Question

2 b) A horizontal disc of radius 9.00 cm rotates about an axis which passes vertically through the center of the disc with angular velocity of 100 r.p.m. A small particle of mass 10g is dropped onto the disc and sticks to the edge of the disc. If the angular velocity is reduced to 90 r.p.m, determine the moment of inertia of the disc about the axis.

Answer 1

Given:

$R=0.09\:\rm m$

$n_1=100\:\rm rpm$

$n_2=90\:\rm rpm$

$m=0.01\:\rm kg$

The law of conservation of energy says

$$In_1=(I+mR^2)n_2$$

Therefore

$$I=mR^2\frac{n_2}{n_1-n_2}\\=0.01*0.09^2\frac{90}{100-90}=7.29*10^{-4}\:\rm kg\cdot m^2$$