Question

A horizontal disc rotating freely about a vertical axis through its center makes $90revolutionsperminute.$ A small piece of wax of mass $m$ falls vertically on the disc and sticks to it at a distance $r$ from the axis. If the number of revolutions per minute reduce to $60,$ then the moment of inertia of the disc is?

Answer 1

Let moment of inertia of the disc is $I.$

Initial angular momentum is $L_1=I_1{\omega }_1=90I$

When the mass $m$ sticks at distance $r$, new moment of inertia is

$I_2=I+m{r}^2$

Now, angular moment of inertia is $L_2=(I+m{r}^2){\omega }_2=60(I+m{r}^2)$

From conservation of angular momentum-

$L_1=L_2$

$⟹90I=60I+60m{r}^2$

$⟹30I=60m{r}^2$

$⟹I=2m{r}^2$