NCERT Solutions Class 11 Physics Chapter 6 System of Particles and Rotational Motion

The Class 11 Physics NCERT Solutions Chapter 6 System of Particles and Rotational Motion will help students prepare for Boards, JEE Main, JEE Advanced, NEET, CUET and NDA in 2026-27. Every back-exercise question is solved with full working, so students can check each centre-of-mass calculation, torque balance, moment-of-inertia value, and angular-momentum step one line at a time.

This chapter moves physics from point particles to extended bodies, and it is one of the highest-scoring mechanics chapters in the paper.

• CBSE Weightage: 6 to 8 marks, usually one numerical on moment of inertia plus one short answer on centre of mass or torque.
• Questions solved: all Exercise 6.1 onward, covering centre of mass, torque, angular momentum, moment of inertia and rolling motion.
• Key formulas: torque, angular momentum, moment of inertia of standard bodies, and the parallel and perpendicular axes theorems.

Each solution in this Class 11 Physics NCERT Solutions Chapter 6 System of Particles and Rotational Motion compilation is curated by subject experts, based on the 2026-27 NCERT textbook, and refined against the last five years of CBSE Board, JEE Main and NEET papers.

Centre of Mass and Motion of the Centre of Mass

Every extended body behaves as if its whole mass sits at one point. This point is the centre of mass. The chapter opens here because it lets students treat a large object as a single particle. The NCERT numericals ask students to find this point for a set of particles and for uniform shapes.

• System of particles: the centre of mass position is the mass-weighted average, R = (Σm_i r_i) / M.
• Uniform bodies: for a ring, disc, rod or sphere, the centre of mass sits at the geometric centre.
• Motion of the centre of mass: it moves as if the total external force acts on the total mass at that one point.

A key result the solutions use again and again is that the centre of mass moves only under an external force; internal forces cancel out. Questions such as Exercise 6.1 ask students to locate the centre of mass of common shapes. The Class 11 Physics NCERT Solutions Chapter 6 System of Particles and Rotational Motion explain the symmetry argument first, then confirm it with the formula, so no step is skipped.

Torque and Angular Momentum of a Rigid Body

To turn a body, you need a torque, not just a force. Torque is the rotational version of force, and angular momentum is the rotational version of linear momentum. These two ideas run through most of the numericals in Class 11 Physics Chapter 6 solutions, so the answers define each one in words before using it.

• Torque: τ = r × F, the turning effect of a force about an axis. Its size is r F sinθ.
• Angular momentum: L = r × p for a particle, and L = I ω for a rigid body about a fixed axis.
• Link between them: torque equals the rate of change of angular momentum, τ = dL/dt.

The relation τ = dL/dt is the rotational form of Newton's second law. The direction of torque and angular momentum is set by the right-hand rule, and the solutions state this direction on its own line so students do not lose the sign. Numericals here often give a force and a position vector and ask for the torque about the origin.

Conservation of Angular Momentum in Rotational Motion

When the net external torque on a body is zero, its angular momentum stays constant. This is the conservation of angular momentum, and it explains many real effects, from a spinning skater to a diver curling up. The NCERT solutions apply it to problems where the moment of inertia changes while the body spins.

The exam favourite here is the equation I₁ ω₁ = I₂ ω₂. If the moment of inertia halves, the angular speed doubles, because their product cannot change. The solutions show this step clearly so students can reproduce it in the exam without guessing which quantity rises.

Moment of Inertia and Radius of Gyration

The moment of inertia measures how hard it is to change a body's rotation. It plays the same role in rotation that mass plays in straight-line motion. Its value depends on both the mass and how that mass is spread about the axis. The radius of gyration is a neat way to state this spread as a single distance.

The radius of gyration k is defined by I = M k², so it is the distance from the axis at which the whole mass could sit to give the same moment of inertia. Numericals like Exercise 6.7 give the mass and shape and ask for the moment of inertia or the radius of gyration. The Class 11 Physics NCERT Solutions Chapter 6 System of Particles and Rotational Motion quote the standard formula, then substitute numbers on a separate line.

Theorems of Parallel and Perpendicular Axes

Standard formulas give the moment of inertia about an axis through the centre. To shift to any other axis, the chapter gives two theorems. These two results save huge effort in the exam, and almost every moment-of-inertia numerical uses at least one of them.

• Parallel axes theorem: I = I_cm + M d², where d is the distance between the two parallel axes.
• Perpendicular axes theorem: for a flat body, I_z = I_x + I_y, where z is perpendicular to the plane.

The perpendicular axes theorem works for planar bodies only, such as a ring or a disc, while the parallel axes theorem works for any body. Always add the term M d² when moving away from the centre of mass, never subtract it. Questions such as Exercise 6.9 ask students to combine both theorems, and the NCERT Solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion show which theorem to apply at each step.

Rolling Motion and Equilibrium of a Rigid Body

Rolling motion joins straight-line motion and rotation in one problem, so it is the natural end of the chapter. A rolling body both moves forward and spins, and the two motions are linked by the contact condition. The chapter also sets the rules for a body that is not moving at all, called equilibrium.

• Rolling without slipping: the contact point is momentarily at rest, so v = R ω links speed and spin.
• Total kinetic energy: a rolling body has both translational and rotational energy, ½M v² + ½I ω².
• Equilibrium of a rigid body: needs both zero net force and zero net torque, so the body neither moves nor turns.

For equilibrium, both conditions must hold together. A body can have zero net force yet still rotate, so the torque condition is separate and must be checked on its own. The rolling numericals compare a ring, disc and sphere down the same incline, and the solutions explain why the sphere reaches the bottom first. This ranking is a favourite in JEE Main and NEET objective papers.

Exercise-wise Breakdown for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

The NCERT back-exercise splits into clear groups. Use this map to plan which answers to practise first for the 2026-27 boards. Every group links to the full solved set.

Solving the groups in this order builds the skills in the same sequence the chapter teaches them. Start with centre of mass, then torque, then moment of inertia, because each group uses the one before it.

Common Mistakes Students Make in the System of Particles and Rotational Motion Chapter

These slips happen while writing or calculating the answer, not because the concept is unclear. Each one costs 1 to 3 marks in the CBSE paper, so the solved answers point them out at the exact step.

Student Feedback on the System of Particles and Rotational Motion Chapter

Practice Questions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

Once the solved answers are clear, test yourself on the full question set. The practice page has every NCERT question with a step-by-step Solution and an Expert Solution behind a click.

Other System of Particles and Rotational Motion Class 11 Physics Resources

NCERT Solutions for Class 11 Physics: All Chapters

Jump to the solutions for any other Class 11 Physics chapter below.

FAQs on Class 11 Physics Chapter 6 System of Particles and Rotational Motion NCERT Solutions