Answer to Question #206577 in Mechanical Engineering for Md nazmul hassan

Question

Question No. 18: (a) In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. If the arm rotates at 150 r.p.m in the anticlockwise direction about the centre of the gear A which is fixed, determine the speed of gear B. If the gear A instead of being fixed, makes 300 r.p.m in the clockwise direction, what will be the speed of gear B?

(b) Prove that the resultant unbalanced force is minimum when half of the reciprocating masses are balanced by rotating masses.

Answer 1

Given: T_A = 36 ; T_B = 45 ; N_C = 150 r.p.m.

(anticlockwise)

The gear train is shown in Fig. 13.7.

By Tabular method

Speed of gear B when gear A is fixed

Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,

y = + 150 r.p.m.

Also the gear A is fixed, therefore

x + y = 0 or x = – y = – 150 r.p.m.

∴ Speed of gear B,

(y-x)(T_a/T_b) = 150 + 150 × (36/45) = 270 r.p.m. (anticlockwise)

Speed of gear B when gear A makes 300 r.p.m. clockwise

Since the gear A makes 300 r.p.m.clockwise, therefore from the fourth row of the table,

x + y = – 300 or x = – 300 – y = – 300 – 150 = – 450 r.p.m.

∴ Speed of gear B,

(y-x)(Ta/Tb) = 150 + 450 × (36/45)

= 510 r.p.m. (anticlockwise).