--- title: "Hartnell governor spring force and stiffness" subject: "Theory of Mechanism and Machine II" chapter: 4 key: "semester_7/theory_of_mechanism_and_machine_ii:hartnell-governor-spring-stiffness" source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/hartnell-governor-spring-stiffness site: BCT Engineering --- # Hartnell governor spring force and stiffness Theory of Mechanism and Machine II · Chapter 4: Governors - Asked in: year not given, 8 marks - Video: [Numerical Problem on Wilson Hartnell Governor](https://www.youtube.com/watch?v=FjcqJpPCm2Y) — Neeraj Gautam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/hartnell-governor-spring-stiffness - Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-4 ## Question In a Hartnell governor, each ball has a mass of 2.5 kg. The ball arm of each bell crank lever is 120 mm and is vertical in the mean position; the sleeve arm is 100 mm and is horizontal. The radius of rotation of the balls is 100 mm at the lowest equilibrium speed of 300 rpm and 140 mm at the highest equilibrium speed of 330 rpm. Neglecting the weight of the balls, the levers and the sleeve, and the obliquity of the arms, find the spring force at the two extreme positions, the stiffness of the spring, and the initial compression. ## Short answer Centrifugal forces are 246.7 N at 300 rpm (r=100 mm) and 418.0 N at 330 rpm (r=140 mm). Spring forces S=2Fc(x/y): S₁=592 N, S₂=1003 N. Sleeve lift=(0.14-0.10)(100/120)=33.3 mm. Stiffness=(1003-592)/0.0333=12.33 N/mm. Initial compression=S₁/stiffness=48.0 mm. ## Answer **Data:** $m = 2.5$ kg, ball arm $x = 120$ mm, sleeve arm $y = 100$ mm, $r_1 = 0.1$ m, $r_2 = 0.14$ m, $N_1 = 300$ rpm, $N_2 = 330$ rpm. ```text ball (m) x = 120 O------+ fulcrum | y = 100 +--> sleeve --> spring force S ``` #### Centrifugal force on each ball $$ \omega_1 = \frac{2\pi \times 300}{60} = 31.42\ \text{rad/s}, \qquad \omega_2 = \frac{2\pi \times 330}{60} = 34.56\ \text{rad/s} $$ $$ F_{c1} = m\omega_1^2 r_1 = 2.5 \times (31.42)^2 \times 0.1 = 246.7\ \text{N} $$ $$ F_{c2} = m\omega_2^2 r_2 = 2.5 \times (34.56)^2 \times 0.14 = 418.0\ \text{N} $$ #### Spring force Taking moments about the fulcrum of the bell crank (per ball, the sleeve force is $S/2$ for each of the two levers): $$ F_c \times x = \frac{S}{2}\times y \Rightarrow S = \frac{2F_c\,x}{y} $$ $$ S_1 = \frac{2 \times 246.7 \times 120}{100} = 592\ \text{N}, \qquad S_2 = \frac{2 \times 418.0 \times 120}{100} = 1003\ \text{N} $$ #### Sleeve lift and stiffness The sleeve moves in proportion to the ball movement: $$ \text{Lift} = (r_2 - r_1)\frac{y}{x} = (0.14 - 0.10)\times\frac{100}{120} = 0.0333\ \text{m} = 33.3\ \text{mm} $$ $$ s = \frac{S_2 - S_1}{\text{lift}} = \frac{1003 - 592}{0.0333} = 12329\ \text{N/m} = 12.33\ \text{N/mm} $$ #### Initial compression $$ \delta_0 = \frac{S_1}{s} = \frac{592}{12329} = 0.0480\ \text{m} = 48.0\ \text{mm} $$ **Answer:** $S_1 = 592$ N; $S_2 = 1003$ N; spring stiffness $= 12.33$ N/mm; initial compression $= 48.0$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the centrifugal force on each ball found? A. m w^2 r at each extreme position B. m g r C. m w r D. m w^2 / r Answer: A. F_c1 = 2.5 x 31.42^2 x 0.1 = 246.7 N and F_c2 = 418.0 N. (Key point: Centrifugal force) 2. Taking moments about the bell crank fulcrum gives the spring force as: A. S = 2 F_c x / y B. S = F_c y / x C. S = 2 F_c y / x D. S = F_c x / 2y Answer: A. F_c x = (S/2) y so S = 2 F_c x / y: S1 = 592 N, S2 = 1003 N. (Key point: Spring force from moments) 3. The sleeve lift is related to the ball movement by: A. (r2 - r1) y / x B. (r2 - r1) x / y C. r2 - r1 D. (r2 + r1) y / x Answer: A. The sleeve moves in proportion to the lever arms: 0.04 x 100/120 = 33.3 mm. (Key point: Sleeve lift) 4. The spring stiffness and initial compression are found from: A. s = (S2 - S1)/lift and compression = S1/s B. s = S2/lift and compression = S2/s C. s = (S2 + S1)/lift and compression = lift D. s = S1 x lift Answer: A. s = 411/0.0333 = 12.33 N/mm, and initial compression is S1/s = 48 mm. 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