(7)
Z = lt. Z1/t = −(α − β) M′Q2 sin θ cos θ − ξQ sin θ = [ −(α − β) M′U + ξ ] V.
(8)
Now suppose the cylinder is free; the additional forces acting on the body are the components of kinetic reaction of the liquid
| −αM′ ( | dU | − VR ), −βM′ ( | dV | + UR ), εC′ | dR | , |
| dt | dt | dt |
(9)
so that its equations of motion are
| M ( | dU | − VR ) = −αM′ ( | dU | − VR ) − (α − β) M′VR, |
| dt | dt |
(10)
| M ( | dV | + UR ) = −βM′ ( | dV | + UR ) − (α − β) M′UR − ξR, |
| dt | dt |