Suppose we take a bowling ball and spin it such that it does not wobble or roll away. (This is to say that the axis of rotation is fixed and \(v_{CM}=0\).) We know that this ball must contain some \(KE\) because it has masses (namely, atoms and molecules) which are in motion with tangential, translational velocities about the fixed axis at a given moment. But if we attempt to calculate this kinetic energy using the formula for translational kinetic energy, we just end up with \(KE_{tr}=\frac{1}{2}M_{ball}v_{CM}^2=1/2M_{ball}(0)^2=0\). This occurs because although the bowling ball is spinning, its center-of-mass is not moving. In our derivation for translational kinetic energy, we made the idealization that the object could be regarded as a point-mass thereby disregarding the motions of the individual particles composing the system. We also disregarded how the mass was distributed in the system.

For example, imagine pushing two objects, say a ball and a box (the particles and molecules are clearly distributed in space differently in these two examples, where in one system they are distributed in such a way as to form a sphere whereas in the other to form a cube), such that both objects are equally massive and they acquire the same velocity. In this scenario, if we use the equation for translational kinetic energy to calculate their \(KE_{tr}\)s, we obtain the same value because we ignore how the mass is distributed in space. Imagine each object is composed of the same number of molecules say \(kN_A\) (they’re just distributed in space differently); you could imagine that had these two objects been pushed across frictionless planes and that after the force had been applied (where in “after” I mean the force is no longer being applied) both objects assume a fixed velocity, take a snap shot at that particular instance where their velocities are the same: you’ll notice that for the box there are \(kN_A\) molecules with translational linear velocities, but no tangential velocity at all associated with the rotation about a fixed, wobbling or un-wobbling, axis; for the sphere, however, each molecule is moving with that same translational velocity, but additionally they are also moving with a tangential velocity abut an axis. (Imagine the bowling ball, the axis is more or less parallel to the floor and all the molecules “spin” and go around that axis.) Thus we can deduce that the sphere clearly will have more kinetic energy than the box; but when we calculate the \(KE_{tr}\), we get the same result. Therefore, we must take into account some missing kinetic energy of the system.

To think about a way for how to calculate this “missing kinetic energy,” imagine at that “snapshot instance” you took away all of the sphere’s translational kinetic energy motion—imagine you took away all of its translational or linear, forward-direction velocity. There would only be that tangential velocity left to account for. Another perhaps simpler way to think about it is imagine you took a sphere or bowling ball and spun it with your hands such that it rotated around a fixed axis and “stayed put” in a fixed position and it didn’t roll away. This would be completely analogous to that previous example where you took the “snapshot instance” of the bowling ball and imagined you took away its translational motion; the ball wouldn’t be moving “forward” anymore and one would calculate \(KE_{tr}=\frac{1}{2}m_{CM}v_{CM}^2=1/2m_{CM}(0)^2=0\) (CM not moving!), and it would just be spinning around where each molecule or particle only had tangential velocity. Anyways, let’s go back to bowling ball example which is far less confusing to think about.

Let’s think about a way to calculate this “missing kinetic energy.” Picture in your mind as many molecules as you can imagine making up the ball (so not many!)—as close to \(kN_A\) as possible. If you imagine trying to make a picture or diagram at a particular snapshot “instant,” you could draw a vector corresponding to the tangential velocity of each molecule and give each the labeling of \(\vec{v}_1, vec{v}_2\text{, ... ,}\vec{v}_{kN_A}\). You could (in this diagram) also draw lines for each molecule say, \(r_1, r_2\text{, ... ,}r_{kN_A}\), from the molecule to the axis of rotation such that when these lines intersect with the axis of rotation, they are perpendicular to that axis. Now, in this picture, we clearly can see \(kN_A\) molecules each with a tangential velocity say \(\vec{v}_{t1}, \vec{v}_{t2}\text{, ... ,}\vec{v}_{tkN_A}\) (or just \(\vec{v}_{ti}\).) Therefore, we can clearly see each molecule will have a kinetic energy of \(\frac{1}{2}m_1v_{t1}^2, \frac{1}{2}m_2v_{t2}^2\text{, … ,}\frac{1}{2}m_{kN_A}v_{tkN_A}^2\). Now, to find the kinetic energy for the total mass of the whole system (the bowling ball), we add up the kinetic energy for each point-mass (each molecule, one by one) until we have accounted for the kinetic energy of every molecule and thus the total mass of the whole bowling ball. Writing this, we have

$$\frac{1}{2}m_1v_{t1}^2+\frac{1}{2}m_2v_{t2}^2\text{, +…+ ,}\frac{1}{2}m_{kN_A}v_{tkN_A}^2=\frac{1}{2}\sum_{i=1}^{kN_A} m_iv_{ti}^2.\tag{1}$$

This equation represents the amount of kinetic energy associated with the rotation of an object which is composed of a finite number of particles. As made explicit in the example where we imagined pushing a block and sphere (both having the same mass) with an equal force so that they acquired the same velocity and then, we imagined taking away the “forward motion” of the ball so that it only had its “spinning motion” left. Perhaps the better example, we imagined spinning the bowling ball with our hands so that it “stayed put” in a fixed position where the center-of-mass didn’t move which is why we got the calculation \(KE_{tr}=\frac{1}{2}m_{CM}v_{CM}^2=\frac{1}{2}m_{CM}(0)^2=0\). Even though the ball is “staying put,” its molecules are still spinning around and we can calculate this “missing \(KE\)” using \(\frac{1}{2}m_1v_{t1}^2+\frac{1}{2}m_2v_{t2}^2\text{, +…+ ,}\frac{1}{2}m_{kN_A}v_{tkN_A}^2\). My point is this: the kinetic energy associated with the ball’s rotation is independent of the ball’s translational motion. We know this from our examples. We can have a ball spinning round-and-round in a fixed position with \(KE_{tr}=0\) and \(KE_{missing}=\text{ some amount}\) and we can vary how fast or slowly the ball spins thereby varying \(KE_{missing}\) while keeping the ball in a fixed position thereby keeping \(KE_{tr}=0\). Similarly, we can push the box with weak or strong forces making it obtain either slow or fast velocities (which varies \(KE_{tr}\)) but at the same time we can push it each time so that the box doesn’t rotate and \(KE_{missing}=0\) the whole time. Let us call this “missing” kinetic energy, rotational kinetic energy, and represent it by \(KE_{rot}\). (Another commonly used name is angular kinetic energy.) This name makes sense because it is the kinetic energy associated with the object’s rotation and nothing else. (One important note is that we have not introduced a new form of energy. \(KE_{rot}=\frac{1}{2}\sum_{i=1}^{kN_A} m_iv_{ti}^2\) is merely the sum of a bunch of translational kinetic energies. It is merely the kinetic energy associated with the rotation of the object, it is not a new form of energy.)

The rotational inertia of an object composed of a finite number, \(n\), of particles (one way of modeling a bowling ball) is given by \(I=\sum_{i=1}^{n} m_ir_{ti}^2\). Using this relation and \(v_t=rω\), we can make the appropriate substitution in order to obtain

$$KE_{rot}=\frac{1}{2}Iω^2.\tag{2}$$

Even if we let the number of particles approach infinity, we could make the analogous substitutions into Equation (1) to get Equation (2). Equation (2), therefore, describes the rotational kinetic energy of a system of \(n\) particles or of an object composed of countless atoms and molecules which is modeled as a continuous mass distribution containing an infinite number of particles.

This article is licensed under a CC BY-NC-SA 4.0 license.