or, putting m = a2ω, so that the vortex velocity is due to an angular velocity ω at a radius a,
(σ + ρ) dU/dt + 2ρωV = 0,
(6)
(σ + ρ) dV/dt − 2ρωU + (σ-ρ) g = 0.
(7)
Thus with g = 0, the cylinder will describe a circle with angular velocity 2ρω/(σ + ρ), so that the radius is (σ + ρ) v/2ρω, if the velocity is v. With σ = 0, the angular velocity of the cylinder is 2ω; in this way the velocity may be calculated of the propagation of ripples and waves on the surface of a vertical whirlpool in a sink.
Restoring σ will make the path of the cylinder a trochoid; and so the swerve can be explained of the ball in tennis, cricket, baseball, or golf.
Another explanation may be given of the sidelong force, arising from the velocity of liquid past a cylinder, which is encircled by a vortex. Taking two planes x = ± b, and considering the increase of momentum in the liquid between them, due to the entry and exit of liquid momentum, the increase across dy in the direction Oy, due to elements at P and P′ at opposite ends of the diameter PP′, is
| ρ dy (U − Ua2r−2 cos 2θ + mr−1 sin θ) (Ua2r−2 sin 2θ + mr−1 cos θ) + ρ dy ( −U + Ua2r−2 cos 2θ + mr−1 sin θ) (Ua2r−2 sin 2θ − mr−1 cos θ) = 2ρdymUr−1 (cos θ − a2r−2 cos 3θ), |
(8)