# Calculate rotational energy of a tri-atomic linear molecule

For linear molecules with two atoms, I understand that the rotational energy is: $$\frac{1}{2} \mu \omega^2R^2$$ Where $$\mu$$ is the reduced mass and $$R$$ the bond length.

However, consider a rotating asymmetric linear molecule of 3 atoms with the heaviest atom being at one of the two edges. The rotating axis will therefore be slightly off from the middle atom. The consequence of this is that the middle atom is very close to the rotating axis. Its distance to it would be not negligible compared to its own radius. This means that different parts of the middle atom have significant different velocities because these parts have different distances to the rotating axis.

How is the rotating energy of such a molecule calculated?

The general formula for a rigid linear molecule is $$E_{r}=\frac12 I \omega^2$$ where $$I$$ is the tensor of inertia. In the particular case of a diatomic molecule the inertia tensor is, as you write $$I=\mu R^2$$. So, if you have a triatomic molecule you just have to use its inertia moment (calculated with respect to the rotation axis) in the general formula.
• Do you mean that; $$E_r = \frac{1}{2}I_1\omega^2 + \frac{1}{2}I_2\omega^2 +\frac{1}{2}I_3\omega^2$$ Where $I_n$ is the inertia tensor for each of the 3 atoms according to $I_n = m_n \cdot R_n^2$? – Phy Jun 18 at 0:15
• Yes, your formula is correct. but there is only one inertia moment for the system. you have $E_r = I \frac{\omega^2}{2}$, with $I=\sum\limits_{n=1,3}m_n r_n^2$. Here $r_n$ are the distances from each atom to the rotation axis – Ebe Stignani Jun 18 at 7:19
• Thanks. How about if the rotation axis is exactly in the middle, such that the middle atom is rotating around its own axis and its distance $r_n = 0$. What would the kinetic energy of the molecule then be? – Phy Jun 18 at 16:19