Answer :
Given data:
* The radius of the wheel is r = 0.33 m.
* The rotational inertia of the wheel is,
[tex]I=0.8kgm^2[/tex]* The mass of the wheel and rider is m = 75 kg.
Solution:
The rotational kinetic energy of the wheel is,
[tex]K_r=\frac{1}{2}I\omega^2[/tex]where,
[tex]\omega\text{ is the angular velocity of the wheel}[/tex]The linear velocity of the wheel in terms of the angular velocity is,
[tex]\begin{gathered} v=r\omega \\ \omega=\frac{v}{r} \end{gathered}[/tex]Thus, the rotational kinetic energy in terms of the linear velocity is,
[tex]\begin{gathered} K_r=\frac{1}{2}I\times(\frac{v}{r})^2 \\ K_r=\frac{1}{2}\frac{Iv^2}{r^2} \end{gathered}[/tex]For the two wheels of the bicycle, the rotational kinetic energy is,
[tex]\begin{gathered} K_r=2\times\frac{1}{2}\frac{Iv^2}{r^2} \\ K_r=\frac{Iv^2}{r^2} \end{gathered}[/tex]The linear kinetic energy of the bicycle is,
[tex]K_t=\frac{1}{2}mv^2[/tex]Thus, the total kinetic energy of the bicycle is,
[tex]\begin{gathered} K=K_t+K_r \\ K=\frac{1}{2}mv^2+\frac{Iv^2}{r^2} \end{gathered}[/tex]The ratio of rotational kinetic energy and total kinetic energy is,
[tex]\begin{gathered} \frac{K_r}{K}=\frac{\frac{Iv^2}{r^2}}{\frac{1}{2}mv^2+\frac{Iv^2}{r^2}} \\ \frac{K_r}{K}=\frac{2I}{mr^2+2I}\times\frac{\frac{1}{2}\times\frac{v^2}{r^2}}{(\frac{1}{2}\times\frac{v^2}{r^2})} \\ \frac{K_r}{K}=\frac{2I}{mr^2+2I} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \frac{K_r}{K}=\frac{2\times0.08}{75\times(0.33)^2+2\times0.08} \\ \frac{K_r}{K}=\frac{0.16}{8.17+\text{0}.16} \\ \frac{K_r}{K}=0.019 \end{gathered}[/tex]Thus, the rotational kinetic energy of the wheels is 0.019 times the total kinetic energy of the bicycle.