physics question

Question

a flywheel in a large factory stores energy in the form of rotational kinetic energy. assume that a flywheel is a uniform iron cylinder with a mass of 1200 kg and a radius of 2.3 m. In addition, the flywheel is embedded with four smaller pieces of lead each placed 1.8 m from the central axis. Each of these lead inclusions has a mass of 30 kg. when the flywheel is operating at 200 rad/s, how much energy does it store?

the answer is 2.8 x 10^8 J but I don't know how to get to it.

Answer 1

The flywheel stores energy by maintaining its rotation. The kinetic energy of a solid cylindrical flywheel of uniform weight is given by

K.E. = 1/2*I*ω^2, where

I = moment of inertia about the rotating axis

ω = angular velocity.

For a uniform solid cylinder, rotating around the cylindrical axis, the moment of inertia is given by

I_flywheel = 1/2*M*r² where

M = Mass of the cylinder

r = radius of the cylinder

This flywheel has an additional 4 embedded weights of 30Kg each at a distance of 1.8m from the rotating axis. Each of these weights also store energy due to rotation around the central axis. The general formula for kinetic energy of an object rotating about an axis is the same as above. Only the moment of inertia of the embedded mass is different.

I_emb = m*d², where

I_emb is the Inertia of each embedded mass

m = 30Kg is the embedded mass

d = 1.8m is the distance of the mass from the rotating axis.

Use this equation to calculate the K.E of one embedded mass. Then you can all four together to obtain the total K.E. of the embedded masses.

Finally, you can obtain the total energy stored by

Total K.E. = K.E. of Flywheel + 4* K.E.of each Embedded Mass

E = 1/2 * (1/2*M*r²) * ω² + 4 * 1/2 * (md²) * ω²

Hope this gives you sufficient guidance to carry out the calculation.