The distance of fulcrum at each bell crank lever is $80\ mm$ from the axis of rotation of governor. The extreme radii of rotation at balls are $75\ mm$ and $1125\ mm$. The maximum equilibrium speed is 5% greater than the mini equilibrium speed which is $360\ rpm$, Find, neglecting the obliquity of arms, initial compression of spring and equilibrium speed corresponding to the radius of rotation of $100\ mm$.

$X = 100 \ mm$

$m = 1\ kg$

$Y = 50 \ mm$

$r = 80 \ mm$

$r_1 = 75 \ mm$

$r_2 = 112.5 \ mm$

$N_2 = (1 + 5 \text{%}) N_1 \ = \ 1.05 N_1 \ = \ 1.05 \times 360 \ = \ 378 \ r.p.m.$

$N_1 = 360 \ rpm$

$N_2 = 378 \ rpm$

$w_1 = 37.69 \ rpm$

$w_2 = 39.58 \ rpm$

$FC_1 \ = \ mr_1 \ w_1^2$

$= 1 \times 0.075 \times 37.09^2$

$FC_1 = 106.54 \times 10^3\ N$

$FC_2 = mr_2 \ w_2^2$

$= 1 \times 0.1125 \times 39.58^2$

$Fc_2 = 176.23 \times 10^3 N$

For 1st position:

$\sum M_o = 0 \curvearrowright +$

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

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

$\frac{S_1}{2} (0.050) – 106.54 \times 10^3 (0.1) = 0$

$S_1 = 426 \ N$

For 2nd position:

$\sum M_o = 0 \curvearrowright$ +

$(\frac{Mg + S_2}{2}) (Y) \ – \ Fc_2 (X) = 0$

$\frac{S_2}{2} (0.05) \ – \ 176.23 \times 10^3 (0.1) = 0$

$S_2 = 705 \ N$

Centrifugal force @ any radius of rotation = $Fc_1 + (Fc_2 – Fc_1)[ \frac{r – r_1}{r_2 – r_1}]$ (NOTE)

$h = \frac{Y}{X} (r_2 – r_1)$

$= \frac{0.05}{0.1} (0.1125 – 0.0 75)$

$H = 0.01875 \ m$

$S = \frac{S_2 – S_2}{h}$

$S = 14.89 \ N/mm$

$\delta = \frac{S_1}{S} = 28.6 mm$

Centrifugal force @ any radius of rotation = $Fc_1 + (Fc_2 – Fc_1) [ \frac{r – r_1}{r_2 – r_1}]$

$= 106.54 \times 10^3 + (176.23 – 106.54) \times 10^3 \times [ \frac{0.1 – 0.075}{0.1125 – 0.075}]$

$Fc \ @ \ 0.1 \ m = 141.385 \ N$

$Fc = mw^2r$

$153 = 1 \times w^2 \times 0.1$

$w = 39.11 \ rad/sec$

$N = \frac{39.11 \times 60}{2 \pi } = 373 \ r.p.m.$