Physics - Rotation Question with Solution | TestHub

For disk, moment of inertia about the diameter, ${\mathrm{I}}_{\mathrm{d}}=\frac{{\mathrm{mr}}^2}{4}$ and moment of inertia about the tangential axis $\begin{array}{rl}& {\mathrm{I}}_{\text{disc }}={\mathrm{I}}_{\mathrm{d}}+{\mathrm{mr}}^2\\ & {\mathrm{I}}_{\text{disc }}=\frac{{\mathrm{mr}}^2}{4}+{\mathrm{mr}}^2\\ & {\mathrm{I}}_{\text{disc }}=\frac{5{\mathrm{mr}}^2}{4}\end{array}$ Let the radius of gyration of disc is ${\mathrm{k}}_{\text{disc }}$. $\begin{array}{rl}{\mathrm{I}}_{\text{disc }}& ={\mathrm{mk}}_{\text{disc }}^2\\ \frac{5{\mathrm{mr}}^2}{4}& ={\mathrm{mk}}_{\text{disc }}^2\\ {\mathrm{k}}_{\text{disc }}^2& =\sqrt{\frac{5}{4}\mathrm{r}}\end{array}$ For ring, moment of inertia about the diameter ${\mathrm{I}}_{\mathrm{a}}^{\prime }={\mathrm{mr}}^2/2$ and moment of inertia about the tangential axis ${\mathrm{I}}_{\text{ring }}={\mathrm{I}}_{\mathrm{a}}^{\prime }+{\mathrm{mr}}^2=\frac{{\mathrm{mr}}^2}{2}+{\mathrm{mr}}^2=\frac{3}{2}{\mathrm{mr}}^2$ Let the radius of gyration of ring ${\mathrm{k}}_{\text{ring }}$. $\begin{array}{rl}{\mathrm{I}}_{\text{ring }}& ={\mathrm{mk}}^2\text{ ring }\\ \frac{3}{2}{\mathrm{mr}}^2& ={\mathrm{mk}}^2\text{ ring }\\ {\mathrm{k}}_{\text{ring }}& =\sqrt{\frac{3}{2}}\mathrm{r}\\ \frac{{\mathrm{k}}_{\text{disc }}}{{\mathrm{k}}_{\text{ring }}}& =\frac{\sqrt{\frac{5}{4}\mathrm{r}}}{\sqrt{\frac{3}{2}}\mathrm{r}}=\sqrt{\frac{5}{6}}\end{array}$