Torque to rotate truncated cone

Reading of the figure: the cone has a full apex angle of $60^\circ$ (half-angle $\alpha=30^\circ$). The wetted frustum extends from 40 mm to 100 mm measured along the axis from the (imaginary) apex: 40 mm from the apex to the small end, and a frustum height of 60 mm. The gap measured normal to the surface is $t=0.2$ mm.

Data: $\mu=2$ poise $=0.2$ Pa·s, $N=100$ rpm, so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s.

    \____________/   large end, z2 = 100 mm from apex
     \          /
      \        /     frustum height z2 - z1 = 60 mm
       \______/      small end, z1 = 40 mm from apex
        \    /
         \  /
          \/  apex (imaginary), 60 deg
    oil gap t = 0.2 mm between cone and casing

Take an element at axial distance $z$ from the apex:

• radius $r=z\tan\alpha$
• slant length $ds=\dfrac{dz}{\cos\alpha}$
• shear stress $\tau=\mu\dfrac{\omega r}{t}$

$$dT=\tau\,(2\pi r\,ds)\,r=\frac{2\pi\mu\omega}{t}\,\frac{r^{3}}{\cos\alpha}\,dz$$ $$T=\frac{2\pi\mu\omega\tan^{3}\alpha}{t\cos\alpha}\int_{z_1}^{z_2}z^{3}dz=\frac{\pi\mu\omega\tan^{3}\alpha}{2\,t\cos\alpha}\left(z_2^{4}-z_1^{4}\right)$$

Substituting ($\tan30^\circ=0.5774$, $\cos30^\circ=0.8660$, $z_1=0.04$ m, $z_2=0.10$ m):

$$T=\frac{\pi(0.2)(10.472)(0.5774)^{3}}{2(0.0002)(0.8660)}\left(0.10^{4}-0.04^{4}\right)=0.356\ \text{N·m}$$

Answer: $T\approx0.356\ \text{N·m}$. (If the figure dimensions differ, use the same formula with the correct $z_1$ and $z_2$.)