Question

Common Data Questions:
A planetary gear train has four gears and one carrier. Angular velocities of the gears are ${\omega }_1,{\omega }_2,{\omega }_3,\text{and}{\omega }_4$, respectively. The carrier rotates with angular velocity ${\omega }_5$

What is the relation between the angular velocities of Gear 1 and Gear 4 ?

• A. $\frac{{\omega }_1-{\omega }_5}{{\omega }_4-{\omega }_5}=6$
• B. $\frac{{\omega }_4-{\omega }_5}{{\omega }_1-{\omega }_5}=6$
• C. $\frac{{\omega }_1-{\omega }_2}{{\omega }_4-{\omega }_5}=-\left(\frac{2}{3}\right)$
• D. $\frac{{\omega }_2-{\omega }_5}{{\omega }_4-{\omega }_5}=\frac{8}{9}$

Answer 1

The correct option is A $\frac{{\omega }_1-{\omega }_5}{{\omega }_4-{\omega }_5}=6$

Arm	Gears	1	2	3	4
0	No,or teeth	15	45	20	40
0	Velocities	$N_1$	$\frac{-15N_1}{45}$	$\frac{-15N_1}{45}$	$\frac{+20\times 15N_1}{40\times 45}$
+Y	Velocities after adding velocity of arm "y"	$N_1+y$	$\frac{-15N_1}{45}+y$	$\frac{-15N_1}{45}+y$	$\frac{+20\times 15N_1}{40\times 45}+y$

$\therefore {\omega }_1=N_1+y$

${\omega }_2=\frac{-15N_1}{45}+y=\frac{-N_1}{3}+y$

${\omega }_3=\frac{-15N_1}{45}+y=\frac{-N_1}{3}+y$

${\omega }_4=\frac{20\times 15\times \times N_1}{45\times 40}+y=\frac{N_1}{6}+y$

$\therefore \frac{{\omega }_1-{\omega }_5}{{\omega }_4-{\omega }_5}=\frac{N_1+y-y}{\frac{N_1}{6}+y-y}=\frac{N_1}{N_{1/6}}=6$

$\frac{{\omega }_1-{\omega }_5}{{\omega }_4-{\omega }_5}=6$