Question

A flywheel (considered as a uniform ring) of mass 60 kg, radius 25 cm is making 240 revolutions per minute. Its kinetic energy is:
A. $40{\pi ^2}J$
B. $80{\pi ^2}J$
C. $120{\pi ^2}J$
D. $160{\pi ^2}J$

Answer 1

Hint: Before we solve this problem, we need to understand the concept of velocity in circular path. When an object is moving in a circular path of radius r, the rate of change of the angular displacement or radians is given by the quantity called angular frequency, represented as $\omega$.
The velocity of the object in the circular path is the product of the radius and the angular frequency.
Velocity, $v = r\omega$

Complete step-by-step solution:
Let us understand the definition of flywheel.
The flywheel is a mechanical device which is used to store the kinetic energy. The kinetic energy is stored in a huge wheel of very high mass with continuous rotation. It also acts as an accumulator wherein the stored kinetic energy in the flywheel can be used, whenever needed and necessary.
One of the most common applications of flywheel is the areas where we need to smoothen out the power output of an energy source, that produces energy in pulses or regular intervals, such as internal combustion engines in the automobiles.
The kinetic energy is given by the formula –
$E = \dfrac{1}{2}m{v^2}$
In this case, we have a rotating flywheel whose radius and angular frequency, $\omega$ is given. So, we have to substitute for the velocity as,
$V = r\omega$
Substituting, we get –
 $E = \dfrac{1}{2}m{r^2}{\omega ^2}$
Given in the problem,
Mass of the flywheel, m = 60 kg
Radius of the flywheel, $r = 25cm = 0 \cdot 25m$
Rotational frequency of the flywheel, n = 240 rpm = $\dfrac{{240}}{{60}} = 4rps$
Angular frequency in terms of radians, $\omega =2\pi n=2\pi \times 4=8\pi rad-{{s}^{-1}}$
Substituting these values, we get –
$E = \dfrac{1}{2}m{r^2}{\omega ^2}$
$\Rightarrow E = \dfrac{1}{2} \times 60 \times 0 \cdot {25^2} \times {8^2}{\pi ^2}$
$\Rightarrow E = \dfrac{{240}}{2}{\pi ^2} = 120{\pi ^2}J$

Hence, the correct option is Option C.

Note: Along with the function of smoothing out the power, the flywheel, additionally, acts as surge protector. This means that when the power output in the engine fluctuates suddenly, the jerk is not felt at the output because the flywheel absorbs the surge power and smooths it out.