Question

A Flywheel of mass 2 kg has radius of gyration 0.2 m. Calculate the K.E. of rotation when it makes
5 rev/sec.​

Answer 1

solution:

Rotational kinetic energy = 0.5Iω²

= 0.5 × mk² × ω²

= 0.5 × 2kg ×

${ (0.2m)}^{2} \:$

× (5 × 2 π rad/s)²

= 39.47

(note : here there may be mistake in calculation so I will suggest you that please calculate yourself)

Answer 2

The K.E. of rotation of a flywheel is 39.46 Joules.

Explanation:

Mass of the flywheel, m = 2 kg

The radius of gyration, r = 0.2 m

Angular velocity of the flywheel, $\omega=5\ rev/s=31.41\ rad/s$

The kinetic energy of rotation of a flywheel is given by :

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

$K=\dfrac{1}{2}\times m\times r^2\times \omega^2$

$K=\dfrac{1}{2}\times 2\times (0.2)^2\times (31.41)^2$

K = 39.46 Joules

So, the K.E. of rotation is 39.46 Joules. Hence, this is the required solution.

Learn more,

Moment of inertia

https://brainly.in/question/6469310