Question

A uniform circular disc with its plane horizontal is rotating about a vertical axis passing through its centre at a speed of $180r.p.m$. A small piece of wax of mass $1.9g$ falls vertically on the disc and sticks to it at a distance of $25cm$ from the axis. If the speed of rotation is now reduced by $60r.p.m$., calculate moment of inertia of the disc:

Answer 1

Given,

Current speed $=180r.p.m$

Mass of wax $=1.9g$

$Distance=25cm$

Speed reduces to $60r.p.m$

So,

$I\omega =I\eta =constant$

Where $I$ is the moment of inertia, $\omega$ is the angular velocity and $\eta$ is the angular frequency.

Let the moment of inertia of disc be $I$, When mass $1.9g$ being added. Then by parallel axis theorem, So final moment of inertia with $60r.p.m$

$(I+m{r}^2)$

$I\times 180=(I+m{r}^2)\times 60$

$I\times 180=I\times 60+1.9\times 625\times 60$

Where, $m=1.9g,r=25cm$

$120I=71250$

$I=\frac{71250}{120}=593.75kg{m}^2$