Relative Velocity And Acceleration - Statics and Dynamics...

Question 1

In a four-bar linkage, the fixed ground link is O₁O₂. Crank O₁A rotates with constant angular velocity $\omega_1 = 10\,\text{rad/s}$ CCW and has length 0.2 m. The coupler AB has length 0.5 m. The follower O₂B has length 0.3 m. At the instant of interest, O₁A is horizontal (pointing right), AB makes 30° above horizontal (pointing up-right), and O₂B is vertical (pointing upward). The velocity of A is therefore $\mathbf{v}_A = (0)(\hat{i}) + (2)\hat{j} = 2\hat{j}\,\text{m/s}$ (since $v_A = \omega_1 \cdot |O_1A| = 2\,\text{m/s}$ upward).

Using the velocity loop equation $\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}_{AB}\times\mathbf{r}_{B/A}$ , combined with the constraint that B moves perpendicular to O₂B (i.e., B's velocity is horizontal at this instant), find $\omega_{AB}$ (angular velocity of the coupler AB).

$\omega_{AB} = -\frac{4}{\sqrt{3}}\,\text{rad/s} \approx -2.31\,\text{rad/s}$ (CW), obtained by enforcing the constraint that $v_{By} = 0$ and solving the $\hat{j}$ -component of the relative velocity equation. (correct answer)
$\omega_{AB} = \frac{4}{\sqrt{3}}\,\text{rad/s} \approx 2.31\,\text{rad/s}$ (CCW), obtained by the same procedure but with a sign error in the direction of $\mathbf{r}_{B/A}$ relative to AB's orientation.
$\omega_{AB} = -4\,\text{rad/s}$ (CW), obtained by using only the $\hat{i}$ -component of the relative velocity equation and ignoring the constraint on $v_{By}$ , which leads to an incorrect scalar equation.
$\omega_{AB} = \frac{2}{0.5} = 4\,\text{rad/s}$ (CCW), obtained by dividing the magnitude of $\mathbf{v}_A$ by the length of the coupler without accounting for the angle between the velocity and the coupler direction.

Explanation: When analyzing four-bar linkage velocities, your strategy should be to write the vector loop equation, express each cross product in component form, then apply kinematic constraints to isolate unknowns — one equation per unknown. Here, the coupler AB makes 30° above horizontal, so

\mathbf{r}_{B/A} = 0.5\cos30°\,\hat{i} + 0.5\sin30°\,\hat{j} = \frac{\sqrt{3}}{4}\hat{i} + \frac{1}{4}\hat{j}

m. With

\boldsymbol{\omega}_{AB} = \omega_{AB}\hat{k}

, the cross product gives

\boldsymbol{\omega}_{AB}\times\mathbf{r}_{B/A} = \omega_{AB}(-\frac{1}{4}\hat{i} + \frac{\sqrt{3}}{4}\hat{j})

. Since O₂B is vertical, B can only move horizontally, meaning

v_{By} = 0

. Applying the

\hat{j}

-component of

\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}_{AB}\times\mathbf{r}_{B/A}

0 = 2 + \omega_{AB}\cdot\frac{\sqrt{3}}{4}

, which yields

\omega_{AB} = -\frac{8}{\sqrt{3}} \cdot \frac{1}{2} = -\frac{4}{\sqrt{3}} \approx -2.31\,\text{rad/s}

. The negative sign confirms clockwise rotation. This is answer A, the correct choice. Answer B reaches the same magnitude but positive, reflecting a sign error — typically from reversing

\mathbf{r}_{B/A}

to point from B to A instead of A to B, which flips the cross product components incorrectly. Answer C uses only the

\hat{i}

-component without enforcing

v_{By} = 0

, leaving one equation with two unknowns (

\omega_{AB}

and

v_{Bx}

) and guessing a value — a fundamental setup error. Answer D simply divides

|\mathbf{v}_A|

by the coupler length, ignoring both the angle of the coupler and the kinematic constraint on B's motion entirely. Your key takeaway: always apply the geometric constraint on B's velocity direction before solving — it's what makes the system determinate.

Statics and Dynamics Quiz: Relative Velocity And Acceleration

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