# The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its...

## Question:

The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its angular velocity decreases from 650 rev/min to 520 rev/min. What moment of inertia is required?

## Rotational kinetic energy

When the body moves about the fixed axis of rotation then it possesses the rotational kinetic energy. It is analogous to the translational kinetic energy in the linear motion. The flywheel is a device used to store the rotational kinetic energy during the peak period and it supplies that energy when it is required.

Rotational kinetic energy {eq}=\dfrac {1}{2}I \omega^2 {/eq}

Where;

- I is the moment of inertia
- {eq}\omega {/eq} is the angular velocity of the body

## Answer and Explanation:

**Given data **

- Kinetic energy supplied by the flywheel {eq}\Delta KE=500\: joules {/eq}

- Initial angular velocity of flywheel {eq}\omega_i =650\: rpm {/eq}

- {eq}\omega_i=650\frac {rev} {min} \times \frac {1\:min}{60\: sec} \times 2\pi\frac {rad} {rev} =68.07\: rad/sec. {/eq}

- Final angular velocity flywheel {eq}\omega_f =520\: rpm {/eq}

- {eq}\omega_f=520\frac {rev} {min} \times \frac {1\:min}{60\: sec} \times 2\pi\frac {rad} {rev} =54.45\: rad/sec. {/eq}

- I is the moment of inertia of the flywheel

A rotating flywheel possesses rotational kinetic energy and when it supplies that energy to gasoline engine then it's rotational kinetic energy decreases. Therefore the energy supplied by the flywheel is equal to the change in rotational kinetic energy of the flywheel.

{eq}\Delta KE=\dfrac {1}{2}I\omega_f^2-\dfrac {1}{2}I\omega_i^2 {/eq}

{eq}\Delta KE=\dfrac {1}{2}I(\omega_f^2-\omega_i^2) {/eq}

{eq}I=\dfrac {2\times \Delta KE} {(\omega_f^2-\omega_i^2)} {/eq}

{eq}I=\dfrac {2\times (-500)}{54.45^2-68.07^2} {/eq}

{eq}I=\dfrac {1000}{1668.72} {/eq}

{eq}\boxed {I=0.6 \:kgm^2} {/eq}

Therefore required a moment of inertia of fly is **0.6 kgm^2**

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