Moment of inertia flywheel problem

[DWill]

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

 

[Doc Al]

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.)

 

[alphysicist]

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.

 

[foxjwill]

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.