What is the Moment of Inertia of a Flywheel?

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SUMMARY

The moment of inertia of a flywheel can be calculated using the equation ½Iω² = 22.0 kJ, where ω is the angular speed in radians per second. In this case, the angular speed of 14.5 revolutions per second converts to approximately 91.1 radians per second. Substituting these values into the equation yields a moment of inertia of 0.00531 kg·m². This calculation emphasizes the importance of unit conversion in physics problems.

PREREQUISITES
  • Understanding of angular motion and moment of inertia
  • Familiarity with the work-energy principle
  • Ability to convert between revolutions per second and radians per second
  • Knowledge of SI units, specifically joules and kilograms
NEXT STEPS
  • Study the work-energy theorem in rotational dynamics
  • Learn about the properties of different types of flywheels
  • Explore the relationship between torque and moment of inertia
  • Investigate practical applications of flywheels in energy storage systems
USEFUL FOR

Students studying physics, mechanical engineers, and anyone interested in the principles of rotational motion and energy storage systems.

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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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