The arms of partner governor are each $250 \ mm$ long and pivoted on the governor axis. The mass of each ball is $5 \ kg$ and mass of central sleeve is $30 \ kg$. The radius of rotation of ball is $150 \ mm$ when sleeve begins to rise reaches a value of $800 \ mm$ for minimum speed. Determine range of speed of the governor.

If the friction of sleeve is $20 \ N$ load. Determine how the speed range is modified.

NOTE: Effects of friction.

1. When radius of rotation is minimum, then speed is minimum and sleeve moves downward therefore frictional force acts upward hence negative sign.

2. When radius of rotation is maximum, then speed is maximum and sleeve moves upward. Hence frictional force acts downward. Hence positive sign.

$h = \frac{g}{w^2} [ 1 + \frac{Mg \pm f}{2mg} (1 + q)]$

SOLUTION:

$M = 5 \ kg$

$M = 30 \ kg$

$r_1 = 150 \ mm$

$r_2 = 200 \ mm$

$\alpha = \beta$

$\therefore q = 1$

Case 1: Without friction:

A] When $r_1 = 150 \ mm$, then $N_1 = ?$

$\therefore$ $h_1 = \sqrt{ 250^2 – 150^2} = 0.2 \ m$

$h_1 = \frac{895}{N_1^2} [ 1 + \frac{M}{2m} (1 + q)]$

$0.2 = \frac{895}{N_1^2} [ 1 + \frac{30}{2 \times 5} (1 + 1)]$

$N_1 = 176.98 \ rpm$

B] When $r_2 = 200 \ mm$, then $N_2$ = ?

$h_2 = \sqrt{250^2 – 200^2} = 0.15 \ m$

$\therefore$ $h_2 = \frac{895}{N_2^2} ( 1 + \frac{M}{2m} (1 + q)]$

$0.15 = \frac{895}{N_2^2} [ 1 + \frac{30}{2 \times 5} (1 + 1)]$

$N_2 = 204.36$ rpm.

Range of speed = $N_2 – N_1 = 204.36 \ – \ 176.98$

$= 27.38 rpm$

Case 2: With friction:

A] When $r_1 = 150 \ mm$, then $N_1 = ?$

$h_1 = \frac{895}{N_1^2} [ 1 + \frac{Mg – f}{2 mg} (1 + q)]$

$0.2 = \frac{895}{N_1^2} [ 1+ \frac{30 \times 9.81 – 20 \times 2}{2 \times 5 \times 9.81}]$

$N_1 = 171.76 \ rpm.$

B] When $r_2 = 300 \ mm$, then $N_2$ = ?

$h_2 = \frac{895}{N_2^2} [ 1+ \frac{Mg + F}{2 mg} (1 + q)]$

$0.15 = \frac{895}{N_2^2} [ 1 + \frac{ 30 \times 9.81 + 20 \times 2}{2 \times 5 \times 9.81}]$

$N_2 = 210.236 \ rpm.$

Range $= 210.236 \ – \ 171.76$

$= 38.48 \ rpm.$