Question

A horizontal disc rotating freely about a vertical axis makes $100rpm.$ A small piece of wax of mass, $10g$ 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.

Answer 1

$I_1{\omega }_1=I_2{\omega }_2$
$\therefore I_1\left(100\right)=(I_1+(10\left)\right(9{)}^2\left]\right(90)$
or $I_1+810=1.11I_1$
$\therefore I_1=7290g.c{m}^2$
$=7.29\times {10}^{-4}kg-m^2$