A conical body turns in a container, as shown in Fig., at constant speed 11 rad/s. A uniform 0.01-in film of oil with viscosity 3.125 × 10 –7 lb.s/in 2 separates the cone from the container. What torque is required to maintain this motion, if the cone has a 2-in radius at its base and is 4 in tall?
Text Solution
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Sol. Consider the conical surface first (r/2 = z/4, r = z/2). The stress on this element is τ = µ(dv/dx) = µ(rꞷ/0.01) = (3.125 × 10 –7 ) [(z/2) (11)/0.01] = 1.719 × 10 –4 z. The area of the strip shown is dA = 2 π r ds = (2 π z/2) [dz/(4/ 
)] = 3.512z dz. The torque on the strip is dT = τ (dA)(r) = (1.719 × 10 –4 z) (3.512z dz) (z/2) = 3.019 × 10 –4 z 3 dz.
T 1 = 
dz = 3.019 × 10 –4 
= 0.01932 in.lb
Next consider the base:
dF f = τ dA, τ = µ(rꞷ/0.01) = (3.125 × 10 –7 ) [(r) (11)/0.01] = 3.438 × 10 –4 r, dF f = (3.438 × 10 –4 r) (r dθ dr) = 3.438 × 10 –4 r 2 dθ dr, dT 2 = (3.438 × 10 –4 r 2 dθ dr) (r) = 3.438 × 10 –4 r 3 dθ dr.
T 2 = 
dθ dr = (3.438 × 10 –4 ) (2 π ) 
= 0.00864 in.lb
T tot = 0.01932 + 0.00864 = 0.280 in.lb
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