Question

In a spring loaded Hartnell type governor, the extreme radii of rotation of the balls are 80 mm and 120 mm. The ball arm and the sleeve arm of the bell crank lever are equal in length. The mass of each ball is 2 kg. If the speeds at the two extreme positions are 400 and 420 r.p.m., find: the change of forces that exert on sleeve

Answer 1

Detailed answer of above asked question is solved as below.

- Gyen r = 80mm V2 - 120mm N = 400 rpm N2 = 420 rpm m = 2 kg. Fo. ball ballarm armg governor axis V At extreme radii of rotation both arm are equally inclined, because of equal length of ball arm and sleeve arm, so that x 4x both at extreme position, are equal X = Y. sleeve am у → Angular speed W, = 2IN, 2 TX 400 = 41.88 ng radis W2 = 211 N2 2 TT X 420 43.9822 radis. - centrifugal force at 'w' Due to ball mass centrifugal force is exerted during rotation Fei= m (w.)?y, = 2X (41.8849 2 x 0.08 Fc, = 280.4353 N Fen= mc w 2) 2 2 = 2x(43.9822) 2X0.12 Fcp = 4 64.264 ) N > let Spring force exerted on sleeve at 'wl. change of the force exerted on sleeve is Sq-Si = 2 ( Fc2 - FC, ) + 1 2 (464-9641 - 280.47353) ::'(X=Y) Sosi 367.0576 N - Change of force exerted on sleeve is 367.0576 N. (Ans)