Page 36 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 2.3 Energy Method 23

Solution: Assume that the system is vibrating harmonically with amplitude 8 from its static
equilibrium position. The maximum kinetic energy is
7max = + >(''1^)1
'
L ' J max
The maximum potential energy is the energy stored in the spring, which is
^max ~ 2^(''2^)max
Equating the two, the natural frequency is
krl
J + mrl
The student should verify that the loss of potential energy of m due to position
r^6 is canceled by the work done by the equilibrium force of the spring in the position
(? = 0.

Example 2.3-2
A cylinder of weight w and radius r rolls without slipping on a cylindrical surface of
radius 7?, 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 r<>= R6.
l

Solution: In determining the kinetic energy of the cylinder, it must be noted that both
translation and rotation take place. The translational velocity of the center of the
cylinder is {R - r)6, whereas the rotational velocity is (<^ - ¿) = (R/r —1)6, be­

cause (j) = (R/r)6 for no slipping. The kinetic energy can now be written as

where (w/gXr^/2) is the moment of inertia of the cylinder about its mass center.
The potential energy referred to its lowest position is
U —>v(7? - r)(l - cos 6)

which is equal to the negative of the work done by the gravity force in lifting the
cylinder through the vertical height (i? - rXl - cos 8).

Figure 2.3-2.