moment of inertia of flywheel experiment viva questions
Answers
The flywheel of a large circular disc will be fitted with a strong projecting axel on both sides. So the axel will be mounted on ball bearings for the fixed support. The cord will be loosely looped around the peg at weight hanger on end carriers.
The axle will be mounted through ball bearings with fixed ends and support the small mass which it will be attached the wheel by a string. So the hook can easily be detached from the axle. The length and height of a line will be chosen to the ground of axle.
Where m is the mass of weight to a hanging rings, And them descends though height "h" to lose the potential energy.
P.E. of mass = K.E. of mass m + K.E. of wheel + work done to overcome the friction.
mgh = 1/2 mv² + 1/2 Imω² + n₁ F
n₂F = 1/2Iω²
F = 1/2n₂Iω²
By substituting equation 2 for F in equation 1, For the expression in moment of inertia as,
mgh = 1/2 mv²ω² + 1/2 Imω² + n₁ 1/2n₂Iω²
I = 2mgh-mr²ω²/ω²(1+n₁/n₂)
So the average angular velocity is ω/2
ω/2 = 2πn₂/t
ω = 4πn₂/t
