Given: m = 1 kg, x = 100 mm = 0.1 m, y = 50 mm = 0.05 m, r = 80 mm = 0.08 m, r1 = 75 mm = 0.075 m, r2 = 112.5 mm = 0.1125 m, N1 =360 rpm, ω1 = 2πx 360/60 = 37.7 rad/sec.
Since the maximum equilibrium speed is 5% greater than minimum equilibrium speed,
ω2 = 1.05 x 37.7 = 39.6 rad/sec
w.k.t Centrifugal force at the minimum equilibrium speed
Fcl = m (ω)2 r1 =1(37.7)2 x 0.075
Fcl = 106.6 N
and centrifugal force at the maximum equilibrium speed
Fcl = m (ω)2r2 = 1(39.6)2 x 0.1125
Fc2 = 176.4 N
(i) Initial compression of the spring:
Let, s1 = spring force corresponding to ω1
s2 = spring force corresponding to ω2
Since the obliquity of arms is neglected, therefore for minimum equilibrium position,
For maximum equilibrium position,
w.k.t. Lift of the sleeve,
and stiffness of the spring
s = s2 - s1/h = 705.6 - 426.4/0.01875
s = 14890 N/m
s = 14.89 N/mm
(i) Initial compression of the spring
= s1/s2 = 426.4/14.89
= 28.6 mm
(ii) Equilibrium speed corresponding to radius of rotation r = 100 mm
Let, N = Equilibrium speed in rpm
Since the obliquity of the arms is neglected. Therefore centrifugal force at any instant,