The rotational kinetic energy is, KE=\(\frac{1}{2} I\omega^{2}\)
The moment of inertia of the ring is, I =MR²
\(I=9\times 3^{2}=9\times 9 = 81 kg m^{2}\)
The angular speed of the ring is,
\(\omega=240 rpm= \frac{240\times 2\pi}{60}\)rad s^-1
KE=\(\frac{1}{2}\times 81 \times (\frac{240\times 2 \pi}{60})^{2}=\frac{1}{2}\times 81 \times (8\pi)^{2}\)
KE=\(\frac{1}{2}\times 81 \times 64 \times (\pi)^{2}=2592 \times (\pi)^{2}\)
KE = 25920 J  \(\because (\pi)^{2}=10\)
KE = 25.9250 kJ

Speed of the cyclist, v = 20 m s^-1
Angle of bending with vertical, θ = 30°
Equation for angle of bending, \(tan \theta=\frac{v^{2}}{rg}\)
Rewriting the above equation for radius r = \(\frac{v^{2}}{tan \theta g}\)
Substituting,  \(r=\frac{(20)^{2}}{(tan 30^{\theta}\times 10)}=\frac{20\times 20}{(tan 30^{\theta})\times 10}=\frac{400}{(\frac{1}{\sqrt{3}}\times10)}\)
\(r=(\sqrt{3})\times 40=1.732 \times 40\)
r = 69.28 m

m₁ = 10 kg
m₂ = 5 kg
\(\overrightarrow { { r }_{ 1 } } = (-3\vec{i}+2\vec{j}+4\vec{k})m\)
\(\overrightarrow {{r}_{2}}=(-3\vec{i}+6\vec{j}+5\vec{k})\)m
\(\overrightarrow {{r}}=\frac{m_1\overrightarrow{r}_{1}+m_2\overrightarrow{r}_{2}}{m_1+m_2}\)
\(\overrightarrow{r}=\frac{10(-3\hat{i}+2\hat{j}+4\hat{k})+5(3\hat{i}+6\hat{j}+5\hat{k})}{10+5}\)
=\(\frac{-30\hat{i}+20\hat{j}+40\hat{k}+15\hat{i}+30\hat{j}+25\hat{k}}{15}=\frac{-15\hat{i}+50\hat{j}+65\hat{k}}{15}\)
\(\overrightarrow{r}=(-\hat{i}+\frac{10}{3}\hat{j}+\frac{13}{3}\hat{k})m\)
The center of mass is located at position \(\overrightarrow {r}\).

For all the four objects namely the ring, disc, hollow sphere and solid sphere, the radii of gyration K are R,\(\sqrt{\frac{1}{2}R},\sqrt{\frac{2}{3}R},\sqrt{\frac{2}{5}R}\). With numerical values the radius of gyration K are 1R, 0.707R, 0.816R, 0.632R respectively. The expression for time taken for rolling has the radius of gyration K in the numerator as per equation. 
\(t=\sqrt{\frac{2h(1+\frac{K^{2}}{R^{2}})}{8 sin ^{2}\theta}}\)     \(V=\sqrt{\frac{2gh}{[1+\frac{K^{2}}{R^{2}}]}}\)
The one with least value of radius of gyration K will take the shortest time to reach the bottom of the inclined plane. The order of objects reaching the bottom is first, solid sphere second, disc third, hollow sphere and last, ring.

The moment of inertia of a disc about an axis passing through the center of mass and perpendicular to the disc is, \(I=\frac{1}{2}MR^{2}\)
In terms of radius of gyration, I = MK²
Hence, \(Mk^{2}=\frac{1}{2}MR^{2} ; K^{2}=\frac{1}{2}R^{2}\)
\(K=\frac{1}{\sqrt{2}}R\ or\ K=\frac{1}{1.414}R\ or\ K=(0.707)R\)
From the case of a rod and also a disc, we can conclude that the radius of gyration of the rigid body is always a geometrical feature like length, breadth, radius or their combinations with a positive numerical value multiplied to it.