Question

A flywheel has moment of inertia $4kg - {m^2}$ and has kinetic energy of $200J$. Calculate the number of revolution it makes before coming to rest if a constant opposing couple of $5N - m$ is applied to the flywheel

Answer 1

Hint We are given with the moment of inertia of the wheel, its kinetic energy and are also given with the value of the opposing couple and are asked to calculate the number of revolutions of the wheel before coming to rest. Thus, we will apply our fundamentals of revolution. Thus, we will take into consideration the angular velocity of the flywheel.
Formulae Used
${E_K} = \dfrac{1}{2}I{\omega ^2}$
Where, ${E_K}$ is the kinetic energy of the object, $I$ is the moment of inertia of the object and $\omega $ is the angular velocity of the object.
$\tau = I\alpha $
Where, $\tau $ is the torque, $I$ is the moment of inertia of the object and $\alpha $ is the angular acceleration of the object.
${\omega _f}^2 - {\omega _i}^2 = 2\alpha \theta $
Where, ${\omega _f}$ is the final angular velocity of the object, ${\omega _i}$ is the initial angular velocity of the object, $\alpha $ is the angular acceleration of the object and $\theta $ is the angular displacement of the object.

Step By Step Solution
Here,
The kinetic energy of the flywheel is, ${E_K} = 200J$
The moment of inertia of the flywheel is, $I = 4kg - {m^2}$
Now,
To calculate the initial angular velocity${\omega _i}$, we apply the formula,
${E_K} = \dfrac{1}{2}I{\omega ^2}$
Thus, we get
${E_K} = \dfrac{1}{2}I{\omega _i}^2$
Substituting the values, we get
$200 = \dfrac{1}{2}\left( 4 \right){\omega _i}^2$
Finally, we get
${\omega _i}^2 = 100rad/s$
Now,
The torque $\tau $ of the flywheel will be equal but opposite to the opposing couple on the wheel as it is the only force due to which the flywheel is rotating.
Thus,
$\tau = - 5N - m$
Thus,
To calculate the angular acceleration $\alpha $ of the flywheel we apply the formula
$\tau = I\alpha $
Substituting the values and calculating, we get
$\alpha = \dfrac{{ - 5}}{4}rad/{s^2}$
Now,
As the wheel comes to rest.
Thus, the final angular velocity of the flywheel will be ${\omega _f} = 0$
Thus,
We apply the formula,
${\omega _f}^2 - {\omega _i}^2 = 2\alpha \theta $
Substituting the values and calculating, we get
$\theta = 40rad$
Thus,
Number of revolutions before coming to rest,
$n = \dfrac{\theta }{{2\pi }} = \dfrac{{40}}{{2\pi }}$
Finally, we get
$n = 6.4$

Note In the question, the opposing couple is the only force in response of which the flywheel is rotating. Thus, we have considered the torque to be equal and opposite of the opposing torque. But if there was any other force for consideration, then we have to take all the forces available for rotation into our consideration.