Answer:
![E_k = M\Omega^2\left((3)/(40)R^2 + (3)/(160)h^2\right)](https://img.qammunity.org/2021/formulas/physics/college/4wttypn4mk42y5nn19qqgdye81pi7v6vda.png)
Step-by-step explanation:
The (rotational) kinetic energy of the right uniform circular cone is as the following:
![E_k = I\omega^2/2](https://img.qammunity.org/2021/formulas/physics/college/xtm8j629d1wwwte0vmd1c8g0j1mf9rknq7.png)
where
is the angular speed, I is the moments of inertia about the axis of rotation, which goes through the center of mass:
![I = M\left((3)/(20)R^2 + (3)/(80)h^2\right)](https://img.qammunity.org/2021/formulas/physics/college/e7iphxwoqt5cxh9lekizprdtqxjvzghe3d.png)
Therefore:
![E_k = M\Omega^2\left((3)/(40)R^2 + (3)/(160)h^2\right)](https://img.qammunity.org/2021/formulas/physics/college/4wttypn4mk42y5nn19qqgdye81pi7v6vda.png)