Tham khảo tài liệu 'dynamics of mechanical systems 2009 part 11', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 482 Dynamics of Mechanical Systems Equation shows that the small disturbance gets larger and larger. Thus 0 n is an unstable equilibrium position. In the next several sections we will use the foregoing technique to explore the stability of several other mechanical systems. A Particle Moving in a Vertical Rotating Tube Consider the system consisting of a particle free to move in the smooth interior of a vertical rotating tube as depicted in Figure we considered the kinematics and dynamics of this system in Section . If the angular speed Q of the tube is prescribed the system has one degree of freedom represented by the angle 0. From Eq. we see that the equation of motion is then 0 - Q2 sin 0 cos 0 g r sin 0 0 where r is the tube radius. If the particle P has reached an equilibrium position 0 will be zero. The equilibrium angle will then satisfy the equation -Q2 sin 0 cos 0 g r sin 0 0 It is readily seen that the solutions to this equation are 0 01 0 0 02 n and 0 03 cos 1 g rQ2 Thus there are three equilibrium positions. In the following paragraphs we consider the stability of each of these. Case 1 0 01 0 Consider first the equilibrium position 0 0. By introducing a small disturbance 0 about 0 0 we have 0 0 0 FIGURE A vertical rotating tube with a smooth interior surface and containing a particle P. Stability 483 Then to the first order in 0 sin0 and cos0 may be approximated as sin 0 sin 0 0 0 and cos 0 cos 0 0 1 By substituting from Eqs. and into we obtain 0 g r Q2 0 0 By referring to the solutions of Eqs. and we see that if g r - Q2 is positive the solution of Eq. may be expressed in terms of trigonometric functions and thus will be bounded and stable. Alternatively if g r - Q2 is negative the solution of Eq. will be expressed in terms of exponential or hyperbolic functions and thus will be unbounded and unstable. Hence the equilibrium .