Answer:
The rotational kinetic energy is 19271.07 Joules.
Explanation:
Given that,
Mass of the motorcycle, m = 14 kg
Angular velocity of the wheel, [tex]\omega=130\ rad/s[/tex]
Inner radius is 0.270 m, [tex]r_1=0.27\ m[/tex]
Outer radius, [tex]r_2=0.3\ m[/tex]
The rotational kinetic energy of the wheel is given by :
[tex]K=\dfrac{1}{2}I\omega^2[/tex]
I is moment of inertia of the wheel, [tex]I=m(r_1^2+r_2^2)[/tex]
So,
[tex]K=\dfrac{1}{2}m(r_1^2+r_2^2)\omega^2\\\\K=\dfrac{1}{2}\times 14\times ((0.27)^2+(0.3)^2)\times (130)^2\\\\K=19271.07\ J[/tex]
So, the rotational kinetic energy is 19271.07 Joules.
