--- title: "Instantaneous centres and Kennedy's theorem" subject: "Theory of Mechanism and Machine I" chapter: 7 key: "semester_6/theory_of_mechanism_and_machine_i:kennedy-theorem-instantaneous-centres-four-bar" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/kennedy-theorem-instantaneous-centres-four-bar site: BCT Engineering --- # Instantaneous centres and Kennedy's theorem Theory of Mechanism and Machine I · Chapter 7: Kinematic Analysis of Mechanisms - Asked in: year not given, 8 marks - Video: [2.3. Kennedy's Theorem | Instantaneous Centre Method | Velocity Analysis | KOM | TOM](https://www.youtube.com/watch?v=PuFkgd_s5W4) — MKS TUTORIALS by Manoj Sir - 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/kennedy-theorem-instantaneous-centres-four-bar - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-7 ## Question Define instantaneous centre of rotation. Write the formula for the number of instantaneous centres in a mechanism and classify them. State and prove Kennedy's theorem and use it to locate all instantaneous centres of a four-bar mechanism. ## Short answer Instantaneous centre: point common to two links with zero relative velocity. Number N = n(n-1)/2; for four-bar, 6 ICs. Types: fixed, permanent, neither. Kennedy's theorem: three bodies' ICs are collinear, proved by requiring equal velocity direction at the IC. In a four-bar ABCD, I13 lies at intersection of AB and CD; I24 at intersection of BC and AD. ## Answer #### Instantaneous centre (IC) For two bodies in relative plane motion, the **instantaneous centre** is the point common to both that has no relative velocity at that instant. Each body appears to rotate about it relative to the other. It is found at the intersection of the perpendiculars to the velocity directions of two points. #### Number of ICs For a mechanism of $n$ links, each pair of links has one IC: $$ N = \frac{n(n-1)}{2} $$ A four-bar mechanism ($n = 4$) has $N = 6$ ICs. #### Types - **Fixed IC**: its location is fixed in space (for example, the pivot of a link on the frame). - **Permanent IC**: its location moves, but it is always at the pin joint between two moving links (for example, the pin between crank and coupler). - **Neither fixed nor permanent**: it is not at a joint and its position changes as the configuration changes, for example the IC between the coupler and the frame. #### Kennedy's theorem (three-centre theorem) *Any three bodies in relative plane motion have three instantaneous centres, and they lie on a straight line.* **Proof.** Let the three bodies be 1, 2 and 3, and take body 1 as fixed, so that body 2 rotates about $I_{12}$ and body 3 about $I_{13}$. Assume $I_{23}$ is not on the line joining $I_{12}$ and $I_{13}$, and call it point $P$ (a point of both body 2 and body 3). The velocity of $P$ as a point of body 2 is perpendicular to $I_{12}P$, and as a point of body 3 it is perpendicular to $I_{13}P$. These two directions are different (since the lines $I_{12}P$ and $I_{13}P$ are different), but at the IC $I_{23}$ the velocities of the two coincident points must be identical in magnitude and direction. This is possible only if $P$ lies on the line through $I_{12}$ and $I_{13}$, where the two perpendicular directions coincide. Hence all three ICs are collinear. #### Four-bar mechanism ABCD (links: 1 = AD frame, 2 = AB, 3 = BC, 4 = CD) **Obvious ICs (pin joints):** $I_{12}$ at A, $I_{23}$ at B, $I_{34}$ at C, $I_{14}$ at D. **Remaining two by Kennedy's theorem:** - $I_{13}$: lies on the line $I_{12}I_{23}$ (AB extended) and on $I_{14}I_{34}$ (DC extended). It is the intersection of AB and DC produced. - $I_{24}$: lies on $I_{23}I_{34}$ (BC extended) and on $I_{12}I_{14}$ (AD extended). It is the intersection of BC and AD produced. ```text I13 * \ . \ B---C \ / \ A-----------D ---- * I24 ``` Use: $\omega_2 \cdot I_{12}I_{24} = \omega_4 \cdot I_{14}I_{24}$ (both equal the speed of $I_{24}$), and $\omega_2 \cdot I_{12}I_{23} = \omega_3 \cdot I_{13}I_{23}$ (both equal the speed of the pin B). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many instantaneous centres does a four-bar mechanism have? A. 4 B. 8 C. 6 D. 12 Answer: C. N = n(n-1)/2 = 4x3/2 = 6. (Key point: Number of ICs) 2. What does Kennedy's theorem state? A. Three bodies in relative plane motion have three instantaneous centres lying on a straight line B. Any two bodies have two ICs C. The IC of three bodies is always at a pin joint D. All ICs of a mechanism lie on one circle Answer: A. Three bodies in relative plane motion have three ICs that are collinear. (Key point: Kennedy's theorem) 3. In four-bar ABCD (frame AD), where is I13? A. Intersection of BC and AD produced B. At pin B C. At the midpoint of BC D. Intersection of AB and DC produced Answer: D. I13 lies on I12I23 (line AB) and on I14I34 (line DC), so it is where AB and DC meet when produced. (Key point: Locating I13) 4. What is a permanent IC? A. One that is fixed in space B. One whose position moves but is a pin joint between two moving links C. One that exists only at dead centres D. One that is always at infinity Answer: B. A permanent IC changes place as the mechanism moves, but it is a pin joint between two moving links, like crank-coupler. 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