/*! 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

\(\mathrm{At} t=0\) a grinding wheel has an angular velocity of \(24.0 \mathrm{rad} / \mathrm{s}\) It has a constant angular acceleration of \(30.0 \mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{~s}\). From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev \(/\) min to 200 rev \(/ \min\) in 4.00 s. (a) Find the angular acceleration in rev/s \(^{2}\) and the number of revolutions made by the motor in the 4.00 s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

A sphere with radius \(R=0.200 \mathrm{~m}\) has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(\rho=3.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}-\left(9.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{4}\right) r .\) (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

A pulley on a frictionless axle has the shape of a uniform solid disk of mass \(2.50 \mathrm{~kg}\) and radius \(20.0 \mathrm{~cm}\). A \(1.50 \mathrm{~kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has \(4.50 \mathrm{~J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Three small blocks, each with mass \(m,\) are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point onefourth of the length from one end.
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