Question

For the epicyclic gear arrangement shown in the figure, ${\omega }_2=100$ rad/s clockwise (cw) and ${\omega }_{\text{arm}}=80$ rad/s counter clockwise (ccw). The angular velocity ${\omega }_5$ (in rad/s) is

• A. 0
• B. 70 cw
• C. 140 ccw
• D. 140 cw

Answer 1

The correct option is C 140 ccw

Method - I:

		Compound gear
Gear	2	3	4	5
Teeth	20	24	32	80
Speed	+N	$\frac{-20N}{24}$	$\frac{-20N}{24}$	$\frac{-20N}{24}\times \frac{32}{80}$
	+N	$\frac{-5N}{6}$	$\frac{-5N}{6}$	$\frac{-5N}{6}\times \frac{2}{5}$
Adding speed of arm in clockwise direction	N+y	$\frac{-5N}{6}+y$	$\frac{-5N}{6}+y$	$\frac{-N}{3}+y$

Given,

$N+y=+100(c.w)$

$y=-80$

$\therefore N-80=100$

$N=180$

$\therefore {\omega }_s=\frac{-N}{3}+y=\frac{-180}{3}-80$

$=-60-80=-140\text{rad/s}$

Method -II:

Relative velocity method:

Given :
${\omega }_2=100\text{rad/s (cw)}$

${\omega }_a=80\text{rad/s (ccw)}=-80\text{rad/s}$

$\frac{{\omega }_5-{\omega }_a}{{\omega }_2-{\omega }_a}=-\frac{T_2}{T_3}\times \frac{T_4}{T_5}$

$\frac{{\omega }_5-(-80)}{100-(-80)}=\frac{-20}{24}\times \frac{32}{80}$

$\frac{{\omega }_5+80}{180}=-\frac{1}{3}$

${\omega }_5=-140\text{rad/s}=140\text{rad/s (ccw)}$