Question

The arm OA of an epicyclic gear train shown in figure revolves counterclockwise about O with an angular velocity of 4 rad/s. Both gears are of the same size. The angular velocity of gear C, if the sun gear B is fixed

• A. 4 rad/s
• B. 8 rad/s
• C. 10 rad/s
• D. 12 rad/s

Answer 1

The correct option is B 8 rad/s

Method -I:

Gear	B	C
No.of teeth	$T_B$	$T_B=T_C$
Speed N in rpm	+N Positive sign means clockwise	$\frac{-N{T}_B}{T_C}=-N$
Speed after adding speed of arm, let speed of arm be +y clockwise	N+y	-N+y

Given :
$y=-4\text{[counter clockwise direction]}$

$N+y=0\text{[sun gear B is fixed]}$

$\therefore$ $N=-y$

$=-1\{-4\}=+4\text{rad/s}$

$\text{Speed of gear}C=-4-4=-8\text{rad/s}$

Method II:
Relative velocity method:

Given:
${\omega }_a=4\text{rad/sec (negative sign CCW)}$

$T_B=T_C,{\omega }_B=0,{\omega }_c=?$

$\frac{{\omega }_C-{\omega }_a}{{\omega }_B-{\omega }_a}=\frac{-T_B}{T_C}$

$\frac{{\omega }_c-(-4)}{0-(-4)}=-1$

${\omega }_C+4=-4$

${\omega }_C=-8\text{rad/s}$

$=8\text{rad/s}$