Question

A spring loaded governor of the Hartnell type has arms of equal length. The masses rotate in a circle of 130 mm diameter when the sleeve is in the mod position and the ball arms are vertical. The equilibrium speed for this position is 450 r.p.m., neglecting friction. The maximum sleeve movement is to be 25 mm and the maximum variation of speed taking in account the friction to be 5 percent of the mid position speed. The mass of the sleeve is 4 kg and the friction may be considered equivalent to 30 N at the sleeve. The power of the governor must be sufficient to overcome the friction by one percent change of speed either way at mid position. Determine, negelecting obliquity effect of arms;

1. The value of each rotating mass.

2. The spring stiffness in N/mm, and

3. The initial compression of spring.

Answer 1

Given x = y ; d = 130 mm or r = 65 mm = 0.065 m; N = 450 r.p.m. or  ω = 2π x 450/60 = 47.23 rad/s ; h = 25 mm = 0.025 m ; M = 4 kg ; F = 30 N

1. Value of each roaring mass

Let  m = Value of each rotating mass in kg, and

S = Spring force on the sleeve at mid position in newtons.

Since the change of the speed at mid position to overcome friction is 1 per cent either way(i.e. ± 1%), therfore 

Minimum speed at mid position, 

ω = ω - 0.01ω = 0.99ω = 0.99 x 47.13 = 46.66 rad/s

and Maximum speed at mid position,

ω₂ = ω + 0.01ω =  1.01ω = 1.01 x 47.13 = 47.6 rad/s

Centrifugal force at the minimum speed,

F_C1  = m(ω₁)² r = m(46.66)² 0.065 = 141.5 mN

and centrifugal force at the maximum speed,

F_C2  = m(ω₂)² r = m(47.6)² 0.0065 = 147.3 mN

We know that for minimum speed at mid- position,

and for maximum speed at mid- position,

From equation (i) and (ii),

m = 5.2 kg

2. Spring stiffness in N/mm

Let s = Spring stiffness in N/mm.

Since the maximum variation of speed, considering friction is ± 5% of the mid-position speed, therefore,

Minimum speed considering friction,

ω₁' = ω - 0.05ω = 0.95 ω = 0.95  x 47.13 = 44.8 rad/s

and maximum speed considering friction,

ω₂' = ω +  0.05ω = 1.05 ω = .105 x  47.13 = 49.5 rad/s

we know that minimum radius of rotation considering friction,

and maximum radius of rotation considering friction,

Centrifugal force at the minimum speed considering friction,

F_c1' = (ω₁')²r₁ = 5.2(44.8)² 0.0525 = 548 N

and centrifugal force at the maximum speed considering friction,

F_c2' = (ω₁')²r₂ = 5.2(49.5)² 0.0775 = 987 N

S₁ = Spring force at minimum speed considering friction, and 

S₂ = Spring force at maximum speed considering friction.

We know that for minimum speed considering friction,

and for maximum speed considering friction,

We know stiffness of the spring,

3. Initial compression of the spring

We know that initial compression of the spring