Two wheels A and C are connected by a belt B as shown in figure. The radius of C is three times the radius of A. What would be the ratio of the rotational inertias $\left({\mathrm{I}}_{\mathrm{C}}/{\mathrm{I}}_{\mathrm{A}}\right)$ if both the wheels have the same rotational kinetic energy?

As the belt does not slip, ${\mathrm{v}}_{\mathrm{P}}={\mathrm{v}}_{\mathrm{Q}}$

i.e., ${\mathrm{r}}_{\mathrm{A}}{\mathrm{\omega }}_{\mathrm{A}}={\mathrm{r}}_{\mathrm{C}}{\mathrm{\omega }}_{\mathrm{C}} [\text{ as }\mathrm{v}=\mathrm{r\omega }]$              …(i)

According to the given problem if ${\mathrm{r}}_{\mathrm{A}}=\mathrm{r}, {\mathrm{r}}_{\mathrm{C}}=3\mathrm{r}$, so Eq. (i) becomes

${\mathrm{\omega }}_{\mathrm{A}}=3{\mathrm{\omega }}_{\mathrm{C}}$                      …(ii)

If both the wheels have the same rotational kinetic energy, then

$\frac{1}{2}{\mathrm{I}}_{\mathrm{A}}{\mathrm{\omega }}_{\mathrm{A}}^2=\frac{1}{2}{\mathrm{I}}_{\mathrm{C}}{\mathrm{\omega }}_{\mathrm{C}}^2$

$\frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{C}}}={\left[\frac{{\mathrm{\omega }}_{\mathrm{C}}}{{\mathrm{\omega }}_{\mathrm{A}}}\right]}^2={\left[\frac{1}{3}\right]}^2=\frac{1}{9}$