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Moment of inertia flywheel problem

  • Thread starter DWill
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
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.)
 
  • #3
alphysicist
Homework Helper
2,238
1
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.
 
  • #4
354
0
I'm not sure why you plugged in [tex]2\pi\frac{520}{650}[/tex] for [tex]\omega[/tex], but the basic idea here is that the change in kinetic energy is -500 J (because it's decreasing). So, you have to plug in to the equation
[tex]\Delta K = \frac{1}{2}I\omega^2-\frac{1}{2}I\omega_0^2.[/tex]​

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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