Expression for kinetic energy of gas per molecule The average kinetic energy (KE) per molecule of a gas is $\frac{3}{2}kT$. While finding this we do   
$$ \text{ Average KE} =\frac{1}{2} M \frac{1}{N}\sum v^2=\frac{3}{2}kT$$
But why do we not add rotational kinetic energy here?
 A: The expression you quote is for a ideal monatomic gas, and we get $C_v = 3/2$ for the three degrees of freedom. For ideal diatomic gases we do indeed have to count rotational degrees of freedom and we get $C_v = 5/2$. See the Wikipedia article on ideal gases for more info.
A: For any Ideal Gas:-


*

*PV = 2/3E

*PV = (γ-1)U
where U = E + R (Internal energy),
E = 1/2mvᵣₘₛ² (avg. translational Kinetic energy), 
& R is rotational Kinetic energy.
For eg:   A diatomic gas with γ = 7/5 has PV = 2/3E  &  PV = (7/5 - 1)U.
Therefore, 2/3E = 2/5U
OR E = 3/5U
CONCLUSION #1:  Translational kinetic energy of a diatomic gas makes up 3/5th of it's internal energy.
From  equation of state, PV = NKᵦT
Therefore, NKᵦT = 2/3E
OR E = 3/2NKᵦT
CONCLUSION #2: Avg. Translational Kinteic Energy Per Molecule (i.e. E/N = 3/2KᵦT) is related only to Temperature & is independent of pressure, volume or nature of gas.
FINAL CONCLUSION: Temperature depends on Avg. TRANSLATIONAL kinetic energy in ideal cases, whether mono, di or poly-atomic.
Thanks!!
