Velocity and acceleration analysis; instantaneous centres

Why it matters

Inertia forces, bearing loads and the torque a mechanism needs all depend on the accelerations of its links, and accelerations come only after velocities are known. Velocity and acceleration analysis is therefore the bridge between the geometry of a mechanism and its dynamics: piston speed, connecting-rod swing, rubbing velocity at a bearing and the inertia force on a wiper arm all come from it.

Key ideas

Relative velocity on a rigid link. For two points A and B on the same rigid link, the distance AB cannot change, so the velocity of B relative to A is perpendicular to AB and equal in magnitude to ω·AB, where ω is the angular velocity of the link. Velocities are vectors: v_B = v_A + v_BA is a vector sum, and you may add magnitudes only when the vectors are parallel.

Velocity polygon (diagram). Start from a pole o that represents all fixed points. Draw known velocities to scale, then draw lines of known direction (perpendicular to links, along slider guides) until they intersect. The polygon gives every absolute and relative velocity in the mechanism. The velocity image of a link is similar to the link itself, which gives the velocity of any third point on it (for example the centre of mass of a connecting rod).

Instantaneous centre (IC). At any instant, a link in plane motion moves as if it were rotating about one point, its instantaneous centre with respect to the frame. That point has zero velocity at that instant, and every other point on the link moves perpendicular to the line joining it to the IC with speed ω times that distance. Its acceleration is not zero, so the IC cannot be used for acceleration analysis. If the IC is at infinity, the link is in instantaneous translation (ω = 0) and all its points have the same velocity.

Locating ICs.

• A pin joint is the IC of the two links it joins.
• A slider on a straight guide: the IC lies at infinity perpendicular to the guide.
• Pure rolling contact: the IC is the contact point.
• Sliding contact (cam, gear tooth): the IC lies on the common normal at the contact point.
• Kennedy's (three-centres) theorem: the three ICs of any three links lie on one straight line. Use it, with a circle diagram, to find the ICs that are not obvious.
• A mechanism with n links has N = n(n − 1)/2 ICs; a four-bar has 6, of which 4 are at the pins (primary) and 2 are found by Kennedy's theorem.

Acceleration on a rigid link. The acceleration of B relative to A has two components:

• centripetal (radial), ω²·AB, directed from B towards A;
• tangential, α·AB, perpendicular to AB, where α is the angular acceleration of the link. An acceleration polygon is built the same way as the velocity polygon. When a point slides along a rotating link, a third term, the Coriolis component 2·ω·v, appears; that is the next topic.

Rubbing velocity. At a pin joining links 1 and 2 of pin radius r_p, the rubbing velocity is r_p·(ω₁ − ω₂) if they turn in the same sense and r_p·(ω₁ + ω₂) if they turn in opposite senses. It sets friction power loss and wear at the bearing.

Slider-crank results. For crank radius r, rod length l, n = l/r, crank speed ω (constant) and crank angle θ from inner dead centre, the connecting rod swings with angular velocity ω_c and angular acceleration α_c given below. At θ = 90° the rod has zero angular velocity but maximum angular acceleration.

Formulas

v_BA = ω·AB (perpendicular to AB)

• v_BA: velocity of B relative to A on one rigid link (m/s); ω: angular velocity of the link (rad/s); AB: distance (m).

v_P = ω·IP

• v_P: velocity of any point P of a link (m/s); IP: distance from the link's instantaneous centre I to P (m); ω: link angular velocity (rad/s). Velocities only.

a_BA = √[(ω²·AB)² + (α·AB)²], with a_n = ω²·AB and a_t = α·AB

• a_n: centripetal component towards A (m/s²); a_t: tangential component perpendicular to AB (m/s²); α: link angular acceleration (rad/s²).

N = n(n − 1) / 2

• N: number of instantaneous centres; n: number of links (including the frame).

ω_c = ω·cos θ / √(n² − sin²θ) and α_c = −ω²·sin θ·(n² − 1) / (n² − sin²θ)^(3/2)

• ω_c, α_c: angular velocity (rad/s) and angular acceleration (rad/s²) of the connecting rod; ω: constant crank speed (rad/s); n = l/r. Exact for a slider-crank with constant crank speed; the minus sign shows that α_c opposes the direction of ω_c growth near θ = 90°.

v_rub = r_p·(ω₁ ± ω₂)

• v_rub: rubbing velocity at a pin (m/s); r_p: pin radius (m); use + when the links rotate in opposite senses, − when in the same sense.

Worked examples

Example 1 (standard: IC method on a slider-crank). Crank r = 100 mm, rod l = 400 mm, crank speed 20 rad/s anticlockwise, θ = 45° from inner dead centre. Find the rod angular velocity, the piston velocity and the rubbing velocity at a 50 mm crank pin.

1. Put the crank centre O at the origin and the line of stroke along x. Crank pin A = (r cos θ, r sin θ) = (70.71, 70.71) mm.
2. Piston pin B on the x-axis: x_B = 70.71 + √(400² − 70.71²) = 70.71 + 393.70 = 464.41 mm.
3. IC of the rod (I₁₃): intersection of the crank line OA extended with the perpendicular to the line of stroke through B. Since OA is at 45°, I = (464.41, 464.41) mm.
4. IA = √[(464.41 − 70.71)² + (464.41 − 70.71)²] = 556.78 mm; IB = 464.41 mm.
5. v_A = ω·r = 20 × 0.1 = 2.0 m/s.
6. ω_c = v_A / IA = 2.0 / 0.55678 = 3.59 rad/s (clockwise, opposite to the crank).
7. v_B = ω_c·IB = 3.592 × 0.46441 = 1.67 m/s.
8. Check: ω_c = 20 × cos 45° / √(16 − 0.5) = 14.142 / 3.937 = 3.59 rad/s.
9. Crank and rod turn in opposite senses, so v_rub = 0.025 × (20 + 3.59) = 0.59 m/s.

Example 2 (GATE level: accelerations at θ = 90°). Same engine (r = 0.1 m, l = 0.4 m, ω = 20 rad/s constant). Find the angular acceleration of the rod and the piston acceleration when θ = 90°.

1. n = 4, sin θ = 1, cos θ = 0, so ω_c = 0: the rod is momentarily not rotating.
2. |α_c| = ω²·(n² − 1) / (n² − 1)^(3/2) = ω² / √(n² − 1) = 400 / √15 = 103.3 rad/s².
3. Crank pin acceleration: only centripetal, ω²r = 400 × 0.1 = 40 m/s² towards O (perpendicular to the line of stroke).
4. Piston acceleration from the exact geometry, x = r cos θ + √(l² − r² sin²θ), gives at θ = 90°: |a_P| = ω²r² / √(l² − r²) = 400 × 0.01 / √0.15 = 4 / 0.3873 = 10.33 m/s², directed away from the crank centre.
5. The approximate formula a_P = ω²r(cos θ + cos 2θ / n) gives ω²r(0 − 1/4) = −10.0 m/s², the same direction within 3%.

Example 3 (quick IC count). A six-link Stephenson chain has N = 6 × 5 / 2 = 15 instantaneous centres.

Common mistakes

• Adding velocity magnitudes (v_B = v_A + ωr) when the vectors are not parallel. Add vectors, or use a polygon.
• Using the instantaneous centre to find accelerations. The IC has zero velocity, not zero acceleration.
• Drawing the centripetal component the wrong way. It always points from the moving point towards the reference point on the same link.
• Forgetting that a slider's IC with the frame is at infinity perpendicular to the guide, not at the slider.
• Using v = ωr for the piston at a general crank angle. The obliquity term is missing.
• Rubbing velocity: using ω₁ − ω₂ when the links rotate in opposite senses.

For GATE ME

Questions ask for the number of ICs, the location of a specific IC (Kennedy's theorem), the angular velocity of a coupler or connecting rod from the IC method, the velocity of a slider at a given crank angle, rubbing velocity at a pin, and the magnitude of an acceleration from its centripetal and tangential parts. Practise the slider-crank and four-bar at a few positions with both the IC method and the polygon, and check one against the other.

Quick check

1. How many ICs does a four-bar mechanism have?
2. Where is the IC of a slider with respect to its straight guide?
3. Point B on a link 0.5 m from fixed pivot A; ω = 4 rad/s, α = 2 rad/s². Magnitude of the acceleration of B?
4. At which crank angle is the connecting-rod angular velocity zero?
5. Can the IC be used to find accelerations?

Answers: 1. Six. 2. At infinity, perpendicular to the guide. 3. √(8² + 1²) = 8.06 m/s². 4. At θ = 90° and 270°. 5. No, only velocities.