Question

In a spring loaded governor of the Hartnell type, the mass of each ball is 1 kg, length of vertical arm of the ball crank lever is 100 mm and that of the horizontal arm is 50 mm. the distance of fulcrum of each ball crank lever is 80 mm from the axis of rotation of the governor. The extreme radii of rotation of the ***** are 75 mm and 112.5 mm. the maximum equilibrium speed is 5% greater than the minimum equilibrium speed which is 360 rpm. Find, neglecting obliquity of arms, initial compression of the spring and equilibrium speed corresponding to the radius of rotation of 100 mm.

Answer 1

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

• = s2 - s1/h = 705.6 - 426.4/0.01875

• = 14890 N/m

• = 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,