A centrifugal clutch is to transmit 25 kW at 1200 r.p.m. There are four shoes. The speed at which the engagement begins is 0.5 times the running speed. The inside radius of the pulley rim is 100 mm and the centre of gravity of the shoe lies at 75 mm from the centre of the spider. Coefficient of friction may be taken as 0.3. Determine mass of the shoes.

Correct Answer: 1.866 kg
Given : P = 25 kW = 25 × 103 W ; N = 1200 r.p.m. or ω = 2 π × 1200/60 = 125.66 rad/s ; n = 4 ; R = 100 mm = 0.1 m ; r = 75 mm = 0.075 m ; µ = 0.3 ; ω1 = 0.5 ω. ω1 = 0.5 ω = 0.5 x 125.66 = 62.83 rad/s Power transmitted = Tω = T x 125.66 T = 198.95 N-m Pc = mω2r = m x 125.662 x 0.075 = 1184.28 m N Ps = mω12r = m x 62.832 x 0.075 = 296.07 m N Frictional force acting tangentially on each shoe, F = µ(Pc–Ps) = 0.3 (1184.28 m – 296.07 m) = 266.46m N Torque transmitted (T), 198.95 = n.F.R = 4 × 266.46 m × 0.1 = 106.584 m m = 1.866 kg.