Consider a badminton racket with length scales as shown in the figure.

If the mass of the linear and circular portions of the badminton racket are same (M)  and the mass of the threads are negligible. The moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{\mathrm{r}}{2}$ distance from the end A of the handle will be ........ Mr² .

The given situation is shown in the following figure

The moment of inertia of the racket about an axis perpendicular to the handle and in the plane of ring at, $\frac{\mathrm{r}}{2}$ distance from end A of the handle is given by

$\begin{array}{l}\mathrm{I}={\mathrm{I}}_{\text{due to handle }}+{\mathrm{I}}_{\text{due to disc }} ....\left(\mathrm{i}\right)\\ {\mathrm{I}}_{\text{due to handle }}=\frac{1}{12}\mathrm{M}(6\mathrm{r}{)}^2+\mathrm{M}{\left(\frac{5\mathrm{r}}{2}\right)}^2\\ =\frac{1}{12}\mathrm{M}36{\mathrm{r}}^2+\frac{25{\mathrm{Mr}}^2}{4}\\ =3{\mathrm{Mr}}^2+\frac{25}{4}{\mathrm{Mr}}^2\\ ={\mathrm{Mr}}^2\left(\frac{37}{4}\right)\\ =\frac{37}{4}{\mathrm{Mr}}^2\\ {\mathrm{I}}_{\text{due to disc }}=\frac{{\mathrm{Mr}}^2}{2}+\mathrm{M}{\left(\frac{13\mathrm{r}}{2}\right)}^2\\ =\frac{{\mathrm{Mr}}^2}{2}+\frac{169{\mathrm{Mr}}^2}{4}\\ =\frac{171{\mathrm{Mr}}^2}{4}\\ \mathrm{I}={\mathrm{I}}_{\text{due to handle }}+{\mathrm{I}}_{\text{due to disc }}\\ =\frac{37}{4}{\mathrm{Mr}}^2+\frac{171{\mathrm{Mr}}^2}{4}\\ =\frac{208}{4}{\mathrm{Mrr}}^2\\ =52{\mathrm{Mr}}^2\end{array}$