# Advanced Dynamics by Donald T. Greenwood

By Donald T. Greenwood

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**Example text**

Now suppose that the system is not in static equilibrium, implying that Fi + Ri = 0 for at least one particle. 254) i=1 since sufﬁcient degrees of freedom remain. 252) we conclude that a virtual displacement can always be found that results in δW = 0 if the system is not in static equilibrium. Thus, if δW = 0 for all possible δrs, the system must be in static equilibrium; this is the sufﬁcient condition. We have assumed a catastatic system. It is possible that a particular system that is not catastatic could, nevertheless, have a position of static equilibrium if a jt and ∂ xk /∂t are not both identically zero, but are equal to zero at the position of static equilibrium.

148) 26 Introduction to particle dynamics where K = Gm 0 m. 673 × 10−11 N · m2 /kg2 and we have used units of Newtons, meters, and kilograms. 148) is a statement of Newton’s law of gravitation where each attracting body is regarded as a particle. 150) r and we note that the gravitational potential energy is generally negative, but goes to zero as r → ∞. V (r ) = − Linear spring A commonly encountered form of potential energy is that due to elastic deformation. As an example, consider a particle P which is attached by a linear spring of stiffness k to a ﬁxed point O, as shown in Fig.

For the case of holonomic constraints, the conﬁguration point moves in a reduced space of (n − m) dimensions since it must remain on each of m constraint surfaces, that is, on their common intersection. Thus, certain regions of n-dimensional conﬁguration space are no longer accessible. By contrast, for nonholonomic constraints, it is the differential motions which are constrained. Since the differential equations representing these nonholonomic constraints are not integrable, there are no ﬁnite constraint surfaces in conﬁguration space and there is no reduction of the accessible region.