--- title: "Link CD 6 rad/s, point E and link AB" subject: "Applied Mechanics (Dynamics)" chapter: 4 key: "semester_3/applied_mechanics_dynamics:link-cd-6rads-velocity-e-link-ab-gpm" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/link-cd-6rads-velocity-e-link-ab-gpm site: BCT Engineering --- # Link CD 6 rad/s, point E and link AB Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies - Asked in: 2080 Baishakh, 8+2 marks - Video: [Dynamics 16-76| If link CD is rotating at wCD = 5 rad/s, determine the angular velocity of link AB](https://www.youtube.com/watch?v=bZ15_E6_Gao) — Learning by Teaching - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/link-cd-6rads-velocity-e-link-ab-gpm - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4 ## Question If link CD has an angular velocity of $\omega_{CD} = 6$ rad/s, determine the velocity of point E on link BC and the angular velocity of link AB at the instant shown. Explain General plane motion (GPM). [Figure: link AB inclined at $30^\circ$ pinned at A; link BC horizontal with midpoint E (0.3 m on each side); link CD vertical, 0.6 m long, pinned at D; $\omega_{CD} = 6$ rad/s.] ## Short answer From relative velocity analysis, v_E = 4.76 m/s, ω_AB = 6.00 rad/s counter-clockwise. General plane motion is the combination of translation and rotation; for any two points, v_B = v_A + ω × r_B/A. ## Answer Assumed layout (from the figure): $D$ and $A$ are fixed pins at the same level; $CD$ is vertical (0.6 m) with $C$ at the top; $BC$ is horizontal (0.6 m) with $B$ to the left of $C$ and $E$ its midpoint; $AB$ rises at $30^\circ$ from $A$ to $B$, so $AB = 0.6/\sin30^\circ = 1.2$ m. Take $\omega_{CD}=6$ rad/s counter-clockwise, $x$ to the right and $y$ up (reverse all senses if $CD$ turns the other way). #### Step 1: velocity of C $$ \vec v_C = \omega_{CD}\hat k\times\vec r_{C/D} = 6\hat k\times0.6\hat j = -3.6\,\hat i\ \text{m/s} $$ #### Step 2: direction of v_B Link $AB$ rotates about $A$, so $\vec v_B \perp AB$. With $\vec v_B = (v_{Bx}, v_{By})$ and $AB$ along $(\cos30^\circ, \sin30^\circ)$: $$ v_{Bx}\cos30^\circ + v_{By}\sin30^\circ = 0 $$ Link $BC$ is horizontal, so $\omega_{BC}\hat k\times\vec r_{B/C}$ is vertical and $v_{Bx} = v_{Cx} = -3.6$ m/s. Hence $$ v_{By} = \frac{3.6\cos30^\circ}{\sin30^\circ} = 6.24\ \text{m/s}, \qquad v_B = \sqrt{3.6^2 + 6.24^2} = 7.20\ \text{m/s} $$ #### Step 3: link BC $$ \vec v_B = \vec v_C + \omega_{BC}\hat k\times(-0.6\hat i) \;\Rightarrow\; v_{By} = -0.6\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.39\ \text{rad/s}\ \text{(clockwise)} $$ #### Step 4: point E $$ \vec v_E = \vec v_C + \omega_{BC}\hat k\times(-0.3\hat i) = -3.6\hat i + (-10.39)(-0.3)\hat j = (-3.60\,\hat i + 3.12\,\hat j)\ \text{m/s} $$ $v_E = 4.76$ m/s, directed 40.9° above the horizontal, towards the left. #### Step 5: angular velocity of AB $$ \omega_{AB} = \frac{v_B}{AB} = \frac{7.20}{1.2} = 6.00\ \text{rad/s}\ \text{(counter-clockwise)} $$ **Answer: $v_E = 4.76$ m/s; $\omega_{AB} = 6.00$ rad/s counter-clockwise.** #### General plane motion **General plane motion (GPM)** is the motion of a rigid body in which every particle stays in a plane parallel to a fixed plane, and the motion is neither pure translation nor pure rotation about a fixed axis. It is the sum of (i) a translation of a chosen reference point and (ii) a rotation of the body about that point. For two points $A$ and $B$ of the same rigid body: $$ \vec v_B = \vec v_A + \vec\omega \times \vec r_{B/A}, \qquad \vec a_B = \vec a_A + \vec\alpha \times \vec r_{B/A} - \omega^2 \vec r_{B/A} $$ Here $v_{B/A} = \omega\, r_{B/A}$ is perpendicular to $AB$, and $a_{B/A}$ has a tangential part $\alpha r$ and a normal part $\omega^2 r$ towards $A$. **Examples** - A wheel rolling on the road: the centre translates while the wheel rotates about the centre. - The connecting rod of an engine: end $A$ (crank pin) moves on a circle, end $B$ (piston) moves in a straight line. - A ladder sliding down a wall. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is w_BC found from the geometry (vB perpendicular to AB, vBx = vCx = -3.6 m/s)? A. 10.39 rad/s clockwise B. 6 rad/s C. 7.2 rad/s ccw D. 3.6 rad/s Answer: A. vBy = 6.24 m/s so w_BC = 6.24/0.6 = 10.39 rad/s clockwise. (Key point: Angular velocity of BC) 2. What is vE? A. 4.76 m/s B. 3.6 m/s C. 6.24 m/s D. 7.20 m/s Answer: A. vE = vC + w_BC x rE/C = (-3.6 i + 3.12 j) m/s, magnitude 4.76 m/s. (Key point: Velocity of E) 3. How is w_AB found? A. w_AB = vB / AB = 7.20 / 1.2 = 6 rad/s ccw B. w_AB = vB / BC = 12 rad/s C. w_AB = w_CD D. w_AB = 10.39 rad/s Answer: A. AB rotates about fixed A, so w = vB / AB with AB = 0.6/sin30 = 1.2 m. (Key point: Angular velocity of AB) 4. General plane motion of a rigid body is A. only translation B. only rotation about a fixed axis C. translation of a reference point plus rotation about it D. motion in three dimensions only Answer: C. GPM is neither pure translation nor fixed-axis rotation, but the sum of both, e.g. a rolling wheel. 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