/*! 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 32 A spinning wheel on a fireworks ... [FREE SOLUTION] | 91Ó°ÊÓ

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A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.00 \(\mathrm{rad} / \mathrm{s}^{2} .\) Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of \(-25.0 \mathrm{rad} / \mathrm{s} .\) While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Short Answer

Expert verified
The time required is 12.5 seconds.

Step by step solution

01

Identify and List Knowns

We are given the following pieces of information:- Angular acceleration, \( \alpha = -4.00 \, \mathrm{rad/s^2} \)- Final angular velocity, \( \omega_f = -25.0 \, \mathrm{rad/s} \)- Angular displacement, \( \theta = 0 \, \mathrm{rad} \)We are asked to find the time, \( t \), for this change to occur.
02

Use the Angular Displacement Formula

Since the angular displacement \( \theta = 0 \), we can use the formula for angular displacement:\[ \theta = \omega_{ ext{initial}} t + \frac{1}{2} \alpha t^2 \]Substitute \( \theta = 0 \) and rearrange:\[ 0 = \omega_{ ext{initial}} t + \frac{1}{2} (-4.00) t^2 \]
03

Solve for Initial Angular Velocity

Since the above equation simplifies to \( 0 = \omega_{ ext{initial}} t - 2t^2 \), divide by t (assuming \( t eq 0 \)):\[ \omega_{ ext{initial}} = 2t \]This expression will help us link \( \omega_{ ext{initial}} \) and \( t \).
04

Relate Angular Velocities Using Acceleration

Use the relationship between angular velocities and angular acceleration:\[ \omega_f = \omega_{ ext{initial}} + \alpha t \]Substitute \( \omega_f = -25.0 \), \( \alpha = -4.00 \), and \( \omega_{ ext{initial}} = 2t \):\[ -25.0 = 2t - 4.00 t \]Which simplifies to:\[ -25.0 = -2t \]
05

Solve for Time t

From the equation \( -25.0 = -2t \), solve for \( t \):\[ t = \frac{-25.0}{-2} = 12.5 \]Thus, the time required for the angular velocity to change is 12.5 seconds.

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Key Concepts

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

Angular Acceleration
Angular acceleration is the rate at which the angular velocity of an object changes with time. It can be thought of as the rotational counterpart of linear acceleration. In this exercise, the angular acceleration is given as \( \alpha = -4.00 \, \mathrm{rad/s^2} \). This negative value indicates that the wheel is slowing down, much like how negative linear acceleration (often called deceleration) describes a slowing car.

When dealing with angular acceleration, you can expect:
  • Units in radians per second squared (\( \mathrm{rad/s^2} \)).
  • A direction: counterclockwise or clockwise, which is determined by the sign. Here, a negative value suggests deceleration or a shift in the clockwise direction.
  • An effect on angular velocity, affecting how quickly an object spins or comes to a halt.
Understanding how the acceleration affects the wheel helps us solve for other rotational dynamics, such as the time or angular displacement needed for changes in rotation to occur.
Angular Velocity
Angular velocity refers to the speed of rotation of an object and the direction in which it is rotating. It is often one of the first-known quantities when analyzing rotational motion. In our problem, the wheel's angular velocity changes from an unknown initial value to a final value of \( \omega_f = -25.0 \, \mathrm{rad/s} \).

It's important to note:
  • Angular velocity is measured in radians per second (\( \mathrm{rad/s} \)).
  • The negative sign indicates the direction of rotation, suggesting counterclockwise rotation here was decelerating to a halt, similar to velocity pointing to the opposite direction of motion.
  • In the context of angular acceleration, anyone studying rotational dynamics needs to find relations between initial and final angular velocities to compute unknowns like time.
Through the relationship \( \omega_f = \omega_{\text{initial}} + \alpha t \), we connect angular velocities easily to time, settling on the need to determine the time \( t \) given known values for \( \omega_f \) and \( \alpha \).
Angular Displacement
Angular displacement looks at the change in the orientation of an object as it rotates about an axis. Unlike linear displacement, angular displacement measures the rotation angle when a body revolves around a circular path. An interesting twist in the exercise is that the angular displacement equals zero, despite changes in velocity and acceleration.

Angular displacement characteristically features:
  • Measured in radians (\( \mathrm{rad} \)).
  • Positive or negative signs indicating direction, though here centered on being zero signifies focus on the initial and final positions being the same.
  • Used in equations such as \( \theta = \omega_{\text{initial}} t + \frac{1}{2} \alpha t^2 \) to tie back to other rotational elements.
In this scenario, a displacement of zero hints at a similar spot after completing an interval, resembling the motion of a ball tossed up that returns to its starting point after slowing down and halting. Understanding this gives key insights into rotational kinematics and aids in determining the period of time taken for these observed rotational changes.

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Most popular questions from this chapter

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is \(8.3 \mathrm{m}\) above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.6 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

Suppose you are driving a car in a counterclockwise direction on a circular road whose radius is \(r=390 \mathrm{m}\) (see the figure). You look at the speedometer and it reads a steady \(32 \mathrm{m} / \mathrm{s}\) (about \(72 \mathrm{mi} / \mathrm{h}\) ). Concepts: (i) Does an object traveling at a constant tangential speed (for example, \(\left.v_{\mathrm{T}}=32 \mathrm{m} / \mathrm{s}\right)\) along a circular path have an acceleration? (ii) Is there a tangential acceleration \(\overrightarrow{\mathbf{a}}_{\mathrm{T}}\) when the angular speed of an object changes (e.g., when the car's angular speed decreases to \(4.9 \times 10^{-2} \mathrm{rad} / \mathrm{s}\) )? Calculations: (a) What is the angular speed of the car? (b) Determine the acceleration (magnitude and direction) of the car. (c) To avoid a rear-end collision with the vehicle ahead, you apply the brakes and reduce your angular speed to \(4.9 \times 10^{-2} \mathrm{rad} / \mathrm{s}\) in a time of 4.0 s. What is the tangential acceleration (magnitude and direction) of the car?

The sun has a mass of \(1.99 \times 10^{30} \mathrm{kg}\) and is moving in a circular orbit about the center of our galaxy, the Milky Way. The radius of the orbit is \(2.3 \times 10^{4}\) light-years ( 1 light-year \(\left.=9.5 \times 10^{15} \mathrm{m}\right)\), and the angular speed of the sun is \(1.1 \times 10^{-15} \mathrm{rad} / \mathrm{s} .\) (a) Determine the tangential speed of the sun. (b) What is the magnitude of the net force that acts on the sun to keep it moving around the center of the Milky Way?

An auto race takes place on a circular track. A car completes one lap in a time of \(18.9 \mathrm{s}\), with an average tangential speed of \(42.6 \mathrm{m} / \mathrm{s}\). Find \((\mathrm{a})\) the average angular speed and (b) the radius of the track.

A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from 3.00 to 5.00 rev/s while rotating through one-half of a revolution. How much time does this maneuver take?

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