Question

The epicyclic gear train consists of the sun gear A which is in mesh with the planet gear B that rotating at 39 rad/s. This gear has an inner hub C which is fixed to B and in mesh with the fixed ring gear R. If the connecting link DE pinned to B and C is rotating about the pin at E, determine the angular velocity of the sun gear (wa).

Answer 1

Solution - The detailed solution is given below, please find the attached solution -

page no 1

Solution ? Anuming a point P to be the contact point between the fixed re ring hub c, Hene, Up=0 gear R and the Vo = 25.2 m/s The diagram is shown as ^ Applying Vp=0 olm qy. the Relative velocity equation - Jo = Ep + WB x Tole -ODIP 1P' Oppp x WB : Up=0 0.5m The relocity of Dil D is defined as- VD - WDEX TOE 36 * * (0.74) UD = 252 VD - 25.2 m/s (1) as- The position rectar from P to D is defined 8b/p= - al put in equation (1), weget- 25.23 0 + (-WB) FX (-0-1) o

page no 2

25.2 ] = 0.1 WB į equating j component- 25.2 = 0.1 WB W8 = 252 rad I see which is known as Assuming angular relocity of the planet gear B. another point p' to be contact point between gear A and B, applying the relative velocity equation- Up = JP + WB x 8P/P (2) The relocity of P is defined as- Jp =0 The position rector from P to p' is defined as- (0.1+0.4) Spive - 8p/p = - 0.5 i Hence from equation (2) Up' = 0+ (-252 ) Ã X (-0.5) ūp'= 126 m/sec. (7). 2)

page no 3

The angular relocity of the sun gear A can then be calculated, WA = Up! YA WA = 196 0-3 WA = 420 Radis Hence, the angular relocity of the sun gear A is 420 Rolls (3)

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