Answer:
[tex]\omega=2.85*10^{13}\frac{rad}{s}[/tex]
Explanation:
The translational kinetic energy depends on the mass and speed of the body, as follows:
[tex]K_T=\frac{mv^2}{2}\\K_T=\frac{5.30*10^{-26}kg(1.49*10^3\frac{m}{s})^2}{2}\\\\K_T=1.18*10^{-19}J[/tex]
While rotational kinetic energy depends on the moment of inertia and the angular velocity of the body, as follows:
[tex]K_R=\frac{I\omega^2}{2}(1)[/tex]. We know that:
[tex]K_R=\frac{2}{3}K_T(2)[/tex]
Replacing (1) in (2):
[tex]\frac{I\omega^2}{2}=\frac{2}{3}K_T\\\\\omega=\sqrt{\frac{4}{3}\frac{K_T}{I}}\\\omega=\sqrt{\frac{4}{3}\frac{1.18*10^{-19}J}{1.94*10^{-46}kg\cdot m^2}}\\\omega=2.85*10^{13}\frac{rad}{s}[/tex]
