Velocity and acceleration analysis of planar mechanisms

Relative velocity and acceleration, instantaneous centres and Kennedy's theorem, the Coriolis component and exact and approximate slider-crank results.

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Why it matters

Inertia forces, bearing loads, cam and gear tooth loads, and the turning-moment diagram of an engine all start from the accelerations of moving links. A designer who gets the velocity and acceleration of a piston or a slotted lever wrong gets every downstream force wrong. In production machines (shapers, presses, packaging linkages) this analysis also decides cutting speed and cycle time.

Key ideas

Relative velocity on a rigid link. For two points A and B on the same rigid link, the distance AB never changes, so B can only move relative to A at right angles to AB: v_BA = ω·AB, perpendicular to AB. The absolute velocity is the vector sum v_B = v_A + v_BA. A velocity polygon (diagram) draws these vectors from a common pole: fixed points sit at the pole, and the velocity image of each link is similar to the link itself.

Instantaneous centre (I-centre). At any instant a link in plane motion turns about one point that has zero velocity relative to the reference link. Every point's velocity is then v = ω × (distance from the I-centre), perpendicular to the line joining the point to the I-centre.

  • A pin joint is the I-centre of the two links it joins.
  • A slider on a straight guide has its I-centre at infinity, perpendicular to the guide.
  • Pure rolling: the I-centre is the contact point.
  • Kennedy's theorem (three-centres): the three I-centres of any three links lie on a straight line.
  • A mechanism of n links has N = n(n − 1)/2 I-centres. I-centres give velocities quickly but not accelerations: the I-centre generally has a non-zero acceleration.

Relative acceleration on a rigid link. B relative to A has two components:

  • centripetal (radial) a^r_BA = ω²·AB = v_BA²/AB, directed from B towards A,
  • tangential a^t_BA = α·AB, perpendicular to AB. The acceleration polygon adds these to a_A. A crank turning at constant speed has only the centripetal component at its pin.

Coriolis component. When a point (a slider block) moves along a link that is itself rotating, its acceleration has an extra term 2ω·v_rel, perpendicular to the slot. Its direction is found by rotating the relative-velocity vector by 90° in the sense of ω. It appears in the crank and slotted lever, Whitworth quick-return and oscillating-cylinder mechanisms, and it is absent when the slot does not rotate (Scotch yoke, ordinary slider on a fixed guide).

Slider-crank analytical results. With crank radius r, rod length l, ratio n = l/r, crank angle θ from inner dead centre and rod angle φ (sin φ = sin θ / n), the exact and approximate piston results below are what GATE expects; the approximate forms drop terms of order 1/n³.

Formulas

v = ω·r — v velocity of a point (m/s), ω angular velocity (rad/s), r distance from the centre of rotation or I-centre (m).

ω = 2πN/60 — N in rev/min, ω in rad/s.

a_r = ω²·r = v²/r and a_t = α·r — α angular acceleration (rad/s²); total a = √(a_r² + a_t²).

a_c = 2·ω·v_rel — Coriolis acceleration (m/s²), ω angular velocity of the guide/slot (rad/s), v_rel sliding velocity along it (m/s).

v_P = ω·r·sin(θ + φ)/cos φ ≈ ω·r·(sin θ + sin 2θ/(2n)) — piston velocity (m/s).

a_P ≈ ω²·r·(cos θ + cos 2θ/n) — piston acceleration (m/s²), crank at constant ω.

ω_c = ω·cos θ/√(n² − sin²θ) — angular velocity of the connecting rod (rad/s).

N = n(n − 1)/2 — number of instantaneous centres of an n-link mechanism.

Worked examples

Example 1 (standard). A slider-crank engine has crank r = 0.1 m, connecting rod l = 0.4 m and turns at N = 300 rev/min. For θ = 30° from inner dead centre, find the piston velocity, the angular velocity of the rod and the piston acceleration.

  1. ω = 2πN/60 = 2π × 300/60 = 31.42 rad/s; n = l/r = 4.
  2. sin φ = sin 30°/4 = 0.125 → φ = 7.18°.
  3. v_P = ω·r·sin(θ + φ)/cos φ = 31.42 × 0.1 × sin 37.18°/cos 7.18° = 1.914 m/s. Check with the approximate form: 31.42 × 0.1 × (0.5 + 0.866/8) = 1.911 m/s, within 0.2 %.
  4. ω_c = ω·cos θ/√(n² − sin²θ) = 31.42 × 0.866/√(16 − 0.25) = 6.86 rad/s.
  5. a_P ≈ ω²·r·(cos θ + cos 2θ/n) = 987.0 × 0.1 × (0.866 + 0.5/4) = 97.8 m/s².
  6. v_P ≈ 1.91 m/s, ω_c ≈ 6.86 rad/s, a_P ≈ 97.8 m/s²

Example 2 (GATE level). In a crank and slotted lever mechanism, at the instant shown the slotted lever turns at ω = 4 rad/s counter-clockwise with zero angular acceleration. The sliding block is 0.3 m from the lever pivot and moves outward along the slot at 1.5 m/s, with no relative acceleration along the slot. Find the Coriolis component and the total acceleration of the block.

  1. Coriolis: a_c = 2ωv_rel = 2 × 4 × 1.5 = 12 m/s², perpendicular to the slot. Rotating the outward velocity 90° counter-clockwise gives its direction: along the direction the lever is turning.
  2. Centripetal: a_r = ω²·r = 4² × 0.3 = 4.8 m/s², along the slot towards the pivot.
  3. Tangential from α: α·r = 0. Relative (sliding) acceleration: 0.
  4. The two non-zero components are perpendicular: a = √(12² + 4.8²) = 12.92 m/s².
  5. a_c = 12 m/s²; total acceleration of the block ≈ 12.9 m/s²

Common mistakes

  • Using v = ω·r for the piston: the piston speed equals the crank-pin speed only when the crank is at 90° to the line of stroke, or approximately when it is perpendicular to the rod.
  • Drawing the centripetal component away from the centre; it always points from the moving point towards the reference point.
  • Forgetting the Coriolis term in slotted-lever problems, or adding it when the guide does not rotate.
  • Using the I-centre to find accelerations; the I-centre has acceleration.
  • Using rev/min in place of rad/s.
  • Adding perpendicular acceleration components arithmetically instead of as vectors.

For GATE PI

Typical questions give a slider-crank or four-bar position and ask for a link angular velocity, the piston velocity or acceleration, the number of I-centres, or the Coriolis component in a quick-return drive. Practise locating I-centres with Kennedy's theorem, the exact and approximate piston formulas, and the direction rule for 2ωv.

Quick check

  1. How many instantaneous centres does a six-link mechanism have?
  2. A crank 50 mm long turns at 120 rev/min. What is the crank-pin velocity?
  3. A block slides at 2 m/s along a link turning at 5 rad/s. What is the Coriolis acceleration?
  4. Where is the I-centre of a slider and its straight fixed guide?

Answers: 1. 15; 2. 0.628 m/s; 3. 20 m/s²; 4. at infinity, perpendicular to the guide.

Interview questions

All Theory of Machines and Machine Design interview questions

Try answering each one aloud before you open it.

  1. 1.What is a planar mechanism in the context of machine design?Concept

    A planar mechanism is a mechanical system where all the components move in parallel planes. This means that the motion of the parts is confined to two dimensions, typically within a single plane. Planar mechanisms are commonly used in machinery where the motion is restricted to a flat surface, such as in linkages and cams.

  2. 2.Explain the difference between velocity and acceleration analysis in planar mechanisms.Concept

    Velocity analysis finds the linear velocity of points and the angular velocity of links at one position, using relative velocity (v_BA = ω·AB, perpendicular to AB) or instantaneous centres. Acceleration analysis needs those velocities first, because each relative acceleration has a centripetal part ω²·AB towards the reference point, a tangential part α·AB, and where a block slides on a rotating link a Coriolis part 2ωv. Accelerations are what you need for inertia forces, bearing loads and the turning-moment diagram.

  3. 3.Why is the relative velocity method used in the analysis of planar mechanisms?Application

    Because two points on a rigid link keep a fixed distance, the velocity of one relative to the other must be perpendicular to the line joining them with magnitude ω times that distance. That one fact lets you build a velocity polygon link by link, starting from the known crank velocity, for any linkage. Unlike the instantaneous-centre method, the same polygon extends directly to the acceleration polygon.

  4. 4.What happens if the acceleration of a component in a planar mechanism is not properly accounted for?Application

    The inertia forces m·a and inertia couples I·α are then wrong, so the bearing reactions, crank torque, shaking forces and stresses used for design are wrong. At high speed inertia loads often exceed the working load, as in the reciprocating parts of an engine, so underestimating them can cause fatigue failure, vibration or loss of contact between a cam and its follower.

  5. 5.Explain how Coriolis acceleration is relevant in the analysis of planar mechanisms.Concept

    When a point slides along a link that is itself rotating, its acceleration includes a Coriolis component of magnitude 2ωv, where ω is the angular velocity of the link and v the sliding velocity. It acts perpendicular to the link, in the direction obtained by rotating the sliding velocity 90° in the sense of ω. It appears in the crank and slotted lever, Whitworth quick-return and oscillating-cylinder mechanisms; leaving it out gives the wrong angular acceleration of the slotted link.

  6. 6.How does the choice of reference frame affect the velocity analysis of a planar mechanism?Application

    Absolute velocities are measured relative to the fixed frame, and relative velocities relative to another moving link; the velocity polygon uses both, since v_B = v_A + v_BA. Velocities between two links do not depend on which link is fixed, which is why inversions keep the same relative motion. If the reference link itself rotates, as with a slotted lever, accelerations measured relative to it need the extra Coriolis term.

  7. 7.What is the significance of the instantaneous centre of rotation in planar mechanisms?Concept

    At any instant a link in plane motion turns about a point, its instantaneous centre, whose velocity relative to the reference link is zero; every point then moves at ω times its distance from that centre, perpendicular to the line joining them. An n-link mechanism has n(n − 1)/2 such centres, located with Kennedy's theorem that any three lie on a straight line. The method gives velocities quickly but not accelerations, because the instantaneous centre itself accelerates.

  8. 8.The input crank of a four-bar linkage turns at 2 rad/s about its fixed pivot. What is the velocity of a point on the crank 0.5 m from the pivot?Numerical

    A point on a link turning about a fixed pivot moves on a circle, so v = ω·r = 2 × 0.5 = 1 m/s, perpendicular to the line from the pivot to the point. This applies only to links with a fixed pivot; for a point on the coupler you need the relative velocity method or the coupler's instantaneous centre.

  9. 9.A slider-crank mechanism has a 0.2 m crank turning at 3 rad/s. What is the slider velocity when the crank is at 90° to the line of stroke?Numerical

    In general the slider velocity is v = ω·r·sin(θ + φ)/cos φ, where φ is the rod angle. At θ = 90°, sin(90° + φ) = cos φ, so v = ω·r = 3 × 0.2 = 0.6 m/s exactly. At other crank angles the slider is slower or faster than the crank pin, so v = ω·r must not be used blindly.

  10. 10.What are common errors in velocity and acceleration analysis of planar mechanisms, and how do you avoid them?Application

    Typical errors are drawing the centripetal component away from the centre instead of towards it, leaving out the Coriolis component in a slotted lever, using the instantaneous centre for accelerations, using rev/min instead of rad/s, and adding perpendicular components arithmetically. Avoid them by writing every term as a vector with its direction, checking units, and cross-checking a velocity polygon result with the instantaneous-centre method or the analytical slider-crank formula.

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