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Chapter 10: Problem 13

A flywheel turns through 40 rev as it slows from an angular speed of \(1.5 \mathrm{rad} / \mathrm{s}\) to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions?

Short Answer

Expert verified
(a) Time to stop is approximately 167.55 s. (b) Angular acceleration is approximately -0.00895 rad/s虏. (c) Time for the first 20 revolutions is about 100.8 s.

Step by step solution

01

Convert Revolutions to Radians

First, we need to convert the number of revolutions to radians, as the angular quantities are often measured in radians. We know that 1 revolution is equal to \(2\pi\) radians. So, for 40 revolutions: \(40 \text{ rev} \times 2\pi \text{ rad/rev} = 80\pi \text{ rad}\).
02

Use Equation for Angular Motion (Part a)

We will use the angular motion equation to find the time (\(t\)) it takes to come to rest. The equation is \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\), where \(\theta = 80\pi\), \(\omega_0 = 1.5 \text{ rad/s}\), and \(\omega_f = 0 \text{ rad/s}\). Since it comes to rest, rearrange and solve the formula: \(80\pi = 1.5t + \frac{1}{2}(-\frac{1.5}{t})t^2\). Solving for \(t\), we find \(t \approx 167.55 \text{ s}\).
03

Calculate Angular Acceleration (Part b)

The angular acceleration (\(\alpha\)) can be found using the formula \(\omega_f = \omega_0 + \alpha t\), and since \(\omega_f = 0\), rearrange this to find \(\alpha = \frac{-\omega_0}{t}\). Substituting \(\omega_0 = 1.5 \text{ rad/s}\) and \(t = 167.55 \text{ s}\), \(\alpha = \frac{-1.5}{167.55} \approx -0.00895 \text{ rad/s}^2\).
04

Solve for Time for First 20 Revolutions (Part c)

Using the same angular equation, but for half of the revolutions, \(40\pi\), solve for time \(t_1\): \(\theta = \omega_0 t_1 + \frac{1}{2} \alpha t_1^2\), where now \(\theta = 40\pi\). Rearranging for time, plug in \(\alpha = -0.00895 \text{ rad/s}^2\) and solve: \(40\pi = 1.5t_1 - 0.004475t_1^2\), giving \(t_1 \approx 100.8 \text{ s}\).

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Key Concepts

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

Angular Acceleration
Angular acceleration represents how fast an object's rotational speed changes over time. It's symbolized by the Greek letter \(\alpha\) and typically measured in radians per second squared (rad/s虏). Think of it as analogous to linear acceleration, which changes the velocity of an object moving in a straight line.

When an object like a flywheel is slowing down, its angular acceleration is negative. This is because it's decelerating. The formula to calculate angular acceleration is:
  • \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time taken for the change.
Being aware of angular acceleration can help in predicting the future motion of rotating systems, crucial for mechanical designs and safety checks.
Angular Velocity
Angular velocity is the rate at which an object rotates or spins around a center. It's like the speedometer on a car's dashboard, but instead of telling you how fast the car goes forwards or backwards, it tells you how fast it rotates. Angular velocity is symbolized by \(\omega\) and is measured in radians per second (rad/s).

In circular motion problems, knowing the initial angular velocity is essential. It gives us a starting point to calculate other properties like displacement or acceleration.

Angular velocity can be constant or variable:
  • When constant, the object rotates at a steady rate.
  • When variable, the rate changes, often described by angular acceleration.
This concept is fundamental in understanding rotational motion, akin to straight path speed in linear motion.
Radians
Radians are the standard unit of angular measure in most mathematical contexts. Unlike degrees, which break a circle into 360 parts, radians divide the circle by its radius, yielding about 6.28 parts (or \(2\pi\)). This makes calculations involving circles more straightforward and tidier.

One crucial identity to remember:
  • 1 revolution of a circle equals \(2\pi\) radians.
In physics and engineering, radians reduce complexity, particularly when developing formulas and equations using calculus. It allows for a more intuitive understanding of angular quantities.
Revolutions
Revolutions represent a complete turn around an axis and are often used interchangeably with cycles in everyday language. For rotational motion, knowing the number of revolutions can help determine total angle moved when converted to radians.

Here's how to convert:
  • 1 revolution = \(2\pi\) radians.
This conversion is useful when working with angular quantities, as many equations demand input in radians. Additionally, counting revolutions is handy in mechanical contexts, like determining how many times a wheel turns in a given travel distance, helping bridge tangible experiences with theoretical calculations.

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Most popular questions from this chapter

Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the \(20 \mathrm{~cm}\) mark. (Treat the stick as a thin rod.)

Cheetahs running at top speed have been reported at an astounding \(114 \mathrm{~km} / \mathrm{h}\) (about \(71 \mathrm{mi} / \mathrm{h})\) by observers driving alongside the animals. Imagine trying to measure a cheetah's speed by keeping your vehicle abreast of the animal while also glancing at your speedometer, which is registering \(114 \mathrm{~km} / \mathrm{h}\). You keep the vehicle a constant \(8.0 \mathrm{~m}\) from the cheetah, but the noise of the vehicle causes the cheetah to continuously veer away from you along a circular path of radius \(92 \mathrm{~m}\). Thus, you travel along a circular path of radius \(100 \mathrm{~m} .\) (a) What is the angular speed of you and the cheetah around the circular paths? (b) What is the linear speed of the cheetah along its path? (If you did not account for the circular motion, you would conclude erroneously that the cheetah's speed is \(114 \mathrm{~km} / \mathrm{h},\) and that type of error was apparently made in the published reports.)

A merry-go-round rotates from rest with an angular acceleration of \(1.50 \mathrm{rad} / \mathrm{s}^{2}\). How long does it take to rotate through (a) the first 2.00 rev and (b) the next 2.00 rev?

A disk rotates at constant angular acceleration, from angular position \(\theta_{1}=10.0\) rad to angular position \(\theta_{2}=70.0\) rad in \(6.00 \mathrm{~s}\). Its angular velocity at \(\theta_{2}\) is \(15.0 \mathrm{rad} / \mathrm{s}\). (a) What was its angular velocity at \(\theta_{1} ?\) (b) What is the angular acceleration? (c) At what angular position was the disk initially at rest? (d) Graph \(\theta\) versus time \(t\) and angular speed \(\omega\) versus \(t\) for the disk, from the beginning of the motion \((\) let \(t=0\) then \()\)

A wheel, starting from rest, rotates with a constant angular acceleration of \(2.00 \mathrm{rad} / \mathrm{s}^{2} .\) During a certain \(3.00 \mathrm{~s}\) interval, it turns through 90.0 rad. (a) What is the angular velocity of the wheel at the start of the \(3.00 \mathrm{~s}\) interval? (b) How long has the wheel been turning before the start of the 3.00 s interval?
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