The Curl of a Vector Evaluate the curl of the velocity vector v = vx ex +vy ey for the following two cases (vz = 0 in both instances): (a) vx = −ωy, vy = ωx; (b) vx = - cy / x^2 + y^2, vy = cx / x^2 + y^2. (E5.4.1)

Question

The Curl of a Vector

Evaluate the curl of the velocity vector [latex]\mathbf{v}=v_{x} \mathbf{e}_{x}+v_{y} \mathbf{e}_{y}[/latex] for the following two cases [latex]\left(v_{z}=0\right.[/latex] in both instances):

(a) [latex]v_{x}=-\omega y, \quad v_{y}=\omega x ;[/latex] (b) [latex]v_{x}=-\frac{c y}{x^{2}+y^{2}}, \quad v_{y}=\frac{c x}{x^{2}+y^{2}}[/latex]. (E5.4.1)

Accepted Answer

In both cases, [latex]v_{z}=0[/latex] and the remaining components [latex]v_{x}[/latex] and [latex]v_{y}[/latex] do not depend on the coordinate [latex]z[/latex]; therefore, the [latex]x[/latex] and [latex]y[/latex] components of [latex] abla imes \mathbf{v}[/latex] (which would multiply the unit vectors [latex]\mathbf{e}_{x}[/latex] and [latex]\mathbf{e}_{y}[/latex] ) are --- from Eqn. (5.29) --- both zero. Thus, we are concerned only with the [latex]z[/latex] components, which multiply the unit vector [latex]\mathbf{e}_{z}[/latex].

Case (a)

[latex]\begin{aligned} \operatorname{curl} \mathbf{v} &=\left(\frac{\partial v_{z}}{\partial y}-\frac{\partial v_{y}}{\partial z} ight) \mathbf{e}_{x}+\left(\frac{\partial v_{x}}{\partial z}-\frac{\partial v_{z}}{\partial x} ight) \mathbf{e}_{y}+\left(\frac{\partial v_{y}}{\partial x}-\frac{\partial v_{x}}{\partial y} ight) \mathbf{e}_{z} \ &=\left|\begin{array}{ccc}\mathbf{e}_{x} & \mathbf{e}_{y} & \mathbf{e}_{z} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ v_{x} & v_{y} & v_{z}\end{array} ight|= abla imes \mathbf{v}. \end{aligned}[/latex] (5.29)

[latex]\operatorname{curl}_{z} \mathbf{v}=\frac{\partial v_{y}}{\partial x}-\frac{\partial v_{x}}{\partial y}=\frac{\partial(\omega x)}{\partial x}-\frac{\partial(-\omega y)}{\partial y}=\omega-(-\omega)=2 \omega.[/latex] (E5.4.2)

Thus, the flow represented by the velocity has a vorticity [latex]2 \omega \mathbf{e}_{z}[/latex] and an angular velocity [latex]\omega \mathbf{e}_{z}[/latex]. It corresponds to a forced vortex, discussed further in Section 7.2.

Case (b)

[latex]\begin{aligned}\operatorname{curl}_{z} \mathbf{v} &=\frac{\partial v_{y}}{\partial x}-\frac{\partial v_{x}}{\partial y}=\frac{\partial}{\partial x}\left(\frac{c x}{x^{2}+y^{2}} ight)-\frac{\partial}{\partial y}\left(-\frac{c y}{x^{2}+y^{2}} ight) \&=\frac{c\left(y^{2}-x^{2} ight)}{\left(x^{2}+y^{2} ight)^{2}}-\frac{c\left(y^{2}-x^{2} ight)}{\left(x^{2}+y^{2} ight)^{2}}=0 (E5.4.3)\end{aligned}[/latex]

Thus, the flow represented by the velocity has zero vorticity and zero angular velocity. It corresponds to a free vortex, to be discussed further in Section 7.2.

As can be discovered from Problem 5.12, the above two cases may be examined somewhat more easily in cylindrical coordinates, which is one of the topics of the next section.

Question 5.4: The Curl of a Vector Evaluate the curl of the velocity vecto...

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