A horizontal disc rotating freely about a vertical axis makes 100 rpm. A small piece of wax of mass 10 g falls vertically on the disc and adheres to it at a distance of 9 cm from the axis if the number of revolution per minute is thereby reduced to 90. Calculate the moment of inertia of disc.

Explanation

`I_(1)omega_(1) = I_(2)omega_(2)`
Let the moment of inertia of the disc about the axis of rotation be / and that of the putty be .I. where `I. = mr^(2)`
`I xx 2pi n_(1) = (I + mr^(2)) xx 2pi n_(2)`
`m = 5 xx 10^(-3) kg, r=0.04 m, n_(1) = 300/60 = 5 rps, n_(2) = 240/60 = 4 rps`
`l xx 2pi xx 5 = (I + 5 xx 10^(-3) xx 0.04^(2)) xx 2pi xx 4`
`5l = 4l + 32 xx 10^(-6) , I = 32 xx 10^(-6) kg m^(2)`
`I = 32 xx 10^(-6) kg m^(2)`