Answer:
[tex]E_k = M\Omega^2\left(\frac{3}{40}R^2 + \frac{3}{160}h^2\right)[/tex]
Explanation:
The (rotational) kinetic energy of the right uniform circular cone is as the following:
[tex]E_k = I\omega^2/2[/tex]
where [tex]\omega = \Omega[/tex] is the angular speed, I is the moments of inertia about the axis of rotation, which goes through the center of mass:
[tex]I = M\left(\frac{3}{20}R^2 + \frac{3}{80}h^2\right)[/tex]
Therefore:
[tex]E_k = M\Omega^2\left(\frac{3}{40}R^2 + \frac{3}{160}h^2\right)[/tex]
