22-Mec-A2 Kinematics and Dynamics of Machines · May 2015
Question 3 of 6: Compound planetary gear train — arm speed & relative gear speed
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: EGBC National Exam, 22-Mec-A2 “Kinematics and Dynamics of Machines”, May 2015 (source code 07-Mec-A2). Open-book, 3 hours; six questions of equal value (20 marks each), candidates answer any five. Part A — mechanisms and machine dynamics (Q1–Q4); Part B — mechanical vibration (Q5–Q6). All six are solved in full below.
Reference texts: R. L. Norton, Design of Machinery, 5th ed. (instant centres & mechanical advantage, cam design, four-bar force balancing, epicyclic trains); S. S. Rao, Mechanical Vibrations, 6th ed. (multi-DOF free vibration, modal analysis, impact response); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (epicyclic train-value cross-check).
Given. Input sun gear 1 ($N_1=78$) at $\omega_1 = 300\ \text{rpm}$ ccw drives the compound planet 2–3 ($N_2=16$ meshing gear 1; $N_3=50$ meshing the fixed gear 4, $N_4=36$). The planet is carried by the arm, which is the output. Gear 4 is held fixed to the frame.
Member
Teeth
Role
Gear 1
78
input sun (300 rpm ccw)
Gear 2
16
planet, meshes gear 1
Gear 3
50
planet (on shaft of 2), meshes gear 4
Gear 4
36
fixed reaction gear
Arm
—
carrier = output
Find. (i) $\omega_{arm}$ and its sense; (ii) $\omega_3 - \omega_1$ (gear 3 relative to the input shaft).
Compound planetary train: input sun 1 (78T) → compound planet 2–3 (16T/50T) on the arm → fixed gear 4 (36T). The arm is the output shaft.
Approach. Use the train-value (tabular / superposition) equation $\dfrac{\omega_L - \omega_{arm}}{\omega_F - \omega_{arm}} = e$, where $e$ is the fixed-arm gear ratio from the first gear $F$ to the last gear $L$; apply it first with gear 4 as the fixed member to get the arm, then again from gear 1 to gear 3 to get $\omega_3$.
Fixed-arm train value, gear 1 → gear 4. The path is $1\to2$ (external mesh) then $3\to4$ (external mesh), with gears 2 and 3 on a common shaft ($\omega_2=\omega_3$): $$e = \left(-\frac{N_1}{N_2}\right)\left(-\frac{N_3}{N_4}\right) = \left(-\frac{78}{16}\right)\left(-\frac{50}{36}\right) = +6.771.$$ The two minus signs (external meshes) make $e$ positive.
Solve for the arm (gear 4 fixed, $\omega_4=0$). $$\frac{\omega_4-\omega_{arm}}{\omega_1-\omega_{arm}} = e \;\Rightarrow\; \frac{0-\omega_{arm}}{300-\omega_{arm}} = 6.771.$$ Solving, $\omega_{arm}(e-1) = e\,\omega_1$, so $$\omega_{arm} = \omega_1\,\frac{e}{e-1} = 300\times\frac{6.771}{5.771} = \boxed{352.0\ \text{rpm (ccw)}}$$ Positive — the arm turns the same sense as the input (ccw), slightly faster.
Angular velocity of gear 3. Gear 3 shares the planet shaft with gear 2, so from gear 1 the fixed-arm ratio to gear 3 is $e_{13} = -N_1/N_2 = -78/16 = -4.875$. Then $$\omega_3 = \omega_{arm} + e_{13}(\omega_1-\omega_{arm}) = 352.0 + (-4.875)(300-352.0) = 605.4\ \text{rpm}.$$
Relative to the input shaft. $$\omega_3 - \omega_1 = 605.4 - 300 = \boxed{305.4\ \text{rpm}}$$ i.e. gear 3 spins 305 rpm faster (same sense) than the input shaft it is geared from.
