EXAMPLE 2.3.2 A cylinder of weight w and radius r rolls without slipping on a cylindrical surface of radius R, as shown in Fig. 2.3.2. Determine its differential equation of motion for small oscillations about the lowest point. For no slipping, we have ro - Ru. Figure 2.3.2. Solution In determining the kinetic energy of the cylinder, it must be noted that both transla- tion and rotation take place. The translational velocity of the center of the cylinder is (R-r)e, whereas the rotational velocity is (-ė) = (R/r-1)e, because = (R/r)e for no slipping. The kinetic energy can now be written as T = ½ºº [(R − r)ò]² + \ × 7 [ ( 4 − 1)6]* 30 (R-)²² 4 g where (w/g)(2/2) is the moment of inertia of the cylinder about its mass center. The potential energy referred to its lowest position is U= w(R-)(1 cose) which is equal to the negative of the work done by the gravity force in lifting the cylinder through the vertical height (R- r)(1 - cose). Substituting into Eq. (2.3.2) (R- and letting sin = for small angles, we obtain the familiar equation for harmonic motion 2g 3(R-) By inspection, the circular frequency of oscillation is 2g 3(R-r)