A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 rpm. 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)kg m^(2), find new frequency of rotation of the disc.

A horizontal disc is freely rotating about a vertical axis passing thr

1.	A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 rpm. 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)kg m^(2), find new frequency of rotation of the disc.
Answer» Solution :`I_(1)=` Moment of inertia of the disc `=2xx10^(-4)KGM^(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)` `(n//t)_(1)=100` rpm , `(n//t)_(2)=?` By the principle of conservation of angular momentum , `I_(1)omega_(1)=I_(2)omega_(2)` `I_(1)2PI((n)/(t))_(1)=I_(2)2pi((n)/(t))_(2)` `2xx10^(-4)xx100=2.5xx10^(-4)((n)/(t))_(2)` `2xx110^(-4)xx100=2.5xx10^(-4)((n)/(t))_(2)` `((n)/(t))_(2)=(100xx2)/(2.5)=80` rpm