## 3 Analysis of Torque
### Torque Equation for Curved Disc-Type MR Drives  
To establish the torque equation for curved disc-type MR drives, the MR torque within a single semi-circular gap was analyzed in cylindrical coordinates (\(\rho\), \(\theta\), \(z\)). The equivalent model is shown in Figure 11.  
Taking an infinitesimal arc ds in the semi-circular gap, the shear yield stress \(\mathrm{d}F\) generated by the infinitesimal arc is  
\[\mathrm{d}F=2\pi y\tau\mathrm{d}s=2\pi y\tau r_{\mathrm{a}}\mathrm{d}\theta \tag{5}\]  
where \(y\) is the distance between the infinitesimal arc and the center of rotation, O is the center of the semi-circular arc, \(L_{n}\) is the distance between the center of the n-th arc and the center of rotation, \(\theta\) is the angle of the corresponding infinitesimal arc, and \(r_{\mathrm{a}}\) is the radius of the semi-circular arc.  
Then, the distance between the infinitesimal arc and the center of rotation \(y\) is  
\[y=L_{n}-r_{\mathrm{a}}\cos\theta \tag{6}\]  
The Bingham model is used to describe the rheological properties of the MRF, and its constitutive relationship is  
\[\begin{cases}\tau=\tau_{y}(T,B)\mathrm{sgn}(\dot{\gamma})+\eta\dot{\gamma}& \left|\tau\right|\geq\tau_{y}\\ \dot{\gamma}=0&\left|\tau\right|<\tau_{y}\end{cases} \tag{7}\]  
where \(\eta\) is the zero-field viscosity and \(\dot{\gamma}\) is the shear strain rate of the MRF.  
Bringing Equations (6) and (7) into Equation (5), we can obtain the MR torque generated by the MRF infinitesimal arc as  
\[\mathrm{d}M=2\pi r_{\mathrm{a}}(L_{n}-r_{\mathrm{a}}\cos\theta)^{2}[\tau_{y}(T,B)+\eta\dot{\gamma}]\mathrm{d}\theta=2\pi r_{\mathrm{a}}(L_{n}-r_{\mathrm{a} }\cos\theta)^{2}\bigg{[}\tau_{y}(T,B)+\eta\frac{\mathrm{d}v_{\theta}}{\mathrm{ d}z}\bigg{]}\mathrm{d}\theta \tag{8}\]  
where \(v_{\theta}\) is the shear linear velocity of the MRF infinitesimal arc in the direction of rotation, i.e., \(v_{\theta}=\omega y\).  
Bringing in Equation (8) yields  
\[\begin{split}\mathrm{d}M=2\pi r_{\mathrm{a}}(L_{n}-r_{\mathrm{a} }\cos\theta)^{2}[\tau_{y}(T,B)+\eta y\frac{\mathrm{d}\omega}{\mathrm{d}z}] \mathrm{d}\theta\\ =2\pi r_{\mathrm{a}}(L_{n}-r_{\mathrm{a}}\cos\theta)^{2}[\tau_{y}(T,B)+\eta y\frac{(\omega_{1}-\omega_{2})}{\hbar}]\mathrm{d}\theta\end{split} \tag{9}\]  
Figure 11: Force model of the MRF infinitesimal arc.  
where \(\omega_{1}\) and \(\omega_{2}\) are the input and output shaft angular velocities, respectively. Bringing Equation (6) into Equation (9) yields  
\[\mathrm{d}M=2\pi r_{\mathrm{a}}(L_{n}-r_{\mathrm{a}}\cos\theta)^{2}\bigg{[}\tau_ {y}(T,B)+\eta(\omega_{1}-\omega_{2})\frac{(L_{n}-r_{\mathrm{a}}\cos\theta)}{h} \bigg{]}\mathrm{d}\theta \tag{10}\]  
By integrating Equation (10) over the entire arc interval, the total MR torque transmitted by double-disc curved MR drives can be obtained as  
\[M_{\mathrm{m1}}=\sum_{n=1}^{n_{\mathrm{m}}}2\pi^{2}r_{\mathrm{a}}\tau_{y}(T,B )\Big{(}2L_{n}^{2}+r_{\mathrm{a}}^{2}\Big{)}+\frac{2\pi r_{\mathrm{a}}\eta L_{ n}}{h}(\omega_{1}-\omega_{2})\Big{(}2\pi L_{n}^{2}+3\pi r_{\mathrm{a}}^{2} \Big{)} \tag{11}\]  
where \(n\) is the current number of arcs and \(n_{\mathrm{m}}\) is the total number of arcs.  
Similarly, the total MR torque of the double-disc planar MR drive is [6]  
\[M_{\mathrm{m2}}=2\int_{R_{4}}^{R_{3}}2\pi r^{2}rdr=\frac{4\pi}{3}\Big{(}R_{3} ^{3}-R_{4}^{3}\Big{)}\tau_{y}(T,B)+\frac{\pi}{h}\Big{(}R_{3}^{4}-R_{4}^{4} \Big{)}(\omega_{1}-\omega_{2})\eta \tag{12}\]