To drive a typical car at 40 mph on a level road for one hour requires about 3.2 × 107 J of energy. Suppose we tried to store this much energy in a spinning, solid, uniform, cylindrical flywheel. A large flywheel cannot be spun too fast or it will fracture. If we used a flywheel of diameter 1.2 m and mass 400 kg, what angular speed would be required to store 3.2 × 107 J?

"Angular velocity" is directly related to kinetic energy, that is, the Kinetic energy equation would allow an approximation to the resolution investigated in the problem.

The equation for KE is given by:

[tex]KE = \frac{1}{2} lw ^ 2[/tex]

Now, starting from there towards the Angular equation of kinetic energy, the moment of inertia (i) is used instead of mass (m), and angular velocity (w) instead of linear velocity (V)

That's how we get

[tex]KE_{Angular} = \frac{1}{2} Iw^2[/tex]

calculating the inertia for a solid cylindrical disk, of

m = 400kg

r = 1.2 / 2 = 0.6m

[tex]I_{disk} = \frac{1}{2} mr^2 = (0.5) (400) (0.6)^2 = 72 kgm^2[/tex]

We understand that the total kinetic energy is 3.2 * 10 ^ 7J, like this:

[tex]3.2*10^7 = \frac{1}{2} Iw^2 = (0.5) (72) w^2 = 36w^2[/tex][tex]w^2 = 3.2*10^7 / 36 = 0.0888*10^7 = 88.8*10^4[/tex]

[tex]w = 9.43*10^2 = 943 rad / s[/tex]

Thus,

943 rad / s ≈ 9000 rpm