We found earlier that the rotational kinetic energy of a rigid body in pure rotational motion is $K_{\text{rot}}=\frac{1}{2}I{\omega }^2$. Now we would like to find the kinetic energy of an object that rolls without slipping, a combination of rotational and translation motion.

We begin with the observation that the bottom point in FIGURE 12.41 is instantaneously at rest. Consequently, we can think of an axis through P as an instantaneous axis of rotation. The idea of an instantaneous axis of rotation seems a little far-fetched, but it is confirmed by looking at the instantaneous velocities of the center point and the top point. We found these in Figure 12.40 and they are shown again in Figure 12.41. They are exactly what you would expect as the tangential velocity $v_t=r\omega$ for rotation about P at distances R and 2R.

FIGURE 12.41 Rolling motion is an instantaneous rotation about point P.

From this perspective, the object’s motion is pure rotation about point P. Thus the kinetic energy is that of pure rotation:

$I_{\text{P}}$ is the moment of inertia for rotation about point P. We can use the parallel-axis theorem to write $I_{\text{P}}$ in terms of the moment of inertia $I_{\text{cm}}$ about the center of mass. Point P is displaced by distance $d=R;$ thus

Using this expression in Equation 12.38 gives us the kinetic energy:

We know from the rolling constraint that $R\omega$ is the center-of-mass velocity $v_{\text{cm}}$. Thus the kinetic energy of a rolling object is

In other words, the rolling motion of a rigid body can be described as a translation of the center of mass (with kinetic energy $K_{cm}$) plus a rotation about the center of mass (with kinetic energy $K_{rot}$).