System of Particles and Rotational Motion

2.1 Core concepts

• A rigid body is one in which distances between all pairs of particles remain unchanged; a rigid body fixed at a point or along a line has only rotational motion, otherwise its motion is pure translation or translation + rotation (NCERT §6.1, p. 92–95).
• In pure translation every particle has the same velocity at any instant; in rotation about a fixed axis every particle moves in a circle whose plane is perpendicular to the axis and whose centre lies on the axis (NCERT §6.1.1, p. 93–95).
• Centre of mass of two particles on a line: X = (m₁x₁ + m₂x₂)/(m₁ + m₂); for equal masses the CM is midway between them (NCERT §6.2 Eq. 6.1, p. 95–96).
• CM of n particles: X = Σm_i x_i / M, Y = Σm_i y_i / M, Z = Σm_i z_i / M; in vector form R = Σm_i r_i / M (NCERT §6.2 Eqs. 6.4a–d, p. 96).
• For three equal masses at the vertices of a triangle, the CM coincides with the centroid (NCERT §6.2, p. 96).
• For a continuous body the sums become integrals: R = (1/M)∫ r dm; if the origin is chosen at the CM, ∫ r dm = 0 (NCERT §6.2 Eqs. 6.5a–b, 6.6, p. 97).
• By symmetry, the CM of a homogeneous rod, ring, disc or sphere coincides with its geometric centre (NCERT §6.2, p. 97).
• Motion of the CM: MA = F_ext — the CM moves as if all mass were concentrated there and all external forces acted on it; internal forces (Newton's third-law pairs) sum to zero (NCERT §6.3 Eqs. 6.10–6.11, p. 99). Illustrated by a projectile that explodes mid-air: the CM continues along the original parabolic path (Fig. 6.12, p. 100).
• Linear momentum of a system: P = Σm_i v_i = MV; dP/dt = F_ext; if F_ext = 0 then P = constant (law of conservation of total linear momentum) (NCERT §6.4 Eqs. 6.14–6.18, p. 100–101). Examples: binary stars and radioactive decay of moving radium (Figs. 6.13–6.14, p. 101).
• Vector product: c = a × b with |c| = ab sin θ, c perpendicular to the plane of a and b, direction given by the right-hand screw / right-hand rule; a × b = −b × a; distributive over addition; i × j = k, j × k = i, k × i = j (NCERT §6.5, p. 102–103).
• Angular velocity ω = dθ/dt is a vector along the axis of rotation; for any particle of a rotating rigid body v_i = ω r_i, and in vector form v = ω × r (NCERT §6.6 Eqs. 6.19–6.20, p. 103–105).
• Angular acceleration α = dω/dt; for a fixed axis the vector equation reduces to the scalar α = dω/dt (NCERT §6.6.1 Eqs. 6.21–6.22, p. 105).
• Torque (moment of force) τ = r × F with magnitude τ = rF sin θ = r⊥ F = rF⊥; SI unit N m; dimensions ML²T⁻² (same as energy but a vector) (NCERT §6.7.1 Eqs. 6.23–6.24, p. 106).
• Angular momentum of a particle l = r × p; magnitude l = rp sin θ = r⊥ p; and dl/dt = τ — the rotational analogue of F = dp/dt (NCERT §6.7.2 Eqs. 6.25a, 6.26, 6.27, p. 106–107).
• For a system of particles L = Σl_i and dL/dt = τ_ext (internal torques cancel in pairs by Newton's third law, assuming central forces) (NCERT §6.7 Eqs. 6.28a–b, p. 107–108).
• Conservation of angular momentum: if τ_ext = 0, L = constant (Eq. 6.29a); applied component-wise as L_x, L_y, L_z each constant (NCERT §6.7 Eq. 6.29b, p. 108).
• Mechanical equilibrium of a rigid body requires both translational equilibrium (ΣF_i = 0, Eq. 6.30a) and rotational equilibrium (Στ_i = 0, Eq. 6.30b); coplanar forces reduce these to three scalar conditions (NCERT §6.8, p. 109–110).
• A couple is a pair of equal and opposite forces with different lines of action; net force zero, net torque nonzero — produces rotation without translation (NCERT §6.8, p. 110–111).
• Principle of moments for a lever: d₁F₁ = d₂F₂; mechanical advantage M.A. = F₁/F₂ = d₂/d₁ (NCERT §6.8.1 Eqs. 6.32a–b, p. 111).
• Centre of gravity is the point where the total gravitational torque vanishes; in uniform gravity the CG coincides with the CM (NCERT §6.8.2 Eq. 6.33, p. 112–113).
• Moment of inertia I = Σm_i r_i² is the rotational analogue of mass; rotational kinetic energy K = ½Iω² (NCERT §6.9 Eqs. 6.34–6.35, p. 114). For a thin ring of radius R about its central axis, I = MR²; for a pair of point masses M/2 at ±l/2 on a light rod, I = Ml²/4 (NCERT §6.9, p. 115).
• Radius of gyration k is defined by I = Mk²; e.g. for a rod about its midpoint k² = L²/12, so k = L/√12; for a disc about its diameter k = R/2 (NCERT §6.9, p. 115). Dimensions of I are ML²; SI unit kg m².
• Standard moments of inertia (Table 6.1, p. 116): thin ring about its central axis = MR², ring about a diameter = MR²/2, thin rod about perpendicular axis at midpoint = ML²/12, disc about its central axis = MR²/2, disc about a diameter = MR²/4, hollow cylinder about its axis = MR², solid cylinder about its axis = MR²/2, solid sphere about a diameter = 2MR²/5.
• Kinematics of rotation about a fixed axis (analogous to linear motion with constant acceleration): ω = ω₀ + αt, θ = θ₀ + ω₀t + ½αt², ω² = ω₀² + 2α(θ − θ₀) (NCERT §6.10 Eqs. 6.36–6.38, p. 117).
• Dynamics: work dW = τ dθ, power P = τω, and τ = Iα — Newton's second law for fixed-axis rotation (NCERT §6.11 Eqs. 6.39–6.41, p. 119–120).
• For rotation about a fixed axis L_z = Iω; for bodies symmetric about the axis L = L_z = Iω k̂ (NCERT §6.12 Eqs. 6.42b, 6.42d, p. 121). If I is time-independent then τ = Iα.
• Conservation of angular momentum about a fixed axis: if τ_ext = 0, Iω = constant — illustrated by the spinning skater / swivel-chair experiment (folding arms decreases I, increases ω) and circus acrobats and pirouetting dancers (NCERT §6.12.1 Eq. 6.44, p. 122–123).
• Composite-body trick for CM: when a body is made of simpler shapes whose individual CMs are known (e.g. an L-lamina made of three squares), the CM of the composite is the mass-weighted average of the individual CMs — R = (Σ mᵢ Rᵢ)/(Σ mᵢ); illustrated in NCERT Example 6.2 (p. 98).
• Why the CM motion is so useful: because internal forces cancel out in pairs (Newton's third law), the equation MA = F_ext for the CM is valid regardless of how complicated the internal motion is; this is what lets us describe the trajectory of an exploding shell, a swinging chain, or a rotating-and-translating wheel as a single particle of total mass M (NCERT §6.3, p. 99).
• Scalar vs vector product distinction: a · b = ab cos θ is a number; a × b = ab sin θ n̂ is a vector. The scalar product is commutative; the vector product is anti-commutative. Both are distributive over addition (NCERT §6.5, p. 102).
• Angular vs ordinary frequency in rotation: for uniform rotation, the magnitude of ω equals 2π × revolution-frequency. When NCERT specifies ω in rad s⁻¹ but a problem gives rpm, students must convert: ω (rad s⁻¹) = 2π × (rpm/60) — exactly the conversion used in Example 6.11 (p. 117).

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig. 6.4 (p. 94): Rigid body rotating about the z-axis; each particle P_i describes a circle of radius r_i in a plane perpendicular to the axis with centre C_i on the axis.
• Fig. 6.8 (p. 97): Symmetry argument for the CM of a thin uniform rod — element at +x is balanced by an equal element at −x, so ∫ x dm = 0.
• Fig. 6.11 (p. 98): CM of an L-shaped lamina decomposed into three unit squares — the worked example for treating composite bodies as a sum of CMs.
• Fig. 6.12 (p. 100): Exploding projectile — CM continues along the original parabolic path because the explosion is an internal force.
• Fig. 6.15 (p. 102): Right-hand-screw and right-hand rules for the direction of a × b.
• Fig. 6.17(b) (p. 104): ω directed along the axis, v = ω × r tangential to the circle.
• Fig. 6.18 (p. 106): τ = r × F perpendicular to the r–F plane, direction by right-hand rule.
• Fig. 6.23 (p. 111): Lever with fulcrum, load arm d₁, effort arm d₂ — pictorial form of the principle of moments.
• Fig. 6.25 (p. 112): Locating the CG of an irregular lamina by suspending it from two points and finding the intersection of plumb lines.
• Fig. 6.32 (p. 122): Spinning girl on a swivel chair / acrobat — visual of Iω = constant.
• Table 6.1 (p. 116): Moments of inertia of standard shapes — this is the single most fact-checked table for CUET.
• Table 6.2 (p. 119): Side-by-side comparison of translational and rotational quantities — every analogy in this chapter is captured here.

2.4 Common confusions / NTA trap points

• CM ≠ centroid in general. Equal masses at the vertices of a triangle give CM = centroid; unequal masses (Example 6.1) do not (p. 97–98). NTA loves to plant "centroid" as a distractor.
• CG vs CM. They coincide only in uniform gravity; the CM depends only on mass distribution, the CG also on g (p. 112–113).
• Direction of τ and L. Both are perpendicular to the plane of (r, F) or (r, p) respectively — students often guess "along r" or "along F". Use right-hand rule.
• a × b ≠ b × a. Vector product is anti-commutative (a × b = −b × a). Scalar product is commutative; do not mix them up (p. 102).
• Table 6.1 axes. Disc about its central axis is MR²/2 but about a diameter is MR²/4; ring about its central axis is MR² but about a diameter is MR²/2 (p. 116). The axis matters.
• τ_ext = 0 ⇒ L conserved, not ω conserved. When a skater folds her arms, I decreases and ω increases so that Iω stays constant — many students wrongly think ω is conserved (p. 122–123).
• Couple. ΣF = 0 does not imply Στ = 0; a couple has zero net force but nonzero net torque (p. 110, Points to Ponder 5).
• K_rot = ½Iω² uses ω in rad s⁻¹. If you forget to convert rpm → rad s⁻¹ you will be off by a factor of (2π/60)² ≈ 0.011 — a classic numerical trap.
• CM of L-shape composite. Treat each rectangular sub-piece as a point mass located at its own CM; the overall CM is the mass-weighted average (NCERT Ex. 6.2). Wrongly taking the geometric centre of the whole L is a common error.
• τ = Iα applies for fixed axis only. If the axis is itself accelerating or non-principal, the relation needs the general τ = dL/dt. Most NCERT problems are fixed-axis and the simpler form suffices.
• Angular momentum L = r × p; magnitude r p sin θ. A common slip is to write |L| = rp regardless of angle — wrong when the velocity is along (parallel to) r, in which case L = 0.

2.5 Key formulas table