/*! 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

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At \(t=0\), a flywheel has an angular velocity of \(4.7 \mathrm{rad} / \mathrm{s}\), 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
(a) 44.18 rad, (b) 5.02 s, (c) 35.22 s, (d) -0.63 s, (e) 33.32 s

Step by step solution

01

Find the maximum angle theta_max

The maximum angle will be achieved when the angular velocity becomes zero. We use the equation \( \omega^2 = \omega_0^2 + 2\alpha\theta \) where \( \omega_0 = 4.7 \mathrm{rad/s} \), \( \alpha = -0.25 \mathrm{rad/s}^2 \), and \( \omega = 0 \) (since the angular velocity is zero). Substituting the values, we get:\[0 = (4.7)^2 + 2(-0.25)\theta_{\max}.\]Solving for \( \theta_{\max} \), we find:\[\theta_{\max} = \frac{(4.7)^2}{0.5} = 44.18 \mathrm{rad}.\]
02

Find the first time at which theta is half of theta_max

We need the time \( t_1 \) when \( \theta = \frac{1}{2} \theta_{\max} = 22.09 \mathrm{rad} \). Using the equation \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), we set \( 22.09 = 4.7t + \frac{1}{2}(-0.25)t^2 \) and simplify to:\[4.7t - 0.125t^2 = 22.09.\]Rearranging gives:\[0.125t^2 - 4.7t + 22.09 = 0,\]solving this quadratic equation using the quadratic formula yields \( t_1 \approx 5.02 \) seconds.
03

Find the second time at which theta is half of theta_max

The second time \( t_2 \) corresponds to the second solution of step 2's quadratic equation. This time, substituting the negative root from the quadratic formula (due to symmetry of the parabolic path):\[t_2 = \frac{4.7 + \sqrt{4.7^2 - 4 \times 0.125 \times 22.09}}{2 \cdot 0.125} \approx 35.22 \text{ seconds}.\]
04

Find the negative time when theta is 10.5 rad

To find a negative time \( t_{neg} \) when \( \theta = 10.5 \) rad, set up the equation \( 10.5 = 4.7t_{neg} + \frac{1}{2}(-0.25)t_{neg}^2 \):\[0.125t_{neg}^2 - 4.7t_{neg} + 10.5 = 0.\]Solving the quadratic equation using the negative root yields \( t_{neg} \approx -0.63 \) seconds.
05

Find the positive time when theta is 10.5 rad

For the positive time \( t_{pos} \), we find the positive solution of the quadratic from step 4:\[t_{pos} = \frac{4.7 + \sqrt{4.7^2 - 4 \cdot 0.125 \cdot 10.5}}{2 \cdot 0.125} \approx 33.32 \text{ seconds}.\]

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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 the rate at which an object spins around a center or a particular axis. It's like how fast a Ferris wheel spins when you’re on it. This speed is measured in radians per second (\( \text{rad/s} \)), with the radians representing the angle through which the wheel rotates. When we say our flywheel has an angular velocity of \( 4.7 \text{ rad/s} \), it means that every second, the wheel turns \( 4.7 \) radians. You can think of radians as a way to measure angles, much like degrees. There are \( 2\pi \) radians in a full turn (about \( 6.28 \) radians), so our flywheel is spinning fairly quickly to the naked eye.
  • An understanding of angular velocity helps us predict how fast or slow something spins.
  • It's an essential parameter in rotational kinematics, covering the math of motion for rotating bodies.
Angular velocity changes can tell us much about the dynamics of rotating systems.
Angular Acceleration
Angular acceleration measures how quickly the angular velocity changes with time. Imagine sitting on a merry-go-round: if it speeds up or slows down, you’re experiencing angular acceleration. For our flywheel, it has a constant angular acceleration of \(-0.25 \text{ rad/s}^2\).What does this negative sign mean? It indicates a decrease in the angular velocity, showing the flywheel is slowing down. Just as cars can accelerate and decelerate, so can rotating objects. Angular acceleration is the measure of a change in angular velocity.
  • An important factor in determining when an object will stop or start moving faster/slower.
  • If acceleration is negative as in our example, it acts to reduce the velocity over time.
Mastering this concept can help predict when spinning objects reach certain speeds or stop.
Quadratic Equations
Quadratic equations are polynomials where the highest exponent of the variable is 2. They appear frequently when analyzing both linear and rotational motions. Think of them like algebra puzzles used to find key points in motion.Our exercise involved solving a quadratic equation to find when the flywheel’s reference line achieves half its maximum rotation (\( \theta = \frac{1}{2} \theta_{\max} \)) and when it reaches \( 10.5 \text{ rad} \). The standard form of a quadratic equation is:\[ ax^2 + bx + c = 0 \]where \( a, b, \text{ and}\ c \) are constants. Using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]we solve these equations to find time values:
  • Gives two solutions often to depict different pathways (either time going forward or backward).
  • Helps us find critical time points in a motion like when an object stops moving.
Quadratics can initially seem complex but offer valuable insights once decoded.
Graphing Motion
Graphing motion in rotational systems can visually represent how position and speed change over time. The \( \theta \) versus \( t \) graph we are asked to draw demonstrates how the flywheel's angle changes from the start.Key aspects to graph include:
  • The starting angular velocity (\( 4.7 \text{ rad/s} \)) which sets the initial slope or steepness of the graph.
  • The influence of angular acceleration (\(-0.25 \text{ rad/s}^2\)) which curbs the slope towards a flat line at maximum angle (\( \theta_{\max} \)).
By marking each calculated time and corresponding angle, the graph becomes a comprehensive story of rotational dynamics:
  • Early on, drastic changes confirm the initial speed of the wheel.
  • Gradual flattening shows deceleration.
  • Positive or negative times illustrate the symmetry and eventual reversal in the rotation.
Graphing these interactions provides valuable understanding and clarity on motion progression.

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Most popular questions from this chapter

A \(0.50 \mathrm{~kg}\) meter stick can rotate around an axis perpendicular to the stick. Find the difference in the stick's rotational inertia about the rotation axis if that axis is initially at the point marked "40 \(\mathrm{cm}\) " and then at the point marked " \(10 \mathrm{~cm} . "\)

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period \(T\) of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of \(T=0.033 \mathrm{~s}\) that is increasing at the rate of \(1.26 \times 10^{-5} \mathrm{~s} / \mathrm{y}\). (a) What is the pulsar's angular acceleration \(\alpha ?\) (b) If \(\alpha\) is constant, how many years from now will the pulsar stop rotating? (c) The pulsar originated in a supernova explosion seen in the year 1054 . Assuming constant \(\alpha\), find the initial \(T\).

A merry-go-round rotates from rest with an angular acceleration of \(1.20 \mathrm{rad} / \mathrm{s}^{2}\). How long does it take to rotate through (a) the first \(2.00\) rev and (b) the next \(2.00\) rev?

A disk, with a radius of \(0.25 \mathrm{~m}\), is to be rotated like a merrygo- round through 800 rad, starting from rest, gaining angular speed at the constant rate \(\alpha_{1}\) through the first 400 rad and then losing angular speed at the constant rate \(-\alpha_{1}\) until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed \(400 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the least time required for the rotation? (b) What is the corresponding value of \(\alpha_{1} ?\)

A gyroscope flywheel of radius \(2.62 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is \(2760 \mathrm{rev} / \mathrm{min}\). (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

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