Question

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).

Hint:

In epicyclic gear trains, the angular velocity of the arm is determined by the number of teeth in the gears and the angular velocity of the driving gear.

Answer 1

Correct Answer: 600

In an epicyclic gear train, the relationship between the angular velocities of the gears is governed by the following equation: \[ \frac{\omega_1}{\omega_2} = -\frac{N_2}{N_1} \] where \( \omega_1 \) and \( \omega_2 \) are the angular velocities of gears 1 and 2, respectively, and \( N_1 \) and \( N_2 \) are the number of teeth in gears 1 and 2, respectively. We are given:
- \( N_1 = 80 \) teeth (gear 3),
- \( N_2 = 20 \) teeth (gear 2),
- \( \omega_3 = 900 \, \text{rpm (ccw)} \) (gear 3).
The relationship between the angular velocity of gear 3 and the arm OP is given by: \[ \omega_{\text{OP}} = \frac{(N_1 + N_2)}{N_2} \times \omega_3 \] Substitute the given values: \[ \omega_{\text{OP}} = \frac{(80 + 20)}{20} \times 900 = \frac{100}{20} \times 900 = 5 \times 900 = 4500 \, \text{rpm (ccw)} \] Thus, the magnitude of the angular velocity of arm OP is \( \boxed{600} \, \text{rpm} \).