Question

The arms of a Hartnell governor are of equal length. At mid-position of the sleeve, the ball arm is vertical and the radius at which the ball rotates is 8.25 cm when the equilibrium speed is 450 rpm. On changing the speed by 1%, the governor is able to overcome the friction at this position. The friction force is assumed to have a constant value of 30 N at the sleeve. The sleeve moves $\pm$ l .6 cm from the mean position. The minimum speed of the governor (including friction) is 428 rpm. The mass of the sleeve is 3.5 kg. The magnitude of the rotating masses,

The initial compression of the spring.

• A. 5.41 cm
• B. 4.63 cm
• C. 6.65 cm
• D. 8.25 cm

Answer 1

The correct option is B 4.63 cm


$\omega =\frac{2\pi \times 450}{60}=47.123rad/sec$

$f=30N$

At equilibrium friction force = 0

At r = 8.25 cm, f = 0

Considering mid-position only and assuming that the total load, on sleeve at this position, due to compression of spring is $F_s$

$m\times \frac{8.25}{100}\times {\omega }^2=\left(\frac{Mg+F_s}{2}\right)...\left(1\right)$

$m\times \frac{8.25}{100}\times (1.01\omega {)}^2=\left(\frac{Mg+F_s+f}{2}\right)...(2)$

Equation (2)- equation (1)

$m\times \frac{8.25}{100}\times \left[\right(1.01{\omega }^2)-{\omega }^2]=\frac{f}{2}=\frac{30}{2}=15$

$m\times \frac{8.25}{100}\times \left[\right(1.01\times 47.123{)}^2-{47.123}^2]=15$

$\Rightarrow m=4.0735kg$

(i) Magnitude of rotating masses,

m = 4.0735 kg

$F_s=Springforce$

$4.0735\times \frac{8.25}{100}\times {47.123}^2=\left(\frac{3.5\times 9.81+F_s}{2}\right)$

$F_s=1458.232N$

Considering extreme positions and the corresponding speeds, when allowing for friction,

For lowest speed

$m{r}_1{\omega }_1^2=\frac{MG+F_{s1}-f}{2}$

$r_1=r-1.6=8.25-1.6=6.65cm$

${\omega }_1=\frac{2\pi \times 428}{60}=44.82rad/sec$

$4.0735\times \frac{6.65}{100}\times {44.82}^2=\left(\frac{3.5\times 9.81+F_{s1}-30}{2}\right)$

$F_{s1}=1084N$

(ii) Spring stiffness,

$k=\frac{F_s-F_{s1}}{1.6}=\frac{1458.232-1084}{1.6}=233.895N/cm$

Initial compression ofspring

$=\frac{F_{s1}}{k}=\frac{1084}{233.895}=4.63cm$