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Chapter 10: Problem 90

The flywheel of an engine is rotating at \(25.0 \mathrm{rad} / \mathrm{s}\). When the engine is turned off, the flywheel slows at a constant rate and stops in \(20.0 \mathrm{~s}\). Calculate (a) the angular acceleration of the flywheel, (b) the angle through which the flywheel rotates in stopping, and (c) the number of revolutions made by the flywheel in stopping.

Short Answer

Expert verified
The angular acceleration is \(-1.25 \, \text{rad/s}^2\), the flywheel rotates through \(250 \, \text{rad}\), and it makes approximately \(39.79\) revolutions.

Step by step solution

01

Calculate Angular Acceleration

To find the angular acceleration \(\alpha\), we use the formula \( \alpha = \frac{\Delta \omega}{\Delta t} \). Here the initial angular velocity \(\omega_i\) is given as \(25.0 \, \text{rad/s}\), the final angular velocity \(\omega_f\) is \(0 \, \text{rad/s}\), and the time \(\Delta t\) is \(20.0 \, \text{s}\). Plugging in the values, we get:\[ \alpha = \frac{0 - 25.0 \, \text{rad/s}}{20.0 \, \text{s}} = -1.25 \, \text{rad/s}^2. \]
02

Calculate the Angle Rotated

The angle \(\theta\) through which the flywheel rotates can be found using the formula \( \theta = \omega_i \cdot t + \frac{1}{2} \alpha t^2 \). Substituting the known values \(\omega_i = 25.0 \, \text{rad/s}\), \(\alpha = -1.25 \, \text{rad/s}^2\), and \(t = 20.0 \, \text{s}\), we have:\[ \theta = 25.0 \, \text{rad/s} \times 20.0 \, \text{s} + \frac{1}{2} \times (-1.25 \, \text{rad/s}^2) \times (20.0 \, \text{s})^2. \]Simplifying,\[ \theta = 500 \, \text{rad} - 250 \, \text{rad} = 250 \, \text{rad}. \]
03

Convert Angle to Revolutions

To convert the angle in radians to revolutions, we use the fact that one revolution equals \(2\pi\) radians. Thus, the number of revolutions \(N\) is given by \(N = \frac{\theta}{2\pi}\). Using \(\theta = 250 \, \text{rad}\), we find:\[ N = \frac{250 \, \text{rad}}{2\pi} \approx 39.79. \]

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Key Concepts

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

Angular Acceleration
Angular acceleration is basically how fast the rotational speed of an object is changing.
It's like when you hit the brakes in a car, and the speed decreases. Similarly, the flywheel here is slowing down.
For a more technical understanding, we use the formula for angular acceleration \( \alpha = \frac{\Delta \omega}{\Delta t} \). Here, \( \alpha \) represents angular acceleration. \( \Delta \omega \) is the change in angular velocity, and \( \Delta t \) is the time period this change occurs over.
The units for angular acceleration are radians per second squared (\( \text{rad/s}^2 \)). In the exercise, the initial speed is 25 radians per second, and it takes 20 seconds to stop. By substituting these values into the formula, you find \( \alpha = -1.25 \, \text{rad/s}^2 \).
  • The negative sign indicates slowing down, or deceleration, as the rotational motion comes to a halt.
  • Constant angular acceleration implies that the flywheel's rotational speed decreased at a steady rate.
Angular Displacement
Angular displacement refers to the angle through which a point or line has been rotated in a specified sense about a specified axis.
For the flywheel, it's the total angle it moves through while it's stopping.
In this exercise, we use the formula \( \theta = \omega_i \cdot t + \frac{1}{2} \alpha t^2 \) to calculate the angle.
Here, \( \theta \) is the angular displacement. By plugging in the values \( \omega_i = 25.0 \, \text{rad/s} \), \( \alpha = -1.25 \, \text{rad/s}^2 \), and \( t = 20.0 \, \text{s} \), you find out that the flywheel rotates through an angle of \( 250 \, \text{rad} \).
  • This displacement is the change from the initial to the final position on a circular path.
  • The formula accommodates initial motion and the effect of any acceleration (or deceleration in this case) over time.
Rotational Motion
Rotational motion is when an object spins around an axis.
In our situation, the flywheel exhibits rotational motion.
A key point is that while linear motion is about objects moving along a path, rotational motion deals with spinning around an axis.
This core concept involves various elements like angular velocity, angular displacement, and angular acceleration which are central to understanding this type of motion. To grasp the idea of rotational motion in terms of revolutions, we translate radians into complete turns of the wheel. Since 1 complete revolution equals \( 2\pi \) radians, we utilize \( N = \frac{\theta}{2\pi} \) to convert the angular displacement into total revolutions.
Here, with \( \theta = 250 \, \text{rad} \), it turns out to be approximately 39.79 revolutions.
  • This conversion helps to relate the scenario to a more tangible measure, like counting the turns.
  • Understanding rotational motion involves appreciating how angular measures link to real world rotations.

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Most popular questions from this chapter

The angular position of a point on the rim of a rotating wheel is given by \(\theta=4.0 t-3.0 t^{2}+t^{3},\) where \(\theta\) is in radians and \(t\) is in seconds. What are the angular velocities at (a) \(t=2.0 \mathrm{~s}\) and \((\mathrm{b}) t=4.0 \mathrm{~s} ?\) (c) What is the average angular acceleration for the time interval that begins at \(t=2.0 \mathrm{~s}\) and ends at \(t=4.0 \mathrm{~s} ?\) What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

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A drum rotates around its central axis at an angular velocity of \(12.60 \mathrm{rad} / \mathrm{s} .\) If the drum then slows at a constant rate of \(4.20 \mathrm{rad} / \mathrm{s}^{2}\) (a) how much time does it take and (b) through what angle does it rotate in coming to rest?
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