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Chapter 10: Problem 76

Starting from rest at \(t=0\), a wheel undergoes a constant angular acceleration. When \(t=2.0 \mathrm{~s}\), the angular velocity of the wheel is \(5.0 \mathrm{rad} / \mathrm{s}\). The acceleration continues until \(t=20 \mathrm{~s}\), when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=40 \mathrm{~s}\) ?

Short Answer

Expert verified
The wheel rotates through 1500 radians from \( t=0 \) to \( t=40 \) s.

Step by step solution

01

Identify the Given Information

We are given that the initial angular velocity \( \omega_0 = 0 \) rad/s (starting from rest), the angular velocity at \( t = 2.0 \) s is \( \omega = 5.0 \) rad/s, and the angular acceleration is constant. We also know that the acceleration stops at \( t = 20 \) s.
02

Calculate Angular Acceleration

To find the angular acceleration \( \alpha \), we use the formula \( \omega = \omega_0 + \alpha t \). Substituting the values, \( 5.0 = 0 + \alpha \times 2.0 \), we solve for \( \alpha \) to get \( \alpha = 2.5 \, \text{rad/s}^2 \).
03

Calculate Angular Displacement During Acceleration

The angular displacement from \( t = 0 \) to \( t = 20 \) s can be calculated using \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \). Substituting the known values, \( \theta = 0 + \frac{1}{2} \times 2.5 \times 20^2 \). This gives \( \theta = 500 \, \text{radians} \).
04

Determine Final Angular Velocity

The final angular velocity at \( t = 20 \) s is found using \( \omega = \omega_0 + \alpha t \). Substitute to get \( \omega = 0 + 2.5 \times 20 \). Thus, \( \omega = 50 \, \text{rad/s} \).
05

Calculate Angular Displacement Without Acceleration

From \( t = 20 \) s to \( t = 40 \) s, the wheel moves with constant angular velocity \( \omega = 50 \, \text{rad/s} \). The angular displacement \( \theta = \omega t = 50 \times 20 \). Thus, \( \theta = 1000 \, \text{radians} \).
06

Calculate Total Angular Displacement

The total angular displacement from \( t = 0 \) s to \( t = 40 \) s is the sum of displacements during acceleration and constant velocity. Total \( \theta = 500 + 1000 = 1500 \, \text{radians} \).

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Key Concepts

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

Angular Acceleration
Angular acceleration (\( \alpha \),) refers to the rate at which angular velocity changes over time. It's similar to linear acceleration but applies to rotating objects.
  • In this problem, angular acceleration is constant, meaning it's the same throughout the period it's active.
  • You can calculate it using the formula: \( \omega = \omega_0 + \alpha t \).
  • Here, \( \omega \) is the final angular velocity,\( \omega_0 \) is the initial angular velocity,and \( t \) is time.
In the example, the wheel starts from rest, so \( \omega_0 = 0 \).After 2 seconds, \( \omega = 5 \, \text{rad/s} \).Solving the equation with these values gives \( \alpha = 2.5 \, \text{rad/s}^2 \).This means for every second, the wheel's angular velocity increases by \( 2.5 \, \text{rad/s} \).Understanding this helps in predicting how the wheel will behave over time.
Angular Velocity
Angular velocity (\( \omega \),) describes how fast an object rotates. Unlike linear velocity, it measures rotation in terms of radians per second.
  • The relationship to angular acceleration is direct: as the wheel accelerates, its angular velocity increases.
  • Initially, when the wheel is at rest, the angular velocity is zero.
  • But by 2 seconds, it increases to \( 5 \, \text{rad/s} \),thanks to the constant angular acceleration.
The angular velocity continues to increase as long as there is angular acceleration. At \( t = 20 \, \text{s} \),the angular velocity reaches \( 50 \, \text{rad/s} \).After this point, the acceleration ceases, but the wheel maintains this constant angular velocity because nothing is slowing it down. This constant rate is essential in figuring out the total rotation after the acceleration stops.
Angular Displacement
Angular displacement (\( \theta \),) is the total angle through which a point or line, that rotates about a center or axis, has moved in a specified sense. It provides insight into how far the wheel has rotated over time.
  • First, calculate the angular displacement under acceleration using \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \).Clearly, because \( \omega_0 = 0 \),this simplifies things.
  • During the first 20 secondswhen the wheel is accelerating, the angular displacement is \( 500 \, \text{radians} \).
  • After that, for the next 20 seconds,the wheel rotates at a constant speed of \( 50 \, \text{rad/s} \).
  • This translates to another \( 1000 \, \text{radians} \).
To find the total angular displacement from start to finish, combine these displacements, leading to \( 1500 \, \text{radians} \).This shows how much the wheel has turned in total.
Kinematics
Kinematics explores the motion of objects without considering forces. It applies to linear motions and rotational ones like in our wheel problem.
  • Key equations in rotational kinematics directly parallel those for linear motion, but with angular variables.
  • The primary equations used here include the angular displacement formula \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \),and \( \omega = \omega_0 + \alpha t \),which find parallels in linear motion equations.
  • Kinematics in this instance allows us to predict how the wheel behaves under constant angular acceleration and then at a constant angular velocity once the acceleration stops.
By using the given values and substituting them into these well-defined equations, we can determine aspects like how far the wheel has rotated (\( \theta \)),how angular velocity changes (\( \omega \)),and over what time frames these take place. Thus, kinematics provides a powerful tool to understand rotational motion comprehensively.

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Most popular questions from this chapter

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 thin spherical shell has a radius of \(1.90 \mathrm{~m}\). An applied torque of \(960 \mathrm{~N} \cdot \mathrm{m}\) gives the shell an angular acceleration of \(6.20 \mathrm{rad} / \mathrm{s}^{2}\) about an axis through the center of the shell. What are (a) the rotational inertia of the shell about that axis and (b) the mass of the shell?

A flywheel with a diameter of \(1.20 \mathrm{~m}\) is rotating at an angular speed of 200 rev/min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute- squared) will increase the wheel's angular speed to 1000 rev/min in \(60.0 \mathrm{~s}\) ? (d) How many revolutions does the wheel make during that \(60.0 \mathrm{~s}\) ?

A bicyclist of mass \(70 \mathrm{~kg}\) puts all his mass on each downwardmoving pedal as he pedals up a steep road. Take the diameter of the circle in which the pedals rotate to be \(0.40 \mathrm{~m}\), and determine the magnitude of the maximum torque he exerts about the rotation axis of the pedals.

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bumps in the groove run into the stylus, causing it to oscillate. The equipment converts those oscillations to electrical signals and then to sound. Suppose that a record turns at the rate of \(33 \frac{1}{3}\) rev/min, the groove being played is at a radius of \(10.0 \mathrm{~cm}\), and the bumps in the groove are uniformly separated by \(1.75 \mathrm{~mm}\). At what rate (hits per second) do the bumps hit the stylus?
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