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Homework Statement
Consider an imaginary circular disc, of radius R, whose arbitrary orientation is described by the unit vector, [tex] \vec {n} [/tex], perpendicular to the plane of the disc. Define the component, in the direction [tex] \vec {n} [/tex], of the angular velocity, [tex] \vec {\Omega} [/tex], at a point in the fluid by [tex] \vec {\Omega}. \vec {n} = \lim_{R \rightarrow 0}[\frac {1}{2 \pi R^2} \oint_C \vec {u}.dl] [/tex], where C denotes the the boundary (rim) of the disc. Use Stokes' theorem, and the arbitrariness of [tex] \vec {n} [/tex], to show that [tex] \vec {\Omega}= \frac {1}{2} \vec {\omega}[/tex], where [tex] \vec {\omega} = \nabla * \vec {u} [/tex] is the vorticity of the fluid at R=0. [This definition is based on a description applicable to the rotation of solid bodies. Confirm this by considering [tex] \vec {u} = \vec {U} + \vec {\Omega}* \vec{r} [/tex], where [tex] \vec {U} [/tex] is the translational velocity of the body, [tex] \vec {\Omega} [/tex] is its angular velocity and [tex] \vec {r} [/tex] is the position vector of a point relative to a point on the axis of rotation.]
Homework Equations
Stokes' theorem : [tex] \oint_c u.dl = \iint_S (\nabla * u) .n ds [/tex]
The Attempt at a Solution
Either 1.:
C is boundary of [tex] x^2 + y^2 = R^2 [/tex]. Parametrically x= Rcost, y = Rsint
dx = -Rsintdt, dy = Rcostdt
L.H.S. of Stokes becomes [tex] \oint_c udx + vdy + wdz [/tex]
= [tex] \oint_C -uRSintdt + vRCostdt[/tex]
=[tex] \int_{0}^{2 \pi} -uRSint dt + vRCost dt [/tex]
=[tex] uRCost + vRSint \right]_{0}^{2 \pi}[/tex]
= -uR - uR
=-2uR
multiply term outside integral
=[tex]-\frac{u}{\pi R}[/tex]
Or 2:
[tex] \frac {1}{2} (\nabla * \vec {u}). \vec {n} = \lim_{R \rightarrow 0}[\frac {1}{2 \pi R^2} \iint_S (\nabla * u) .n ds ] [/tex], ...
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