Class 11 Physics Chapter 6 System of Particles and Rotational Motion — Formulas & Key Points

The centre of mass (COM) is the point where the entire mass of a system can be considered concentrated for translational motion analysis. For a system of n particles with masses m₁, m₂,… at positions r₁, r₂,…, the position vector of COM is R_cm = (Σ mᵢ rᵢ) / M, where M = Σ mᵢ is the total mass. In Cartesian coordinates, x_cm = (Σ mᵢ xᵢ)/M, y_cm = (Σ mᵢ yᵢ)/M, z_cm = (Σ mᵢ zᵢ)/M. For continuous bodies, replace summation by integration: R_cm = (∫ r dm) / M. The velocity of COM is v_cm = (Σ mᵢ vᵢ) / M, and acceleration a_cm = F_ext / M, where F_ext is the net external force. Internal forces cancel in pairs (Newton's third law), so only external forces change the COM motion. For symmetric, uniform-density objects—rectangle, circle, sphere, cylinder—the COM lies at the geometric centre. For composite bodies, treat each part as a point mass at its own COM, then apply the discrete-particle formula.

Moment of inertia I quantifies rotational inertia—resistance to angular acceleration—and depends on mass distribution relative to the axis. For a point mass m at perpendicular distance r from the axis, I = m r². For a system of particles, I = Σ mᵢ rᵢ². For continuous bodies, I = ∫ r² dm. The radius of gyration k is defined by I = M k², where M is total mass; k is the RMS distance of mass from the axis. The table below lists standard moments of inertia for common shapes (mass M, relevant dimension given). Always check which axis is specified—axis through COM, along length, perpendicular, etc. Parallel axis theorem: I = I_cm + M d², where I_cm is MOI about an axis through COM, d is perpendicular distance between the two parallel axes. Perpendicular axis theorem (for planar laminae only): I_z = I_x + I_y, where x, y lie in the plane and z is perpendicular.

Torque τ is the rotational analogue of force, causing angular acceleration. Mathematically, τ = r × F (vector cross product), where r is the position vector from the axis to the point of application of force F. Magnitude: τ = r F sinθ, where θ is the angle between r and F. The perpendicular (lever arm) distance from the axis to the line of action of F is r⊥ = r sinθ, so τ = r⊥ F. Direction: determined by the right-hand rule—curl fingers from r toward F; thumb points along τ. Positive torque produces counter-clockwise rotation (conventionally), negative torque clockwise. For multiple forces, net torque τ_net = Σ τᵢ. Relation to angular acceleration: τ_net = I α, where I is moment of inertia and α is angular acceleration (rad/s²). Units: N·m (not Joule, even though dimensionally equivalent). Always choose the axis carefully; torque depends on the choice of pivot/axis.

Angular momentum L measures rotational motion quantity. For a particle: L = r × p, where p = m v is linear momentum. Magnitude: L = m v r sinθ. For a rigid body rotating about a fixed axis: L = I ω, where I is moment of inertia and ω is angular velocity (rad/s). Relation to torque: τ = dL/dt (rotational analogue of F = dp/dt). Conservation of angular momentum: If τ_ext = 0 (no net external torque), then L is constant. This principle explains ice-skater spin-up (reducing I increases ω), planetary motion (Kepler's second law), and collisions involving rotation. For systems of particles, total L = Σ Lᵢ. Direction: right-hand rule—curl fingers in direction of rotation; thumb points along L. Units: kg·m²/s or J·s. In problems involving collisions (a rotating disc colliding with another disc, person jumping onto a turntable), equate L_initial = L_final and solve for unknown ω or I.

Rotational motion about a fixed axis mirrors linear kinematics. Angular displacement θ (rad), angular velocity ω = dθ/dt (rad/s), angular acceleration α = dω/dt (rad/s²). For constant α, the SUVAT analogues are: ω = ω₀ + α t; θ = ω₀ t + (1/2) α t²; ω² = ω₀² + 2 α θ. Relation to linear quantities: s = r θ (arc length), v = r ω (tangential velocity), a_t = r α (tangential acceleration), a_c = ω² r = v²/r (centripetal acceleration). Total acceleration a = √(a_t² + a_c²). Torque-angular acceleration relation: τ = I α. Work done by torque: W = ∫ τ dθ; for constant τ, W = τ θ. Rotational kinetic energy: KE_rot = (1/2) I ω². Power: P = τ ω. These formulas are essential for solving problems on rotating wheels, pulleys, flywheels, and gyroscopes in CBSE Class 11 Physics Chapter 6.

Rolling motion combines translation of the centre of mass and rotation about the COM. For a body (radius R) rolling on a surface, the point of contact is instantaneously at rest. Condition for pure rolling (no slipping): v_cm = ω R. If this holds, kinetic energy KE_total = KE_trans + KE_rot = (1/2) M v_cm² + (1/2) I_cm ω². Substitute ω = v_cm/R and I_cm = M k² to get KE = (1/2) M v_cm² [1 + k²/R²]. For a solid sphere, I_cm = (2/5) M R², so KE = (7/10) M v_cm². For a disc/cylinder, I_cm = (1/2) M R², so KE = (3/4) M v_cm². Acceleration down an incline (angle θ): a = g sinθ / [1 + I_cm/(M R²)] = g sinθ / [1 + k²/R²]. Friction provides the torque needed for rolling; for pure rolling on an incline, static friction f = (M g sinθ) / [1 + M R²/I_cm]. If the incline is frictionless, the body slips, and only translational motion occurs.

Rigid Body: a body with definite shape where distance between any two particles remains constant under applied forces. Centre of Mass: the point representing mean position of mass distribution; moves as if all external forces act there. Moment of Inertia (I): rotational inertia, depends on mass distribution and axis, analogous to mass in linear motion. Radius of Gyration (k): defined by I = M k²; the RMS distance of mass from axis. Torque (τ): turning effect of force, τ = r × F. Angular Momentum (L): rotational momentum, L = I ω for rigid bodies. Pure Rolling: motion where contact point is instantaneously at rest, v_cm = ω R. Couple: pair of equal, opposite, parallel forces producing rotation without translation. Parallel Axis Theorem: I = I_cm + M d². Perpendicular Axis Theorem (planar laminae only): I_z = I_x + I_y. These terms appear frequently in NCERT Class 11 Physics Chapter 6 numerical and theory questions.

Students often lose marks in CBSE board exams and JEE by mixing sign conventions, forgetting to convert units, or misapplying formulas. (1) Sign of torque: choose one direction as positive consistently; reversing mid-problem causes sign errors. (2) Confusing radius of gyration k with actual radius R: I = M k² does not mean k = R. (3) Mixing degrees and radians: always use radians in ω, α, θ formulas; 1 revolution = 2π rad; rpm to rad/s multiply by 2π/60. (4) Using wrong MOI formula: check axis carefully (through centre, end, diameter, perpendicular, etc.). (5) Forgetting parallel axis theorem: when axis shifts from COM, add M d². (6) Misapplying perpendicular axis theorem to 3D bodies: valid only for planar laminae. (7) In rolling problems, assuming v_cm = ω R even when slipping occurs; verify pure rolling condition. (8) Writing torque units as Joules instead of N·m. (9) Ignoring direction in vector quantities L and τ. (10) Not converting cm to m or gram to kg before substitution.

Mnemonics help recall the multitude of formulas in System of Particles and Rotational Motion. (1) MOI decreases as mass concentrates near axis: Ring > Disc > Sphere. Remember 'RDS descending': Ring (MR²) → Disc (½MR²) → Sphere (⅖MR²). (2) Rod through centre vs end: 'Centre is one-third of end' — (1/12)ML² vs (1/3)ML². (3) Parallel axis theorem: 'I = I_cm + M d²' — 'MOI grows with distance squared'. (4) Pure rolling: 'v = ωR' — 'Velocity equals omega times Radius'. (5) Torque direction: Right-Hand Rule — 'Curl fingers from r to F, thumb = τ'. (6) Energy in rolling: 'Total KE = ½Mv² (1 + k²/R²)' — 'Add rotational fraction'. (7) Acceleration on incline: 'a = g sinθ / (1 + k²/R²)' — 'Bigger k² means slower a' (ring slowest). Write these on a flashcard and revise daily before CBSE practicals or term exams.

Example 1 (Centre of Mass): Three particles of masses 1 kg, 2 kg, 3 kg are at (0,0), (1,0), (0,1) m. Find COM. Solution: x_cm = (1×0 + 2×1 + 3×0)/(1+2+3) = 2/6 = 1/3 m; y_cm = (1×0 + 2×0 + 3×1)/6 = 3/6 = 0.5 m. COM at (1/3, 0.5) m. Example 2 (Torque & Angular Acceleration): A disc of mass 2 kg, radius 0.4 m is acted on by a tangential force 5 N at the rim. Find α. Solution: I = (1/2)×2×(0.4)² = 0.16 kg·m². τ = F×r = 5×0.4 = 2 N·m. α = τ/I = 2/0.16 = 12.5 rad/s². Example 3 (Rolling Motion): A solid sphere of mass 5 kg, radius 0.2 m rolls down a 30° incline. Find acceleration. Solution: I_cm = (2/5)MR² ⇒ k² = (2/5)R². a = g sinθ/(1 + k²/R²) = 10×0.5/(1 + 2/5) = 5/(7/5) = 25/7 ≈ 3.57 m/s². These examples mirror typical CBSE Class 11 Physics board questions.

Use this condensed summary 24 hours before your CBSE Physics exam or JEE mock. Centre of Mass: R_cm = (Σ m r)/M; v_cm = (Σ m v)/M; a_cm = F_ext/M. MOI: I = Σ m r²; rod (centre) (1/12)ML², (end) (1/3)ML²; disc (1/2)MR²; ring MR²; solid sphere (2/5)MR²; hollow sphere (2/3)MR². Parallel axis: I = I_cm + Md². Perpendicular axis (2D): I_z = I_x + I_y. Torque: τ = r F sinθ = I α. Angular momentum: L = I ω; τ = dL/dt; conserved if τ_ext = 0. Rotational KE: (1/2)I ω². Rolling: v_cm = ωR; KE = (1/2)Mv² (1 + k²/R²); incline a = g sinθ/(1 + k²/R²). Kinematics: ω = ω₀ + αt; θ = ω₀t + (1/2)αt²; ω² = ω₀² + 2αθ. Units: [I] = kg·m²; [τ] = N·m; [L] = kg·m²/s; [ω] = rad/s; [α] = rad/s². Keep this box on your phone wallpaper or print it for quick glances during revision.

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