= Solution
In a <parallel-plate rheometer>, a point at radius $r$ on the upper plate moves at speed $\Omega r$. The thin-gap approximation therefore gives the local <shear rate>
$$
\dot\gamma(r)=\frac{\Omega r}{h}.
$$
Define the rim value
$$
\dot\gamma_R=\frac{\Omega R}{h},
\qquad
\dot\gamma(r)=\dot\gamma_R\frac rR.
$$
An annulus of radius $r$ and width $dr$ has area $2\pi r\,dr$; its tangential force is $\tau(r)2\pi r\,dr$, and its moment arm is $r$. The measured <torque> is consequently
$$
T=2\pi\int_0^R\tau(\dot\gamma(r))r^2\,dr.
$$
For a <generalized Newtonian fluid>, $\tau=\eta(\dot\gamma)\dot\gamma$. Changing the integration variable from $r$ to $\dot\gamma$ gives
$$
\boxed{
T=\frac{2\pi R^3}{\dot\gamma_R^3}
\int_0^{\dot\gamma_R}\eta(\dot\gamma)\dot\gamma^3\,d\dot\gamma}.
$$
Thus the requested kernel is
$$
f(\dot\gamma,R,\dot\gamma_R)
=\frac{2\pi R^3\dot\gamma^3}{\dot\gamma_R^3}.
$$
Multiplying by $\dot\gamma_R^3$ and taking a <derivative> with respect to $\dot\gamma_R$ turns the upper-limit contribution of the <integral> into the rim viscosity:
$$
\frac{d}{d\dot\gamma_R}\left(T\dot\gamma_R^3\right)
=2\pi R^3\eta(\dot\gamma_R)\dot\gamma_R^3.
$$
Therefore
$$
\boxed{
\eta(\dot\gamma_R)
=\frac{3T+\dot\gamma_R\,dT/d\dot\gamma_R}
{2\pi R^3\dot\gamma_R}
=\frac{3T+\Omega\,dT/d\Omega}
{2\pi R^3\dot\gamma_R}}.
$$
A measured torque curve and its slope hence determine $\eta$ over the range of imposed rim shear rates.
The experimental law implies
$$
\dot\gamma_R=\frac{\Omega R}{h}=\alpha(T-T_c),
\qquad
T=T_c+\frac{\dot\gamma_R}{\alpha}.
$$
Substitution gives
$$
\boxed{
\eta(\dot\gamma)
=\frac{3T_c}{2\pi R^3\dot\gamma}
+\frac{2}{\pi\alpha R^3}}.
$$
The graph is a decreasing rectangular hyperbola with a positive high-rate plateau $2/(\pi\alpha R^3)$ and a $1/\dot\gamma$ divergence at the origin. Equivalently,
$$
\tau=\eta\dot\gamma
=\frac{3T_c}{2\pi R^3}
+\frac{2}{\pi\alpha R^3}\dot\gamma,
$$
so the inferred material is a <Bingham plastic> with yield stress $3T_c/(2\pi R^3)$.
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