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Hartnell governor spring force and stiffness

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.

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.

Data: m=2.5m = 2.5 kg, ball arm x=120x = 120 mm, sleeve arm y=100y = 100 mm, r1=0.1r_1 = 0.1 m, r2=0.14r_2 = 0.14 m, N1=300N_1 = 300 rpm, N2=330N_2 = 330 rpm.

        ball (m)        x = 120
          O------+  fulcrum
                 |  y = 100
                 +--> sleeve --> spring force S

Centrifugal force on each ball

ω1=2π×30060=31.42 rad/s,ω2=2π×33060=34.56 rad/s\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} Fc1=mω12r1=2.5×(31.42)2×0.1=246.7 NF_{c1} = m\omega_1^2 r_1 = 2.5 \times (31.42)^2 \times 0.1 = 246.7\ \text{N} Fc2=mω22r2=2.5×(34.56)2×0.14=418.0 NF_{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/2S/2 for each of the two levers):

Fc×x=S2×y⇒S=2Fc xyF_c \times x = \frac{S}{2}\times y \Rightarrow S = \frac{2F_c\,x}{y} S1=2×246.7×120100=592 N,S2=2×418.0×120100=1003 NS_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:

Lift=(r2−r1)yx=(0.14−0.10)×100120=0.0333 m=33.3 mm\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=S2−S1lift=1003−5920.0333=12329 N/m=12.33 N/mms = \frac{S_2 - S_1}{\text{lift}} = \frac{1003 - 592}{0.0333} = 12329\ \text{N/m} = 12.33\ \text{N/mm}

Initial compression

δ0=S1s=59212329=0.0480 m=48.0 mm\delta_0 = \frac{S_1}{s} = \frac{592}{12329} = 0.0480\ \text{m} = 48.0\ \text{mm}

Answer: S1=592S_1 = 592 N; S2=1003S_2 = 1003 N; spring stiffness =12.33= 12.33 N/mm; initial compression =48.0= 48.0 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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