Question

The arms of a Hartnell governor are of equal length. At midposition 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 $\pm$ 1.6 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,

• A. 4.075 kg
• B. 2.6725 kg
• C. 2.175 N
• D. 3.425 N

Answer 1

The correct option is A 4.075 kg


Mass of sleeve, M = 3.5 kg

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

$f=30N$

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)

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

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

$\Rightarrow m=4.0735kg$