Question

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

According to angular momentum conservation:

$$I_1n_1=(I_1+I)n_2,\\\space\\ I_1=\frac{In_2}{n_1-n_2}=\frac{mr^2n_2}{n_1-n_2},\\\space\\ I_1=\frac{0.01·0.09^2·90}{100-90}=7.29·10^{-4}\text{ kg}·\text{m}^2.$$