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uFy=ybsinptyOl 1ulFAϕFigure 2.10 Calculation scheme in the case of changing cutting conditionsThe general solution of the differential equation (2.4) for the declination angle ofthe blank taking into account the initial conditionsMF F uϕ = ϕ C t C tptr r1 + ϕ 2 = 1 cos ω + 2 sin ω +sin + − t +2 2J ( p ) kl 2ω −kl(2.5)*Fal* 2Fauω* Fau*+sin ω t +cos ω t −t sin ω t2 *222 *2J ( ω − ω ) 2 *2J ( ω − ω ) J ( ω − ω )where*F 2FuωC1= ϕ0− r −aklJ2 *2( ω −ω) 2, C2 =& 0−ϕωωJMpF u*F lω+r a−2 2 2 2 *2( ω − p ) ωklωJ( ω − ω )(2.6)From Eq. (2.5) it is easy to find the displacement y = ϕland the rotation speed νof the blank end in relation to initial conditions y 0 and v 0 .2.4. Dynamical model with one degree of freedom with dampingIt has been so far assumed that no frictional resisting forces act on the systems infree vibrations, but actually, in practice, such forces are always present and will intime damp out the vibrations. Usually, in the study of steady-state vibration, thevalues of the components, which determine free damping vibrations, are reduced.However, it is impossible to achieve it in this instance, because the operatingconditions in the cutting are changed due to surface roughness. Dynamical modelswithout damping are more precisely described in Paper II.The differential equations of forced vibrations with damping varying with thevelocity in much the same way as in Eq. (2.4):**( l −ut) + F sinωt ⋅l− F ⋅utsinωtJo& 2 2ϕ&+ βϕ&l + kylϕ = M y sin pt + Fraa , (2.7)where β is the damping factor.The general solution of Eq. (2.7) represents free vibrations:ϕ 1 = exp( − nt)( C1cosω1t+ C2sinω1t) , (2.8)2where 2n = β l / Jois the natural frequency of free damping vibrations.The values of the damping factor for the lathe are assumed n = (0.06…0.08) ω(Hakansson, 1999). In practical problems it can be assumed with sufficient accuracythat small resisting forces do not affect the frequencies, thus ω 1 ≈ ω. A particularsolution of Eq. (2.7), depending on the foundation vibrations and disturbing force25