/*! 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] | 91Ó°ÊÓ

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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
It takes 9 revolutions to stop the blade from an initial speed of \( 3\omega_{1} \).

Step by step solution

01

Understand Initial Problem Conditions

The initial problem states that a blade comes to rest from an initial angular speed \( \omega_{1} \) in 1.00 revolution. This implies an angular displacement of \( \theta = 2\pi \) radians. The acceleration is constant.
02

Use Angular Kinematics for First Situation

The kinematic equation \( \omega^{2} = \omega_{0}^{2} + 2\alpha \theta \) applies. For the first situation, with final angular speed \( \omega = 0 \), \( \omega_{0} = \omega_{1} \), and \( \theta = 2\pi \): \( 0 = \omega_{1}^2 + 2\alpha (2\pi) \). This simplifies to \( \alpha = -\frac{\omega_{1}^2}{4\pi} \).
03

Determine Angular Displacement for New Condition

Given \( \omega_{3} = 3\omega_{1} \), substitute into the equation \( \omega^2 = \omega_{0}^2 + 2\alpha \theta \) for the new initial speed. \( 0 = (3\omega_{1})^2 + 2(-\frac{\omega_{1}^2}{4\pi})\theta \) simplifies to \( \theta = 9(2\pi) \), because \( 9\omega_{1}^2 = \frac{\omega_{1}^2}{2\pi}(2\theta) \).
04

Calculate Number of Revolutions

With \( \theta = 18\pi \) for \( \omega_{3} = 3\omega_{1} \), convert the angular displacement to revolutions. Since \( 1 \) revolution equals \( 2\pi \) radians, it would take \( \frac{18\pi}{2\pi} = 9 \) revolutions to bring the blade to rest from \( 3\omega_{1} \).

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Key Concepts

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

Angular Displacement
Angular displacement measures how much an object has rotated from its initial position. It is vital in understanding rotational motion. The displacement is typically measured in radians or revolutions, where one full revolution equals \( 2\pi \) radians. In the context of rotating systems like a mower blade, knowing the angular displacement helps determine how much rotation occurs before coming to a stop or changing speeds.

For instance, if a blade starts at a certain angular speed and is brought to rest, the angular displacement tells us the complete rotation angle covered during this process. In the example provided, the blade has an angular displacement of \( 2\pi \) radians, which equates to one full revolution before stopping.

Understanding angular displacement allows us to relate angular speeds with the rotations completed while stopping or changing velocities.
Constant Acceleration
Constant acceleration in rotational motion refers to a steady change in angular velocity over time. This is analogous to constant linear acceleration but applied to rotating systems. When a system experiences constant acceleration, its angular velocity changes at a uniform rate, simplifying the analysis of its motion.

With constant acceleration, the angular velocity (\( \omega \)) can be described by:
  • Initial angular speed (\( \omega_0 \))
  • Time (\( t \))
  • Angular acceleration (\( \alpha \))
In rotational motion problems, knowing the constant acceleration is necessary to predict future positions and velocities of rotating objects. In our problem, a constant negative acceleration brings the blade to rest, showing a uniform decrease in speed until stopping.

Using kinematic equations, we can find unknown variables like how many revolutions it takes for an object to come to a stop, using known initial speeds and constant acceleration.
Kinematic Equations
Kinematic equations form the foundation of analyzing both linear and angular motion. In angular motion, these equations help relate angular displacement, velocity, and acceleration. The scenario discussed uses the angular kinematic equation:\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]Here:
  • \( \omega \) is the final angular velocity
  • \( \omega_0 \) is the initial angular velocity
  • \( \alpha \) is the angular acceleration
  • \( \theta \) is the angular displacement
This equation is pivotal in problems where an object's angular velocity needs to change, such as bringing a mower blade to rest. By setting the final velocity to zero, we find angular displacement needed given initial angular speeds and constant acceleration.

In the exercise, given a constant acceleration and an initial speed, this equation helps calculate the number of revolutions the blade completes before stopping. It simplifies the process of understanding and predicting rotational behaviors under specific conditions.

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Most popular questions from this chapter

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center. (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 20.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During the 20.0 -rad angular displacement of part (b), the kinetic energy of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J} .\) What is the moment of inertia of the wheel about the rotation axis?

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b .\) Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

A uniform disk with radius \(R=0.400 \mathrm{m}\) and mass 30.0 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 \(\theta(t)=\) \((1.10 \mathrm{rad} / \mathrm{s}) t+\left(8.60 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2} .\) 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 \(\mathrm{rev}\) ?

An airplane propeller is rotating at 1900 rpm (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 \(35^{\circ} ?\)

A wheel of diameter 40.0 \(\mathrm{cm}\) starts from rest and rotates with a constant angular acceleration of 3.00 \(\mathrm{rad} / \mathrm{s}^{2} .\) At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\text { rad }}=\omega^{2} r\) and \((b)\) from the relationship \(a_{\text { rad }}=v^{2} / r\) .

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