Given, ω1=60rpm=60×602π=2πrad/sω2=360rpm=360×602π=12πrad/s Change in kinetic energy, ΔK.E484⇒I=21I(ω22−ω12)=21×I(ω2+ω1)(ω2−ω1)=14π×10π484×2=0.7 kg⋅m2 (where, the symbols have their usual meanings)
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:
NEET UG 2022 — Physics Mechanics
An energy of 484 J is spent in increasing the speed of a flywheel from 60rpm to 360rpm. The moment of inertia of the flywheel is:
Held on 30 Apr 2022 · Verified 9 Jul 2026.
Options
- A
0.7 kg−m2
- B
3.22 kg−m2
- C
30.8 kg−m2
- D
0.07 kg−m2
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