/*! 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 A safety device brings the blade... [FREE SOLUTION] | 91影视

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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 with three times the initial speed will come to rest after completing 9 revolutions.

Step by step solution

01

Setup the known parameters

Given that the initial angular speed is \(\omega_{1}\), and the blade comes to rest (i.e final angular speed, \(\omega_{2}= 0\)) in 1.00 revolution (which is \(\Theta_{1} = 2 \pi\) radians). The problem also mentions the blade coming to rest from an initial angular speed which was thrice as great (\(\omega_{3}\) = 3\(\omega_{1}\)). We need to find \(\Theta_{3}\), the angular displacement when the blade has the initial speed of \(\omega_{3}\)
02

Use the angular kinematic equation

The first scenario gives us the angular deceleration (\(\alpha_{1}\)). We can find this by using the angular kinematic equation \(\omega_{2}^{2} = \omega_{1}^{2} + 2\alpha_{1}\Theta_{1}\). From this, we can determine the angular deceleration, \(\alpha_{1} = \(\frac{\omega_{2}^{2} - \omega_{1}^{2}}{2\Theta_{1}}\)
03

Substitute in the known values

Substituting given values, we get \(\alpha_{1} = \(\frac{0^{2} - \omega_{1}^{2}}{2(2 \pi)}\). Simplifying, we get \(\alpha_{1} = -\frac{\omega_{1}^{2}}{4 \pi}\)
04

Find the angular displacement (\(\Theta_{3}\)) for \(\omega_{3}\)

Next, we use the same angular kinematic equation to find the angular displacement for the blade that rotates with an initial speed of \(\omega_{3}\). Here, \(\omega_{2} = 0\) (comes to rest), \(\omega_{1} = \omega_{3}\), \(\alpha_{1} = \alpha_{3}\). The equation becomes \(0^{2} = \omega_{3}^{2} + 2\alpha_{3}\Theta_{3}\). Solving for \(\Theta_{3}\) gives us \(\Theta_{3} = \frac{\omega_{3}^{2} - 0^{2}}{2\alpha_{3}}\)
05

Substitute \(\omega_{3}\), \(\alpha_{3}\) into the equation

Replacement of given values gives, \(\Theta_{3} = \frac{(3\omega_{1})^{2} - 0^{2}}{2(-\frac{\omega_{1}^{2}}{4\pi})}\)
06

Evaluate \(\Theta_{3}\)

Solving the equation yields \(\Theta_{3} = -\frac{9\omega_{1}^{2}}{-\frac{\omega_{1}^{2}}{2\pi}}\). Simplifying, we get \(\Theta_{3} = 18\pi\). Note that we are seeking the magnitude only, so the negative sign indicates only that the blade moves in the opposite direction of its initial motion. We can convert \(\Theta_{3}\) back into revolutions by recognizing that \(2\pi rad = 1 rev\), so \(18\pi rad = 9\) revolutions. So, it will take 9 revolutions to stop the blade with three times the initial speed.

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Key Concepts

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

Angular Acceleration
Angular acceleration, denoted as \(\alpha\), is a measure of how quickly an object's rotational speed changes. Just like linear acceleration pertains to changes in velocity, angular acceleration corresponds to changes in angular velocity (\(\omega\)). It is defined as the rate of change of angular velocity with respect to time.

Mathematically, angular acceleration is given by \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) represents the time interval over which the change occurs. In our mower blade scenario, the blade's angular acceleration is constant and it's calculated by using the change in angular speed \(\omega_{1}\) to \(\omega_{2} = 0\) over one revolution. The negative sign signifies that this is actually an angular deceleration, as the blade is slowing down.
Angular Displacement
Angular displacement is the angle in radians through which a point or line has been rotated in a specified sense about a specified axis. It is the measure of the angle that an object has moved through in its circular path and is usually denoted by \(\Theta\). One revolution corresponds to an angular displacement of \(2\pi\) radians.

In the context of our problem, the angular displacement \(\Theta_{1}\) for the first situation is one complete revolution, or \(2\pi\) radians, and we are tasked with finding the angular displacement \(\Theta_{3}\) when the blade is rotating at three times the initial speed. Understanding how to relate angular displacement to revolutions is key to converting the final answer from radians back to revolutions, which is more intuitive for many.
Kinematic Equations
Kinematic equations allow us to predict the future state of an object's motion鈥攕uch as position, velocity, and acceleration鈥攂ased on its current state and assuming constant acceleration. In angular motion, these equations relate angular velocity, angular acceleration, and angular displacement.

The angular kinematic equation that we use in this problem is \(\omega_{2}^{2} = \omega_{1}^{2} + 2\alpha\Theta\). In the exercise, we used the equation twice: once to find the angular deceleration with the initial parameters, and again to discover the angular displacement needed for the blade to stop when starting with a higher initial speed. Each application of this equation provides insight into the rotatory motion, illustrating the direct connection between the angular quantities.

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Most popular questions from this chapter

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of \(0.200 \mathrm{rev} / \mathrm{s}^{2}\). The design specifications call for it to have a rotational kinetic energy of \(240 \mathrm{~J}\) after it has turned through 30.0 revolutions. What should be the moment of inertia of the flywheel about its rotation axis?

The earth is approximately spherical, with a diameter of \(1.27 \times 10^{7} \mathrm{~m} .\) It takes 24.0 hours for the earth to complete one revolution. What are the tangential speed and radial acceleration of a point on the surface of the earth, at the equator?

A uniform disk with radius \(R=0.400 \mathrm{~m}\) and mass \(30.0 \mathrm{~kg}\) rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=(1.10 \mathrm{rad} / \mathrm{s}) t+\left(6.30 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\gamma-\beta t^{2}, \quad\) where \(\quad \gamma=5.00 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.800 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_{z}\) at \(t=3.00 \mathrm{~s}\) and the average angular acceleration \(\alpha_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=3.00 \mathrm{~s}\). How do these two quantities compare? If they are different, why?

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 \(/ \mathrm{s},\) 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 \(0.330 \mathrm{~m}\).
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