The steady motions of a disc on an absolutely rough plane

The Routh theorem for mechanical systems with unknown first integrals

“…The explicit expression of this critical value is very complicated and we omit it here. The results obtained in [11,12] are in completely agreement with bifurcation diagrams presented in [9,10] and here.…”

“…The stable region is outside this ellipse and the unstable region is within it. Thus we can give a geometric interpretation for conditions of existence and stability of stationary motions of a disk [11,12]. Obviously, the stationary motions of a disk corresponding to the points of the hyperbola lying outside the ellipse are stable (Figures 6 and 7).…”

“…Application of Theorem 2 for the investigation of stability of stationary motions (4) of the system (1) is connected with the investigation of eigenvalues of matrix (7) and requires the knowledge of function ( , ) and matrix Γ( , ) only; the explicit form of first integrals (11) (and also matrices ( ) and = ( ,˙, )) is not required. Note also that condition (6) for search the stationary values of integral (2) on fixed levels of integrals (11) does not require the knowledge of explicit form of these integrals. Moreover, equations ( 6) determine stationary motions (4) of system (1) even in the case, when these integrals are absent (i.e., when equations ( 8) are failed).…”

“…Example 2. Let us consider a particular case of the previous example, when the rotationally symmetric body, moving on a fixed perfectly rough plane, is a circular disk of a radius [9][10][11][12]. Let be the mass of a disk, 1 = 2 and 3 = 2 2 are its moments of inertia.…”

“…Conditions of existence (22) and stability (25) of stationary motions (16) of a disk have been analyzed in [11,12]. In particular, it was shown [11,12] that the stationary motions (16) of a disk are stable (independently of 1 and 2 ) for any satisfying the condition In particular, for a homogeneous disk ( = 1/4) we have For other values of stationary motions (16) will be stable if the absolute value of 1 exceeds a certain critical value. The explicit expression of this critical value is very complicated and we omit it here.…”