Class 11 Physics · NCERT Chapter 6
~23 min readSystem of Particles and Rotational Motion Class 11 Physics Notes
Complete, exam-ready notes on the system of particles and rotational motion: centre of mass, torque and equilibrium, angular momentum and its conservation, moment of inertia with the theorems of parallel and perpendicular axes, rotational kinetic energy and rolling motion — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.
What is torque, and how is it related to angular momentum?
Torque τ = r × F is the rotational analogue of force, and angular momentum L = r × p is the rotational analogue of linear momentum. Just as F = dp/dt, torque equals the rate of change of angular momentum: τ = dL/dt. With no external torque, angular momentum is conserved.
Centre of Mass
Centre of mass
The centre of mass of a system of particles is the point where the entire mass of the system may be considered to be concentrated for translational purposes. The position vector of the CM of particles of masses m₁, m₂, … at positions r₁, r₂, … is R = (Σmᵢrᵢ)/(Σmᵢ).
The velocity and acceleration of the CM are V = dR/dt and A = d²R/dt². The net external force acting on the system equals the total mass times the acceleration of the CM: F_ext = M·A. Internal forces cannot change the motion of the centre of mass.
- For a symmetric object (rod, ring, disc, sphere), the CM lies at its geometric centre.
- The CM may lie outside the body — e.g. a hollow ring's CM is at its centre even though there is no material there.
- The CM of two particles divides the line joining them in the inverse ratio of their masses (closer to the heavier particle).
The CM follows a simple path
MCQ Questions
1 Mark Questions
- 1Define the centre of mass of a system of particles.
- 2Write the expression for the position vector of the centre of mass of two particles of masses m₁ and m₂.
- 3Where is the centre of mass of a uniform ring located?
2 Mark Questions
- 1State the rule for locating the centre of mass of two particles of unequal masses.
- 2Show that internal forces cannot change the motion of the centre of mass.
- 3Two bodies of masses 2 kg and 8 kg are placed at the ends of a light rod 1 m long. Find the position of their centre of mass from the 8 kg mass.
3 Mark Questions
- 1Derive the expression for the position vector of the centre of mass of a system of n particles and show it behaves like a point mass under a net external force.
- 2Three particles of masses 1 kg, 2 kg and 3 kg are placed at the vertices of an equilateral triangle of side 1 m. Find the position of the centre of mass.
Torque and Translational Equilibrium
Torque
Torque is the rotational analogue of force — it measures the tendency of a force to rotate a body about a pivot. Its magnitude is τ = rF sinθ, where r is the perpendicular distance of the line of action of the force from the pivot (the moment arm) and F sinθ the perpendicular component of the force.
Torque is a vector perpendicular to both r and F, with direction given by the right-hand rule. A body in equilibrium must have zero net force (no translation) and zero net torque about any point (no rotation): ΣF = 0 and Στ = 0. This gives the three conditions that let us analyse ladders, beams and seesaws.
- The turning effect is stronger when the force is applied farthest from the pivot.
- Forces through the pivot (moment arm = 0) produce zero torque.
- A couple is two equal, opposite, parallel forces that produce pure rotation with moment τ = F·d.
Choosing the pivot
MCQ Questions
1 Mark Questions
- 1Define torque. Write its SI unit.
- 2State the conditions for translational equilibrium of a rigid body.
- 3What is a couple?
2 Mark Questions
- 1A force of 10 N acts at the end of a rod of length 0.5 m, perpendicular to it. Find the torque about the pivot at the other end.
- 2Explain why the effectiveness of a wrench is increased by making its handle longer.
- 3State the conditions for equilibrium of a rigid body under the action of several forces.
3 Mark Questions
- 1A uniform rod of weight 40 N and length 2 m is pivoted at one end and held horizontal by a force at the other end. Find the force needed to balance the rod.
- 2Derive the torque due to a force F applied at a point whose position vector is r, and show τ is perpendicular to both r and F.
Angular Momentum and Its Conservation
Angular momentum
For a particle, the angular momentum about a point is L = r × p = r × (mv). Its magnitude is L = rmv sinθ = mvr (for r ⊥ v). For a rigid body rotating about a fixed axis, L = Iω.
The rotational analogue of Newton's second law is τ = dL/dt. When the net external torque is zero, angular momentum is conserved: L = constant. This is why an ice skater speeds up by pulling in her arms — reducing I increases ω while Iω stays the same.
- Angular momentum is conserved for a system with no external torque.
- A planet orbiting the Sun sweeps equal areas in equal times — a direct application of L conservation.
- The direction of L is along the axis of rotation (right-hand rule).
Spinning faster by pulling in
MCQ Questions
1 Mark Questions
- 1Define angular momentum of a particle about a point.
- 2Write the relation between torque and angular momentum.
- 3State the law of conservation of angular momentum.
2 Mark Questions
- 1Show that τ = dL/dt follows from Newton's second law for a particle.
- 2A diver curls into a tight ball during a dive then straightens out on entry. Explain the change in angular speed using angular momentum conservation.
- 3State the law of conservation of angular momentum and give two examples where it applies.
3 Mark Questions
- 1Prove that the angular momentum of a particle is conserved when the net torque on it is zero, and relate it to Kepler's second law of equal areas.
- 2A man stands at the centre of a rotating platform and holds weights with arms stretched out. Explain what happens when he pulls the weights in, using conservation of angular momentum.
Moment of Inertia and Theorems
Moment of inertia
Moment of inertia I is the rotational analogue of mass — it measures a body's resistance to a change in its rotational motion. For a particle of mass m at distance r from the axis, I = mr²; for a system I = Σmr². The moment of inertia depends on the mass distribution and the choice of the axis.
The radius of gyration k is the distance at which the entire mass of the body may be assumed concentrated to give the same moment of inertia: I = Mk². Standard results include a thin ring (MR²), a solid disc (½MR²), a solid sphere (2/5 MR²) and a thin rod about its centre (ML²/12).
Theorems of moment of inertia
Parallel axes theorem: the moment of inertia about any axis equals that about a parallel axis through the CM plus M times the square of the perpendicular distance between them, I = I_CM + Md². Perpendicular axes theorem (for a lamina): the moment of inertia about an axis perpendicular to the lamina equals the sum of the moments of inertia about two mutually perpendicular axes in the plane meeting at that point, I_z = I_x + I_y.
Remember the thin disc
MCQ Questions
1 Mark Questions
- 1Define moment of inertia. Write its SI unit.
- 2State the theorem of parallel axes.
- 3Define the radius of gyration.
2 Mark Questions
- 1State and prove the theorem of perpendicular axes for a plane lamina.
- 2A ring and a disc of the same mass and radius rotate about their central axes. Compare their moments of inertia.
- 3Write the moment of inertia of a thin uniform rod of length L about an axis through its centre, perpendicular to its length, and state the parallel-axes theorem.
3 Mark Questions
- 1State and prove the theorem of parallel axes, using it to find the moment of inertia of a thin rod of length L and mass M about an axis through one end.
- 2Find the moment of inertia of a uniform disc of mass M and radius R about a tangent in its plane, using the theorems of moment of inertia.
Rotational Kinetic Energy and Rolling
Rotational kinetic energy
The kinetic energy of a rotating rigid body is KE_rot = ½Iω², the analogue of ½mv². When a body rolls without slipping, its motion is a combination of translation of the CM and rotation about the CM, so the total kinetic energy is KE = ½Mv² + ½Iω².
For pure rolling motion (no slipping), the point of contact is instantaneously at rest, giving the rolling condition v = rω. The work done in rotation is W = τθ, and the rotational power is P = τω, matching W = Fs and P = Fv in translation.
- Rolling without slipping requires v = rω at the contact point.
- A body rolling down an incline converts potential energy into both translational and rotational KE.
- Objects with the smallest moment-of-inertia factor reach the bottom of a slope first (sphere beats cylinder beats ring).
Why the sphere wins the race
MCQ Questions
1 Mark Questions
- 1Write the expression for the rotational kinetic energy of a rigid body.
- 2State the rolling condition for a body moving without slipping.
- 3Give the expression for the total kinetic energy of a rolling body.
2 Mark Questions
- 1A flywheel of moment of inertia 2 kg m² rotates at 300 rpm. Find its rotational kinetic energy.
- 2Explain why a solid sphere rolls down an incline faster than a ring of the same mass and radius.
- 3Derive the total kinetic energy of a rolling body as ½Mv² + ½Iω².
3 Mark Questions
- 1A solid sphere rolls down an inclined plane of height h starting from rest. Obtain the velocity of its centre of mass at the bottom.
- 2State and prove that the total kinetic energy of a rolling body equals ½Mv²(1 + k²/R²), where k is its radius of gyration.
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q6.1–Q6.17)
- 6.1) Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
- 6.2) In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10⁻¹⁰ m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
- 6.3) A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
- 6.4) Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
- 6.5) Show that a·(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
- 6.6) Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components pₓ, pᵧ and p_z. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
- 6.7) Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
- 6.8) A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig. 6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
- 6.9) A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
- 6.10) Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
- 6.11) A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
- 6.12) (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
- 6.13) A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
- 6.14) To maintain a rotor at a uniform angular speed of 200 rad s⁻¹, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
- 6.15) From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
- 6.16) A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
- 6.17) The oxygen molecule has a mass of 5.30 × 10⁻²⁶ kg and a moment of inertia of 1.94 × 10⁻⁴⁶ kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Answers at a glance
- 1Q6.1 → All four lie at the centre of symmetry: the geometric centre (cylinder: mid-point of the axis). No, the CM need not lie inside the body.
- 2Q6.2 → CM ≈ 1.24 Å from the hydrogen nucleus (about 0.035 Å from the chlorine nucleus).
- 3Q6.3 → The CM of the system continues to move with speed V.
- 4Q6.4 → Area = ½ |a||b| sin θ = ½ |a × b|.
- 5Q6.5 → Volume = base area × height = |b × c| × [a·(b × c)/|b × c|] = a·(b × c).
- 6Q6.6 → lₓ = y p_z − z pᵧ; lᵧ = z pₓ − x p_z; l_z = x pᵧ − y pₓ. In the x-y plane only l_z remains.
- 7Q6.7 → L = m v d, fixed in magnitude and direction (⊥ to the plane), independent of the reference point.
- 8Q6.8 → The centre of gravity is 0.72 m from the left end.
- 9Q6.9 → Each front wheel: 3675 N; each back wheel: 5145 N.
- 10Q6.10 → The solid sphere acquires the greater angular speed (J = (2/5)MR² is smaller than MR²).
- 11Q6.11 → KE = 3125 J; L = 62.5 kg m² s⁻¹.
- 12Q6.12 → (a) ω₂ = 100 rev/min. (b) KE rises to 2.5 times — the increase comes from the muscular work the child does in folding his arms.
- 13Q6.13 → α = 25 rad s⁻²; linear acceleration of the rope = 10 m s⁻².
- 14Q6.14 → P = 36 kW.
- 15Q6.15 → CG is at distance R/6 from the centre of the disk, on the side opposite the hole.
- 16Q6.16 → Mass of the metre stick = 66 g.
- 17Q6.17 → Average angular velocity ≈ 6.75 × 10¹² rad s⁻¹.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1Define the centre of mass of a system of particles.
- 2Define torque and give its SI unit.
- 3State the unit of angular momentum.
- 4Define the radius of gyration.
2 Mark Questions
- 1Two particles of masses 2 kg and 3 kg are 1 m apart. Find the position of their centre of mass. (Ans: 0.4 m from the 2 kg body)
- 2A force of 10 N is applied tangentially to the rim of a wheel of radius 0.5 m. Find the torque about the centre. (Ans: τ = Fr = 5 N m)
- 3Find the moment of inertia of a disc of mass 2 kg and radius 0.1 m about its own axis. (Ans: I = ½MR² = 0.01 kg m²)
3 Mark Questions
- 1Derive the expression for the moment of inertia of a rod of mass M and length L about an axis through its centre perpendicular to its length. (Ans: I = ML²/12)
- 2State and prove the theorem of parallel axes for the moment of inertia of a rigid body.
- 3Obtain the relation L = Iω for the angular momentum of a rigid body rotating about a fixed axis. (Ans: sum of mᵢ rᵢ² × ω)
5 Mark Questions
- 1State and prove the theorem of parallel axes and apply it to find the moment of inertia of a thin rod about an axis through one end perpendicular to its length. (Ans: I = ML²/3)
- 2Explain the equilibrium conditions of a rigid body (translational and rotational), and derive the work–energy theorem for rotation using torque and angular displacement.
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.
Covers all five topics: centre of mass; torque and translational equilibrium; angular momentum and its conservation; moment of inertia and theorems; and rotational kinetic energy and rolling. Arranged topic by topic.
Higher-difficulty set with multi-topic integration problems (centre of mass, torque and couples, angular momentum conservation, moment of inertia, rolling bodies) and full worked answers. Includes an answer key.
Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.
Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Centre of mass
Torque
Angular momentum
Torque–angular momentum
Moment of inertia
Parallel axes
Perpendicular axes
Rotational KE
Rolling
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- CM may lie outside the body; internal forces cannot change its motion.
- Torque τ = rF sinθ; SI unit N m.
- No external torque → angular momentum conserved: I₁ω₁ = I₂ω₂.
- Standard values: ring MR², disc ½MR², solid sphere 2/5 MR², rod about centre ML²/12.
- Parallel axes: I = I_CM + Md².
- Rolling condition v = rω; rolling KE = ½Mv² + ½Iω².
- Solid sphere beats disc beats ring down a slope.
Practice Numericals
Step-by-step worked examples
Standard exam-level problems solved with clear conceptual breakdowns.
Centre of mass of two particles
Step-by-Step Solution
- 1x_CM = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2×0 + 3×2)/5 = 6/5 = 1.2 m.
- 2y_CM = 0. The centre of mass is at (1.2, 0) m, closer to the 3 kg mass.
Final Answer
(1.2, 0) m
Torque about a pivot
Step-by-Step Solution
- 1τ = rF sinθ = 0.4 × 12 × sin30° = 0.4 × 12 × 0.5 = 2.4 N m.
- 2The turning effect is 2.4 N m about the pivot.
Final Answer
2.4 N m
Rolling sphere down an incline
Step-by-Step Solution
- 1Energy conservation: mgh = ½mv² + ½Iω², with I = 2/5 mR² and v = Rω.
- 2mgh = ½mv² + ½(2/5 mR²)(v²/R²) = ½mv² + (1/5)mv² = (7/10)mv².
- 3v² = (10/7)gh, so v = √(10gh/7).
Final Answer
v = √(10gh/7)
FAQ
Frequently asked questions
Why can the centre of mass lie outside the body?
The centre of mass is a weighted average of positions, not necessarily a point occupied by matter. For a hollow ring it lies at the geometric centre where there is no material — internal forces and even the shape of the body do not require the CM to be inside.
What is the difference between linear and angular momentum conservation?
Linear momentum is conserved when the net external force is zero; angular momentum is conserved when the net external torque is zero. They are the translational and rotational statements of the same idea — no external influence means the corresponding momentum stays constant.
Why does an ice skater spin faster when pulling in her arms?
With no external torque, angular momentum L = Iω is conserved. Pulling arms in reduces the moment of inertia I, so the angular speed ω must increase to keep L constant — she spins faster.
What is the perpendicular axes theorem and when does it apply?
For a plane lamina, the moment of inertia about an axis perpendicular to the lamina equals the sum of moments of inertia about two perpendicular axes in the lamina meeting at the same point: I_z = I_x + I_y. It applies only to planar (lamina) bodies.
Why does a sphere roll down an incline faster than a ring?
Rolling divides the potential energy between translation and rotation. The sphere has the smallest moment-of-inertia factor (2/5) so it stores less energy in rotation, leaving more for translational acceleration — it reaches the bottom first.
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