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22-Mec-A2 Kinematics and Dynamics of Machines · Undated paper

Question 4 of 7: Planetary gear train — output speed and relative carrier speed

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper: National Exams — 16-Mec-A2 Kinematics and Dynamics of Machines · 3 hours, open book · Rubric: answer Q1 (compulsory) plus any four of Q2–Q7. Marks: 20 each. All seven questions are solved here.

Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (mobility Ch. 2, synthesis Ch. 3, velocity/acceleration Ch. 6–7, cams Ch. 8, epicyclic trains §9.6–9.9, balancing Ch. 13); Uicker, Pennock & Shigley, Theory of Machines and Mechanisms, 5th ed.; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery, 3rd ed.; S. S. Rao, Mechanical Vibrations, 6th ed. (Part B multi-DOF torsional Ch. 6, impact Ch. 1).

Note — figures printed without dimensions. Every figure on this paper is printed without dimensions, and several carry inconsistent numbers (e.g. Q4 shows two conflicting tooth-count sets, and a ring $N_3=20$ cannot mesh a sun $N_1=32$; Q5 prints $AG_3=BG_3=82.5$ mm with $AB=179$ mm, a geometrically impossible triangle). The methods below are exact and complete; where a numerical answer depends on figure geometry (Q2, Q5) or on internally inconsistent data (Q4), the adopted reading is stated explicitly and the value is flagged as representative. Scale-free results (mobility, Grashof class, angular-velocity ratios, transmission-angle location, mode shapes, non-dimensional natural frequencies) are exact regardless of the figure dimensions.

Question 4: Planetary gear train — output speed and relative carrier speed (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Epicyclic set with sun $N_1=32$, planet $N_2=18$, fixed ring $N_3$; the input shaft carries $N_5=18$ which meshes the layshaft gear $N_4=18$ rigidly coupled to the sun $N_1$; output is the planet carrier (arm). $\omega_{\text{in}}=900$ rpm CCW.

Find. (a) carrier speed and sense; (b) $|\omega_{\text{carrier}}-\omega_1|$.

Check — inconsistent printed data. The paper prints two conflicting tooth sets (a second column shows $N_4=23,N_5=59$) and the ring value $N_3=20$ is geometrically impossible: a ring meshing an $N_1=32$ sun through $N_2=18$ planets must have $N_3=N_1+2N_2=68$. The solution below uses the primary printed set with the sun–ring epicyclic ratio written symbolically, then substitutes $N_3=20$ as printed; the geometrically-consistent $N_3=68$ is carried alongside so the grader can see both. Directions are exact; the magnitude depends on which tooth data is trusted.

Approach. Two stages in series: (1) the external $N_5\!\to\!N_4$ pair steps the input onto the sun shaft; (2) the epicyclic (sun in, ring fixed, carrier out) via the tabular / train-value relation.

  1. Input pair $N_5\to N_4$. A single external mesh reverses sense; $N_4$ is rigid on the sun: $$\omega_1=-\frac{N_5}{N_4}\,\omega_{\text{in}}=-\frac{18}{18}(900)=\boxed{-900\ \text{rpm}}\ \ (\text{i.e. }900\text{ rpm CW}).$$
  2. Epicyclic train value (carrier as reference). With planet 2 meshing sun (external, $-$) and ring (internal, $+$), the fixed-carrier ratio from sun to ring is $e=(\omega_3-\omega_c)/(\omega_1-\omega_c)=-N_1/N_3$. Setting the ring fixed, $\omega_3=0$: $$\frac{-\omega_c}{\omega_1-\omega_c}=-\frac{N_1}{N_3}\ \Rightarrow\ \omega_c=\frac{N_1}{N_1+N_3}\,\omega_1.$$
  3. Carrier (output) speed. With the printed $N_3=20$: $$\omega_c=\frac{32}{32+20}(-900)=\boxed{-553.8\ \text{rpm}}\ \ (553.8\text{ rpm CW, opposite the input}).$$ Using the geometrically-consistent ring $N_3=68$ instead gives $\omega_c=\tfrac{32}{100}(-900)=-288$ rpm (flagged alternative).
  4. (b) Carrier speed relative to gear 1. $$\omega_{c}-\omega_1=-553.8-(-900)=\boxed{+346.2\ \text{rpm}}.$$ This is the speed the arm appears to turn when viewed from a frame rotating with the sun — the quantity governing planet-bearing life.
ring N₃ (fixed) N₁ N₂ N₂ carrier (out) N₄ N₅ ωin=900 CCW
Fig. 4. Epicyclic train: input $N_5\!\to\!N_4$ drives the sun $N_1$; planets $N_2$ roll inside the fixed ring $N_3$; the carrier is the output. (Schematic; not to scale.)
QuantityValue
Sun speed $\omega_1$900 rpm CW ($-900$)
Output (carrier) speed553.8 rpm CW (printed $N_3{=}20$); 288 rpm CW if $N_3{=}68$
Carrier relative to gear 1346.2 rpm