Answer:
[tex]\frac{K_{rot}} {K_T} = 0.019[/tex]
Explanation:
We know that Rotational Kinetic Energy is given by,
[tex]K_{rot}=\frac{1}{2}Iw^2[/tex]
Where I is the inertia and w the angular velocity. We know as well that[tex]w= \frac{v}{r}[/tex]
The we can replace,
[tex]K_{rot} = \frac{1}{2}I(\frac{v}{r})^2[/tex]
[tex]K_{rot} = \frac{1}{2}I(\frac{v^2}{r^2})[/tex]
That equation are perfect for the 4 wheel, however we need a similar expresion for the ycycle
[tex]K_{tran} = \frac{1}{2}mv^2[/tex]
The total energy can be expressed as follow,
[tex]K_T = K_{rot}+K{tran}[/tex]
[tex]K_T=2(\frac{1}{2}I\frac{v^2}{r^2})+\frac{1}{2}mv^2[/tex]
Whit this expression is easy to find te ratio of rotational kinetic energy, we need to make the relation between [tex]K_{rot} \Rightarrow K_T[/tex]
[tex]\frac{K_{rot}} {K_T} = \frac{2I}{2I+mr^2}[/tex](You only need to put the previous values and simplify)
Substituting [tex]I=0.08kg.m^2, m=79kg and r=0.32m[/tex]
We have
[tex]\frac{K_{rot}} {K_T} = \frac{2I}{2I+mr^2} \\\frac{K_{rot}} {K_T} = \frac{2(0.08)}{2(0.08)+(79)(32)}[/tex]
[tex]\frac{K_{rot}} {K_T} = 0.019[/tex]
