Kinematics and kinetics of rigid bodies in plane motion

Why it matters

Gears, wheels, cranks, connecting rods, cams and robot links all move in a plane while translating and rotating at the same time. To size bearings, motors and brakes you must know both how the parts move (kinematics) and which forces and torques that motion needs (kinetics). Slider-crank mechanisms, rolling wheels and pulleys appear in almost every machine.

Key ideas

Kinematics vs kinetics. Kinematics describes motion (position, velocity, acceleration) without asking what causes it. Kinetics relates the motion to the forces and moments acting.

Types of plane motion of a rigid body.

• Translation: every line in the body keeps its direction; all points have the same velocity and acceleration (rectilinear or curvilinear).
• Rotation about a fixed axis: points move on circles about the axis; every point has the same ω and α but its speed grows with radius.
• General plane motion: translation plus rotation, for example a rolling wheel or a connecting rod. It can always be treated as translation with a reference point plus rotation about that point.

Relative velocity. For two points A and B on the same rigid body, v_B = v_A + ω × r_B/A. The relative velocity is perpendicular to the line AB and of magnitude ω·AB. All points share the same ω.

Instantaneous centre of rotation (ICR). At any instant a body in plane motion has a point of zero velocity (possibly outside the body). Every point then moves as if rotating about it: speed = ω × distance from the ICR, direction perpendicular to that line. Find it where the normals to two known velocity directions meet. For a wheel rolling without slipping, the ICR is the contact point. The ICR has zero velocity but not, in general, zero acceleration, so it cannot be used for accelerations.

Relative acceleration. a_B = a_A + α × r_B/A − ω² r_B/A: a tangential part α·AB perpendicular to AB and a centripetal part ω²·AB directed from B toward A.

Rolling without slipping. Contact point has zero velocity relative to the ground: v_G = ω r and a_G = α r for a round body of radius r. If slipping occurs, these do not hold and friction equals μk N.

Kinetics (Newton–Euler equations). For a rigid body in plane motion with mass centre G:

• ΣF = m a_G (the mass centre moves as if all mass and all external force were there);
• ΣM_G = I_G α (moments about G);
• for rotation about a fixed axis O, also ΣM_O = I_O α. Moments may be taken about any point P if you include the moment of m·a_G: ΣM_P = I_G α + (r_G/P × m a_G). This is D'Alembert's principle: add inertia force −m a_G at G and inertia couple −I_G α, then use statics.

Moment of inertia. I = Σ m r², the rotational counterpart of mass. Radius of gyration k: I = m k². Parallel-axis theorem: I_O = I_G + m d².

Formulas

ω = ω0 + α t, θ = ω0 t + ½ α t², ω² = ω0² + 2 α θ (constant α)

• ω in rad/s, α in rad/s², θ in rad, t in s.

v = ω r, a_t = α r, a_n = ω² r = v²/r, a = √(a_t² + a_n²) (point at radius r on a body rotating about a fixed axis)

v_P = ω · (distance from ICR to P)

ΣF = m a_G, ΣM_G = I_G α

• F in N, m in kg, a in m/s², M in N·m, I in kg·m².

I_G = ½ m r² (solid disc or cylinder), I_G = m r² (thin ring), I_G = m L²/12 (slender rod about centre), I_end = m L²/3 (rod about end), I_G = (2/5) m r² (solid sphere)

I_O = I_G + m d² (parallel-axis theorem; d = distance between the axes, m)

a = g sin θ / (1 + k²/r²) (body rolling without slipping down an incline of angle θ)

Worked examples

Example 1 (standard, kinematics). A wheel of radius 0.4 m rolls without slipping on level ground; its centre moves at 6 m/s. Find the angular velocity and the speeds of (a) the top point and (b) the front point level with the centre.

1. Rolling: ω = v_G / r = 6 / 0.4 = 15 rad/s.
2. ICR is the contact point.
3. (a) Top point is 2r = 0.8 m from the ICR: v = 15 × 0.8 = 12 m/s, horizontal, forward.
4. (b) Front point is √(0.4² + 0.4²) = 0.566 m from the ICR: v = 15 × 0.566 = 8.49 m/s, at 45° to the horizontal (forward and down).

Answer: ω = 15 rad/s; top point 12 m/s; front point 8.49 m/s.

Example 2 (GATE level, kinetics). A solid cylinder (m = 20 kg, r = 0.3 m) is released from rest on a 30° incline and rolls without slipping. Find its acceleration, the friction force, the least μs that allows rolling, and its speed after 5 m along the incline. Take g = 9.81 m/s².

1. FBD: weight mg, normal N, friction f up the slope at the contact.
2. Along the slope: m g sin θ − f = m a.
3. Moments about G: f r = I_G α = ½ m r² (a/r), so f = ½ m a.
4. Combine: m g sin θ = (3/2) m a, so a = (2/3) g sin θ = (2/3) × 9.81 × 0.5 = 3.27 m/s².
5. Friction: f = ½ × 20 × 3.27 = 32.7 N.
6. Normal: N = m g cos θ = 20 × 9.81 × 0.866 = 169.9 N; rolling needs f ≤ μs N, so μs ≥ 32.7/169.9 = 0.192 (equivalently tan θ/3).
7. Speed: v = √(2 a s) = √(2 × 3.27 × 5) = 5.72 m/s. Check by energy: m g h = ¾ m v² with h = 2.5 m gives v = √(4 × 9.81 × 2.5 / 3) = 5.72 m/s. ✓

Answer: a = 3.27 m/s², f = 32.7 N, μs,min = 0.192, v = 5.72 m/s.

Common mistakes

• Using the ICR to find accelerations; it is valid for velocities only.
• Writing ΣM = I α about a point that is neither G nor a fixed axis, without the m·a_G term.
• Forgetting the centripetal component ω²r when asked for the total acceleration of a point.
• Assuming friction equals μN in rolling; it is whatever is needed for rolling, up to μs N.
• Mixing rpm and rad/s: ω = 2πN/60.
• Using I = ½ m r² for a ring or hollow pipe.

For GATE PI

Expect questions on velocities in slider-crank and four-bar linkages by ICR or relative velocity, rolling bodies on inclines (which reaches the bottom first: sphere, cylinder or ring), angular acceleration of pulleys with hanging masses, and moments of inertia with the parallel-axis theorem. Practise drawing separate FBDs for each body and writing one force and one moment equation per body.

Quick check

1. A wheel of radius 0.5 m rolls without slipping at 4 rad/s. What is the speed of its centre?
2. Where is the ICR of a wheel rolling without slipping?
3. A point 2 m from a fixed axis has ω = 3 rad/s and α = 4 rad/s². What is its total acceleration?
4. Which reaches the bottom of an incline first when rolling from rest: a solid sphere or a thin ring?

Answers: 1. 2 m/s. 2. At the contact point with the ground. 3. a_t = 8 m/s², a_n = 18 m/s², total ≈ 19.7 m/s². 4. The solid sphere (smaller k²/r², so larger acceleration).