## 2 Basic Equations and Assumptions for Cone-and-Plate Rheometer  
A schematic diagram of cone-and-plate rheometers is shown in Figure 1. The circular disk is motionless. The cone rotates with angular velocity \(\omega\). The process has an axis of symmetry. End and inertia effects are neglected. Then, the material flow between the disk and cone can be approximated by the material flow between two coordinate surfaces, \(\theta=\pi/2\) and \(\theta=\pi/2-\beta=\theta_{0}\), of a spherical coordinate system \((r,\ \theta,\ \varphi)\) whose axis \(\theta=0\) coincides with the axis of symmetry of the process (Figure 2). It is evident that the solution is independent of \(\varphi\). Those are typical assumptions used for describing the flow in the cone-and-plate rheometer [16]. Let \(\sigma_{rr}\), \(\sigma_{\theta\theta}\), \(\sigma_{\varphi\varphi}\), \(\sigma_{r\theta}\), \(\sigma_{rp}\), and \(\sigma_{\theta\varphi}\) be the stress components referred to the spherical coordinate system. It is assumed that  
\[\sigma_{rr}=\sigma_{\theta\theta}=\sigma_{\varphi\varphi}=\sigma,\ \sigma_{r\theta}=\sigma_{rp}=0. \tag{1}\]  
Figure 1: Cone-and-plate rheometer – notation. Figure 2: Illustration of the boundary value problem. Thus the associated flow rule is satisfied if (12) is satisfied. Substituting (10) into (9), and taking into account (12), one can get  
\[\xi_{eq}=\xi_{\theta\varphi}. \tag{13}\]  
Turning to the yield criterion (7), this criterion, (1) and (3) combine to give  
\[\sigma_{\theta\varphi}=k_{0}\Phi\bigg{(}\frac{\xi_{eq}}{\xi_{0}}\bigg{)}. \tag{14}\]  
Using (13) this equation can be rewritten as  
\[\sigma_{\theta\varphi}=k_{0}\Phi\bigg{(}\frac{\xi_{\theta\varphi}}{\xi_{0}} \bigg{)} \tag{15}\]  
giving the relationship between the shear stress and shear strain rate in the spherical coordinate system. The viscosity is usually defined as (in our nomenclature)  
\[\eta=\frac{\sigma_{\theta\varphi}}{\xi_{\theta\varphi}}. \tag{16}\]  
In the case of materials with vanishing viscosity, \(\eta\to 0\) as \(\xi_{\theta\varphi}\rightarrow\infty\). Then, it follows from (15) and (16) that  
\[\frac{\Phi\Big{(}\xi_{\theta\varphi}/\xi_{0}\Big{)}}{\xi_{\theta\varphi}/\xi_ {0}}\to 0 \tag{17}\]  
as \(\xi_{\theta\varphi}\rightarrow\infty\). Henceforward, attention is focused on the class of functions that are represented as  
\[\Phi\bigg{(}\frac{\xi_{\theta\varphi}}{\xi_{0}}\bigg{)}=\frac{k_{s}}{k_{0}}+A \bigg{(}\frac{\xi_{\theta\varphi}}{\xi_{0}}\bigg{)}^{\alpha}+o\bigg{[}\bigg{(} \frac{\xi_{\theta\varphi}}{\xi_{0}}\bigg{)}^{\alpha}\bigg{]} \tag{18}\]  
as \(\xi_{\theta\varphi}\rightarrow\infty\). Here \(k_{s}\), \(A\) and \(\alpha\) are constant. It is evident that this class is large enough for any function \(\Phi\Big{(}\xi_{\alpha\varphi}\Big{)}\) used in applications. It follows from the representation (18) that (17) is satisfied if  
\[\sigma_{rr}=\sigma_{\theta\theta}=\sigma_{\varphi\varphi}=\sigma,\ \sigma_{r\theta}=\sigma_{rp}=0. \tag{1}\]0  
Since \(\Phi\) is a monotonically increasing function of its argument, \(A>0\) if \(\alpha>0\) and \(A<0\) if \(\alpha<0\). The function \(\Phi\) is independent of the shear strain rate if \(\alpha=0\). The response of such materials is independent of the speed of loading. An example is the rigid perfectly plastic material [30]. This special case is not considered in the present paper. If \(\alpha>0\), then the term \(k_{s}/k_{0}\) in (18) is negligible as \(\xi_{\theta\varphi}\rightarrow\infty\). If \(\alpha<0\), then \(k_{s}\) is the maximum possible shear stress that may appear in the material. Equation (18) can be solved for \(\xi_{\theta\varphi}\) to give  
\[\frac{\xi_{\theta\varphi}}{\xi_{0}}=\left\{