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All the arms of a porter governor are $178mm$long and are hinged @ a distance of $38mm$ from the axis of rotation. The mass of each ball is $1.15 kg$ and mass of each sleeve is $20 kg.$
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The governor sleeve begins to rise @ $280 \ mm$ when the links are @ angle of $30°$ to the vertical. Assuming the friction force to be constant, Determine the minimum and maximum speed of rotation when the inclination of the arms of the vertical is $45°$. $m = 1.15 \ kg$

$M = 20 \ kg$

$N = 280 \ rpm$ @ $30°$

F = constant

$N_{min} = \ ? \ @ \ 45^0$

$N_{max} = \ ? \ @ \ 45^0$

$sin \ 30° = \frac{a}{178}$

$a = 89 \ mm$

$\therefore r = a + 38$

$= 89 + 38$

$r = 127 \ mm$

$tan \ 30° = \frac{r}{h} = \frac{127}{h}$

$h = 219.97 \ mm$

$0.21997 = \frac{895}{280^2} [ 1 + \frac{20 \times 9.81 \pm f}{2 \times 1.15 \times 9.81} (1+1)]$

$\pm F = 9.9 \ N$

$sin \ 45° = \frac{a}{178}$

$a = 125.865 \ mm$

$r = a + 38$

$= 125.865 + 38$

$r = 163.865 \ mm$

$tan \ 45° = \frac{r}{h^*} = \frac{163.865}{h^* = h^{**}}$

$h^* = h^{**} =163.86$

$h^* = \frac{895}{N^2_{min}} [1 + \frac{Mg – f}{2mg} (1 + q)]$

$0.1638 = \frac{895}{N_{min}^2} [ 1+ \frac{ 20 \times 9.81 – 9.9}{2 \times 1.15 \times 9.81} (1+1)]$

$N_{min} = 309.28 \ rpm$

$h^{**} = \frac{895}{N_{max}^2} [1 + \frac{Mg + f}{2mg} (1+q)]$

$0.1638 = \frac{895}{N^2_{max}} [ 1 + \frac{20 \times 9.81 + 9.9}{2 \times 1.15 \times 9.81} (1+1)]$

$N_{max} = 324.91 \ rpm$

Range $= 324.41 \ – \ 309.28 = 15.13 \ r.p.m.$