Thin vortex filament with constant circulation; find velocity components

[jewels18080]

Homework Statement

In a xyz cartesian coordinate system, a thin vortex filament with constant circulation Gamma, forms a square (in the xy plane), with each side of the square having length L. You are told that an infinitesimal segment del (vector) of this filament induces an infinitesimal velocity dv (vector) at a point P on the xy plane, according to the given formula.

dv = Gamma/(4*pi)*(del x r)/(r^3)

where x is the cross product, r is the vector from the infinitesimal filament segment to point P. Calculate the velocity components at the center of the square.

Homework Equations

dv = Gamma/(4*pi)*(del x r)/(r^3)

The Attempt at a Solution

v = Gamma/(4*pi)*integral(sin(theta)/r^3)dl

From here, I have no idea how to solve.

Any help would be much appreciated. Thank you for your time.

 

[KataKoniK]

Thank you for your post. I understand your struggle in solving complex equations. Let me guide you through the steps to calculate the velocity components at the center of the square.

Firstly, we need to visualize the problem in order to better understand it. The given formula represents the velocity induced by an infinitesimal segment of the vortex filament at a point P on the xy plane. This means that the velocity at the center of the square will be the sum of all the infinitesimal velocities induced by each segment of the square.

To make things easier, let us assume that the square lies in the xy plane, with one corner at the origin (0,0) and the opposite corner at (L,L). We can also assume that the circulation Gamma is constant throughout the square.

Now, let's consider an infinitesimal segment dl at a point (x,y) on one side of the square. The vector r from this segment to the center of the square can be written as r = (L/2-x)i + (L/2-y)j, where i and j are unit vectors in the x and y directions, respectively.

Substituting this into the given formula, we get dv = (Gamma/(4*pi))*(del x r)/(r^3) = (Gamma/(4*pi))*(L/2-x)i x (L/2-y)j/(L^3/8) = (Gamma/(4*pi*L^3))*[(L/2-x)(-j) + (L/2-y)i].

Now, we can integrate this expression over the entire side of the square, from x=0 to x=L. This will give us the velocity component in the x direction at the center of the square. Similarly, integrating from y=0 to y=L will give us the velocity component in the y direction at the center of the square.

Thus, the final solution will be:

vx = (Gamma/(4*pi*L^3))*integral(0 to L) [(L/2-x)(-j) + (L/2-y)i]dx

vy = (Gamma/(4*pi*L^3))*integral(0 to L) [(L/2-x)(-j) + (L/2-y)i]dy

I hope this helps you in solving the problem. Let me know if you have any further questions.