CBSE • Class 11 • Physics
System of Particles and Rotational Motion
Centre of mass, torque, angular momentum, and rotational dynamics.
Chapter 6
Verified Curriculum Topic
What is System of Particles and Rotational Motion?
Centre of mass, torque, angular momentum, and rotational dynamics.
System of Particles and Rotational Motion matters because it connects theory, equations, and real physical behaviour. At Class 11 level, students are typically expected to explain concepts precisely, apply laws correctly, and interpret numerical or experimental questions with confidence.
Study System of Particles and Rotational Motion now
Summary
The One Thing
The motion of a system or rotating body can be analysed by separating translational motion of its centre of mass from rotational motion about an axis. Torque, moment of inertia and angular momentum provide the rotational counterparts of force, mass and linear momentum, while conservation laws simplify systems with no external force or torque.
Reactions, Processes and Experiments
| What happens | Equation or process | What you observe | Type |
|---|---|---|---|
| The position of the centre of mass of two particles is determined from their masses and position vectors. | rcm = (m1r1 + m2r2)/(m1 + m2) | The centre of mass lies closer to the particle with the greater mass. | Centre-of-mass calculation |
| The position of the centre of mass of a system is determined from all particle masses and position vectors. | rcm = (Σmi ri)/(Σmi) | The result represents the mass-weighted average position of the system. | Centre-of-mass calculation |
| The centre-of-mass velocity is determined from the velocities of the particles. | vcm = (Σmi vi)/(Σmi) | Internal motions may cancel, leaving the velocity of the system’s centre of mass. | Translational motion of a system |
| The centre-of-mass acceleration is determined by the net external force. | acm = Fext/M | Internal forces do not alter the motion of the centre of mass; only external forces do so. | Newtonian motion of a system |
| The total linear momentum of a system is related to the centre-of-mass velocity. | P = Mvcm | The total momentum is the vector sum of the momenta of all particles. | Linear momentum |
| The total linear momentum of an isolated system remains unchanged. | For an isolated system, total linear momentum remains constant. | Momentum before and after an interaction is equal. | Conservation of linear momentum |
| A force produces a turning effect about a point or axis. | τ = r × F | The turning effect depends on the position vector and the applied force. | Torque |
| The magnitude of torque depends on the angle between the position vector and the force. | τ = rF sinθ = F × perpendicular distance | Torque is maximum when the force is perpendicular to the position vector and zero when its line of action passes through the axis. | Torque |
| Angular momentum is determined for a particle relative to an origin or axis. | L = r × p | Angular momentum depends on the position vector and linear momentum. | Angular momentum |
| The magnitude of a particle’s angular momentum is determined by the angle between position and momentum vectors. | L = rp sinθ | Angular momentum is zero when the position vector and momentum are parallel. | Angular momentum |
| A rigid body rotating about a fixed axis has angular momentum proportional to angular velocity. | L = Iω | A larger moment of inertia gives greater angular momentum at the same angular velocity. | Rotational dynamics |
| Net torque produces angular acceleration. | τnet = Iα | A larger moment of inertia makes a body more resistant to changes in rotational motion. | Rotational form of Newton’s second law |
| The moment of inertia of a collection of particles depends on their masses and distances from the axis. | I = Σmi ri² | Particles farther from the axis contribute more strongly to rotational inertia. | Moment of inertia |
| The moment of inertia about an axis parallel to an axis through the centre of mass is determined using the distance between the axes. | I = Icm + Md² | Moving the axis away from the centre of mass increases the moment of inertia. | Parallel-axis theorem |
| The moment of inertia of a plane lamina about an axis perpendicular to its plane is related to the moments about two perpendicular axes in the plane. | Iz = Ix + Iy | The perpendicular-axis moment equals the sum of the two in-plane moments. | Perpendicular-axis theorem |
| Rotational kinetic energy is associated with angular motion. | Krot = 1/2 Iω² | Rotational kinetic energy increases with moment of inertia and with the square of angular velocity. | Rotational kinetic energy |
| A constant torque does work through an angular displacement. | W = τθ | Work increases with torque and angular displacement. | Rotational work |
| Torque produces rotational power. | P = τω | Power increases when either torque or angular velocity increases. | Rotational power |
| Torque acting over a time interval changes angular momentum. | angular impulse = τΔt = change in angular momentum | The change in angular momentum equals the angular impulse. | Angular impulse |
| Angular momentum remains unchanged when the net external torque is zero. | Li = Lf | Angular speed can change when mass distribution changes, while total angular momentum remains constant. | Conservation of angular momentum |
| A body rolls without slipping through simultaneous translational and rotational motion. | vcm = Rω | The point of contact is instantaneously at rest relative to the surface. | Pure rolling |
| The linear and angular accelerations of a body rolling without slipping are related. | acm = Rα | Translational acceleration and angular acceleration remain linked by the radius. | Rolling motion |
| The kinetic energy of a rolling body contains translational and rotational components. | K = 1/2 Mvcm² + 1/2 Icmω² | Both motion of the centre of mass and rotation contribute to total kinetic energy. | Rolling kinetic energy |
| The moment of inertia of a thin ring about its central axis is determined by its mass and radius. | I = MR² | All mass is effectively distributed at the same radius from the axis. | Standard moment of inertia |
| The moment of inertia of a disc or solid cylinder about its central axis is determined by its mass and radius. | I = 1/2 MR² | The value is less than that of a thin ring with the same mass and radius because some mass lies closer to the axis. | Standard moment of inertia |
| The moment of inertia of a solid sphere about a diameter is determined by its mass and radius. | I = 2/5 MR² | The mass is distributed throughout the sphere relative to the diameter. | Standard moment of inertia |
| The moment of inertia of a hollow sphere about a diameter is determined by its mass and radius. | I = 2/3 MR² | The distribution of mass produces a different rotational inertia from a solid sphere. | Standard moment of inertia |
| The moment of inertia of a rod about an axis through its centre is determined by its mass and length. | I = 1/12 ML² | The rod has a smaller moment of inertia about its centre than about an end. | Standard moment of inertia |
| The moment of inertia of a rod about an axis through one end is determined by its mass and length. | I = 1/3 ML² | Moving the axis from the centre to the end increases the moment of inertia. | Standard moment of inertia |
| The centre of mass of a uniform symmetrical body is located at its geometrical centre. | The centre of mass of a uniform symmetrical body lies at its geometrical centre. | The geometrical centre is the balance point for the uniform symmetrical body. | Centre of mass |
| In uniform circular motion, angular speed remains constant while the direction of linear velocity changes. | In uniform circular motion, angular speed is constant but the direction of linear velocity changes continuously. | The speed is constant, but the velocity is not constant because its direction changes. | Circular motion |
| Rotational equilibrium requires no net external torque. | The net external torque on a body is zero, so its angular acceleration is zero. | The body has no angular acceleration. | Rotational equilibrium |
| Complete equilibrium requires both translational and rotational balance. | For equilibrium, both the net external force and the net external torque must be zero. | Neither the centre of mass accelerates nor the body undergoes angular acceleration. | Mechanical equilibrium |
Key Terms
- System of Particles: A collection of two or more particles considered together for studying their motion and interactions.
- Centre of Mass: The point at which the entire mass of a system may be considered concentrated for analysing translational motion.
- Centre of Mass of Two Particles: For masses
m1andm2at positionsr1andr2, the centre-of-mass position isrcm = (m1r1 + m2r2)/(m1 + m2). - Centre of Mass of a System: For particles with masses
miand position vectorsri,rcm = (Σmi ri)/(Σmi). - Linear Momentum: The product of mass and velocity; for a system, total momentum is the vector sum of the momenta of all particles.
- Torque: The turning effect of a force about a point or axis, given by
τ = r × F, with magnitudeτ = rF sinθ. - Moment Arm: The perpendicular distance from the axis of rotation to the line of action of the force.
- Angular Momentum: The rotational equivalent of linear momentum; for a particle,
L = r × p, and for a rigid body rotating about a fixed axis,L = Iω. - Moment of Inertia: The rotational inertia of a body, defined for particles as
I = Σmi ri²and dependent on the mass distribution relative to the axis. - Radius of Gyration: The distance
kfrom the axis at which the entire mass could be assumed concentrated to produce the same moment of inertia;I = Mk². - Angular Displacement: The angle through which a body rotates, usually measured in radians.
- Angular Velocity: The rate of change of angular displacement, given by
ω = dθ/dt. - Angular Acceleration: The rate of change of angular velocity, given by
α = dω/dt. - Rigid Body: An ideal body in which the distance between every pair of particles remains constant during motion.
- Rotational Equilibrium: A state in which the net external torque on a body is zero, so its angular acceleration is zero.
- Rolling Motion: Combined translational and rotational motion in which a body rolls without slipping when
vcm = Rω.
Easily Confused
- Force and torque: Force governs translational motion, whereas torque is the turning effect of a force and governs rotational motion.
- Mass and moment of inertia: Mass measures resistance to translational acceleration, whereas moment of inertia measures resistance to angular acceleration and depends on mass distribution relative to the axis.
- Linear momentum and angular momentum: Linear momentum is
p = mv, whereas angular momentum isL = r × porL = Iωfor a rigid body about a fixed axis. - Centre of mass and geometrical centre: The centre of mass is determined by mass distribution; it coincides with the geometrical centre only for a uniform symmetrical body.
- Rotational equilibrium and complete equilibrium: Rotational equilibrium requires zero net external torque, whereas complete equilibrium requires both zero net external force and zero net external torque.
- Angular speed and linear velocity in uniform circular motion: Angular speed remains constant, but the direction of linear velocity changes continuously.
- Pure rolling and rotation alone: Pure rolling combines translation and rotation and satisfies
vcm = Rω; rotation alone does not require this condition. - Internal and external forces: Internal forces cannot change the motion of the centre of mass, whereas only external forces can do so.
- Thin ring and disc or solid cylinder: For the same mass and radius, a thin ring has
I = MR², whereas a disc or solid cylinder hasI = 1/2 MR². - Rod about its centre and rod about one end: The moments of inertia are
I = 1/12 ML²andI = 1/3 ML², respectively.
What Gets Asked
- Calculate the centre of mass of two particles or of a system using
rcm = (m1r1 + m2r2)/(m1 + m2)orrcm = (Σmi ri)/(Σmi). Marks are lost by failing to use mass-weighted positions. - Determine the motion of a system’s centre of mass using
vcm = (Σmi vi)/(Σmi),P = Mvcmoracm = Fext/M. The key slip is attributing a change in centre-of-mass motion to internal forces. - Calculate torque using
τ = r × Forτ = rF sinθ, and identify when torque is maximum or zero. Marks are lost by using the distance to the point of application instead of the perpendicular distance to the line of action. - Apply rotational dynamics through
τnet = Iα, rotational kinetic energyKrot = 1/2 Iω², workW = τθor powerP = τω. The common error is substituting mass for moment of inertia. - Use the parallel-axis theorem, perpendicular-axis theorem and standard moments of inertia. Marks are lost by selecting the wrong axis or confusing the values for a thin ring, disc or solid cylinder, sphere or rod.
- Apply conservation of angular momentum using
Li = Lfwhen the net external torque is zero, including cases in which mass distribution or angular speed changes. The key condition must not be omitted. - Analyse rolling motion using
vcm = Rω,acm = RαandK = 1/2 Mvcm² + 1/2 Icmω². Marks are lost by including only translational or only rotational kinetic energy.
Flashcards
Quick quiz
What does the centre of mass represent for a system of particles?
Save this & unlock the full study pack
Create a free account to save System of Particles and Rotational Motion, get the complete set of notes, flashcards, quizzes, mind maps, and mock exams, and track your progress across Physics.
Sign up free — save & unlock everythingKey ideas to master
- Explain the core principle behind System of Particles and Rotational Motion in clear scientific language.
- Use the correct equations, symbols, and units when solving numerical questions.
- Interpret diagrams, graphs, or experiments linked to the topic.
- Connect conceptual understanding with the final answer instead of memorising formulas alone.
Common exam prompts
- State the law, principle, or definition behind System of Particles and Rotational Motion precisely.
- Apply the relevant equation to a short numerical problem with correct units.
- Explain a diagram, graph, or experiment related to System of Particles and Rotational Motion.
- Distinguish between conceptual understanding and memorised formula use in this chapter.
How to study System of Particles and Rotational Motion effectively
Step 1
Start with a clear summary
Generate a concise summary first so you can see the core idea, the main vocabulary, and the chapter structure before going deeper.
Step 2
Turn it into active recall
Use flashcards and a short quiz to test whether you can reproduce the ideas in your own words instead of only recognising them.
Step 3
Ask the tutor where you are weak
Use AI Tutor for step-by-step explanations, simpler language, and one-question checks whenever part of the chapter still feels unclear.
Quick answers students usually need
What is System of Particles and Rotational Motion in CBSE Class 11 Physics?
Centre of mass, torque, angular momentum, and rotational dynamics.
How should I study System of Particles and Rotational Motion effectively?
Start with a concise summary, then move into notes, flashcards, and a short quiz. Use AI Tutor when you need a simpler explanation, a worked example, or a quick oral check on the part that still feels unclear.
What can Study Buddy generate for System of Particles and Rotational Motion?
From this verified topic path, Study Buddy can generate summaries, detailed notes, flashcards, quizzes, mind maps, and follow-up tutor explanations that stay aligned with the selected curriculum branch.
Generate Your Study Pack
Get AI-generated notes, flashcards, quizzes, and mind maps for System of Particles and Rotational Motion. All content is curriculum-aligned and tailored to Class 11 level.
More Topics in Physics
Physical quantities, SI units, significant figures, and dimensional analysis.
Kinematics of one-dimensional motion, velocity, and acceleration.
Vectors, projectile motion, and uniform circular motion.
Newton's laws, momentum, equilibrium, and common forces.
Work-energy theorem, power, collisions, and conservation ideas.
Useful next links for this topic
Back to all Physics topics
Compare this chapter with the rest of the subject and open the next verified topic path directly.
Browse the full Class 11 library
Jump back to the grade hub if you need to switch subjects or revise another chapter next.
Audio study podcast
Review laws, definitions, and explanation chains while away from your desk.
Mind map generator
Map out concepts, formulas, and linked units across the chapter.