An epicyclic gear train is shown schematically in the adjacent figure. The sun gear 2 on the input shaft is a 20 teeth external gear. The planet gear 3 is a 40 teeth external gear. The ring gear 5 is a 100 teeth internal gear. The ring gear 5 is fixed and the gear 2 is rotating at 60 rpm ccw (ccw = counterclockwise and cw = clockwise)

The arm 4 attached to the output shaft will rotate at

10 rpm ccw

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10 rpm cw

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12 rpm cw

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12 rpm ccw

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Solution

The correct option is A 10 rpm ccw
Method I:

Arm- 4 Arm- 2 Arm- 3 Arm- 5

0 +x $- \frac{T_{2}}{T_{3}} x$ $- \frac{T_{2}}{T_{3}} x \times \frac{T_{3}}{T_{5}}$

+y +y +y +y

y x+y $y - \frac{T_{2}}{T_{3}} x$ $y - \frac{T_{2}}{T_{5}} x$

Gear 5 is fixed
$y - \frac{T_{2}}{T_{5}} x = 0$

$y = \frac{20}{100} x$

$y = 0.2 x$

$x + y = - 60$

[Gear 2 rotates counterclockwise]

$x + 2 x = - 60$

$1.2 x = - 60$

$x = - \frac{60}{1.2} = - 50 rpm$

$y = - 10 rpm$

Method II:

Relative velocity method:
Given: $T_{2} = 20, T_{3} = 40, T_{5} = 100, N_{5} = 0$

$N_{2} = 60 rpm (C C W) = - 60 rpm$

$\frac{N_{2} - N_{4}}{N_{5} - N_{4}} = \frac{- T_{5}}{T_{3}} \times \frac{T_{3}}{T_{2}} = \frac{- T_{5}}{T_{2}}$

$\frac{- 60 - N_{4}}{0 - N_{4}} = \frac{- 100}{20} = - 5$

$- 60 - N_{4} = + 5 N_{4}$

$6 N_{4} = - 60$

$N_{4} = - 10 rpm = 10 rpm (ccw)$

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