Why isn't rotation included when calculating the average energy of a monatomic gas?

It should be noted that it has been assumed that atoms have no rotational or internal degrees of freedom. This is in fact untrue. For example, atomic electrons can exist in excited states and even the atomic nucleus can have excited states as well. Each of these internal degrees of freedom are assumed to be frozen out due to their relatively high excitation energy. Nevertheless, for sufficiently high temperatures, these degrees of freedom cannot be ignored. In a few exceptional cases, such molecular electronic transitions are of sufficiently low energy that they contribute to heat capacity at room temperature, or even at cryogenic temperatures. One example of an electronic transition degree of freedom which contributes heat capacity at standard temperature is that of nitric oxide (NO), in which the single electron in an anti-bonding molecular orbital has energy transitions which contribute to the heat capacity of the gas even at room temperature.

However, my thought is that if the moment of inertia is very small, it just means that the particle would be spinning extremely fast in order to reach an average energy of kT/2 for each rotational axis.

Greetings, this is my first post, though I have been reading these forums for a while.

I understand that the average energy of each degree of freedom in a thermodynamic system in equilibrium is kT/2. My textbook says that for a monatomic gas particle, the only degrees of freedom that count are movement in three dimensional space, so the average energy of such a particle is 3kT/2.

My question is, why don't rotations contribute toward the average energy? My textbook suggests that this is because the moment of inertia is vanishingly small. However, my thought is that if the moment of inertia is very small, it just means that the particle would be spinning extremely fast in order to reach an average energy of kT/2 for each rotational axis.

sanbyakuman