The sleeve arms and the ball arms are $80 \ mm$ and $120 \ mm$ respectively. The levers are pivoted @ $120 \ mm$ from the governor axis and mass of each ball is $2.5 \ kg$. The law arms are parallel to the governor axis @ lowest equilibrium speed.

Determine:

1] Loads on the spring.

2] Lowest and highest equilibrium speeds.

3] Stiffness of spring.

Solution:

$N_1 \ = \ 290 \ r.p.m.$

$w_1 = \frac{2 \pi \times 290}{60} \ = \ 30.36 \ rad/sec$

$N_2 = 310 \ r.p.m.$

$H = 15 \ mm$

$Y = 80 \ mm$

$X = 120 \ mm$

$r = 120 \ mm$

$m = 2.5 \ kg$

$Fc_1 \ = \ mr_1 \ w_1^2 \ = \ 2.5 \times \ 0.120 \times \ (30.36)^2$

$Fc_1 \ = \ 276.51 \ N$

1st Position:

$\sum M_0 = 0 \ \curvearrowright$ +

($\frac{Mg+S_1}{2}) \ (Y) \ – \ Fc_1 \ (x) \ = \ 0$

{As sleeve mass M is not given, $\therefore M = 0$ }

$\frac{S_1}{2} \ (Y) \ – \ Fc_1 \ (x) \ = \ 0$

$\frac{S_1}{2} \ (0.08) \ – \ 276.57 (0.12) \ = \ 0$

$S_1 = 829.71 \ N$

$N_2 = 310\ rpm$

$w_2 = \frac{2 \pi \times 310}{60} = 32.46\ rad/sec$

$fc_2 = mr_2 \ w_2^2$

$= 2.5 \times 0.1425 \times 32.46^2$

$fc_2 = 375.36N$

for $‘r_2’$ :

$h = \frac{Y}{X} (r_2 – r_1)$

$0.015 = \frac{0.08}{0.12} \ (r_2 – 0.12)$

$r_2 \ = \ 0.1425 \ m$

$\sum M_o \ = \ 0 \curvearrowright$ + . . . . {Neglecting obliquity effect}

$(\frac{Mg + S_2}{2}) \ (y) \ – \ FC_2 \ (x) \ = \ 0$

$\frac{S_2}{2} \ (y) \ – \ FC_2 (x) \ = \ 0$

$\frac{S_2}{2} \ (0.08) \ – \ 375.36 \ (0.12) \ = \ 0$

$S_2 \ = \ 1126.08 \ N$

$\therefore$ $S = \frac{S_2 \ – \ S_1}{h} \ = \ \frac{1126.08 \ – \ 829.71}{0.015}$

$S = 0.01975 \ N/m$