- #1
Wildcat04
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Homework Statement
Equation solved from previous problem:
(g/[tex]\nu[/tex]) = (1/r) d/dr [ r dvz/dr]
[tex]T[/tex] = [tex]\mu[/tex] * dvz/dr
Boundary condtions:
vz (r=a) = vz (r=b) = 0
The Attempt at a Solution
(d/dr)[d/dr r*vz = g*r / [tex]\nu[/tex]
(d/dr) r*vz = gr2/2[tex]\nu[/tex] + A
vz = gr2/6[tex]\nu[/tex] + A + B/r
I am not sure if this is correct or what I am supposed to do with the B.C.s. I know that I can have 2 equations with 2 unknowns ( v = 0 @ r = a and v = 0 @ r = b, unks = A, B)
0 = vz (r=a) = ga2/6[tex]\nu[/tex] + A
A = -ga2 / 6[tex]\nu[/tex]
0 = vz (r=b) = gb2/6[tex]\nu[/tex] - ga2/6[tex]\nu[/tex] + B/r
B = -[rg(b2-a2)] / 6[tex]\nu[/tex]
vz = gr/3[tex]\nu[/tex] - ga/3[tex]\nu[/tex] -[rg(b2-a2)] / 6[tex]\nu[/tex]
I need to take the derivative of vz to get the torque...
However the solution is supposed to be:
[tex]T[/tex]=([tex]\mu[/tex]/r) [ (g(b2-a2)/4[tex]\nu[/tex]) / 2 ln(b/a) - gr2/2[tex]\nu[/tex] ]
Obviously I messed up somewhere in there. Could someone please point out my error so I can get this thing correct?
Thank you very much in advance!
I am sorry for it being a little sloppy...the [tex]\nu[/tex] kept shooting up to the top!