Velocity & Acceleration Analysis of Linkages

Velocity and acceleration analysis is a core GATE ME skill (~1–2 marks) and the Coriolis-acceleration term is a favourite trap. Master the relative-velocity method and instantaneous centres.

Velocity methods

Relative velocity: V_B = V_A + V_(B/A), where V_(B/A) ⊥ the link AB and has magnitude ω·AB. Build a velocity polygon.

Instantaneous centre (I-centre) method: each link has a point of zero velocity at any instant; V = ω·(distance from I-centre). Kennedy's theorem: the three I-centres of any three bodies in relative motion lie on a straight line. Number of I-centres = n(n − 1)/2 for n links.

Acceleration

Point acceleration has two parts along a link: centripetal a_c = ω²·r (toward centre) and tangential a_t = α·r (⊥ link). For a point moving along a rotating link (slider in a slotted lever), add the Coriolis acceleration a_cor = 2·ω·V, directed perpendicular to the sliding velocity in the sense of ω. Total acceleration is the vector sum in an acceleration polygon.

Exam Tricks & Tips

• 🎯 Coriolis acceleration = 2ωV appears only when a body slides along another rotating body (e.g. Whitworth, slotted-lever) — spot the sliding-on-rotating pair.
• 🎯 Number of I-centres = n(n−1)/2; for a four-bar (n=4) that's 6 — locate them using Kennedy's theorem (collinear triples).
• 🎯 Relative velocity V_(B/A) is always perpendicular to the link AB — set directions before magnitudes in the polygon.
• 🎯 Centripetal a_c = ω²r points from the point toward the fixed centre; tangential a_t = αr is perpendicular — never mix their directions.
• 🎯 Velocity of a point = ω × (distance from its link's I-centre) — a shortcut that avoids the full polygon.
• ❌ Common mistake: forgetting the Coriolis term when a slider moves in a rotating slot — omitting 2ωV gives a badly wrong acceleration.

Expected exam pattern

A 1-mark velocity/angular-velocity or "number of I-centres" NAT, and a 2-mark acceleration analysis, frequently featuring the Coriolis component in a quick-return or slotted-lever mechanism.

Quick recap

Relative velocity: V_B = V_A + ω·AB (⊥ AB). I-centres = n(n−1)/2, collinear by Kennedy's theorem. Acceleration: centripetal ω²r + tangential αr; add Coriolis 2ωV when a slider moves on a rotating link. Use polygons.

Cover the answer, recall, then check. 10 cards on linkage kinematics for GATE ME.

Q1. State the relative-velocity relation for two points on a rigid link.
A1. V_B = V_A + V_(B/A), where V_(B/A) is perpendicular to link AB with magnitude ω·AB.

Q2. How many instantaneous centres does an n-link mechanism have?
A2. n(n − 1)/2. For a four-bar mechanism (n = 4), that is 6.

Q3. State Kennedy's theorem.
A3. The three instantaneous centres of any three bodies in relative planar motion lie on one straight line.

Q4. What are the two acceleration components of a point on a rotating link?
A4. Centripetal a_c = ω²r (toward the centre) and tangential a_t = αr (perpendicular to the link).

Q5. When does the Coriolis acceleration appear?
A5. When a point slides along a rotating link (a slider in a rotating slot), e.g. Whitworth or crank-and-slotted-lever mechanisms.

Q6. Give the magnitude and direction of the Coriolis acceleration.
A6. Magnitude 2ωV (ω = link's angular velocity, V = sliding velocity); direction perpendicular to V, in the sense of ω rotated 90°.

Q7. How do you find a point's velocity using an I-centre?
A7. V = ω × (distance of the point from its link's instantaneous centre).

Q8. In a velocity polygon, what sets the direction of each relative-velocity vector?
A8. It is drawn perpendicular to the corresponding link; magnitude follows from ω·(link length).

Q9. Centripetal acceleration direction along a link?
A9. Always directed from the point toward the centre of rotation (radially inward).

Q10. A slider moves at 2 m/s in a slot on a link rotating at 5 rad/s. Coriolis acceleration?
A10. a_cor = 2ωV = 2 × 5 × 2 = 20 m/s².

Once a mechanism moves, designers need the velocity and acceleration of every point — for output speed, inertia forces and vibration. GATE tests the instantaneous-centre and relative-velocity methods, plus the often-forgotten Coriolis term.

Core concept: rigid-body motion means every point's velocity is a rotation about the body's instantaneous centre; relative-velocity and acceleration diagrams then propagate these through the linkage.

Deep explanation

Beginner — velocity basics

For a link rotating at angular velocity ω, a point at radius r has speed v = ωr (perpendicular to r). For two points A and B on the same rigid link:
v_B = v_A + v_(BA), where v_(BA) = ω × r_(BA) is perpendicular to AB with magnitude ω·AB.

Intermediate — instantaneous centre (I-centre) method

The instantaneous centre of a link is the point (possibly off the body) about which it appears to rotate at that instant. Kennedy's theorem: for three bodies in relative motion, their three I-centres are collinear. Number of I-centres = n(n−1)/2. Once located, v = ω × (distance from I-centre), giving output velocity ratios directly without a full diagram.

Advanced — acceleration and Coriolis

Acceleration of B relative to A on a rotating link has two parts:

• Centripetal: a_r = ω²·r (along BA, toward the centre),
• Tangential: a_t = α·r (perpendicular, α = angular acceleration).
  When a point slides along a rotating link (e.g. slotted-lever/quick-return), add the Coriolis acceleration:
  a_cor = 2 ω v,
  directed perpendicular to the sliding velocity, rotated in the direction of ω. Missing this term is the classic error in shaper/Whitworth problems.

Worked example

A slider moves at v = 2 m/s along a link rotating at ω = 5 rad/s. Find the Coriolis acceleration.
a_cor = 2 ω v = 2 × 5 × 2 = 20 m/s², directed perpendicular to the sliding velocity (sense given by rotating v by 90° in the direction of ω). This adds to the centripetal (ω²r) and tangential (αr) components in the full acceleration diagram.

GATE relevance

Velocity/acceleration of linkages is a core TOM topic. The Coriolis term (2ωv) is a favourite exam trap in quick-return mechanisms, and I-centre/Kennedy questions test whether you can find velocity ratios efficiently.

Exam tricks & shortcuts

• Use I-centres when only a velocity ratio is wanted — much faster than a full velocity polygon.
• Number of I-centres = n(n−1)/2; locate the rest with Kennedy's collinearity.
• Include 2ωv Coriolis whenever a slider moves along a rotating member.
• Mnemonic: "Sliding on a spinning link? Add two-omega-vee."

Key formulas

• Velocity: v = ωr; relative velocity v_BA = v_B − v_A.
• Acceleration: centripetal a_c = ω²r (toward centre); tangential a_t = αr.
• Coriolis acceleration: a_cor = 2ωv (slider on rotating link).
• Instantaneous centre: point of zero velocity; number = n(n−1)/2 (Kennedy's theorem).

Velocity/acceleration polygons or the instantaneous-centre method analyse linkage motion; Coriolis appears with sliding on rotating links.

Worked Example

Problem: In a slider-crank mechanism the crank radius is r = 100 mm rotating at ω = 100 rad/s, and the connecting rod length is L = 400 mm (so ratio n = L/r = 4). Find the piston (slider) velocity when the crank angle is θ = 45° from the inner dead centre.

Solution:
The piston velocity for a slider-crank is given (to the standard second-order approximation) by:
V_p = r·ω·[ sinθ + (sin2θ)/(2n) ].

Evaluate the trigonometric terms at θ = 45°:
sinθ = sin45° = 0.7071,
sin2θ = sin90° = 1.

Substitute r = 0.1 m, ω = 100 rad/s, n = 4:
V_p = (0.1)(100)·[0.7071 + 1/(2×4)]
= 10·[0.7071 + 0.125]
= 10 × 0.8321 = 8.32 m/s.

Answer: The piston velocity at θ = 45° is approximately 8.32 m/s.