--- title: "Four-bar angular velocities by IC method" subject: "Theory of Mechanism and Machine I" chapter: 7 key: "semester_6/theory_of_mechanism_and_machine_i:four-bar-velocity-by-ic-method" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/four-bar-velocity-by-ic-method site: BCT Engineering --- # Four-bar angular velocities by IC method Theory of Mechanism and Machine I · Chapter 7: Kinematic Analysis of Mechanisms - Asked in: year not given, 8 marks - Video: [Instantaneous Centre method of Velocity Analysis - Four bar mechanism | Four bar IC method | KTU MOM](https://www.youtube.com/watch?v=znMRW9j0Tlw) — Mohith Dadu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/four-bar-velocity-by-ic-method - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-7 ## Question In a four-bar mechanism ABCD, AD = 300 mm is fixed, AB = 100 mm, BC = 250 mm and CD = 200 mm. The crank AB rotates at 10 rad/s anticlockwise and makes 60 degrees with AD. [Take A as origin and AD along the x-axis. In this position B is at (50.0, 86.6) mm, D at (300, 0) mm and C at (273.7, 198.3) mm.] Using Kennedy's theorem, locate the instantaneous centres I13 and I24 and hence find the angular velocities of the coupler BC and the rocker CD, and the linear velocity of C. ## Short answer Using Kennedy’s theorem, I13 is at intersection of AB and DC, I24 at intersection of BC and AD. Angular velocity of coupler BC is 2.578 rad/s clockwise, rocker CD is 2.916 rad/s anticlockwise, velocity of C is 0.583 m/s. ## Answer #### Instantaneous centres of the four links (1 = AD, 2 = AB, 3 = BC, 4 = CD) Number of ICs $= \dfrac{4 \times 3}{2} = 6$. Four are pin joints: $I_{12} = A\,(0, 0)$, $I_{23} = B\,(50.0, 86.6)$, $I_{34} = C\,(273.7, 198.3)$, $I_{14} = D\,(300, 0)$ mm. By Kennedy's theorem the remaining two are: - $I_{13}$ is the intersection of line AB (through $I_{12}$, $I_{23}$) and line DC (through $I_{14}$, $I_{34}$). - $I_{24}$ is the intersection of line BC (through $I_{23}$, $I_{34}$) and line AD (through $I_{12}$, $I_{14}$). Solving the line equations: $$ I_{13} = (243.9,\ 422.5)\ \text{mm}, \qquad I_{24} = (-123.5,\ 0)\ \text{mm} $$ ```text I13 * (243.9, 422.5) \ . \ C B / \ / \ I24 *--A---------D------ ``` (Not to scale; $I_{24}$ lies on AD produced beyond A.) #### Angular velocity of rocker CD $I_{24}$ is a point common to links 2 and 4, so its velocity is the same whether taken on AB or on CD: $$ \omega_2 \cdot I_{12}I_{24} = \omega_4 \cdot I_{14}I_{24} $$ $$ I_{12}I_{24} = 123.5\ \text{mm}, \qquad I_{14}I_{24} = 300 + 123.5 = 423.5\ \text{mm} $$ $$ \omega_4 = 10 \times \frac{123.5}{423.5} = 2.916\ \text{rad/s} $$ $I_{24}$ lies outside A and D on the same side of both, so $\omega_4$ has the same sense as $\omega_2$: **anticlockwise**. #### Angular velocity of coupler BC $I_{13}$ is the IC of links 1 and 3. The pin B is common to links 2 and 3: $$ v_B = \omega_2 \cdot AB = 10 \times 0.100 = 1.0\ \text{m/s} $$ $$ \omega_3 = \frac{v_B}{I_{13}B} = \frac{1.0}{387.8/1000} = 2.578\ \text{rad/s} $$ B lies between $I_{12}$ and $I_{13}$ on the same line, so the senses of $\omega_2$ and $\omega_3$ are opposite: $\omega_3$ is **clockwise**. #### Velocity of C $$ v_C = \omega_4 \cdot CD = 2.916 \times 0.200 = 0.583\ \text{m/s} $$ perpendicular to CD (check: $v_C = \omega_3 \cdot I_{13}C$ gives the same value). **Answer: $\omega_3 = 2.578$ rad/s (clockwise), $\omega_4 = 2.916$ rad/s (anticlockwise), $v_C = 0.583$ m/s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where is I24 located using Kennedy's theorem? A. Intersection of AB and DC produced B. At pin C C. Intersection of BC and AD produced D. At pin D Answer: C. I24 lies on I23I34 (line BC) and on I12I14 (line AD). (Key point: Locating I24) 2. Which equation gives omega_4? A. omega_2 x I12I24 = omega_4 x I14I24 B. omega_2 x AB = omega_4 x BC C. omega_4 = omega_2 x AD/CD D. omega_2 x I13B = omega_4 x I24C Answer: A. I24 is common to links 2 and 4, so its velocity is the same on both: omega_2 x 123.5 = omega_4 x 423.5. (Key point: Rocker angular velocity) 3. Why is omega_4 in the same sense as omega_2? A. I24 lies between I12 and I14 B. I24 lies outside A and D on the same side of both C. CD is shorter than AB D. The coupler is horizontal Answer: B. When the IC lies outside both centres on the same side, the two links rotate in the same sense. (Key point: Sense of omega_4) 4. How is omega_3 of the coupler found? A. omega_3 = v_C/BC B. omega_3 = omega_2 x AB/BC C. omega_3 = omega_4 x CD/AD D. omega_3 = v_B/I13B, giving 2.578 rad/s clockwise Answer: D. Link 3 rotates about I13, and B is at 387.8 mm from it with v_B = 1.0 m/s; B lies between I12 and I13 so the sense is opposite to the crank. 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