\(\mathrm{I}_{\text {cylinder }}=\frac{1}{12} \mathrm{M}\left(4 \mathrm{R}^2\right)+\frac{1}{4} \mathrm{MR}^2+\mathrm{M}(2 \mathrm{R})^2\)
\(=\frac{67}{12} \mathrm{MR}^2=\mathrm{Mk}_2^2\)
\(\frac{\mathrm{k}_1}{\mathrm{k}_2}=\sqrt{\frac{2}{3} \cdot \frac{12}{67}}=\sqrt{\frac{8}{67}}\)