Question

Energy of $484joule$ is spent in increasing the speed of a flywheel from $60$ RPM to $360$ RPM find the moment of inertia of the flywheel:

• A. $0.7kg{m}^2$
• B. $9.8kg{m}^2$
• C. $2kg{m}^2$
• D. $15kg{m}^2$

Answer 1

The correct option is A $0.7kg{m}^2$

Given E=484J

Initial speed of fly wheel $\omega$i=$\frac{60\ast 2\pi }{60}$ =$2\pi$

final angular velocity of fly wheel $\omega$f =$\frac{360\ast 2\pi }{60}$=$12\pi$

Rotational energy of flywheel ,

K E=$\frac{1I(\omega f^2-\omega i^2)}{2}$

$I=\frac{2\ast 484}{140{\pi }^2}$

=0.7 kg m sq.