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 20 g falls on the disc and sticks to it a distance of 5 cm 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 20 g falls on the disc and sticks to it a distance of 5 cm 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)=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)/(r))_(1)=I_(2)2pi((n)/(t))_(2)` `2xx10^(-4)xx100=2.5xx10^(-4)((n)/(t))_(2)` `((n)/(t))_(2)=(100xx2)/(2.5)=80` rpm.