/*! 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 25 A spinning wheel on a fireworks ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 25

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 the known values

We are given the following values: the angular acceleration \( \alpha = -4.00 \, \mathrm{rad/s}^2 \), the initial angular velocity \( \omega_i \), the final angular velocity \( \omega_f = -25.0 \, \mathrm{rad/s} \), and the angular displacement \( \theta = 0 \).
02

Set up the angular kinematic equation

Use the kinematic equation for angular motion: \[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]. Since \( \theta = 0 \), the equation simplifies to: \[ 0 = \omega_i t + \frac{1}{2} \alpha t^2 \].
03

Solve for initial angular velocity, \( \omega_i \)

Use the equation \( \omega_f = \omega_i + \alpha t \) to solve for \( \omega_i \). Rearrange the equation to get:\[ \omega_i = \omega_f - \alpha t \].
04

Substitute \( \omega_i \) into the angular displacement equation

Substitute \( \omega_i = \omega_f - \alpha t \) into \( 0 = \omega_i t + \frac{1}{2} \alpha t^2 \). The equation becomes:\[ 0 = (\omega_f - \alpha t)t + \frac{1}{2} \alpha t^2 \].
05

Simplify and solve the quadratic equation

The equation \( 0 = (\omega_f - \alpha t)t + \frac{1}{2} \alpha t^2 \) simplifies to:\[ 0 = \omega_f t - \alpha t^2 + \frac{1}{2} \alpha t^2 \].This further reduces to:\[ 0 = \omega_f t - \frac{1}{2} \alpha t^2 \].Reorder to find: \[ \frac{1}{2} \alpha t^2 = \omega_f t \].
06

Factor out \( t \) and solve for \( t \)

Factor \( t \) out of the equation: \[ t(\omega_f - \frac{1}{2} \alpha t) = 0 \].Solve for \( t \) by setting each factor to zero:1. \( t = 0 \) 2. \( \omega_f - \frac{1}{2} \alpha t = 0 \)The second equation gives us: \[ \frac{1}{2} \alpha t = \omega_f \]\[ t = \frac{2 \omega_f}{\alpha} \].
07

Plug in the values to find the time \( t \)

Substitute \( \omega_f = -25.0 \, \mathrm{rad/s} \) and \( \alpha = -4.00 \, \mathrm{rad/s}^2 \) into: \[ t = \frac{2 \times -25.0}{-4.00} \].Calculate to find: \[ t = \frac{-50.0}{-4.00} = 12.5 \, \mathrm{s} \].

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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 of change of angular velocity over time. It tells us how quickly an object is speeding up or slowing down its spin. In this problem, the angular acceleration is given as \(-4.00 \mathrm{rad/s}^2\). The negative sign indicates that the wheel is decelerating.When working with physics, it鈥檚 important to pay attention to signs, as they indicate direction. In angular motion:
  • Positive acceleration might mean speeding up if the rotation is counterclockwise.
  • Negative acceleration might mean slowing down.
This exercise features negative angular acceleration. Therefore, the wheel is slowing down its counterclockwise rotation.
Angular Velocity
Angular velocity is how fast something spins. It measures how much of the angle the rotating object covers in a specific time period, typically in radians per second. In this exercise:- The initial angular velocity isn't explicitly given, but it's something we need to calculate.- The final angular velocity provided is \(-25.0 \mathrm{rad/s}\),and it's in a counterclockwise direction as indicated by the negative sign.An object's angular velocity can change due to angular acceleration, which either speeds up or slows down the rotation.
Angular Displacement
Angular displacement is the change in the angle as an object rotates, and in this problem, it's given as zero. This may sound strange at first, since we're discussing a rotating wheel. The zero angular displacement suggests that, similar to a ball thrown up and coming back to the start point, the wheel rotates and returns to its original position after a period. It's a unique situation of motion where also the initial and final points align perfectly.
Kinematic Equations
The kinematic equations describe how things move. They help us connect displacement, velocity, acceleration, and time in a systematic way. In our problem, we used one such equation uniquely suited for rotational motions:\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]Where:
  • \(\theta\) is angular displacement,
  • \(\omega_i\) is initial angular velocity,
  • \(t\) is time,
  • \(\alpha\) is angular acceleration.
Since angular displacement is zero here, the equation transforms into a quadratic form to solve for time. This simplification helps find unknowns when solving motion-related problems.
Quadratic Equation
In this exercise, solving for time required us to work through a quadratic equation, which surfaces when we have motion equations involving squared terms, like time squared.The equation derived was:\[ t(\omega_f - \frac{1}{2} \alpha t) = 0 \]Rearranging gets us:\[ t = \frac{2 \omega_f}{\alpha} \]Where \(t=0\) is a trivial solution that does not apply in this context.A quadratic equation is key in physics problems where solving for variables requires identifying roots. Often it connects precisely back with physical constraints that simplify understanding complex motion.

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Most popular questions from this chapter

A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes through one of the open spaces between the propeller blades. The vertical drop of the arrow may be ignored. There is a maximum value for the angular speed \(\omega\) of the propeller beyond which the arrow cannot pass through an open space without being struck by one of the blades. (a) If the arrow is to pass through an open space, does it matter if the arrow is aimed closer to or farther away from the axis (see points \(\mathrm{A}\) and \(\mathrm{B}\) in the drawing, for example)? Explain. (b) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow speed \(v ?\) Why? (c) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow length \(L\) ? Justify your answer.

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 \(/ \mathrm{s}\) while rotating through one-half of a revolution. How much time does this maneuver take?

At reviews the approach that is necessary for solving problems such as this one. A motorcyclist is traveling along a road and accelerates for \(4.50 \mathrm{~s}\) to pass another cyclist. The angular acceleration of each wheel is \(+6.70 \mathrm{rad} / \mathrm{s}^{2}\) and, just after passing, the angular velocity of each is \(+74.5 \mathrm{rad} / \mathrm{s}\), where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?

The penny-farthing is a bicycle that was popular between 1870 and 1890 . As the drawing shows, this type of bicycle has a large front wheel and a small rear wheel. On a Sunday ride in the park the front wheel (radius \(=1.20 \mathrm{~m}\) ) makes 276 revolutions. How many revolutions does the rear wheel (radius \(=0.340 \mathrm{~m}\) ) make?

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by whirling it rapidly in a horizontal circle and releasing it at the right moment. Such a slingshot is used to throw a stone from the edge of a cliff, the point of release being \(20.0 \mathrm{~m}\) above the base of the cliff. The stone lands on the ground below the cliff at a point \(X .\) The horizontal distance of point \(X\) from the base of the cliff (directly beneath the point of release) is thirty times the radius of the circle on which the stone is whirled. Determine the angular speed of the stone at the moment of release.
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