The x and y components of a velocity field are given by u = 5y/(x^2 + y^2) and v = (-5x/(x^2 + y^2). Calculate the circulation around a circular path of radius 3m. Let the units of u and v be in m/s.
Consider a velocity field where x and y components of velocity are \(u = 5y/\left( {{x^2} + {y^2}} \right)\) and \(v = – 5x/\left( {{x^2} + {y^2}} \right)\). Calculate circulation around a circular path of radius \(3m\) . Let the units of \(u\) and \(v\) be in \(m/s\).
Let \(x = r\cos \theta ,\,y = r\sin \theta \), therefore \({x^2} + {y^2} = {r^2}\). In polar- coordinates \({V_r} = u\cos \theta + v\sin \theta \) and \({V_\theta } = – u\sin \theta + v\cos \theta \)
Hence,
\(u = \frac{{5y}}{{{x^2} + {y^2}}} = \frac{{5r\sin \theta }}{{{r^2}}} = \frac{{5\sin \theta }}{r}\) , \(v = – \frac{{5x}}{{{x^2} + {y^2}}} = – \frac{{5r\cos \theta }}{{{r^2}}} = – \frac{{5\cos \theta }}{r}\)
\({V_r} = \frac{{5\sin \theta }}{r}\left( {\cos \theta } \right) + \left( { – \frac{{5\cos \theta }}{r}} \right)\sin \theta = 0\) , \({V_\theta } = – \frac{{5\sin \theta }}{r}\sin \theta + \left( { – \frac{{5\cos \theta }}{r}} \right)\cos \theta = – \frac{5}{r}\)
\(V.ds = \left( {{V_r}{e_r} + {V_\theta }{e_\theta }} \right).\left( {dr{e_r} + rd\theta {e_\theta }} \right)\)\( = \left( {{V_r}dr + r{V_\theta }d\theta } \right)\)\( = 0 + r\left( { – \frac{5}{r}} \right)d\theta \)\( = – 5d\theta \)
Therefore, \(\tau = – \oint_c {V.ds} = – \int\limits_0^{2\pi } { – 5d\theta } = 5\int\limits_0^{2\pi } {d\theta } \)\( = 5 \times 2\pi \ = \,10\,{m^2}/s\).
Here, value of circulation, is independent of diameter of circular path.
