Degrees of freedom and temperature I quote the following lines directly from the Wikipedia page titled "Heat capacity":

"...rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, since this energy does not contribute to temperature."

Why doesn't rotational energy contribute to temperature? 
 A: 
For example, rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, since this energy does not contribute to temperature.

This description is misguiding in two ways.
First, the statement that
rotational energy does not contribute to temperature
makes an impression that temperature is a quantity that is closely connected with the translational kinetic energy, but not rotational kinetic energy. But that is false; according to classical theory (applicable when temperatures are high) in thermodynamic equilibrium, all quadratic degrees of freedom, translational and rotational, correspond to kinetic energy $k_BT/2$ on average.
It is only true that rotational energy does not contribute to translational kinetic energy of molecules, since the two energies are exclusive contributions to total kinetic energy.
Second, heat capacity when molecules are allowed to rotate is not higher because rotational energy does not contribute to translational kinetic energy of molecules.
It is higher because for the same temperature, such system has higher energy than system without rotation. This is because there are additional degrees of freedom, to which corresponds additional average kinetic energy.
Equilibrium implies temperature implies average energies of molecules. Value of average kinetic energies of molecules neither implies temperature exists nor implies temperature is only connected to translational kinetic energy.
A: This question is surprisingly difficult to answer and requires clear thinking.  I am not sure that I have got it entirely right myself.  I don't think the usual explanations - which state that the energy in rotational and vibrational motions are also proportional to temperature - solve the problem of why only the translational motion determines temperature.  
If two containers of gases are placed in thermal contact, heat will transfer between them until they are at the same temperature.  Their molecules then have the same average translational kinetic energy.  However, the molecules of one gas could have only translational kinetic energy while the other has vibrational and rotational energy also, but this extra energy is not reflected in the temperature of the gas.  I guess that is why this 'hidden' energy is often called 'internal energy' or - in the similar case of changes of phase - 'latent heat'.  
I think the answer stems from the fact that temperature is defined in terms of heat transfer - specifically by conduction - rather than heat content.  Body A is at a higher temperature than body B if there is a net transfer of heat energy from A to B; A and B are said to be at the same temperature if, when placed in thermal contact, there is no net transfer of heat between them.  [This is not entirely satisfactory because the only way that we know if there is a transfer of heat is by detecting a change in temperature.  So it seems to be a circular definition.] 
Transfer of heat energy by conduction only happens when atoms or molecules move through space and collide with each other and with the walls of a container.  Whatever rotational or vibrational energy a molecule has does not directly help it to move through space and collide with the walls of the container.  
The thermodynamic scale of temperature was originally based on the ideal gas laws relating temperature to the pressure and volume of a gas, and these were related to particle motion when Kinetic Theory was developed.  The pressure which a gas exerts on the walls of a container is caused by the force of molecular collisions, and this depends on the linear momentum of molecules and the frequency of collisions, each of which is proportional to linear velocity.
Measurement of temperature also depends on heat transfer - eg the mercury thermometer.
Thermodynamic temperature is now defined in terms of entropy and energy so that a temperature can be assigned to systems which don't include any translational motion.  The familiar form of temperature which we recognise by touch is often called 'kinetic temperature' because it relates to average (translational) kinetic energy.  
A: The rotation does contribute to the temperature. The heat capacity (at constant volume) is the derivative of the internal energy of the gas with respect to the temperature. If the internal energy increases (by adding more degrees of freedom), the heat capacity increases accordingly. The heat you provide to the system is now equally partitioned over all degrees of freedom, so you need to add more energy to increase the temperature of the gas. 
In a way you could actually say that because the rotation of the molecules contributes to the temperature the heat capacity goes up, as you also have to "heat up" the rotation. You can google for images of graphs that show the heat capacity of rotating and vibrating molecules as a function of the temperature (that is, the energy put into the system). At low temperatures, close to 0 K, the heat put into the system is not enough to populate higher rotational levels and the heat capacity is the same as that for a monoatomic gas. As the temperature increases, you start populating excited rotational states, and the heat capacity jumps to a higher value. Another jump is seen when you start populating the excited vibrational states (and in principle also for the excited electronic states).
A: 
"...rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, since this energy does not contribute to temperature."

The sentence is just wrong. It comes from a Wikipedia article that notes "This article has multiple issues. ..." 
Let's conduct a simple Gedankenexperiment. Given a diatomic gas we magically stop time and remove all the rotational modes of each molecule of the gas. Since we have removed energy the gas must lose energy overall. Now we start the time clock again and let the gas molecules start bumping into each other again and come to a new equilibrium. The molecules will eventually collide at an angle which will result in rotational motion. So does the gas go up in temperature, stay the same, or go down in temperature?
The gas must go down in temperature since some of the energy in the translation and vibrational modes will have been converted to rotational energy in the new equilibrium. 
A: The confusion in all the answers stated above is that somehow people are not making clear distinction between heat capacity, internal energy, temperature and kinetic energy.  The point of the original statement that "...rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, [but] this energy does not contribute to temperature" is that there is energy which does not get realized as a thermodynamic temperature.
In the ideal gas the state variable temperature is only definable in a state of equilibrium.  As such we need to expect that the gas is in equilibrium with the walls of its container.  Of course this is an ensemble average condition, but that's the game of thermodynamics.  
The reason we need to consider equilibrium is two-fold:  First, state variables such as temperature, pressure, etc have no meaning outside of equilibrium and second, the microscopic model on which the definition of temperature is built assumes elastic collisions between molecules and containing vessel.  
This is where the translational kinetic energy comes in.  When a molecule collides with a wall of the vessel (thereby producing pressure on the walls), that pressure is a consequence of momentum transfer.  It is not and cannot be a consequence of angular momentum transfer.  Furthermore, there is no momentum in a vibrational mode of a molecule since by definition they are consequences of internal forces within a molecule which can never change the momentum of a system.
So the bottom line is that an ideal gas can only produce pressure against a vessel due linear momentum transfer which only depends on a molecule's translational kinetic energy.  In this case it will satisfy the ideal gas law with a temperature T that will never depend on rotational, vibrational or any other type of motion or energy.
To suggest that the temperature of a polyatomic gas molecule is harder to change is irrelevant.  That's heat capacity, and of course by the equipartition theorem it will be larger than for a monatomic gas.  Regardless, at the same temperature both the monatomic gas and polyatomic one will have the same mean translational kinetic energy, will produce the same pressure on the walls of a vessel and will satisfy the same ideal gas law.
