Measurement of the moment of inertia of a flywheel
A flywheel of radius R is set up on a horizontal axle of radius r. A string of length h is wrapped round the axle with a mass m tied to the end (Figures 1 and 2). The moment of inertia of the flywheel and axle is I. The flywheel is accelerated by the couple applied by the mass m. The mass is allowed to fall through a height h at which point the string leaves the axle. The velocity of the falling mass at this instant is v and the angular velocity of the flywheel ω.

The potential energy lost by the weight is converted into kinetic energy of the weight, kinetic energy of the flywheel and heat due to friction in the bearings.

If the energy lost per revolution due to friction is E and the flywheel makes n1 revolutions during acceleration, then:

mgh = ½ mv2 + ½ Iω2 + n1E

The flywheel is then allowed to come to rest due to the frictional couple. If it stops after a further n2 revolutions then:

½ Iω2 = n2E

Therefore:

mgh = 1/2 mv2 + 1/2 Iω2 + (n1/n2)1/2 Iω2 = 1/2 mv2 + 1/2 Iω2(1 + n1/n2)

We could convert linear velocity (v) into angular velocity (ω) if we wished using v = Rω.

Now the angular velocity ω at the end of the period of the acceleration is given by:

ω/2 = 2πn1/t

Since ω>/2 is the average angular velocity of the flywheel and 2πn1 is the angular distance covered by any point on it in a time t.

Hence the moment of inertia of the flywheel can be calculated