The oxygen molecule has a total mass of 5.30×10^–26kg and a rotational inertia of 1 .94×10^–46 kgm^2 about an axis through the center perpendicular to the line joining the atoms . Suppose that such a molecule in a gas has a mean speed of 500m/s and that its rotational kinetic energy is two-thirds of its translational kinetic energy.Find its average angular velocity.

Moment of inertia, I = 1.94 × 10^–46 kg m²
Velocity of the oxygen molecule, v = 500 m/s
The separation between the two atoms of the oxygen molecule = 2r
Mass of each oxygen atom = m/2
Hence, moment of inertia I, is calculated as:
(m/2)r² + (m/2)r² = mr²
r = ( I / m)^1/2
(1.94 × 10^-46 / 5.36 × 10^-26 )^1/2  =  0.60 × 10^-10m
It is given that:
KE_rot = (2/3)KE_trans
(1/2) I ω² = (2/3) × (1/2) × mv²
mr²ω² = (2/3)mv²
ω = (2/3)^1/2 (v/r)
= (2/3)^1/2 (500 / 0.6 × 10^-10) = 6.80 × 10¹²rad/s.