Question

A disc of moment of inertia `4kgm^2` is spinning freely at `3rads^(-1)`. A second disc of moment of inertia `2kgm^2` slides down the spindle and they rotate together. (a) What is the angular velocity of the combination ? (b) What is the change in kinetic energy of the system?

Answer 1

(a) Since there are no external torques acting, we may apply the conservation of angular momentum,
`I_(f)omega_(f)=I_(i)omega_(i)rArr(6 k g m^(2))omega_(f)=(4kgm^(2))(3 rads^(-1))`
Thus `omega_(f)=2rads^(-1)`
(b) The kinetic energies before and after the collision are
`K_(i)=1/2I_(i)omega_(i)^(2), K_(f)=1/2I_(f)omega_(f)^(2)=12 J`
The change is `DeltaK=K_(f)-K_(i)=-6 J`
In order for the two discs to spin together at the same rate, there had to be friction between them . the loss in kinetic energy is converted into thermal energy.