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is2.5. Dynamical model with two degrees of freedom2.5.1. Dynamical model with two degrees of freedom without dampingThe differential equations of forced vibrations, caused by the cutting force F,according to the theorem on kinetic moment, are presented in the following form:2J0 & z&− Aω by&+ kzzl= Mzlsinpt,2*J0& y+ Aω b z&+ k y yl = M ylcos pt + Frl1l+ Fal1lcosωt , (2.15)where ω * the angular velocity of blank rotation, 2M mp y l / 2ωy = b and2Mz= mp zbl/ 2 , y b and z b are the amplitudes of the foundation vibrations, k z andk y are spring constants, and A is the moment of inertia of the blank in relation to theaxes of rotation.The general solution of Eqs (2.15) represents free vibrationsy1= a1sin( p1t+ α1)+ a2sin( p2t+ α2) ,(2.16)z1= µ1a1sin( p1t+ α1)+ µ2a2sin( p2t+ α2)where a 1 , a 2 , α 1 and α 2 are constants of integration to be determined from theinitial conditions, µ 1 and µ 2 are ratios of the amplitudes of the two principalmodes of vibrations, p 1 and p 2 are the natural frequencies of vibrations withgyroscopic forces:22 2 22( J l ( k + k ) + A ) ± J l ( k + k )2 2 2 2 4 ⎞( 0 y z + A b ) − 4J0 l k y k z 0. J 0⎛p1,2= ⎜0 y z bω⎟ 5⎝⎠ω (2.17)Our analysis shows that with an increase in the value of ω b the difference betweenthe higher and the lower frequencies, p 1 and p 2 , is increased.It was found that for the first mode with the higher frequency p 1 , the ratio µ 1 waspositive, i.e. their vibrations y 1 and z 1 were in phase or in the so-called directprecession. For the lower frequency p 2 , their vibrations y 2 and z 2 were in theopposite phase or in the so-called reverse precession. In the first mode of vibration,a point of the blank axis moves along the circle in the direction of its own rotation,while in the second mode it moves in the direction opposite to rotation.A particular solution of Eqs (2.8), depending on the disturbing force, representsthe forced vibrations of the system, which are represented as follows:whereB=*y2 = B cos pt + b cosω t + Fr l1/ k yl, (2.18)2 2M z l pA ω b − M y l ( k z l − J o p ),2 2 2 2 2 2 2p A ωb − ( k y l − J 0 p )( k z l − J 0 p )2 2M z l ( k y l − J o p ) − M y l p ω b2 2 2 2 2 2 2( k l − J p )( k l − J p ) − p A ωD = ,z 0 y 0b27