Velocity, Acceleration and Cams

Relative velocity method, instantaneous centres and Kennedy's theorem, Coriolis acceleration, slider-crank kinematics, and cam follower motions and terminology.

Once a mechanism's input speed is known, the speeds and accelerations of all its links follow from geometry. GATE tests three tools: relative velocity, instantaneous centres and the slider-crank formulas, plus the basics of cams.

Relative velocity

For two points A and B on the same rigid link rotating at ω: v_BA = ω × AB, perpendicular to AB. Velocity polygons are drawn from this rule, link by link.

Instantaneous centres (I-centres)

  • A mechanism with n links has n(n − 1)/2 instantaneous centres.
  • Kennedy's theorem: the three I-centres of any three links lie on a straight line.
  • The velocity of a point = ω × (its distance from the link's I-centre relative to the frame).

Acceleration

  • A point on a link rotating about a fixed point has centripetal acceleration ω²r and tangential acceleration αr.
  • Coriolis acceleration appears when a slider moves along a rotating link: 2vω, perpendicular to the link. It is needed in quick-return and slotted-lever mechanisms.

Slider-crank mechanism

For crank radius r, connecting rod length l, n = l/r and crank angle θ from inner dead centre, at crank speed ω:

  • Piston velocity v ≈ ωr (sin θ + sin 2θ/(2n))
  • Piston acceleration a ≈ ω²r (cos θ + cos 2θ/n)
  • Angular velocity of the rod ≈ ω cos θ / n

Cams

A cam turns rotary motion into a set follower motion.

  • Terms: base circle, pitch curve, pressure angle (angle between the follower's path and the normal to the cam at contact), lift (stroke), dwell.
  • Follower motions:
  • Uniform velocity: infinite acceleration at the ends (jerk), so it is used only with modifications.
  • Simple harmonic motion: max velocity = (πh/2)(ω/β); max acceleration = (π²h/2)(ω/β)², where β is the cam angle for the rise.
  • Uniform acceleration and retardation (parabolic).
  • Cycloidal: zero acceleration at the start and end, best for high speeds.

Worked example

A crank 100 mm long turns at 300 rpm; the rod is 400 mm (n = 4). Find the piston velocity at θ = 90°.

  • ω = 2π × 300/60 = 31.42 rad/s.
  • v = 31.42 × 0.1 × (sin 90° + sin 180°/8) = 3.14 m/s.

Practice

GATE-styleNAT · numerical answer

How many instantaneous centres does a six-link mechanism have?

Show answer and explanationAnswer: 15

n(n − 1)/2 = 6 × 5/2 = 15.

GATE-styleNAT · numerical answer

A slider moves at 2 m/s along a link rotating at 5 rad/s. What is the Coriolis acceleration in m/s²?

Show answer and explanationAnswer: 20

2vω = 2 × 2 × 5 = 20 m/s².

GATE-styleNAT · numerical answer

A follower rises 20 mm with SHM during 90° of cam rotation; the cam turns at 10 rad/s. What is the maximum follower velocity in mm/s? (Give 1 decimal place.)

Show answer and explanationAnswer: 200

v_max = (πh/2)(ω/β) = (π × 20/2)(10/(π/2)) = 200 mm/s.

GATE-styleMCQ · one correct answer

Which follower motion has no sudden change in acceleration at the start and end of the rise?

Show answer and explanationAnswer: D

Cycloidal motion has zero acceleration at both ends, so there is no jerk; it suits high-speed cams.