# Moment of inertia flywheel problem

1. Jun 6, 2008

### DWill

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
K = (1/2) * I * w^2

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

3. 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

2. Jun 6, 2008

### Staff: Mentor

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. Jun 6, 2008

### 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.

4. Jun 6, 2008

### 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 in to 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.