Answer:
2
Explanation:
The rotational kinetic energy of the dumbbell is:
[tex]E_r = \frac{1}{2}I\omega^2[/tex]
where I is the moments of inertia of the two point-like system
[tex]I = 2Mr^2 = 2ML^2[/tex]
[tex]E_r = ML^2\omega^2[/tex]
The translation kinetic energy is:
[tex]E_t = \frac{1}{2}Mv^2 = \frac{1}{2}M\omega^2L^2[/tex]
Therefore the ratio of the rotational kinetic energy to the translational (linear) kinetic energy is:
[tex]\frac{E_r}{E_t} = \frac{ML^2\omega^2}{\frac{1}{2}M\omega^2L^2} = 2[/tex]
