Velocity and acceleration analysis; instantaneous centres

Why it matters

Velocities decide the speed of a tool, the slip at a joint and the power through a linkage; accelerations decide the inertia forces that load bearings, shake frames and size motors. Every dynamic force analysis, cam design or balancing calculation starts from a correct velocity and acceleration analysis, so this is the working core of kinematics.

Key ideas

Relative velocity on one rigid link. For two points A and B on the same rigid link, the velocity of B relative to A is perpendicular to AB with magnitude ω·AB, where ω is the link's angular velocity. Velocities are vectors: v_B = v_A + v_BA, and you add them as vectors, never as plain numbers.

Velocity diagram (polygon). Starting from a fixed point (zero velocity), draw each known velocity to scale and close the loop with directions you know (perpendicular to links, or along a slider guide). The triangle formed by the velocity images of three points on a link is similar to the link itself (velocity image), so the velocity of any extra point on the link is found by proportion.

Instantaneous centre (I-centre). The point, on the body or its extension, about which a body is purely rotating at that instant relative to another body; at that point the two bodies have the same velocity. For a body moving relative to the frame, its I-centre has zero velocity, and every point's velocity is v = ω·(distance to the I-centre), perpendicular to the line joining them.

• A pin joint is the I-centre of the two links it joins.
• A slider on a straight guide: the I-centre lies at infinity, perpendicular to the guide.
• Pure rolling contact: the contact point.
• Rolling with sliding (cam, gear teeth): on the common normal at the contact.
• If the I-centre of a link is at infinity, the link is in instantaneous translation (ω = 0, all points same velocity).
• The I-centre moves as the mechanism moves; it gives velocities but not accelerations (its own acceleration is generally not zero).

Kennedy's (three-centres) theorem. The three I-centres of any three bodies in plane motion lie on one straight line. With the obvious centres at pins and sliders, it locates the rest (use a circle diagram to keep track).

Angular velocity ratio theorem. For links 2 and 4 moving relative to frame 1: ω₄ / ω₂ = I₁₂I₂₄ / I₁₄I₂₄. If I₂₄ lies between I₁₂ and I₁₄ the links turn in opposite senses; outside, the same sense.

Acceleration of a point on a link. Relative to another point on the same link, the acceleration has two parts:

• centripetal (radial) ω²·r, directed from the point towards the reference point (along the link);
• tangential α·r, perpendicular to the link. A point sliding along a rotating link also has a Coriolis component (next topic). Acceleration diagrams are drawn after the velocity diagram, because the centripetal terms need the velocities.

Slider-crank analytical results. For an engine with crank radius r, rod length l and n = l/r, the closed-form approximations below are standard (Klein's construction gives the same results graphically).

Formulas

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

N = n(n − 1) / 2 (number of I-centres for n links)

ω₄ / ω₂ = I₁₂I₂₄ / I₁₄I₂₄ (angular velocity ratio theorem)

a_r = ω²·r = v² / r (centripetal, m/s²) ; a_t = α·r (tangential, m/s²) ; a = √(a_r² + a_t²)

v_rub = (ω₁ ± ω₂)·r_pin (rubbing velocity at a pin; minus if the links turn the same way, plus if opposite)

Slider-crank (θ from the dead centre where the slider is farthest from the crank axis, ω constant):

• v_P ≈ ω·r·(sin θ + sin 2θ / (2n)) (piston velocity, m/s)
• a_P ≈ ω²·r·(cos θ + cos 2θ / n) (piston acceleration, m/s²)
• ω_CR = ω·cos θ / √(n² − sin² θ) (connecting-rod angular velocity, rad/s)
• α_CR = −ω²·sin θ·(n² − 1) / (n² − sin² θ)^(3/2) (connecting-rod angular acceleration, rad/s²)

Worked examples

Example 1 (standard): engine slider-crank. Given: r = 150 mm, l = 600 mm, crank speed 300 rpm (constant), θ = 45°.

1. ω = 2πN / 60 = 2π × 300 / 60 = 31.42 rad/s ; n = 600 / 150 = 4.
2. v_P ≈ ω·r·(sin θ + sin 2θ / (2n)) = 31.42 × 0.15 × (0.7071 + 1/8) = 4.712 × 0.8321 = 3.92 m/s (the exact relation gives 3.93 m/s).
3. a_P ≈ ω²·r·(cos θ + cos 2θ / n) = 987.0 × 0.15 × (0.7071 + 0) = 104.7 m/s²
4. ω_CR = ω·cos θ / √(n² − sin² θ) = 31.42 × 0.7071 / √(16 − 0.5) = 22.21 / 3.937 = 5.64 rad/s Answer: v_P ≈ 3.92 m/s, a_P ≈ 104.7 m/s², ω_CR ≈ 5.64 rad/s.

Example 2 (GATE level): four-bar by I-centres, checked by relative velocity. Given (coordinates in mm): A(0, 0) and D(400, 0) fixed; crank AB = 100 mm vertical, B(0, 100); coupler BC = 300 mm horizontal, C(300, 100); rocker CD = √(100² + 100²) = 141.4 mm. Crank turns at ω₂ = 10 rad/s counter-clockwise. Find ω₃, v_C and ω₄.

1. v_B = ω₂·AB = 10 × 0.1 = 1.0 m/s, perpendicular to AB (towards −x).
2. I₁₃ (coupler about frame) lies where line AB (x = 0) meets line DC. Line DC runs from (400, 0) through (300, 100), so at x = 0 it reaches y = 400: I₁₃ = (0, 400).
3. ω₃ = v_B / I₁₃B = 1.0 / 0.300 = 3.33 rad/s (clockwise).
4. I₁₃C = √(300² + 300²) = 424.3 mm → v_C = ω₃·I₁₃C = 3.333 × 0.4243 = 1.414 m/s.
5. ω₄ = v_C / CD = 1.414 / 0.1414 = 10.0 rad/s (counter-clockwise).
6. Check: I₂₄ lies where line AD (y = 0) meets line BC (y = 100); they are parallel, so I₂₄ is at infinity and ω₄ / ω₂ = 1 by the ratio theorem.
7. Check by vectors: v_C = v_B + ω₃ × BC = (−1, 0.3ω₃) and v_C = ω₄ × DC = ω₄(−0.1, −0.1). So ω₄ = 10 rad/s and ω₃ = −3.33 rad/s; v_C = (−1, −1) m/s, magnitude 1.414 m/s. Answer: ω₃ = 3.33 rad/s (CW), v_C = 1.41 m/s, ω₄ = 10 rad/s (CCW).

Common mistakes

• Adding velocity magnitudes as scalars (v_B = v_A + ω·r) when the vectors are not parallel.
• Using the I-centre for accelerations; it is valid for velocities only.
• Drawing the centripetal component away from the centre; it always points towards the reference point.
• Forgetting the centripetal term on a link whose ω is not zero even when α = 0.
• Counting I-centres as n(n − 1) instead of n(n − 1)/2.
• Using rpm directly in ω²r; convert to rad/s first.

For GATE ME

Expect: number and location of I-centres; velocity of a slider or a coupler point using I-centres (often a ladder or a rolling disc); angular velocity ratio of four-bar links; piston velocity or acceleration from the slider-crank approximations; rubbing velocity at a pin. Practise drawing I-centres quickly with Kennedy's theorem, and solving four-bars with the vector loop.

Quick check

1. How many I-centres does a six-link mechanism have?
2. Where is the I-centre of a slider on a fixed straight guide relative to the frame?
3. A link 0.5 m long turns at 4 rad/s with α = 0. What is the centripetal acceleration of one end relative to the other?
4. Two links joined by a 20 mm radius pin turn at 3 rad/s CW and 5 rad/s CCW. Rubbing velocity?
5. At θ = 90° in a slider-crank, what is v_P in terms of ω and r?

Answers: 1. 15. 2. At infinity, perpendicular to the guide. 3. 4² × 0.5 = 8 m/s². 4. (3 + 5) × 0.02 = 0.16 m/s. 5. v_P = ω·r.