Answer to Question #208127 in Mechanical Engineering for Sanajit

Question

An epicyclic gear consists of bevel wheels as shown in Fig. 13.49. The driving pinion A has 20 teeth

and meshes with the wheel B which has 25 teeth. The wheels B and C are fixed together and turn freely

on the shaft F. The shaft F can rotate freely about the main axis X X. The wheel C has 50 teeth and

meshes with wheels D and E, each of which has 60 teeth. Find the speed and direction of E when A

rotates at 200 r.p.m., if

1. D is fixed, and 2. D rotates at 100 r.p.m., in the same direction as A.

In both the cases, find the ratio of the torques transmitted by the shafts of the wheels A and E, the

friction being neglected

Answer 1

The speed of A

$x+y=200 \implies y=200-x$

Part a

D is fixed

$y+x \frac{T_A*T_C}{T_B*T_D}=0 \implies 200-x+x \frac{20*50}{25*60}=0$

$x=600 ; y= 200-600= -400$

Speed of E, $y-x \frac{T_A*T_C}{T_B*T_E}=0 \implies -400-600 \frac{20*50}{25*60}=-800 rpm$

E rotates at 800 rpm in the direction opposite to A

Part b

D rotates at 100 rpm in the same direction of A

$y+x \frac{T_A*T_C}{T_B*T_D}=-100 \implies -x+ \frac{2}{3}x=-100 rpm \implies x= 300$

$y=200-x=-100$

Speed of E

$y-x \frac{T_A*T_C}{T_B*T_E}=-100 -300* \frac{2}{3}= 300$

E rotates at 300 rpm in the direction to A

$T_1 \omega _A=T_0\omega _E \implies \frac{T_1}{T_0}=\frac{\omega _E}{\omega _A}$

Case A

$\frac{T_1}{T_0}=\frac{800}{200}=4$

The ratio of torque transmitted by the shaft of wheel A and E is 4

Case B

$\frac{T_1}{T_0}=\frac{300}{200}=1.5$

The ratio of torque transmitted by the shaft of wheel A and E is 1.5