A unifrom horizontal disc is freely rotating about a vertical axis through its centre at the rate of 240 rpm. A blob of wax of mass 2 g falls on it and sticks to it at 25 cm from the axis. If the frequency of rotation of the disc is reduced by 60 rpm. calculate the moment of inertia of the disc. lt

Explanation

Data :`f_(1) = 240 rpm = 240 //60 "rps,m" m =2 , g= 2xx 10^(3)` kg,
` f_(2) = (240 -60) "rpm" = 180 //60 "rps" = r = 25 cm = 0.25 cm`
Let ` I_(1)` the MI of the disc. Let `I_(2)` be the MI of the disc and the blob.
` I_(2) =I_(1) + mr^(2)`
According to the law of conservation of angular momentu.
` I_(1)omega_(2) = I_(2)omega_(2)`
` I_(1) (2pif_(1)) = (I_(1) + mr^(2)) (2pif_(2))`
` I_(1)f_(1) = (I_(1) + mr^(2)) f_(2)`
` I_(1) (f_(1)-f_(2)) = mr^(2)f_(2)`
`I_(1) = (mr^(2)f)/(f_(1)-f_(2)) = (2xx 10^(-3) xx (0.25)^(2) xx 3)/( 4-3) `
` = 6xx 10^(-3) xx 6.25 xx 10^(-2)`
= `4.75 xx 10^(-4)` kg.`m^(2)`