An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :
- 0.07 kg m2
- 0.7 kg m2
- 3.22 kg m2
- 30.8 kg m2
Answer
(2) 0.7 kg m2
Solution
1 = 2 6060 = 2 rad s-1, 2 = 2 36060 = 12 rad s-1
Work done = change in rotational KE
W = 12I( 22 - 12)
484 = 12I(144 2 - 4 2) = 70 2 I
I = 48470 9.87
= 0.70 kg m2
Answer: (2)