Question

The oxygen molecule, O2, has a total mass of 5.30×10-26 kg and a rotational inertia of 1.94×10-46 kg-m2 about an axis perpendicular to the center of the line joining the atoms. Suppose that such a molecule in a gas has a speed of 1.49×103 m/s and that its rotational kinetic energy is two-thirds (2/3) of its translational kinetic energy.
What then is the molecule's angular speed aboutthe center of mass. (rad/s)

Answer 1

Answer:

$\omega=2.85*10^{13}\frac{rad}{s}$

Explanation:

The translational kinetic energy depends on the mass and speed of the body, as follows:

$K_T=\frac{mv^2}{2}\\K_T=\frac{5.30*10^{-26}kg(1.49*10^3\frac{m}{s})^2}{2}\\\\K_T=1.18*10^{-19}J$

While rotational kinetic energy depends on the moment of inertia and the angular velocity of the body, as follows:

$K_R=\frac{I\omega^2}{2}(1)$. We know that:

$K_R=\frac{2}{3}K_T(2)$

Replacing (1) in (2):

$\frac{I\omega^2}{2}=\frac{2}{3}K_T\\\\\omega=\sqrt{\frac{4}{3}\frac{K_T}{I}}\\\omega=\sqrt{\frac{4}{3}\frac{1.18*10^{-19}J}{1.94*10^{-46}kg\cdot m^2}}\\\omega=2.85*10^{13}\frac{rad}{s}$