/*! 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 whecl on a fircworks ... [FREE SOLUTION] | 91Ó°ÊÓ

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A spinning whecl on a fircworks 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 Known Variables

First, let's identify the known variables given in the problem. The initial angular velocity \( \omega_i \) is unknown, the final angular velocity \( \omega_f \) is \(-25.0 \, \text{rad/s}\), the angular acceleration \( \alpha \) is \(-4.00\, \text{rad/s}^2\), and the angular displacement \( \theta \) is zero.
02

Use Angular Displacement Equation

We use the angular displacement equation: \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \). Given that \( \theta = 0 \), the equation simplifies to: \( 0 = \omega_i t + \frac{1}{2} \alpha t^2 \).
03

Solve for Initial Angular Velocity

Rearrange the equation to find \( \omega_i \): \( \omega_i = -\frac{1}{2} \alpha t \).
04

Use Angular Velocity Equation

We also use the angular velocity equation: \( \omega_f = \omega_i + \alpha t \). Substituting \( \omega_i = -\frac{1}{2} \alpha t \), we have \( \omega_f = -\frac{1}{2} \alpha t + \alpha t \).
05

Simplify to Relate Final Velocity and Acceleration

Simplify to \( \omega_f = \frac{1}{2} \alpha t \). Substitute \( \omega_f = -25.0 \, \text{rad/s} \) and \( \alpha = -4.00 \, \text{rad/s}^2 \), so \( -25.0 = \frac{1}{2} (-4.00) t \).
06

Solve for Time

Calculate \( t \) by rearranging the equation: \( -25.0 = -2.00 t \). Solving gives \( t = 12.5 \, \text{seconds} \).

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Key Concepts

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

Angular Displacement
Angular displacement refers to the angle through which an object has rotated around a fixed point. It is typically measured in radians and is denoted by the Greek letter theta (\(\theta\)). In the context of circular motion, it is the change in the angle as an object moves along a circular path.

For the fireworks display wheel, the problem statement offers a special case where the angular displacement is zero. This means that although the wheel was initially spinning, it returned to its initial angular position after completing its motion.

Understanding angular displacement can be likened to imagining a pendulum swing that returns to its starting point, hence experiencing zero net displacement despite having moved.
Angular Velocity
Angular velocity represents how quickly an object is rotating. It is expressed in radians per second (\(\text{rad/s}\)) and indicates the rate of change of the angular position.

In the fireworks wheel example, the angular velocity changes from an unknown initial value to a final value of \(-25.0\, \text{rad/s}\). The negative sign here suggests that the wheel, which was rotating counterclockwise, has reversed direction to clockwise movement by the end of the given motion.

Angular velocity is crucial in determining how a rotating system is behaving over time and can be calculated using various equations based on the specifics of the problem, such as initial conditions and time elapsed.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. It is measured in radians per second squared (\(\text{rad/s}^2\)) and indicates whether an object is speeding up or slowing down its rotation.

For the fireworks wheel, the angular acceleration is \(-4.00\, \text{rad/s}^2\). The negative value means that the wheel is decelerating in its counterclockwise motion, which eventually leads it to stop momentarily and then reverse its direction.

Understanding this concept is similar to considering a car that's slowing down as you apply brakes if you're driving in a circle. Eventually, this negative acceleration will cause the car to stop and potentially roll back in the opposite direction.
Spinning Motion
Spinning motion refers to the motion of an object rotating around an axis. In the instance of the fireworks wheel, the wheel is initially spinning in a counterclockwise direction.

The spinning motion involves all the previously discussed concepts—angular displacement, velocity, and acceleration. Each of these aspects helps frame the overall rotational behavior of the wheel.

In practical applications, spinning motion is found in numerous real-world scenarios like the spinning of a merry-go-round, the turning of gears, and the movement of celestial bodies. This makes understanding spinning motion and its parameters essential for grasping more complex rotational dynamics.

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Most popular questions from this chapter

A motorcyclist is traveling along a road and accelerates for 4.50 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 wheel 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?

A 220-kg speedboat is negotiating a circular turn (radius 32 m) around a buoy. During the turn, the engine causes a net tangential force of magnitude 550 N to be applied to the boat. The initial tangential speed of the boat going into the turn is 5.0 m/s. (a) Find the tangential acceleration. (b) After the boat is 2.0 s into the turn, find the centripetal acceleration.

A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The angular speed of the child has a constant valuc of 0.250 \(\mathrm{rad} / \mathrm{s}\) . At the instant the child spots the horse, onc-quarter of a turn away, the merry-go-round begins to move (in the direction the child is runing) with a constant angular acceleration of 0.0100 \(\mathrm{rad} / \mathrm{s}^{2}\) . What is the shortest time it takes for the child to catch up with the horse?

A compact disc (CD) contains music on a spiral track. Music is pul onto a CD with the assumption that, during playback, the music will be detected at a constant tangential speed at any point. Since \(v_{\mathrm{T}}=r \omega, \mathrm{a}\) CD rotates at a smaller angular speed for music near the outer edge and a larger angular speed for music neur the inner part of the disc. For music at the outer edge (r 0.0568 m), the angular speed is 3.50 rev/s. Find (a) the constant tangential speed at which music is detected and (b) the angular speed (in rev/s) for music at a distance of 0.0249 m from the center of a CD.

ssm 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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