Mindyard | A schematic of an epicyclic gear train is shown in the figure. The sun (gear 1) and planet (gear 2) are external, and the ring gear (gear 3) is internal. Gear 1, gear 3 and arm OP are pivoted to the ground at O. Gear 2 is carried on the arm OP via the pivot joint at P, and is in mesh with the other two gears. Gear 2 has 20 teeth and gear 3 has 80 teeth. If gear 1 is kept fixed at 0 rpm and gear 3 rotates at 900 rpm counter clockwise (ccw), the magnitude of angular velocity of arm OP is _______

A schematic of an epicyclic gear train is shown in the figure. The sun (gear 1) and planet (gear 2) are external, and the ring gear (gear 3) is internal. Gear 1, gear 3 and arm OP are pivoted to the ground at O. Gear 2 is carried on the arm OP via the pivot joint at P, and is in mesh with the other two gears. Gear 2 has 20 teeth and gear 3 has 80 teeth. If gear 1 is kept fixed at 0 rpm and gear 3 rotates at 900 rpm counter clockwise (ccw), the magnitude of angular velocity of arm OP is ________ rpm (in integer).

Correct Answer :

Correct answer is : 600

T₁ = ?, T₂ = 20, T₃ = 80

N₃ = 900 rpm (ccw), N₁ = 0 rpm (fixed)

Since the module of all mating gears is always the same, hence radius can be expressed as:

r₃ = r₁ + 2r₂

_{$\frac{m T_{3}}{2} = \frac{m T_{1}}{2} + 2. \frac{m T_{2}}{2}$} ⇒ T₃ = T₁ + 2T₂

T₁ = 80 - 40 = 40 teeth

According to the tabular method:

Motions	Arm	Gear 1 (40)	Gear 2 (20)	Gear 3 (80)
Arm fixed, gear1 rotate +x revolutions	0	+ x	$- x \times \frac{40}{20}$	$- x \times \frac{40}{20} \times \frac{20}{80}$
Arm effect is considered	y	y + x	y - 2x	y - (x/2)

It is given that:

N₁ = 0 ⇒ y + x = 0

N₃ = 900 rpm ⇒ y - (x/2) = 900

on solving we get,

3y/2 = 900 ⇒ y = 600 rpm

x = - 600 rpm

Speed of arm OP = y (from table)

Speed of arm is 600 rpm counter - clockwise.

Solution :

The correct answer is 600.

Analysis of the Gear Train from the Figure:
By inspecting the provided schematic diagram, we observe:

The central sun gear, labeled Gear 1, is pivoted at center O and has a speed of 0 rpm (fixed to the ground).
An intermediate planet gear, labeled Gear 2, has 20 teeth (T₂ = 20) and is pivoted at point P on the arm.
The outer ring gear, labeled Gear 3, has 80 teeth (T₃ = 80) and rotates at 900 rpm counter-clockwise (ccw).
The link arm, labeled OP, is pivoted at the center O and carries the planet gear at P.

Step 1: Determine the Number of Teeth on Gear 1
From the geometry of the epicyclic gear train shown in the figure, the outer radius of the ring gear must equal the radius of the sun gear plus the diameter of the planet gear:

$r_{3} = r_{1} + 2 r_{2}$

Since the module ( $m$ ) is identical for all mating gears, the radii are directly proportional to the number of teeth ( $r = \frac{m T}{2}$ ). Substituting this relationship gives:

$T_{3} = T_{1} + 2 T_{2}$

Given that $T_{3} = 80$ and $T_{2} = 20$ :
$80 = T_{1} + 2 (20)$
$T_{1} = 80 - 40 = 40 teeth$

Step 2: Set up the Tabular Method for Motion Analysis
We analyze the rotations of each member using the tabular method. Let counter-clockwise (ccw) rotation be positive (+).

Operation / Condition	Arm OP	Gear 1 (T₁ = 40)	Gear 2 (T₂ = 20)	Gear 3 (T₃ = 80)
1. Arm fixed, Gear 1 rotates by +x	0	+x	$- x \times \frac{T_{1}}{T_{2}} = - 2 x$	$- 2 x \times \frac{T_{2}}{T_{3}} = - \frac{x}{2}$
2. Add rotation +y of the arm	y	y + x	y - 2x	y - x/2

Step 3: Solve for the Speeds using Boundary Conditions
We are given:
1. Gear 1 is fixed:
$N_{1} = y + x = 0 \Rightarrow x = - y$
2. Gear 3 rotates at 900 rpm ccw (positive direction):
$N_{3} = y - \frac{x}{2} = 900$

Substitute $x = - y$ into the equation for Gear 3:

$y - \frac{(- y)}{2} = 900$

$y + \frac{y}{2} = 900$

$\frac{3 y}{2} = 900$

$y = 600 rpm$

The speed of the arm OP is represented by $y$ in the table. Therefore, the magnitude of the angular velocity of the arm OP is 600 rpm (rotating in a counter-clockwise direction).

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