The sun (S') and the planet (P') of an epicyclic gear train shown in the figure have identical number of teeth

If the sun (S') and the outer ring (R) gears are rotated in the same direction with angular speed $ω_{S}$ and $ω_{R}$ respectively, then the angular speed of the arm AB is

\frac{1}{4} ω_{R} + \frac{3}{4} ω_{S}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{4} ω_{R} + \frac{1}{4} ω_{S}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3}{4} ω_{R} - \frac{1}{4} ω_{S}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} ω_{R} - \frac{2}{2} ω_{S}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{3}{4} ω_{R} + \frac{1}{4} ω_{S}$
As per given configuration diagram :

$R_{S} + 2 R_{P} = R_{R}$

$\Rightarrow T_{S} + 2 T_{P} = T_{R} (T_{P} = T_{S})$

$3 T_{P} = T_{R}$

$\Rightarrow 3 T_{P} = T_{R}$

Motion Arm S( $T_{S}$ ) P( $T_{P}$ ) R $(3 T_{P})$

1.Let arm fixed
Sun $\to + x$ rpm
clockwise 0 +x $- x \frac{T_{p}}{T_{p}}$ $- x \frac{T_{p}}{3 T_{p}}$

2.Arm free y (y+x) (y-x) $y - \frac{x}{3}$

$y + x = ω_{S}$

$y - \frac{x}{3} = ω_{R}$

Substract by,

$\frac{4 x}{3} = (ω_{S} - ω_{R})$

$x = \frac{3}{4} (ω_{S} - ω_{R})$

$y = ω_{S} - x = ω_{S} - \frac{3}{4} (ω_{S} - ω_{R})$

$= ω_{S} - \frac{3 ω_{s}}{4} + \frac{3 ω_{R}}{4} = \frac{ω_{s}}{4} + \frac{3 ω_{R}}{4}$

Suggest Corrections

Motion	Arm	S( $T_{S}$ )	P( $T_{P}$ )	R $(3 T_{P})$
1.Let arm fixed Sun $\to + x$ rpm clockwise	0	+x	$- x \frac{T_{p}}{T_{p}}$	$- x \frac{T_{p}}{3 T_{p}}$
2.Arm free	y	(y+x)	(y-x)	$y - \frac{x}{3}$