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

taking wax & disk as a system ....

then there is no external torque acting on this system so angular momentam will remain conserved..

IW = I_fW_{f              ..............1}

finally moment of inertia = I + I_wax

I_wax = mr² =  10(9*10^-2)²                                         (r = 9cm = 9*10^-2m  &   m =10kg )

                = 0.081 kg-m²

I_f = I + 0.081           ..........2

putting 2 in 1

 IW = (I+0.081)W_f

 W/W_f = (1+.081/I)                                 {W = 100rpm &  W_f=90rpm (given)}

 100/90 = 1 + 0.081/I

  I = 0.729kg-m²

Approved