Answer:
5.9*10^-3 Pa*s
Explanation:
The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.
τ = μ * du/dr
u(r) = 0.4/r - 1000*r
The derivative is:
du/dr = -1/r^2 - 1000
On the radius of the inner cylinder this would be
u'(0.02) = -1/0.02^2 - 1000 = -3500
So:
τ = -3500 * μ
We don't care about the sign
τ = 3500 * μ
That is a tangential force per unit of area.
The area of the inner cylinder is:
A = h * π * D
And the torque is
T = F * r
T = τ * A * D/2
T = τ * h * π/2 * D^2
T = 3500 * μ * h * π/2 * D^2
Then:
μ = T / (3500 * h * π/2 * D^2)
μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s
