Simplified view of a Weissenberg rheogoniometer—M. Consider the Weissenberg rheogoniometer with a very shallow cone; thus, referring to Fig. E6.9, β = π/2 − α, where α is a small angle. (a) Without...

Simplified view of a Weissenberg rheogoniometer—M. Consider the Weissenberg rheogoniometer with a very shallow cone; thus, referring to Fig. E6.9, β = π/2 − α, where α is a small angle. (a) Without going through the complicated analysis presented in Example 6.9, outline your reasons for supposing that the shear stress at any location on the cone is: (τθφ)θ=β = ωμ α . (b) Hence, prove that the torque required to hold the cone stationary (or to rotate the lower plate) is: T = 2 3 πωμR4 H . (c) By substituting β = π/2 − α into Eqn. (E6.9.13) and expanding the various functions in power series (only a very few terms are needed), prove that g(β) = 1/(2α), and that Eqn. (E6.9.18) again leads to the expression just obtained for the torque in part (b) above.