To damp oscillations, the pointer of a galvanometer is fixed to a circular disk which turns in a container of oil (see Fig.). What is the damping torque for ꞷ = 0.3 rad/s if the oil has a viscosity of 8 × 10 –3 Pa.s? Neglect edge effects.
Text Solution
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Sol. Assume at any point that the velocity profile of the oil is linear dv/dn = rꞷ/ (0.5/1000) = (r) (0.3)/ (0.5/1000) = 600r; τ = µ(dv/dn) = τ (600r) = (8 × 10 –3 ) (600r) = 4.80r.
The force dF f on dA on the upper face of the disc is then dF f = τ dA = (4.80r) (r dθ dr) = 4.80r 2 dθdr. The torque dT for dA on the upper face is then dT = r dF f = r(4.80r 2 dθ dr) = 4.80r 3 dθdr. The total resisting torque on both faces is
T = 2 
= (9.60) (2 π ) 
= 2.98 × 10 –5 N.m
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