Question

A cycle wheel of radius 0.4m completes a revolution in one second, then the acceleration of the cycle will be:
A) $0.8m/{s^2}$
B) $3.9m/{s^2}$
C) $1.6{\pi ^2}m/{s^2}$
D) $0.4{\pi ^2}m/{s^2}$

Answer 1

Hint: Use the concept of angular frequency. Angular frequency is $2\pi $ times of the frequency of the revolving object. Now, this angular frequency can be used to calculate the angular acceleration. Angular acceleration is directly proportional to the linear acceleration.

Complete step by step solution:
Step 1: The angular frequency $\omega $ is a scalar measure of rotation rate. It refers to the angular displacement per unit time. It is given by $\omega = \dfrac{{2\pi }}{n}$ . where, $n$ is the frequency of the revolving body. We are given that the wheel completes a revolution in one second, therefore $n = 1$ . the angular frequency will be
$\therefore \omega = \dfrac{{2\pi }}{1}$
$ \Rightarrow \omega = 2\pi $ Rad/sec.
Step 2: express the formula for the linear acceleration of a rotation body in terms of angular frequency.
$\Rightarrow a = {\omega ^2}r$ , where $r$ is the radius of the wheel.
Step 3: Now put the values
$\therefore a = {(2\pi )^2} \times 0.4$
$ \Rightarrow a = 4{\pi ^2} \times 0.4$
Multiply both terms on the right-hand side.
$\therefore a = 1.6{\pi ^2}$ $m/{s^2}$

Hence the correct option is Option C.

Note: How the angular speed of the wheel and the acceleration of the cycle is related to each other? To understand this let's take the example of a cycle itself. Suppose a cycle having a wheel of radius r. if the wheel is rotating with some speed and there is some angular acceleration, that is, how fast the angular velocity is changing with time. Now when the cycle is moving on the road the acceleration of the center of the wheel is equal to the acceleration of the cycle. Therefore we can use only the relation between angular speed and acceleration to find the acceleration of the cycle.