In a compound gear train, gear F rotates in clockwise direction with 200 rpm. Find the rpm and direction of gear A. T_A = 20, T_B = 80, T_C = 25, T_D = 50, T_E = 20, T_F = 50

Concept:

\({\rm{speed\;ratio}} = \frac{{{\rm{speed\;of\;driver\;gear}}}}{{{\rm{speed\;of\;driven\;gear}}}} = \frac{{{\rm{teeth\;of\;driven\;gear}}}}{{{\rm{teeth\;of\;driver\;gear}}}}\)

Gears mounted on the same shaft will have the same speed and direction.

The direction of the two mating gears will always be opposite.

Calculation:

Given:

Teeth on gears, TA = 20, TB = 80, TC = 25, TD = 50, TE = 20, TF = 50.

Speed of gear F, NF = 200 rpm (clockwise)

\(\frac{{{{\rm{N}}_{\rm{F}}}}}{{{{\rm{N}}_{\rm{E}}}}} = \frac{{{{\rm{T}}_{\rm{E}}}}}{{{{\rm{T}}_{\rm{F}}}}}\)

\({{\rm{N}}_{\rm{E}}} = \frac{{50}}{{20}} \times 200\)

N_E = 500 rpm (anti-clockwise)

∵  gear E and D are mounted on same shaft

∴ N_D = N_E = 500 rpm  (anti-clockwise)

Now, \(\frac{{{{\rm{N}}_{\rm{D}}}}}{{{{\rm{N}}_{\rm{C}}}}} = \frac{{{{\rm{T}}_{\rm{C}}}}}{{{{\rm{T}}_{\rm{D}}}}}\)

\({{\rm{N}}_{\rm{c}}} = \frac{{50}}{{25}} \times 500\)

N_C = 1000 rpm (clockwise)

∵  gear B and C are mounted on same shaft

∴ N_B= N_C = 1000 rpm (clockwise)

Now, \(\frac{{{{\rm{N}}_{\rm{B}}}}}{{{{\rm{N}}_{\rm{A}}}}} = \frac{{{{\rm{T}}_{\rm{A}}}}}{{{{\rm{T}}_{\rm{B}}}}}\)

\({{\rm{N}}_{\rm{A}}} = \frac{{80}}{{20}} \times 1000\)

N_A = 4000 rpm (anti-clockwise)