A horizontal disc ratating about a vertical axis passing through its centre makes `180` rpm. A small piece of wax of mass `10 g` falls vertically on the disc and stricks to it at a distance reduced to `150 rpm`, calculate moment of inertia of the disc.

Explanation

Here, `n_(1) = 180"rpm" = 3 rps`,
`n_(2) = 150 "rpm" = (5)/(2) rps`
Let `I_(1)` be the moment of inertia of the disc about the given axis.
When mass `m = 10g = 10^(-2) kg` stricks to disc at
`r = 8 cm = 8 xx 10^(-2) m`, then `I_(2) = I_(1) + mr^(2)`.
According to principle of conservation of angular momentum
`I_(2) omega_(2) = I_(1)omega_(1) or (I_(1) + mr^(2)) 2pi n_(2) = I_(1) 2pi n_(1)`
or `I_(1) + mr^(2) = I_(1)(n_(1))/(n_(2)) = I_(1) xx (3)/(5//2) = (6)/(5)I_(1)`
`(6)/(5)I_(1) - I_(1) = mr^(2)`
`I_(1) = 5 mr^(2) = 5 xx 10^(-2) (8 xx 10^(-2))^(2)`
`= 3.2 xx 10^(-4) kg m^(2)`