What is the Moment of Inertia of a Flywheel?

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a flywheel given that it accelerates from rest to an angular speed of 14.5 revolutions per second with a work input of 22.0 kJ. The relevant equation used is ½Iω² = 22.0 kJ, leading to the rearrangement I = (2)(22.0 kJ)/ω². The angular speed is converted to radians per second, resulting in ω = 29π rad/sec. After performing the calculations, the moment of inertia is determined to be approximately 0.00531 kg·m². The discussion also highlights the importance of using SI units correctly, specifically converting 22 kJ to joules.
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Homework Statement


The work done in accelerating a flywheel from rest to an angular speed of 14.5 revolutions per second is 22.0 kJ. What is the moment of inertia of the flywheel?

Homework Equations


½Iω² = 22.0 kJ

The Attempt at a Solution


½Iω² = 22.0 kJ
I = (2)(22.0 kJ)/ω²
ω = (14.5 revs/sec)
= (14.5 revs/sec) × (2π rad/rev)
= 29π = 91.1

I = (2)(22)/(91.1)2
I = 44/8300.3
I = 0.00531
 
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You are using SI units. How many joules are there in 22kJ?
 
Oh, thank you! I hadn't even noticed!
 
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