Kinematics of particles and rigid bodies in plane motion

Why it matters

Kinematics is the geometry of motion: where a point is, how fast it moves and how fast that speed changes, without asking which forces cause it. Stopping distances, wheel slip, piston speed, the acceleration that a passenger or a crankpin bearing feels: all are kinematics first. In kinetics you write ΣF = m·a, and you cannot do that until you know a.

Key ideas

Particle kinematics. Position r(t), velocity v = dr/dt, acceleration a = dv/dt.

• Rectilinear motion: use v = ds/dt, a = dv/dt = v·dv/ds. The constant-acceleration equations apply only when a is constant; otherwise integrate.
• Curvilinear motion in normal–tangential components: the tangential acceleration a_t = dv/dt changes the speed; the normal (centripetal) acceleration a_n = v²/ρ, directed toward the centre of curvature, changes the direction. A car cornering at constant speed still accelerates.
• Projectile motion (no air drag): constant horizontal velocity, constant downward acceleration g.

Rigid body in plane motion. Every point moves in parallel planes. Three types:

• Translation: every line in the body keeps its orientation; all points have the same velocity and acceleration (rectilinear or curvilinear translation, e.g. a coupler of a parallel linkage).
• Rotation about a fixed axis: every point moves on a circle; one angular velocity ω and one angular acceleration α describe the whole body.
• General plane motion: translation plus rotation, e.g. a rolling wheel or a connecting rod. Any such motion can be seen as translation with a reference point plus rotation about it.

Angular quantities are properties of the body. ω and α are the same for every point of a rigid body. That is why one ω describes a whole wheel.

Relative velocity. For two points A and B on the same rigid body, v_B = v_A + ω × r_B/A. The relative velocity of B with respect to A is perpendicular to AB and has magnitude ω·AB. Velocity polygons for mechanisms are drawn from this.

Instantaneous centre (I-centre). At any instant a body in plane motion has a point of zero velocity (possibly at infinity for pure translation). Every point then moves as if rotating about it: v = ω·(distance from I-centre), perpendicular to the line from the I-centre. For a wheel rolling without slip, the contact point is the I-centre, so the top of the wheel moves at twice the centre speed. The I-centre has zero velocity but not, in general, zero acceleration.

Relative acceleration. a_B = a_A + α × r_B/A − ω²·r_B/A. The tangential part α·AB is perpendicular to AB; the centripetal part ω²·AB points from B toward A.

Rolling without slip. v_O = ω·r and a_O = α·r for the centre O of a wheel of radius r. Slip (braking lock-up or wheel spin) breaks this link.

Rotating frames and Coriolis. When a point moves relative to a rotating body (a slider on a rotating slotted link), its acceleration gains a Coriolis component 2·ω·v_rel, perpendicular to v_rel. It appears in quick-return mechanisms and in the slider of a rotary vane pump.

Formulas

v = u + a·t, s = u·t + ½·a·t², v² = u² + 2·a·s

• u, v: initial and final velocities (m/s); a: constant acceleration (m/s²); t: time (s); s: displacement (m). Constant acceleration only.

a_t = dv/dt, a_n = v² / ρ, a = √(a_t² + a_n²)

• ρ: radius of curvature of the path (m).

ω = dθ/dt, α = dω/dt, v = r·ω, a_t = r·α, a_n = r·ω²

• θ (rad), ω (rad/s), α (rad/s²); r: distance from the rotation axis (m). Rotation about a fixed axis.

ω = 2π·N / 60

• N: speed (rev/min).

v_B = v_A + ω × r_B/A, a_B = a_A + α × r_B/A − ω²·r_B/A

• Two points on one rigid body in plane motion.

v_O = r·ω, a_O = r·α

• Wheel of radius r rolling without slip on a fixed surface.

v_p ≈ r·ω·(sin θ + sin 2θ / (2n))

• Piston velocity of a slider-crank; r: crank radius (m); n = l/r, with l the connecting-rod length; θ: crank angle from inner dead centre. Approximation good when n is about 4 or more.

a_c = 2·ω·v_rel

• Coriolis acceleration (m/s²) of a point sliding with velocity v_rel along a body rotating at ω.

Worked examples

Example 1 (standard): braking distance. Given: a car at 72 km/h brakes with constant deceleration and stops in 40 m. Find the deceleration and the stopping time.

1. u = 72 / 3.6 = 20 m/s, v = 0, s = 40 m.
2. v² = u² + 2·a·s: 0 = 400 + 2·a·40, so a = −5 m/s².
3. v = u + a·t: 0 = 20 − 5·t, so t = 4 s. Answer: deceleration 5 m/s² (about 0.51 g), stopping time 4 s. Doubling the speed would make the distance four times larger at the same deceleration.

Example 2 (GATE level): rolling wheel. Given: a wheel of radius r = 0.3 m rolls without slip to the right. Its centre O has velocity 15 m/s and acceleration 3 m/s², both to the right. Find (a) ω and α, (b) the velocity of the top point T, (c) the velocity of the point Q at the front, level with O, (d) the acceleration of the contact point P.

1. ω = v_O / r = 15 / 0.3 = 50 rad/s clockwise; α = a_O / r = 3 / 0.3 = 10 rad/s² clockwise.
2. T is 2r from the I-centre P: v_T = ω·2r = 50 × 0.6 = 30 m/s, horizontal to the right.
3. Q is √2·r = 0.4243 m from P: v_Q = 50 × 0.4243 = 21.21 m/s, perpendicular to PQ, i.e. at 45° below the horizontal, forward. (Check: v_O = 15 m/s forward plus ω·r = 15 m/s downward from rotation gives the same vector.)
4. Contact point P, taking O as reference: a_O = 3 m/s² forward; the tangential term α·r = 3 m/s² is backward at P and cancels it; the centripetal term ω²·r = 2500 × 0.3 = 750 m/s² points from P toward O, upward. Answer: ω = 50 rad/s, α = 10 rad/s²; v_T = 30 m/s; v_Q = 21.2 m/s at 45° below horizontal; a_P = 750 m/s² upward. The contact point has zero velocity but a large acceleration.

Example 3: piston speed. Given: crank radius r = 50 mm, connecting rod l = 200 mm (n = 4), engine speed 3000 rev/min, θ = 45°.

1. ω = 2π × 3000 / 60 = 314.16 rad/s.
2. v_p ≈ r·ω·(sin θ + sin 2θ / (2n)) = 0.05 × 314.16 × (0.7071 + 1/8) = 15.708 × 0.8321 = 13.07 m/s.
3. The exact value from the geometry is 13.10 m/s, so the approximation is within 0.3 %. Answer: v_p ≈ 13.1 m/s.

Common mistakes

• Using v = u + a·t when the acceleration varies (with time, position or speed). Integrate a = v·dv/ds instead.
• Forgetting the normal acceleration on a curved path at constant speed.
• Treating the instantaneous centre as having zero acceleration.
• Using rev/min directly in v = r·ω; convert to rad/s.
• Assuming v = r·ω for a wheel that is slipping or skidding.
• Getting the centripetal direction backwards: it points toward the reference point or the centre of rotation.

For GATE ME

Expect rolling-wheel velocity and acceleration questions, velocity of a point on a link by relative velocity or I-centres, rectilinear motion with variable acceleration, and constant-acceleration problems with vehicles. The Theory of Machines paper builds directly on this (velocity and acceleration diagrams, Coriolis in quick-return mechanisms). Practise vector addition of relative terms with clear sketches.

Quick check

1. A flywheel accelerates uniformly from rest to 600 rev/min in 10 s. Find α.
2. What is the velocity of the top of a car tyre when the car travels at 25 m/s without slip?
3. A car rounds a 50 m radius curve at 15 m/s. What is its normal acceleration?
4. Can the instantaneous centre of a link lie outside the link?

Answers: 1. ω = 62.83 rad/s, α = 6.28 rad/s². 2. 50 m/s. 3. 225 / 50 = 4.5 m/s². 4. Yes; it is often outside the physical link, for example for a connecting rod.