Explanation:
Given:
Vector potential
[tex]A (r) = \frac{u_o}{4*\pi } \int {\frac{K(r')}{r*} } \, da'[/tex]
Where, K = б*v ; r* = sqrt (R^2 + r^2 -2R*r*cos(θ')) ; da' = R^2*sin(θ')*dθ'dΦ'
Solution:
- Velocity of v point a point r' in a rotating rigid body is given by:
v = w x r' =
[tex]\left[\begin{array}{ccc}x&y&z\\wsin(a)&0&wcos(a)\\Rsin(b')cos(c')&Rsin(b')sin(c')&Rcos(b')\end{array}\right][/tex]
- where a = Ψ and b' = θ' and c' = Φ'
v = R*w [-(cos Ψ *sin θ' *sin Φ') x + (cos Ψ *sin θ' *cos Φ' - sin Ψ * cos θ') y
+ (cos Ψ *sin θ' *sin Φ') z ]
- Notice that terms like sin Φ' and cos Φ' contribute to zero:
[tex]\int\ {sin (c')} \, dc' = \int\ {cos (c')} \, dc' = 0[/tex]
- Hence,
[tex]A(r) = - \frac{u_o *R^3*w*sigma*sin(a)}{2}*\int\limits^p_0 {\frac{cos (b')*sin(b') }{\sqrt{R^2 + r^2 -2*r*R*cos(b')} } } \, db' y[/tex]
- Evaluate integral u = cos (b')
[tex]= - \frac{1}{3R^2*r^2} * [ (R^2 + r^2 +rR)*(R - r) - (R^2 + r^2 -rR)*(R + r)][/tex]
- From we can determine two cases when r > R and r < R
Hence,
r < R [tex]A(r) = - \frac{u_o *R*sigma}{3} * (w x r)[/tex]
r > R [tex]A(r) = - \frac{u_o *R^4*sigma}{3r^3} * (w x r)[/tex]
- Reverting back to original coordinate system given in figure 5.45:
r < R [tex]A(r) = - \frac{u_o *R*w*sigma}{3} *r*sin(b) . c[/tex]
r > R [tex]A(r) = - \frac{u_o *R^4*w*sigma}{3r^2} *sin(b) . c[/tex]
Where, b = θ and c = direction along Φ.
Hence, A ( r , θ , Φ )