What is the type of kinetic energy associated with a molecule spinning about its center of mass called?

If we define "temperature" as "a useful number proportional to the mean energy stored in some degree of freedom" (which is not the most fundamental definition of temperature, but it's close in this circumstance), then it's possible for the "translational temperature" and the "spin temperature" of a gas to get out of thermal equilibrium with each other. When this happens, a "thermometer" which couples to the translational degree of freedom will measure the translational temperature, while a thermometer which couples to the spin degree of freedom will measure the spin temperature. In general it takes more cleverness and subterfuge to build a "spin thermometer," which is where the sense of your quote comes from.

A naturally-occurring example of such disequilibrium occurs in diatomic hydrogen, $\rm H_2$. The two protons in a hydrogen molecules must obey Fermi-Dirac statistics, so their wavefunction must be antisymmetric under exchange. The rotational wavefunctions with quantum number $L$ have symmetry $(-1)^L$ under exchange — that is, the $s$-wave and $d$-wave are symmetric, the $p$-wave and $f$-wave are antisymmetric, and so on. To have the total wavefunction antisymmetric, the states with even $L$ must have the two protons in a spin singlet, while the states with odd $L$ must have the two protons in a spin triplet. The energies of the rotational states are

$$ E_L = \frac{15\rm\,meV}{2}L(L+1) $$

Here the energy of the $L=1$ state, a few milli-eV, is consistent with the moment of inertia for a rotor whose mass and size are set by the hydrogen molecule's mass and bond length. There isn't any important correction to the molecular energy due to the proton-proton spin interaction because the proton's magnetic moment is very small. If the hydrogen gas is in thermal equilibrium, the probability of finding a molecule in a state with angular momentum $L$ is proportional to

$$ P(L) \propto (2S+1)(2L+1) e^{-E_L/kT} $$

This is just the ordinary Boltzmann factor $e^{-E_\text{state}/kT}$ and the degeneracy factors for the spin and orbital angular momentum quantum numbers: there are $2S+1=3$ ways to orient a spin triplet, but only $2S+1=1$ way to orient a spin singlet. At room temperature ($kT=25\rm\,meV$), there are nontrivial populations in states with many $L$, so hydrogen's heat capacity is $\frac52R$. Near its triple point ($T \approx 30\,\mathrm K = 2.5\,\mathrm{meV}/k$) the rotational degrees of freedom are "frozen out" as the Boltzmann factor $e^{-E_L/kT}$ becomes smaller and smaller for all the states with $L>0$. The heat capacity of cold hydrogen gas approaches $\frac32R$ as the rotational states become less accessible.

At least, that's what you'd think until you starting making liquid hydrogen.

It turns out that there isn't any internal mechanism for an isolated molecule of $\rm H_2$ to convert from a spin triplet to a spin singlet (or vice-versa), but collisions between molecules allow spin exchange pretty effectively. So hydrogen which has been at room temperature for a long time will have a thermal-equilibrium distribution of $L$. The requirement that protons obey Fermi-Dirac statistics, and the spin-degeneracy factor $(2S+1)$ attached to the Boltzmann factor, conspire to mean that three-quarters of the population will be spin-triplet molecules with $L$ odd. When you cool hydrogen vapor down below its triple point, you actually don't bring all of the molecules into the $L=0$ ground state. Most of your hydrogen has odd $L$ and gets stuck in the $L=1$ state, waiting to collide with another $L=1$ molecule to be able to release its rotational energy. The even-$L$ and odd-$L$ molecules actually behave like two miscible but distinct gases, called "parahydrogen" ($L$ even) and "orthohydrogen" ($L$ odd). They have slightly different heat capacities, because the first excitation in parahydrogen $L=0\to2$ has different energy and degeneracy than the first excitation in orthohydrogen $L=1\to3$. They also have slightly different densities and other thermal properties.

The rate of ortho-to-para conversion in cold hydrogen (where the $L=1$ state should be "frozen out") is proportional to the square of the orthohydrogen density: the orthos have to find each other. When you liquify cold hydrogen vapor its total density increases by a thousandfold, and the rate of orthohydrogen downconversion increases by a factor of a million. Each $L=1\to0$ downconversion releases 15 meV, which happens to be approximately the same the latent heat of vaporization. So what happens when you liquify hydrogen gas from room temperature is

1. Your room-temperature "normal hydrogen" (75% ortho, 25% para) falls below its boiling temperature and condenses into a liquid.
2. The orthohydrogens in the dense liquid undergo collisions with each other more rapidly than they did in the sparse gas. They find each other, have spin-exchange collisions, and convert to parahydrogen.
3. The conversion to parahydrogen releases so much heat that all of your hydrogen boils again!
4. If you have kept your parahydrogen vapor in your cryostat, rather than venting it and replacing it with more normal hydrogen from your bottle outside, you can re-liquify the parahydrogen and it will stay liquid this time.

Any "thermometer" that you attach to this system is going to tell you that hydrogen's triple point is about 33 K and its boiling point at ordinary pressure is about 20 K; ordinary thermometers are not sensitive to the extra heat stored in orthohydrogen's orbital angular momentum degree of freedom. You cannot tell the difference between cold $L=0$ hydrogen gas and cold $L=1$ hydrogen gas using only a thermometer; this statement directly addresses your question.

The problem with the statement you quote, alluded to in the very first sentence of this long answer, is that thermometers don't measure internal energy. A thermometer measures temperature, which is is a system's "willingness to give up its internal energy," to paraphrase Schroeder's excellent intro-thermal textbook. In many systems the temperature and the internal energy are proportional to each other. But it's quite possible for a system with non-interacting degrees of freedom to function at a different temperature for each type of interaction. The most common examples are "spin temperatures," but there are others as well.

It’s plausible to suppose that the greater the velocity of a body, the greater effect it could have on other bodies. This does not depend on the direction of the velocity, only its magnitude. At the end of the seventeenth century, a quantity was introduced into mechanics to explain collisions between two perfectly elastic bodies, in which one body makes a head-on collision with an identical body at rest. The first body stops, and the second body moves off with the initial velocity of the first body. (If you have ever played billiards or croquet, or seen a model of Newton’s Cradle, you have observed this type of collision.) The idea behind this quantity was related to the forces acting on a body and was referred to as “the energy of motion.” Later on, during the eighteenth century, the name kinetic energy was given to energy of motion.

With this history in mind, we can now state the classical definition of kinetic energy. Note that when we say “classical,” we mean non-relativistic, that is, at speeds much less that the speed of light. At speeds comparable to the speed of light, the special theory of relativity requires a different expression for the kinetic energy of a particle, as discussed in Relativity.

Since objects (or systems) of interest vary in complexity, we first define the kinetic energy of a particle with mass m.

We then extend this definition to any system of particles by adding up the kinetic energies of all the constituent particles:

Note that just as we can express Newton’s second law in terms of either the rate of change of momentum or mass times the rate of change of velocity, so the kinetic energy of a particle can be expressed in terms of its mass and momentum (p→=mv→),(p→=mv→), instead of its mass and velocity. Sincev=p/mv=p/m, we see that

K=12m(pm)2=p22mK=12m(pm)2=p22m

also expresses the kinetic energy of a single particle. Sometimes, this expression is more convenient to use than Equation 7.6.

The units of kinetic energy are mass times the square of speed, or kg·m2/s2kg·m2/s2. But the units of force are mass times acceleration, kg·m/s2kg·m/s2, so the units of kinetic energy are also the units of force times distance, which are the units of work, or joules. You will see in the next section that work and kinetic energy have the same units, because they are different forms of the same, more general, physical property.

Because velocity is a relative quantity, you can see that the value of kinetic energy must depend on your frame of reference. You can generally choose a frame of reference that is suited to the purpose of your analysis and that simplifies your calculations. One such frame of reference is the one in which the observations of the system are made (likely an external frame). Another choice is a frame that is attached to, or moves with, the system (likely an internal frame). The equations for relative motion, discussed in Motion in Two and Three Dimensions, provide a link to calculating the kinetic energy of an object with respect to different frames of reference.

Kinetic Energy Relative to Different Frames

A 75.0-kg person walks down the central aisle of a subway car at a speed of 1.50 m/s relative to the car, whereas the train is moving at 15.0 m/s relative to the tracks. (a) What is the person’s kinetic energy relative to the car? (b) What is the person’s kinetic energy relative to the tracks? (c) What is the person’s kinetic energy relative to a frame moving with the person?

Strategy

Since speeds are given, we can use 12mv212mv2 to calculate the person’s kinetic energy. However, in part (a), the person’s speed is relative to the subway car (as given); in part (b), it is relative to the tracks; and in part (c), it is zero. If we denote the car frame by C, the track frame by T, and the person by P, the relative velocities in part (b) are related by v→PT=v→PC+v→CT.v→PT=v→PC+v→CT. We can assume that the central aisle and the tracks lie along the same line, but the direction the person is walking relative to the car isn’t specified, so we will give an answer for each possibility, vPT=vCT±vPCvPT=vCT±vPC, as shown in Figure 7.10.

Figure 7.10 The possible motions of a person walking in a train are (a) toward the front of the car and (b) toward the back of the car.

Solution

1. K=12(75.0kg)(1.50m/s)2=84.4J.K=12(75.0kg)(1.50m/s)2=84.4J.
2. vPT=(15.0±1.50)m/s.vPT=(15.0±1.50)m/s. Therefore, the two possible values for kinetic energy relative to the car are

   K=12(75.0kg)(13.5m/s)2=6.83kJK=12(75.0kg)(13.5m/s)2=6.83kJ

   and

   K=12(75.0kg)(16.5m/s)2=10.2kJ.K=12(75.0kg)(16.5m/s)2=10.2kJ.

3. In a frame where vP=0,K=0vP=0,K=0 as well.

Significance

You can see that the kinetic energy of an object can have very different values, depending on the frame of reference. However, the kinetic energy of an object can never be negative, since it is the product of the mass and the square of the speed, both of which are always positive or zero.

The kinetic energy of a particle is a single quantity, but the kinetic energy of a system of particles can sometimes be divided into various types, depending on the system and its motion. For example, if all the particles in a system have the same velocity, the system is undergoing translational motion and has translational kinetic energy. If an object is rotating, it could have rotational kinetic energy, or if it’s vibrating, it could have vibrational kinetic energy. The kinetic energy of a system, relative to an internal frame of reference, may be called internal kinetic energy. The kinetic energy associated with random molecular motion may be called thermal energy. These names will be used in later chapters of the book, when appropriate. Regardless of the name, every kind of kinetic energy is the same physical quantity, representing energy associated with motion.