Question 4 (paper Question IV) — Four-Bar Linkage Velocity & Acceleration (Part B · Dynamics, 20 marks): 04-BS-3, December 2015, Q4

Question 4 (paper Question IV) — Four-Bar Linkage Velocity & Acceleration (Part B · Dynamics, 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.

Check — the paper does not state an angular acceleration for the driving crank CB; consistent with the standard textbook version of this problem, α_CB = 0 (CB rotates at the constant 2 rad/s stated) is assumed for part (b).

Given. O(0,0), A(0,100), B(175,50), C(250,50) mm (all fixed pivots/pins except A, B which move); ω_CB=2 rad/s CCW (constant, α_CB=0); OA=100 mm, CB=75 mm, OC=250 mm.

Given data
Point	x (mm)	y (mm)
O (fixed pin)	0	0
A	0	100
B	175	50
C (fixed pin)	250	50

Find. ω_OA, ω_AB (part a); α_OA, α_AB (part b).

Figure 4 — four-bar linkage O-A-B-C in the position where CB is horizontal and OA is vertical.

Approach. B moves on a known circle about the fixed pin C, so $\mathbf v_B=\boldsymbol\omega_{CB}\times\mathbf r_{B/C}$ is known directly; equating this to $\mathbf v_B=\mathbf v_A+\boldsymbol\omega_{AB}\times\mathbf r_{B/A}$ with $\mathbf v_A=\boldsymbol\omega_{OA}\times\mathbf r_{A/O}$ gives two scalar equations for the two unknown angular velocities; the identical relative-motion equation applied to acceleration (with the centripetal $-\omega^2\mathbf r$ terms now known) gives the two angular accelerations.

Velocity of B from its circular motion about fixed C. $\mathbf r_{B/C}=(175-250,\,50-50)=(-75,0)\ \text{mm}$. $$\mathbf v_B=\omega_{CB}\,\hat{\mathbf k}\times\mathbf r_{B/C}=(0,-150,0)\ \text{mm/s}$$ (150 mm/s straight down — consistent with CCW rotation carrying the point left of C downward).
Relative-velocity equation, solved for ω_OA, ω_AB. $\mathbf r_{A/O}=(0,100)$, $\mathbf r_{B/A}=(175,-50)\ \text{mm}$. $$\mathbf v_B=\omega_{OA}\hat{\mathbf k}\times\mathbf r_{A/O}+\omega_{AB}\hat{\mathbf k}\times\mathbf r_{B/A}$$ Matching x- and y-components: $-100\,\omega_{OA}+50\,\omega_{AB}=0$ and $175\,\omega_{AB}=-150$, giving $$\omega_{AB}=\boxed{-0.857\ \text{rad/s}}=0.857\ \text{rad/s CW}, \qquad \omega_{OA}=\boxed{-0.429\ \text{rad/s}}=0.429\ \text{rad/s CW}$$
Acceleration of B, with α_CB=0 (constant ω_CB), so only the centripetal term survives: $$\mathbf a_B=-\omega_{CB}^2\,\mathbf r_{B/C}=-2^2(-75,0)=\boxed{(300,\,0)\ \text{mm/s}^2}$$
Relative-acceleration equation, solved for α_OA, α_AB. $$\mathbf a_B=\alpha_{OA}\hat{\mathbf k}\times\mathbf r_{A/O}-\omega_{OA}^2\mathbf r_{A/O}+\alpha_{AB}\hat{\mathbf k}\times\mathbf r_{B/A}-\omega_{AB}^2\mathbf r_{B/A}$$ With $\omega_{OA}^2=0.1837$, $\omega_{AB}^2=0.7347$, matching components gives $175\,\alpha_{AB}+18.37=0$ and $-100\,\alpha_{OA}+50\,\alpha_{AB}-128.57=300$, so $$\alpha_{AB}=\boxed{-0.105\ \text{rad/s}^2}=0.105\ \text{rad/s}^2\text{ CW}, \qquad \alpha_{OA}=\boxed{-4.338\ \text{rad/s}^2}=4.338\ \text{rad/s}^2\text{ CW}$$

Final results
Quantity	Value
ω_OA	0.429 rad/s CW
ω_AB	0.857 rad/s CW
α_OA	4.338 rad/s² CW
α_AB	0.105 rad/s² CW

Topic: Rigid-body kinematics of a four-bar linkage (relative velocity and acceleration).

Key relations: $\mathbf v_B=\mathbf v_A+\boldsymbol\omega\times\mathbf r_{B/A}$; $\mathbf a_B=\mathbf a_A+\boldsymbol\alpha\times\mathbf r_{B/A}-\omega^2\mathbf r_{B/A}$, applied link by link from a point whose motion is already fully known (a fixed-pivot end).

Why this works: every link is rigid, so any two points on it are related by pure rotation about the instantaneous angular velocity/acceleration of that link; chaining the equation from the known crank CB, through the coupler AB, to the known-direction crank OA turns the mechanism into two vector equations (four scalar equations) with exactly four unknowns overall, but the problem is structured so velocity and acceleration each separate into 2-by-2 linear systems.

Common pitfall: forgetting the centripetal term $-\omega^2\mathbf r$ in the acceleration equation (it is often the dominant term, as it is for B here) or mixing up which link's $\omega$ multiplies which $\mathbf r$.

Source: Hibbeler, Engineering Mechanics: Dynamics, 14th ed., Ch. 16 (Planar Kinematics of a Rigid Body § Relative-Motion Analysis: Velocity and Acceleration).

Question 4 of 6: Question 4 (paper Question IV) — Four-Bar Linkage Velocity & Acceleration (Part B · Dynamics, 20 marks)

Question 4 (paper Question IV) — Four-Bar Linkage Velocity & Acceleration (Part B · Dynamics, 20 marks)