--- title: "Acceleration analysis of slider-crank" subject: "Engineering Mechanics" chapter: 5 key: "semester_3/engineering_mechanics:crank-slider-acceleration-numerical" source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/crank-slider-acceleration-numerical site: BCT Engineering --- # Acceleration analysis of slider-crank Engineering Mechanics · Chapter 5: Kinematics of Rigid Bodies - Asked in: year not given, 8 marks - Video: [Lecture 2.5: Acceleration diagram for slider crank mechanism](https://www.youtube.com/watch?v=o4D1Lw3igMs) — icgindia - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/crank-slider-acceleration-numerical - Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-5 ## Question A crank OA of length 150 mm rotates counter-clockwise at a constant angular speed of 20 rad/s. The connecting rod AB is 600 mm long and the slider B moves in a horizontal guide through O. When the crank is at 45° to the line of stroke (crank pin above the line), find the acceleration of the slider B and the angular acceleration of the connecting rod. ## Short answer With constant crank ω = 20 rad/s, v_A = 3 m/s, ω_AB = 3.59 rad/s clockwise. Relative acceleration method gives a_B = 42.67 m/s² toward O and angular acceleration of AB = 69.52 rad/s² counter-clockwise. ## Answer **Given:** $r = OA = 0.15$ m, $l = AB = 0.6$ m, $\omega_{OA} = 20$ rad/s (constant, counter-clockwise), $\theta = 45^\circ$. #### Geometry $\sin\phi = \dfrac{0.15\sin45^\circ}{0.6} = 0.1768$, $\phi = 10.18^\circ$. Horizontal projection of AB: $d = l\cos\phi = 0.5906$ m. Vertical drop from A to B: $0.15\sin45^\circ = 0.1061$ m. #### Velocity analysis $v_A = \omega r = 20 \times 0.15 = 3$ m/s, perpendicular to OA: $\mathbf{v}_A = (-2.121,\ 2.121)$ m/s. B moves horizontally, so the vertical components give $$ \omega_{AB} = \frac{2.121}{0.5906} = 3.592\ \text{rad/s (clockwise)} $$ #### Acceleration analysis Since $\omega_{OA}$ is constant, A has only a normal acceleration toward O: $$ a_A = \omega^2 r = 20^2 \times 0.15 = 60\ \text{m/s}^2,\quad \mathbf{a}_A = (-42.43,\ -42.43)\ \text{m/s}^2 $$ Relative acceleration of B with respect to A has: - normal part $\omega_{AB}^2\,l = (3.592)^2(0.6) = 7.74$ m/s² directed from B toward A, - tangential part $\alpha_{AB}\,l$ perpendicular to AB. Resolve $\mathbf{a}_B = \mathbf{a}_A + \mathbf{a}^n_{B/A} + \mathbf{a}^t_{B/A}$ with B having no vertical acceleration. The vector from A to B is $(d,\ -0.1061)$ m. Vertical components: $$ 0 = a_{Ay} + \alpha_{AB}\,d + \omega_{AB}^2(0.1061) \text{ with } a_{Ay} = -42.43 \;\Rightarrow\; \alpha_{AB} = \frac{42.43 - \omega_{AB}^2(0.1061)}{d} $$ $$ \alpha_{AB} = \frac{42.43 - (12.903)(0.1061)}{0.5906} = 69.52\ \text{rad/s}^2\ \text{(counter-clockwise)} $$ Horizontal direction: $$ a_B = a_{Ax} + \alpha_{AB}(0.1061) - \omega_{AB}^2\,d $$ $$ a_B = -42.43 + (69.52)(0.1061) - (12.903)(0.5906) = -42.67\ \text{m/s}^2 $$ The slider's acceleration is directed toward the crank centre O. **Check (approximate formula):** $a_B \approx -\omega^2 r\left(\cos\theta + \dfrac{\cos2\theta}{n}\right)$ with $n = l/r = 4$ gives $-60(0.7071 + 0) = -42.4$ m/s², which agrees closely. **Answer: a_B = 42.67 m/s² toward O; α_AB = 69.52 rad/s² counter-clockwise.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The crank turns at constant speed, so A has: A. Only normal acceleration omega^2 r = 60 m/s^2 toward O B. Only tangential acceleration C. Zero acceleration D. Normal and tangential acceleration Answer: A. With omega constant there is no tangential part. (Key point: Acceleration of A) 2. The rod's angular velocity is found by equating vertical components, giving about: A. 20 rad/s B. 3.59 rad/s clockwise C. 10 rad/s D. 0 Answer: B. 2.121 / 0.5906 = 3.592 rad/s. (Key point: Rod angular velocity) 3. The relative acceleration of B with respect to A has: A. Normal part omega_AB^2 l toward A and tangential part alpha_AB l perpendicular to AB B. Only a tangential part C. Only a normal part along the guide D. Normal part alpha l Answer: A. B moves relative to A on a circle of radius l about A. (Key point: Relative acceleration terms) 4. Which constraint determines alpha_AB? A. B has no vertical acceleration B. B has no horizontal acceleration C. A has no acceleration D. The rod has no angular velocity Answer: A. B moves in a horizontal guide, so the vertical components sum to zero; then a_B is found from the horizontal components (about 42.7 m/s^2 toward O). (Key point: Using the constraint) ## Related in Engineering Mechanics - [Velocity analysis of slider-crank mechanism](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Acrank-slider-velocity-numerical) — Engineering Mechanics, chapter 5 - [Rod released from horizontal position](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-release-from-horizontal-reaction) — Engineering Mechanics, chapter 6 ## Related in other subjects - [Crank OA 12 rad/s, piston B, constrained motion](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acrank-oa-12rads-piston-constrained-motion) — Applied Mechanics (Dynamics), chapter 4 - [Crank AB 5 rad/s, piston velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acrank-5rads-piston-velocity-coriolis) — Applied Mechanics (Dynamics), chapter 4 - [Slider crank velocity by instantaneous centre method](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aslider-crank-velocity-instantaneous-centre) — Theory of Mechanism and Machine I, chapter 7 - [Engine crank 1800 rpm, connecting rod velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aengine-crank-1800rpm-rod-velocity) — Applied Mechanics (Dynamics), chapter 4