Kinematics of particles and rigid bodies in plane motion

Particle motion, normal and tangential acceleration, translation, fixed-axis rotation and general plane motion with relative velocity, instantaneous centre and Coriolis, with a servo run-up and slider-crank example.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Before you can size a motor or write a motion profile, you need the motion itself: how fast the end effector moves when a joint turns, what acceleration a gripper sees, how a crank's steady rotation becomes a slider's varying speed. Kinematics answers these questions without forces, and the accelerations it gives feed directly into F = m·a and T = I·α for drive sizing. Encoder-based velocity estimation and trajectory planning in mechatronic systems are applied kinematics.

Key ideas

Particle kinematics. Position s(t), velocity v = ds/dt, acceleration a = dv/dt. Also a = v·dv/ds, useful when acceleration is given as a function of position. For constant acceleration the familiar equations apply; for anything else, integrate.

Curvilinear motion, normal and tangential components. A point moving on a curved path has a tangential acceleration a_t = dv/dt (changes the speed) and a normal (centripetal) acceleration a_n = v²/ρ towards the centre of curvature (changes the direction). The total is √(a_t² + a_n²). A point moving at constant speed on a curve still accelerates.

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

  • Translation: every line in the body keeps its direction; all points have the same velocity and acceleration (rectilinear or curvilinear, for example a parallelogram-linkage platform).
  • Rotation about a fixed axis: each point moves on a circle; v = ω·r, a_t = α·r, a_n = ω²·r.
  • General plane motion: translation plus rotation, for example a connecting rod or a rolling wheel. The angular velocity ω and angular acceleration α are the same for every line in a rigid body.

Relative velocity. For two points A and B on the same rigid body, v_B = v_A + v_B/A, where v_B/A has magnitude ω·AB and is perpendicular to AB. This is the basis of velocity diagrams for linkages. For accelerations, a_B = a_A + (a_B/A)_t + (a_B/A)_n with (a_B/A)_t = α·AB perpendicular to AB and (a_B/A)_n = ω²·AB directed from B towards A.

Instantaneous centre (IC) of zero velocity. At any instant a body in plane motion behaves as if rotating about one point I with zero velocity. Find it where the perpendiculars to the velocities of two points meet. Then v = ω·(distance from I). For a wheel rolling without slip on fixed ground, the contact point is the IC, so the centre moves at v = ω·R and the top point at 2v. The IC generally has non-zero acceleration, so never use it for accelerations.

Rotating frames and Coriolis. If a point slides along a link that is itself rotating (a slider on a rotating arm, a quick-return mechanism), its acceleration includes the Coriolis term 2·ω·v_rel, perpendicular to the link.

Units. Angles in radians. Convert rpm with ω = 2πN/60.

Formulas

v = ds/dt, a = dv/dt = v·dv/ds

  • s: position (m); v: velocity (m/s); a: acceleration (m/s²); t: time (s).

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

  • u: initial velocity (m/s). Only for constant acceleration.

ω = ω₀ + α·t, θ = ω₀·t + ½·α·t², ω² = ω₀² + 2·α·θ

  • θ: angle (rad); ω: angular velocity (rad/s); α: angular acceleration (rad/s²). Constant α only.

v = ω·r, a_t = α·r, a_n = ω²·r = v²/r

  • r: distance from the axis (m). Fixed-axis rotation.

v_B = v_A + ω × r_B/A

  • Relative velocity of two points of one rigid body; |ω × r_B/A| = ω·AB, perpendicular to AB.

a_c = 2·ω·v_rel

  • Coriolis acceleration (m/s²) of a point sliding with relative speed v_rel (m/s) along a link rotating at ω.

ω = 2πN / 60

  • N: speed in rev/min.

Worked examples

Example 1 (standard): servo pulley run-up. Given: a servo motor drives a 25 mm radius pulley from rest to 3000 rpm in 0.5 s at constant angular acceleration. Find α, the number of revolutions turned, and the velocity and acceleration components of a point on the rim at t = 0.5 s.

  1. Final speed: ω = 2πN/60 = 2π × 3000/60 = 314.16 rad/s.
  2. α = (ω − ω₀)/t = 314.16/0.5 = 628.3 rad/s².
  3. θ = ½·α·t² = 0.5 × 628.3 × 0.25 = 78.54 rad = 12.5 rev.
  4. Rim speed: v = ω·r = 314.16 × 0.025 = 7.85 m/s (also the belt speed).
  5. a_t = α·r = 628.3 × 0.025 = 15.7 m/s²; a_n = ω²·r = 314.16² × 0.025 = 2467 m/s². Answer: α = 628 rad/s², 12.5 rev, v = 7.85 m/s, a_t = 15.7 m/s², a_n ≈ 2470 m/s². The normal component dwarfs the tangential one, which is why balancing matters at speed.

Example 2 (GATE level): slider-crank velocities. Given: crank OA = 0.1 m rotates anticlockwise at 100 rad/s; connecting rod AB = 0.4 m; slider B moves on the horizontal line through O. Crank angle θ = 45° from the line of stroke (A above the line, B to the right of O). Find the slider velocity and the rod's angular velocity.

  1. Rod angle φ: sin φ = r·sin θ / l = 0.1 × 0.7071/0.4 = 0.1768, so φ = 10.18°, cos φ = 0.9843.
  2. Velocity of A: v_A = ω·r = 100 × 0.1 = 10 m/s, perpendicular to OA.
  3. Relative velocity: v_B = v_A + v_B/A, with v_B horizontal and v_B/A perpendicular to AB. Resolving perpendicular to AB eliminates v_B/A and gives the standard result v_B = ω·r·sin(θ + φ) / cos φ.
  4. v_B = 10 × sin 55.18° / 0.9843 = 10 × 0.8208 / 0.9843 = 8.34 m/s, directed towards O for this sense of rotation.
  5. Resolving vertically: ω_AB = ω·r·cos θ / (l·cos φ) = 10 × 0.7071/(0.4 × 0.9843) = 17.96 rad/s, clockwise. Answer: v_B ≈ 8.34 m/s towards the crank, ω_AB ≈ 18.0 rad/s clockwise. A numerical derivative of the slider position x = r·cos θ + √(l² − r²·sin² θ) gives the same 8.34 m/s.

Common mistakes

  • Using degrees or rpm in v = ω·r. Convert to rad/s first.
  • Applying the constant-acceleration equations when the acceleration varies.
  • Forgetting the centripetal component a_n = ω²·r when asked for the total acceleration of a point.
  • Using the instantaneous centre to find accelerations.
  • Drawing the relative velocity v_B/A along AB instead of perpendicular to it.
  • Assuming every point of a rolling wheel moves at the speed of its centre; the contact point is momentarily at rest and the top moves at twice the centre speed.
  • Leaving out the Coriolis component when a slider moves along a rotating link.

For GATE ME

Expect rectilinear motion with variable acceleration (integrate a(t) or use v·dv/ds), projectiles, rolling without slip, velocity of a point on a link by the instantaneous centre, slider-crank velocity and angular velocity of the rod, and Coriolis acceleration in slotted-lever mechanisms. Practise locating the instantaneous centre quickly and checking answers by a second method.

Quick check

  1. Convert 600 rpm to rad/s.
  2. A wheel of radius 0.3 m rolls without slipping with its centre at 2 m/s. What is the speed of the topmost point?
  3. A point 0.2 m from a fixed axis has ω = 10 rad/s and α = 0. What is its acceleration?
  4. Can the instantaneous centre be used to find accelerations?
  5. A slider moves at 0.5 m/s along an arm rotating at 4 rad/s. What is the Coriolis acceleration?

Answers: 1. 62.8 rad/s. 2. 4 m/s. 3. 20 m/s² towards the axis. 4. No, it generally has non-zero acceleration. 5. 2 × 4 × 0.5 = 4 m/s².

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is the difference between kinematics and dynamics in the context of mechanics?Concept

    Kinematics is the study of motion without considering the forces that cause it. It focuses on parameters like displacement, velocity, and acceleration. Dynamics, on the other hand, deals with the forces and torques that cause motion. It involves studying the relationship between motion and the forces affecting it.

  2. 2.Explain the concept of relative velocity in plane motion.Concept

    Relative velocity is the velocity of one object as observed from another moving object. In plane motion, it is calculated by vectorially subtracting the velocity of the reference object from the velocity of the object being observed. This concept is crucial in understanding how different observers perceive the motion of objects differently.

  3. 3.What is the significance of the instantaneous center of rotation in the analysis of plane motion?Concept

    The instantaneous centre (IC) is the point, on the body or its extension, whose velocity is zero at that instant, so the body's motion at that instant is a pure rotation about it. Then every point's speed is ω times its distance from the IC and its velocity is perpendicular to the line joining it to the IC. You locate it where the perpendiculars to two known velocity directions meet; for a wheel rolling without slip it is the contact point. It is valid for velocities only: the IC itself generally accelerates, so it cannot be used to find accelerations.

  4. 4.Why is the Coriolis acceleration important in the study of rotating systems?Application

    When a point moves with relative velocity v_rel along a body that is itself rotating at ω, its absolute acceleration includes the Coriolis component 2·ω·v_rel, perpendicular to the relative velocity in the plane of motion. It appears in slotted-lever and quick-return mechanisms, a slider on a rotating robot arm, and rotating fluid machinery. Leaving it out gives wrong link accelerations and therefore wrong inertia forces and actuator torques.

  5. 5.What happens to the motion of a rigid body if the net external force acting on it is zero?Application

    Zero net external force means the centre of mass does not accelerate: it stays at rest or moves with constant velocity (Newton's first law applied to the centre of mass). It says nothing about rotation. If the forces form a couple, the net moment is not zero and the body has an angular acceleration even though its centre of mass is unaccelerated. Both ΣF = 0 and ΣM = 0 are needed for a rigid body to have no acceleration of any kind.

  6. 6.How does the choice of reference frame affect the analysis of particle motion?Application

    The choice of reference frame can significantly affect the analysis of particle motion. In an inertial reference frame, Newton's laws of motion are directly applicable. However, in a non-inertial reference frame, additional fictitious forces, such as centrifugal and Coriolis forces, must be considered to accurately describe the motion.

  7. 7.A particle moves in a plane with a velocity given by v = 3i + 4j m/s. What is the magnitude of its velocity?Numerical

    To find the magnitude of the velocity, use the formula: |v| = √(vx² + vy²). Here, vx = 3 m/s and vy = 4 m/s. So, |v| = √(3² + 4²) = √(9 + 16) = √25 = 5 m/s.

  8. 8.A rigid body rotates with an angular velocity of 2 rad/s. If a point on the body is 0.5 m from the axis of rotation, what is its linear velocity?Numerical

    The linear velocity v of a point on a rotating body is given by v = ω·r, where ω is the angular velocity and r is the distance from the axis of rotation. Here, ω = 2 rad/s and r = 0.5 m. So, v = 2 rad/s × 0.5 m = 1 m/s.

  9. 9.Why is it important to consider both translational and rotational motion in the analysis of rigid bodies?Application

    Points on a rigid body in general plane motion have different velocities, because the motion is a translation of a reference point plus a rotation about it: v_B = v_A + ω × r_B/A. A connecting rod or a rolling wheel cannot be described by one velocity or one angular velocity alone. In dynamics this splits into ΣF = m·a_G for the centre of mass and ΣM_G = I_G·α for rotation, and both are needed to find forces and motor torques.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousPrinciple of virtual workNext Impulse, momentum and energy methods; impact