Question

A energy storage system based on a flywheel (a rotating disk) can store a maximum of 4.0 MJ when the flywheel is rotating at 20,000 revolutions per minute. What is the moment of inertia of the flywheel?

Answer 1

Answer:

Moment of inertia of the flywheel, $I=1.82\ kg-m^2$          

Explanation:

Given that,

The maximum energy stored on  flywheel, $E=4\ MJ=4\times 10^6\ J$

Angular velocity of the flywheel, $\omega=20000\ rev/s=2094.39\ rad/s$

We need to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

$E=\dfrac{1}{2}I\omega^2$

I is the moment of inertia of the flywheel

On rearranging we get :

$I=\dfrac{2E}{\omega^2}$

$I=\dfrac{2\times 4\times 10^6}{(2094.39)^2}$

$I=1.82\ kg-m^2$

So, the moment of inertia of the flywheel is $I=1.82\ kg-m^2$. Hence, this is the required solution.

Answer 2

The moment of inertia of the flywheel $I= 1.82\ kg-m^2$

What will be the Moment of inertia of the flywheel?

It is given that,

The maximum energy stored on the flywheel is given as

E=4 MJ=$4\times 10^6\ J$

Angular velocity of the flywheel is

$w=20000\ \dfrac{Rev}{Sec} =2094.39 \frac{rad}{sec}$

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

$E= \dfrac{1}{2} Iw^2$

By rearranging the equation

$I=\dfrac{2E}{w^2}$

$I=1.82\ kg-m^2$

Thus the moment of inertia of the flywheel $I= 1.82\ kg-m^2$

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