Q6.6 Translational and rotational motion (S)
An oxygen molecule, O2, with a total mass of 2.7×10-26 kg, is travelling at 500 ms-1, and at the same time rotating about an axis through its center of mass, and perpendicular to the ‘bond’ (the line between the atoms). If its rotational kinetic energy is 2/3 its translational kinetic energy, find its angular velocity. [ The separation of the two atoms in the molecule is 0.121×10-9 m (3.s.f) ]Hint
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![\[ \frac{2}{3} \times \frac{1}{2}(2m) v^2 = \frac{1}{2} I \omega^2 \]](mastermathpng-0.png)

Solution
First we have to calculate the moment of inertia of the molecule about an axis through its center of mass, perpendicular to the bond. Using Key Point 5.12,
I=2mr2,
where r is half the distance separating the two atoms and m is the mass of one oxygen atom.
Appealing to Key Point 5.13 and using the information provided to relate the kinetic and rotational energies we find
![\[ \frac{2}{3} \times \frac{1}{2}(2m) v^2 = \frac{1}{2} I \omega^2 \]](mastermathpng-0.png)
Then, substituting I=2mr2 we find that
![\[ \omega = \frac{\sqrt{2}}{\sqrt{3}} \frac{v}{r} = 6.8 \times 10^{12}\,rad\,s^{-1} \]](mastermathpng-1.png)