Physics - Rotation Question with Solution | TestHub

$I_1=\frac{1}{3}M{\ell }^2$

$r^2={\ell }^2+{\left(\frac{\ell }{2}\right)}^2={\ell }^2+\frac{{\ell }^2}{4}=\frac{5{\ell }^2}{4}$

$I_2=\frac{1}{12}M{\ell }^2+M{r}^2=\frac{1}{12}M{\ell }^2+\frac{5}{4}M{\ell }^2=\frac{1}{12}M{\ell }^2+\frac{15}{12}M{\ell }^2=\frac{16}{12}M{\ell }^2=\frac{4}{3}M{\ell }^2$

$I=I_1+I_2=\frac{1}{3}M{\ell }^2+\frac{4}{3}M{\ell }^2=\frac{5}{3}M{\ell }^2$

Thus, $\frac{x}{y}=\frac{5}{3}$.

If $x+y=8$, then $x=5$ and $y=3$. Thus x/y = 5/3 in lowest terms, so x = 5 and y = 3. Therefore x + y = 8.