Question

Energy of 484 J is spent in increasing the speed of a flywheel from 60rpm to 360 rpm.Find the MI of the wheel.

Answer 1

application of work energy theorem

Answer 2

Answer:

The moment of inertia of the fly wheel is $0.7\ kg m^2$

Explanation:

The energy by which the fly wheel is rotating is E=484 J

The initial speed of the fly wheel is $\omega _{i}=\dfrac{60\times 2\pi}{60}=2\pi$

The final angular velocity of the fly wheel is $\omega_{f}=\dfrac{360\times 2\pi }{60}\\\omega_{f}=12\pi$

The fly wheel has rotational kinetic energy

So,

$KE_{r}=\dfrac{1}{2} I(\omega_{f}^2-\omega_{i}^2)\\KE_{r}=\dfrac{1}{2} I((12 \pi)^2-(2\pi)^2)\\\\484=\dfrac{1}{2}I(140 \pi^2)\\I=\dfrac{2\times 484}{140 \pi^2}\\I=0.7\ kg m^2$