Question

A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of `100rpm`. A bob of wax of mass `20g` falls on the disc and sticks to it a distance of `5cm` from the axis. If the moment of inertia of the disc about the given axis is `2xx10^(-4) kgm^2`, find new frequency of rotation of the disc.

Answer 1

`I_(1)`=Moment of inertia of disc `=2xx10^(-4)kg m^(2)`
`I_(2)`=moment of inertia of the disc+moment of inertia of the bob of wax on the disc
`=2xx10^(-4)+mr^(2)=2xx10^(-4)+20xx10^(-3)(0.05)^(2)`
`=2xx10^(-4)+0.5xx10^(-4)=2.5xx10^(-4) kgm^(2)`
By the principle of conservation of angular momentum
`I_(1)n_(1)=I_(2)n_(2)rArr2xx10^(-4)xx100=2.5xx10^(-4) n_(2)`
`n_(2)=(100xx2)/2.5=80 rpm`