# A car wheel turning at 0.5 rad/s is brought to rest by the brakes in exactly two revolutions....

## Question:

A car wheel turning at 0.5 rad/s is brought to rest by the brakes in exactly two revolutions. What is the angular acceleration of the wheel?

## Constant Acceleration in Rotational Motion:

In the constant angular acceleration cases, we have the following formulae:

1. {eq}\omega_f=\omega_i+\alpha t {/eq} (Angular Velocity-Time Relation)

2. {eq}\theta=\omega_it+\frac{1}{2}\alpha t^2 {/eq} (Angular Displacement-Time Relation)

3. {eq}{\omega_f}^2={\omega_i}^2+2\alpha\theta {/eq} (Angular Displacement-Angular Velocity Relation)

Where,

• {eq}\omega_f {/eq} is the final angular velocity,
• {eq}\omega_i {/eq} is the initial angular velocity,
• {eq}\alpha {/eq} is the angular acceleration,
• {eq}t {/eq} is the time and
• {eq}\theta {/eq} is the angular displacement.

## Answer and Explanation: 1

Given:

{eq}\omega_i=0.5\ rad/s\\ \omega_f=0\\ \theta=2\times2\pi=4\pi {/eq}

(one revolution is {eq}2\pi\ rad {/eq} )

By using angular displacement-angular velocity relation, the angular acceleration of the body is,

{eq}\displaystyle \alpha=\frac{{\omega_f}^2-{\omega_i}^2}{2\theta}=-10\times10^{-3}\ rad/s^2 {/eq}

Practice Applying Rotational Motion Formulas

from

Chapter 17 / Lesson 15
2.4K

This lesson introduces the reader to rotational motion and gives examples of solving rotational motion problems such as calculating the angular velocity of an object, the rotational motion of a cylinder, and the angular kinetic energy of an object.