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Question

For the epicyclic gear arrangement shown in the figure, ω2=100 rad/s clockwise (cw) and ωarm=80 rad/s counter clockwise (ccw). The angular velocity ω5 (in rad/s) is

A
0
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B
70 cw
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C
140 ccw
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D
140 cw
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Solution

The correct option is C 140 ccw
Method - I:

Compound gear
Gear 2 3 4 5
Teeth 20 24 32 80
Speed +N 20N24 20N24 20N24×3280
+N 5N6 5N6 5N6×25
Adding speed of
arm in
clockwise
direction
N+y 5N6+y 5N6+y N3+y

Given,

N+y=+100(c.w)

y=80

N80=100

N=180

ωs=N3+y=180380

=6080=140rad/s

Method -II:

Relative velocity method:

Given :
ω2=100 rad/s (cw)

ωa=80rad/s (ccw)=80rad/s

ω5ωaω2ωa=T2T3×T4T5

ω5(80)100(80)=2024×3280

ω5+80180=13

ω5=140 rad/s=140rad/s (ccw)

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