linear relationship between ball path radius and controlling force

Question

In a spring controlled governor, mass of each governor ball is 7 kg and moves radially under the action of a controlling force F. If the speed range is 420 RPM to 440 RPM with range of ball path radius r equal to 12.4 cm to 13.2 cm, determine the linear relationship between ball path radius and controlling force. Sketch the graph between F and r. What is the equilibrium speed at r = 12.8 cm ?

Answer 1

Given that

m = 7 kg

at N₁ = 420 rpm, r₁ = 12.4cm

let the relation be     F = ar + b            ---  (1)

F₁ = mr₁$\omega$₁² = 7 x 12.4 x $(\frac{2\times\pi\times 420}{60})^2\times10^{-2}$ 

     =1679.096 N

At N₂=440 rpm, r₂= 13,2 cm

F2= 7 x 13.2 x  $(\frac{2\times\pi\times 440}{60})^2\times10^{-2}$ 

  =1961.7089 N

∴ By substituting these values in eq (1)

      1679.096 = a x 12.4 x 10^–2 + b --- (2)

      1961.7089 = a x 13.2◊10^–2 +b --- (3)

By solving eq(2) and eq(3)

⇒ a = 35326.6125 N/m, b = –2701.4039N

                   F = 35326.613r – 2701.404 N

what is F at r = 12.8 cm ?

F = 35326.613 x 12.8 x 10^-2 - 2701.404

= 1820.4025 N

Finding $\omega$ corresponding to force F

But F=mr$\omega$²

⇒ 7 x 12.8 x 10^–2 x$(\frac{2\times\pi\times N}{60})^2$ =1820.4025 

⇒ N = 430.43 rpm