22-Mec-A2 Kinematics and Dynamics of Machines · December 2014

Question 2 of 6: Four-bar function generator — output velocity & acceleration

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

Notes on this paper

National Exams, December 2014 — 07-Mec-A2, Kinematics and Dynamics of Machines. 3 hours, OPEN BOOK. Six questions of 20 marks each; candidates answer any FIVE. All SIX are solved in full below (Part A — mechanisms and machine dynamics, Q1–Q4; Part B — mechanical vibration, Q5–Q6).

Reference texts: R. L. Norton, Design of Machinery, 5th ed. (instant centres and velocity analysis, four-bar position/velocity/acceleration, planetary gear kinematics, engine balancing); S. S. Rao, Mechanical Vibrations, 6th ed. (single- and multi-degree-of-freedom free and forced response, modal analysis); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (epicyclic train-value cross-check).

Question 2: Four-bar function generator — output velocity & acceleration (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. $r_1=0.50,\ r_2=0.22,\ r_3=0.69,\ r_4=0.48\ \text{m}$; input crank at $\theta_2=30^{\circ}$ measured from the ground line O₂O₄, turning CCW at a constant $\omega_2=20\ \text{rad/s}$ ($\alpha_2=0$).

Find. $\omega_4$ and $\alpha_4$ of the output link at $\theta_2=30^{\circ}$ (open circuit).

Figure 2.1 — the four-bar at $\theta_2=30^{\circ}$ in the open (uncrossed) configuration: ground O₂O₄ ($r_1$), input crank $r_2$ (blue), coupler $r_3$, output rocker $r_4$ (green).

Approach. Close the vector loop for the two unknown angles, then differentiate the loop once for the angular velocities and twice for the angular accelerations, solving each as a $2\times2$ linear system.

Position solution. The loop $r_2e^{i\theta_2}+r_3e^{i\theta_3}=r_1+r_4e^{i\theta_4}$ places pin A at $(0.191,0.110)$ m and O₄ at $(0.500,0)$. Intersecting the coupler circle ($r_3$ about A) with the output circle ($r_4$ about O₄) gives the open-circuit joint B, whence $\theta_3=19.16^{\circ},\ \theta_4=44.51^{\circ}$.
Velocity equations. Differentiating the loop and separating real/imaginary parts, $$-r_2\omega_2\sin\theta_2-r_3\omega_3\sin\theta_3+r_4\omega_4\sin\theta_4=0,$$ $$\phantom{-}r_2\omega_2\cos\theta_2+r_3\omega_3\cos\theta_3-r_4\omega_4\cos\theta_4=0.$$ Solving the pair, $\omega_3=-3.73$ and $\omega_4=+4.03\ \text{rad/s}$ (same sense as $\omega_2$).
Acceleration equations. The second derivative (with $\alpha_2=0$) gives another $2\times2$ system in $\alpha_3,\alpha_4$: $$-r_3\alpha_3\sin\theta_3+r_4\alpha_4\sin\theta_4=r_2\omega_2^2\cos\theta_2+r_3\omega_3^2\cos\theta_3-r_4\omega_4^2\cos\theta_4,$$ $$\phantom{-}r_3\alpha_3\cos\theta_3-r_4\alpha_4\cos\theta_4=r_2\omega_2^2\sin\theta_2+r_3\omega_3^2\sin\theta_3-r_4\omega_4^2\sin\theta_4.$$ The centripetal terms are $r_2\omega_2^2=88.0$, $r_3\omega_3^2=9.61$, $r_4\omega_4^2=7.78\ \text{m/s}^2$; solving gives $\alpha_3=291$ and $\alpha_4=+433\ \text{rad/s}^2$.

Quantity	Value
Coupler / output angles $\theta_3,\theta_4$	19.16°, 44.51°
Coupler angular velocity $\omega_3$	−3.73 rad/s
Output angular velocity $\omega_4$	+4.03 rad/s (CCW)
Coupler angular acceleration $\alpha_3$	291 rad/s²
Output angular acceleration $\alpha_4$	+433 rad/s²

Check. The open (uncrossed) circuit is taken, matching the figure (coupler and rocker meet above the ground line). The crossed circuit would give the mirror branch $\theta_4\approx-44^{\circ}$ with different signs; the open branch is the physically continuous one for a crank sweeping through $30^{\circ}$.

Topic. Analytical position, velocity and acceleration analysis of a planar four-bar linkage (vector-loop / complex-number method).

Key relations. Loop closure $r_2e^{i\theta_2}+r_3e^{i\theta_3}-r_4e^{i\theta_4}-r_1=0$; its first and second time-derivatives give linear systems for $(\omega_3,\omega_4)$ and $(\alpha_3,\alpha_4)$ once $(\theta_3,\theta_4)$ are known.

Why this works. The single loop equation contains two scalar constraints, exactly matching the two unknown link angles (1 DOF plus the driven position). Differentiating keeps the same coefficient matrix (the Jacobian), so velocities and accelerations are each one linear solve — no iteration once the position is fixed.

Common pitfall. Choosing the wrong assembly branch (open vs crossed) flips $\theta_4$ and all downstream signs; and forgetting the $-r_i\omega_i^2$ (centripetal) terms in the acceleration equations, which here dominate the right-hand side. (Norton, Design of Machinery, ch. 6–7.)

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