written 4.8 years ago by |
The mass of each ball is $2 \ kg$ and friction of the sleeve together with the resistance of the operating gear is equal to a load of $25 \ N$ and sleeve. If the limitations inclinations of the upper arms to the vertical are 30° and 40° Find, taking friction into account, range of speed of the governor.
$m = 2kg$
$M = 15 \ kg$
$f = 25 \ N$
A] $\alpha = 30°$
$cos \ 30° = \frac{h_1}{0.2}$
$h_1 = 0.173$
$sin \beta = \frac{r_1}{0.25}$
$= \frac{0.1}{0.25}$
$\beta = 23.58°$
$sin 30° = \frac{r_1}{0.2}$
$r_1 = 0.1m$
$q = \frac{tan \ 23.58}{tan \ 30} = 0.7559$
$0.173 = \frac{895}{N_1^2} [ 1 + \frac{15 \times 9.81 - 25}{2 \times 9.81 \times 2} \times (1 + 0.7531)]$
$N_1 = 182.77 \ rpm$
$\alpha =$ 40°
$cos \ 40° = \frac{h_2}{0.2}$
$h_2 = 0.1532$
$sin \beta = \frac{r_2}{0.250} = \frac{0.1286}{0.250}$
$\beta = 30.946°$
$sin \ 40° = \frac{r_2}{0.2}$
$r_2 = 0.1286$
$q = \frac{tan \ 30.946}{tan \ 40} = 0.7101$
$h_2 = \frac{895}{N_2^2} [ 1+ \frac{Mg + F}{2 mg} (1 + q)]$
$0.1532 = \frac{895}{N_2^2} [ 1 + \frac{15 \times 9.81 + 25}{2 \times 2 \times 9.81} (1 + 0.7101)]$
$N_2 = 222.87 \ rpm$
Range = $N_2 – N_1 = 222.87 – 182.77$
$= 40.1 r.p.m$.