Why does a figure skater spin faster when pulling their arms inward? Why is it easier to open a heavy door by pushing near the outer handle rather than near the hinge? How does a rolling wheel combine translational and rotational motion? System of Particles and Rotational Motion forms Chapter 7 of the CBSE Class 11 Physics syllabus and is one of the highest-weightage topics in JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers Centre of Mass (CM) derivations for discrete and continuous systems, rotational kinematics & dynamics analogies, torque, angular momentum conservation, moment of inertia theorems (parallel & perpendicular axes), master moment of inertia formulas table, pure rolling motion, inclined plane acceleration, and five step-by-step solved entrance exam numericals.
- 1. Centre of Mass (CM): Discrete & Continuous Systems
- 2. Linear vs Rotational Kinematics & Dynamics Analogy
- 3. Torque, Angular Momentum & Conservation Law
- 4. Moment of Inertia (I), Radius of Gyration & Theorems
- 5. Master Moment of Inertia Formulas Table for Rigid Bodies
- 6. Pure Rolling Motion & Acceleration Down an Inclined Plane
- 7. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 8. Frequently Asked Questions (FAQ)
1. Centre of Mass (CM): Discrete & Continuous Systems
The Centre of Mass (CM) of a body or system of particles is the imaginary point at which the whole mass of the system may be assumed to be concentrated for describing its translational motion.
R_cm = (m1 r1 + m2 r2 + ... + mn rn) / M = Σ(mi ri) / M
For 2-particle system: X_cm = (m1 x1 + m2 x2) / (m1 + m2).
If m1 = m2: X_cm = (x1 + x2) / 2 (Midpoint!).
R_cm = (1 / M) ∫ r dm
• Uniform Thin Rod (Length L): X_cm = L / 2.
• Uniform Semi-Circular Ring: Y_cm = (2 R) / π.
• Uniform Semi-Circular Disc: Y_cm = (4 R) / (3 π).
• Solid Hemisphere: Y_cm = (3 R) / 8.
2. Linear vs Rotational Kinematics & Dynamics Analogy
Rotational motion about a fixed axis mirrors translational linear motion with corresponding rotational quantities:
| Physical Parameter | Linear Motion (Translational) | Rotational Motion (About Fixed Axis) | Connecting Relation |
|---|---|---|---|
| Displacement | Linear displacement s (m) | Angular displacement θ (rad) | s = θ · r |
| Velocity | Linear velocity v = ds/dt (m/s) | Angular velocity ω = dθ/dt (rad/s) | v = ω · r (v_vec = ω_vec × r_vec) |
| Acceleration | Linear acceleration a = dv/dt (m/s²) | Angular acceleration α = dω/dt (rad/s²) | a_t = α · r (Tangential) |
| Inertia / Mass | Mass m (kg) | Moment of Inertia I (kg·m²) | I = Σ m_i r_i² = M k² |
| Force / Cause | Force F = m · a (N) | Torque τ = I · α (N·m) | τ_vec = r_vec × F_vec |
| Momentum | Linear Momentum p = m · v (kg·m/s) | Angular Momentum L = I · ω (kg·m²/s) | L_vec = r_vec × p_vec |
| Kinetic Energy | K_trans = ½ m v² (J) | K_rot = ½ I ω² (J) | K_rot = p_rot² / (2I) = L² / (2I) |
3. Torque, Angular Momentum & Conservation Law
τ_vec = r_vec × F_vec = r F sinθ n_hat or τ = I α
2. Angular Momentum (L): Moment of linear momentum about an axis:
L_vec = r_vec × p_vec = r p sinθ n_hat or L = I ω
3. Relation between Torque and Angular Momentum:
τ_ext = dL_vec / dt (Rotational analogue of Newton's 2nd Law!).
L_vec = constant ⇒ I1 ω1 = I2 ω2
Classic Applications:
• Ice Skater / Ballet Dancer: Folding arms decreases Moment of Inertia I, automatically increasing angular speed ω to keep Iω constant.
• Acrobat / Diver: Curling body into a tuck position decreases I, allowing rapid somersault spins.
• Kepler's 2nd Law: Planetary areal velocity is constant because gravitational torque τ_vec = 0.
4. Moment of Inertia (I), Radius of Gyration & Theorems
Moment of Inertia (I) is the measure of rotational inertia of a body about a given axis: I = Σ m_i r_i² = M k² (where k is Radius of Gyration).
I = I_cm + M d²
Applies to 3D bodies of ANY shape!
I_z = I_x + I_y
Applies ONLY to flat 2D planar bodies!
5. Master Moment of Inertia Formulas Table for Rigid Bodies
| Rigid Body & Mass M | Axis of Rotation | Moment of Inertia (I) | Radius of Gyration (k²/R²) |
|---|---|---|---|
| Thin Circular Ring (Radius R) | Perpendicular to plane through center | I = M R² | k² / R² = 1 |
| Thin Circular Ring (Radius R) | About any diameter | I = ½ M R² | k² / R² = 1/2 |
| Circular Disc (Radius R) | Perpendicular to plane through center | I = ½ M R² | k² / R² = 1/2 |
| Circular Disc (Radius R) | About any diameter | I = ¼ M R² | k² / R² = 1/4 |
| Solid Cylinder (Radius R) | Central longitudinal axis | I = ½ M R² | k² / R² = 1/2 |
| Hollow Cylinder (Radius R) | Central longitudinal axis | I = M R² | k² / R² = 1 |
| Solid Sphere (Radius R) | About any diameter | I = ⅖ M R² | k² / R² = 2/5 = 0.4 |
| Hollow Sphere (Radius R) | About any diameter | I = ⅔ M R² | k² / R² = 2/3 = 0.67 |
| Thin Uniform Rod (Length L) | Perpendicular to rod through center | I = (1/12) M L² | k² / L² = 1/12 |
| Thin Uniform Rod (Length L) | Perpendicular to rod through one end | I = (1/3) M L² | k² / L² = 1/3 |
6. Pure Rolling Motion & Acceleration Down an Inclined Plane
Pure Rolling Motion (Rolling without slipping) is a combination of translational motion of the centre of mass and rotational motion about the centre of mass. Condition at contact point: v_cm = ω R.
K_total = K_trans + K_rot = ½ M v_cm² + ½ I ω² = ½ M v_cm² + ½ (M k²) (v_cm / R)².
K_total = ½ M v_cm² (1 + k² / R²)
2. Acceleration down an Inclined Plane (θ):
a = (g sinθ) / (1 + k² / R²)
Comparison of Acceleration down incline:
Solid Sphere (k²/R² = 2/5 = 0.4) > Solid Cylinder / Disc (k²/R² = 0.5) > Hollow Sphere (k²/R² = 0.67) > Ring / Hollow Cylinder (k²/R² = 1).
Solid Sphere reaches the bottom FIRST with maximum acceleration!
7. Solved Entrance Exam Numerical Problems (JEE / NEET)
X_cm = (m1 x1 + m2 x2) / (m1 + m2) = [(2 × 1) + (4 × 4)] / (2 + 4) = (2 + 16) / 6 = 18 / 6 = 3.
Y_cm = (m1 y1 + m2 y2) / (m1 + m2) = [(2 × 2) + (4 × 5)] / (2 + 4) = (4 + 20) / 6 = 24 / 6 = 4.
Distance between diameter and parallel tangent d = R.
By Parallel Axes Theorem: I_tangent = I_cm + M R² = ⅖ M R² + M R² = (7/5) M R².
Given M = 5 kg, R = 0.2 m ⇒ R² = 0.04 m².
I_tangent = (7/5) × 5 × 0.04 = 7 × 0.04 = 0.28 kg·m².
(a) τ = I α ⇒ α = -4 / 0.20 = -20 rad/s².
(b) ω = ω0 + α t ⇒ 0 = 20 - 20 t ⇒ t = 1 s.
(c) θ = ω0 t + ½ α t² = 20(1) - ½(20)(1)² = 20 - 10 = 10 rad.
Revolutions N = θ / (2π) = 10 / (2 × 3.1416) = 10 / 6.283 ≈ 1.59 revolutions.
By Conservation of Angular Momentum: I1 ω1 = (I1 + I2) ω_f.
4 × 10 = (4 + 1) ω_f ⇒ 40 = 5 ω_f ⇒ ω_f = 40 / 5 = 8 rad/s.
1. For Solid Cylinder: k²/R² = 1/2 = 0.5 ⇒ a_solid = (0.5 g) / (1 + 0.5) = 0.5 g / 1.5 = g / 3 = 3.27 m/s².
2. For Hollow Cylinder: k²/R² = 1 ⇒ a_hollow = (0.5 g) / (1 + 1) = 0.5 g / 2 = g / 4 = 2.45 m/s².
Ratio a_solid / a_hollow = (g/3) / (g/4) = 4/3 = 1.33.
Explore Related CBSE Physics & Lab Apparatus Guides
8. Frequently Asked Questions (FAQ)
The Parallel Axes Theorem states that the moment of inertia of a body about any axis equals its moment of inertia about a parallel axis passing through its centre of mass plus the product of its mass and the square of distance between the axes: I = I_cm + M d².
By pulling arms in, mass is brought closer to the rotation axis, decreasing moment of inertia I. Since external torque τ_ext = 0, angular momentum L = I ω is conserved, so angular velocity ω increases.
The condition for pure rolling on a stationary surface is v_cm = ω R, meaning the point of contact between body and surface is instantaneously at rest relative to the surface.
Because a solid sphere has a smaller radius of gyration ratio (k²/R² = 0.4) than a hollow sphere (k²/R² = 0.67), resulting in a larger linear acceleration a = (g sinθ) / (1 + k²/R²).
Radius of Gyration (k) is the perpendicular distance from the axis of rotation to a point where the entire mass of the body could be concentrated without altering its moment of inertia: I = M k² ⇒ k = √(I / M).
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