A flywheel has a 135-kg solid cylindrical rotor with radius 30 cm and spins at 31,000 rpm. How much energy does it store?

To calculate the energy stored in the flywheel, we need to use the formula for rotational kinetic energy:

E = (1/2) I ω^2

Where:
E is the energy stored in the flywheel
I is the moment of inertia of the flywheel
ω is the angular velocity of the flywheel

The moment of inertia of a solid cylindrical rotor can be calculated using the formula:

I = (1/2) m r^2

Where:
m is the mass of the rotor
r is the radius of the rotor

Given:
m = 135 kg
r = 30 cm = 0.3 m
ω = 31,000 rpm = (31,000/60) rad/s = 516.67 rad/s

First, let's calculate the moment of inertia:

I = (1/2) m r^2
I = (1/2) * 135 kg * (0.3 m)^2
I = 6.075 kg·m^2

Now, let's calculate the energy stored in the flywheel:

E = (1/2) I ω^2
E = (1/2) * 6.075 kg·m^2 * (516.67 rad/s)^2
E = 0.5 * 6.075 kg·m^2 * 267,223.89 rad^2/s^2
E = 805,180.34 Joules

Therefore, the flywheel stores approximately 805,180.34 Joules of energy.