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Chapter 9: Problem 17

A safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)

Short Answer

Expert verified
The blade will take 18 revolutions to come to rest from an angular speed of \( \omega_3 = 3\omega_1 \).

Step by step solution

01

Understand the Problem

We need to find the number of revolutions required for the mower's blade to come to rest from an angular speed three times greater while undergoing the same constant angular deceleration.
02

Define Known Variables

Initially, the blade stops in 1.00 revolution from angular speed \( \omega_1 \). Therefore, the angular acceleration \( \alpha \) is constant. We are given \( \omega_3 = 3 \omega_1 \).
03

Use Kinematic Equation for Angular Motion

The kinematic equation for angular motion is \( \omega^2 = \omega_0^2 + 2\alpha\theta \), where \( \omega \) is the final angular velocity, \( \omega_0 \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( \theta \) is the angular displacement.
04

Apply Initial Condition Equation

For the first condition: \( 0 = \omega_1^2 + 2\alpha\theta_1 \), where \( \theta_1 = 1.00 \, \text{revolution} \). Solve for \( \alpha \): \( \alpha = -\frac{\omega_1^2}{2} \).
05

Solve for New Condition

Using the same equation for the new initial speed \( \omega_3 \) and displacement \( \theta_2 \): \( 0 = (3\omega_1)^2 + 2\alpha\theta_2 \). Substitute \( \alpha \): \( 9\omega_1^2 = -\frac{\omega_1^2}{2} \theta_2 \).
06

Solve for \( \theta_2 \)

Solve the equation \( 9\omega_1^2 = -\frac{\omega_1^2}{2} \theta_2 \) for \( \theta_2 \). Simplifying, we get \( \theta_2 = 18 \), meaning it takes 18 revolutions.

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Key Concepts

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

Angular Acceleration
In the realm of rotational dynamics, angular acceleration (\( \alpha \)) measures how quickly an object speeds up or slows down as it rotates. It describes the rate of change of angular velocity over time. Just like linear acceleration defines how quickly something moves faster or slower in a straight line, angular acceleration does the same but in a circular path.

In practical terms, when you spin a merry-go-round or stop it, you are causing angular acceleration.
  • Positive angular acceleration: when an object spins faster.
  • Negative angular acceleration (or deceleration): when it slows down.
In this exercise, the mower's blade undergoes constant angular deceleration since it slows from a speed to a halt in a defined span (1 revolution initially). This constant nature of \( \alpha \) simplifies calculations, allowing us to use kinematic equations to predict outcomes like total revolutions for given conditions.
Kinematic Equations
Kinematic equations in rotational motion are similar to those for linear motion. They allow us to predict unknown motion variables when others are known. For angular motion, the basic kinematic equation we utilize is:\[\omega^2 = \omega_0^2 + 2\alpha\theta\]Here, \( \omega \) is the final angular speed, \( \omega_0 \) is initial angular speed, \( \alpha \) is angular acceleration, and \( \theta \) is angular displacement.
  • This equation helps relate the initial and final speeds to the angular displacement and acceleration.
  • It鈥檚 particularly useful in scenarios involving constant acceleration.
Understanding each term can help solve problems like the mower's blade, where the speed is adjusted. We know the initial speed, calculate the angular acceleration, and check how displacement changes if speed changes.
Angular Displacement
Angular displacement (\( \theta \)) is the angle a rotating object sweeps out over time. Unlike linear displacement, which is measured in meters, angular displacement is typically measured in radians or revolutions.
In our exercise, one complete revolution was linked to bringing the mower blade to a halt from its initial speed. Here, \( \theta = 1.00 \) revolution. We wanted to find the angular displacement for a speed three times greater.
  • Angular displacement calculates how far an object has rotated.
  • It plays a crucial role in determining time in rotational problems.
In our scenario, switching from \( \omega_1 \) to \( 3\omega_1 \) increases angular displacement to \( 18 \) revolutions, showcasing how initial speed amplifies travel distance during deceleration.
Angular Speed
Angular speed (\( \omega \)) describes how fast an object spins around an axis. The unit here is radians per second (rad/s) or revolutions per unit time. It is similar to linear speed but in a circular course.
In our exercise, the initial angular speed is given as \( \omega_1 \), and we explore what happens when it triples to \( \omega_3 = 3\omega_1 \).
  • Initial angular speed: Starting speed of an object (mower's blade in our case).
  • Final angular speed: Speed at which an object ends, often zero when stopping.
Tripling the starting speed notably affects the number of revolutions needed to stop with the same angular acceleration. Understanding this relationship highlights how changes in speed impact rotational behaviors in systems like power tools or vehicles.

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Most popular questions from this chapter

An airplane propeller is rotating at 1900 TPm (rev/min). a) Compute the propeller's angular velocity in rad/s. (b) How many seconds does it take for the propeller to turn through \(35^{\circ}\) ?

The flywheel of a gasoline engine is required to give up 500 \(\mathrm{J}\) of kinetic energy while its angular velocity decreases from 650 rev /min to 520 rev/min. What moment of inertia is required?

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 \(\mathrm{cm}\) . If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 \(\mathrm{m} / \mathrm{s} ?\) The rear wheel has radius 0.330 \(\mathrm{m} .\)

Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7} \mathrm{m}\) deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 \(\mathrm{mm}\) and 58.0 \(\mathrm{mm}\) , respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 \(\mathrm{m} / \mathrm{s}\) . (a) What is the angular speed of the CD when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a \(\mathrm{CD}\) is 74.0 min. What would be the length of the track on such a maximum-duration \(\mathrm{CD}\) if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum-duration CD during its 74.0 -min playing time? Take the direction of rotation of the dise to be positive.

A twirler's baton is made of a slender metal cylinder of mass \(M\) and length \(L\) . Each end has a rubber cap of mass \(m,\) and you can accurately treat each cap as a particle in this problem. Find the total moment of inertia of the baton about the usual twirling axis (perpendicular to the baton through its center).
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