Moment of inertia flywheel problem

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SUMMARY

The discussion centers on calculating the moment of inertia required for a flywheel in a gasoline engine, given a change in kinetic energy of 500 J as the angular velocity decreases from 650 rev/min to 520 rev/min. The equation used is K = (1/2) * I * w^2, where K represents kinetic energy, I is the moment of inertia, and w is the angular velocity. The correct approach involves calculating the change in kinetic energy using both initial and final angular velocities, leading to a required moment of inertia of 0.600 kg m², rather than the incorrect 0.337 kg m² initially calculated.

PREREQUISITES
  • Understanding of angular velocity in revolutions per minute (rev/min)
  • Familiarity with the kinetic energy formula K = (1/2) * I * w^2
  • Knowledge of how to convert angular velocity from rev/min to radians per second
  • Basic principles of energy conservation in rotational dynamics
NEXT STEPS
  • Learn how to convert angular velocity from rev/min to radians per second
  • Study the concept of rotational kinetic energy and its applications
  • Explore the relationship between moment of inertia and angular velocity in different mechanical systems
  • Investigate energy conservation principles in rotational motion
USEFUL FOR

This discussion is beneficial for physics students, mechanical engineers, and anyone involved in the design and analysis of rotating systems, particularly in automotive applications.

DWill
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Homework Statement


The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its angular velocity decreases from 650 rev/min to 520 rev/min. What moment of inertia is required?


Homework Equations


K = (1/2) * I * w^2

K = kinetic energy
I = moment of inertia
w = angular velocity


The Attempt at a Solution


I thought that since the engine gives up 500 J of KE (so change of 500?) I can plug in 500 for K, and the final angular velocity for w, to solve for I.

500 = (1/2) * I * ((520)/60) * 2pi)^2
I = .337 kg m^2

The correct answer is 0.600 kg m^2. What am I doing wrong here? thanks
 
Physics news on Phys.org
500 J is not the KE at one speed but the change in KE as the speeds change. (Use both speeds and set the difference in KE equal to 500 J.)
 
Hi DWill,

The quantity 500 J is the change in kinetic energy. However, you only have the final kinetic energy on the right side of your equation.
 
I'm not sure why you plugged in 2\pi\frac{520}{650} for \omega, but the basic idea here is that the change in kinetic energy is -500 J (because it's decreasing). So, you have to plug into the equation
\Delta K = \frac{1}{2}I\omega^2-\frac{1}{2}I\omega_0^2.​

edit: I just find it really funny that we all just responded with the exact same thing at pretty much the same time.
 

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