# 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}

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Chapter 17 / Lesson 15This 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.