \[\omega = \omega_0 + \alpha t = 0\]where \(\omega_0 = 4.7 \mathrm{rad}/\mathrm{s}\) is the initial angular velocity and \(\alpha = -0.25 \mathrm{rad}/\mathrm{s}^2\) is the angular acceleration. Solve for time \(t\):\[t = \frac{-\omega_0}{\alpha} = \frac{-4.7}{-0.25} = 18.8 \text{ seconds}\]Substitute into the rotational kinematic equation:\[\theta_{\max} = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 = 0 + 4.7 \cdot 18.8 + \frac{1}{2}(-0.25)(18.8)^2 = 44.18 \text{ radians}\]
02

Find θ When θ = (1/2)θ_max

To find when \(\theta = \frac{1}{2}\theta_{\max}\), calculate:\[\theta = \frac{1}{2} \times 44.18 = 22.09 \text{ radians}\]Use the rotational kinematic equation:\[\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2\]Substitute values:\[22.09 = 0 + 4.7 t - 0.125 t^2\]Rearranging gives a quadratic:\[-0.125 t^2 + 4.7 t - 22.09 = 0\]Solve this quadratic using the quadratic formula:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = -0.125\), \(b = 4.7\), \(c = -22.09\):\[t = \frac{-4.7 \pm \sqrt{4.7^2 - 4(-0.125)(-22.09)}}{2(-0.125)}\]Calculate the roots to get the first and second times.
03

Calculate (b) First Time: t_1

Solving the quadratic gives:\[t = \frac{-4.7 + \sqrt{4.7^2 - 4 \times (-0.125) \times (-22.09)}}{2 \times (-0.125)} = 4.5 \text{ seconds (approximately)}\]
04

Compute (c) Second Time: t_2

Continuing from the Step 2 quadratic, calculate the second time:\[t = \frac{-4.7 - \sqrt{4.7^2 - 4 \times (-0.125) \times (-22.09)}}{2 \times (-0.125)} = 35.0 \text{ seconds (approximately)}\]
05

Determine Times for θ = 10.5 rad

Use the equation again for \(\theta = 10.5\):\[10.5 = 0 + 4.7 t - 0.125 t^2\]Reforming the quadratic gives:\[-0.125 t^2 + 4.7 t - 10.5 = 0\]Solve using the quadratic formula:\[t = \frac{-4.7 \pm \sqrt{4.7^2 - 4(-0.125)(-10.5)}}{2(-0.125)}\]Find negative and positive times from roots.
06

Calculate (d) Negative Time and (e) Positive Time

Solving the quadratic formula from Step 5 gives:For negative time:\[t = \frac{-4.7 - \sqrt{4.7^2 - 4 \times (-0.125) \times (-10.5)}}{2 \times (-0.125)} = -1.5 \text{ seconds (approximately)}\]For positive time:\[t = \frac{-4.7 + \sqrt{4.7^2 - 4 \times (-0.125) \times (-10.5)}}{2 \times (-0.125)} = 65.0 \text{ seconds (approximately)}\]

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.

Kinematics
Kinematics describes the motion of objects without considering the forces that cause this motion. In rotational motion, where angular motion takes place, we are particularly interested in quantities like angular displacement, angular velocity, and angular acceleration.
It's the foundation for understanding how objects move in circular paths.
In this exercise, we have a flywheel beginning with an angular velocity and a constant angular acceleration. We use kinematic equations to determine how its position (angular displacement) changes over time. These equations are:
  • The angular displacement equation: \[ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \]
  • The angular velocity equation: \[ \omega = \omega_0 + \alpha t \]
Understanding these equations helps solve questions related to how a rotating object changes its motion parameters over time.
Angular Velocity
Angular velocity represents how fast an object rotates or spins. It's similar to linear velocity but in the context of circular motion. It's typically denoted by the symbol \(\omega\) and measured in radians per second.
Initially, the flywheel in our problem has an angular velocity of 4.7 rad/s. As time progresses, this velocity is adjusted by any angular acceleration present.
To determine when the flywheel stops accelerating positively, we set angular velocity to zero (\(\omega = 0\)) and solve for the time using the angular velocity formula:
  • \[ \omega = \omega_0 + \alpha t = 0 \]
By calculating time \(t\), we find that the flywheel stops after 18.8 seconds. This knowledge allows us to explore further changes in the flywheel's motion.
Angular Acceleration
Angular acceleration measures how the rate of angular velocity changes over time. It's analogous to linear acceleration in straight-line motion and is denoted by \(\alpha\), measured in radians per second squared.
In this problem, the flywheel has a constant angular acceleration of -0.25 rad/s², indicating the angular velocity is decreasing over time.
To predict motion, we need to calculate how this acceleration impacts angular velocity and displacement:
  • The kinematic equation, \(\omega = \omega_0 + \alpha t\), allows us to see when the flywheel's motion reverses due to negative acceleration.
  • The equation for angular displacement, \(\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2\), determines how far the flywheel turns.
Understanding angular acceleration is essential for predicting rotational dynamics and examining how objects adjust their spin over time.
Quadratic Equations
Quadratic equations are polynomials of degree two, often taking the form \(ax^2 + bx + c = 0\). In this exercise, they emerge naturally when applying kinematic equations to find time \(t\).
To solve for times when certain angular displacements occur, such as when the flywheel reaches half its maximum angle, we rearrange the angular displacement equation into a quadratic form:
  • \(-0.125 t^2 + 4.7 t - 22.09 = 0\)
By using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]we solve for \(t\), discovering critical points in the flywheel's motion. Quadratic equations help pinpoint two distinct times when specific motion criteria are met, illustrating the utility of algebra in rotational kinematics.

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) What is the angular speed \(\omega\) about the polar axis of a point on Earth's surface at latitude \(40^{\circ} \mathrm{N} ?\) (Earth rotates about that axis.) (b) What is the linear speed \(v\) of the point? What are (c) \(\omega\) and \((\mathrm{d}) v\) for a point at the equator?

Calculate the rotational inertia of a wheel that has a kinetic energy of \(24400 \mathrm{~J}\) when rotating at 602 rev/min.

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by \(\theta=0.40 e^{2 t}\), where \(\theta\) is in radians and \(t\) is in seconds. Consider a point on the object that is \(4.0 \mathrm{~cm}\) from the axis of rotation. At \(t=0\), what are the magnitudes of the point's (a) tangential component of acceleration and (b) radial component of acceleration?

The flywheel of an engine is rotating at \(25.0 \mathrm{rad} / \mathrm{s}\). When the engine is turned off, the flywheel slows at a constant rate and stops in \(20.0 \mathrm{~s}\). Calculate (a) the angular acceleration of the flywheel, (b) the angle through which the flywheel rotates in stopping, and (c) the number of revolutions made by the flywheel in stopping.

Two uniform solid spheres have the same mass of \(1.65 \mathrm{~kg}\), but one has a radius of \(0.226 \mathrm{~m}\) and the other has a radius of \(0.854 \mathrm{~m}\). Each can rotate about an axis through its center. (a) What is the magnitude \(\tau\) of the torque required to bring the smaller sphere from rest to an angular speed of \(317 \mathrm{rad} / \mathrm{s}\) in \(15.5 \mathrm{~s}\) ? (b) What is the magnitude \(F\) of the force that must be applied tangentially at the sphere's equator to give that torque? What are the corresponding values of (c) \(\tau\) and (d) \(F\) for the larger sphere?

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.