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Class 11 Physics · NCERT Chapter 6

System 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.

ChapterSystem of Particles & Rotational MotionClassClass 11SubjectPhysicsBoardCBSEExamsCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

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).

R⃗=m1r⃗1+m2r⃗2+⋯m1+m2+⋯=∑mir⃗i∑mi\vec{R} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \cdots}{m_1 + m_2 + \cdots} = \frac{\sum m_i\vec{r}_i}{\sum m_i}
Centre of mass of a system of particles

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

When a projectile or a system of connected parts is in flight, the centre of mass continues along the original parabolic path even if the pieces fly apart — internal forces cannot alter the CM's trajectory.

MCQ Questions

1The centre of mass of a system of particles —

2The motion of the centre of mass is governed only by —

3Two particles of masses 1 kg and 3 kg are 40 cm apart. The centre of mass is located —

4The centre of mass of a uniform rod lies —

1 Mark Questions

  1. Define the centre of mass of a system of particles.
  2. Write the expression for the position vector of the centre of mass of two particles of masses m₁ and m₂.
  3. Where is the centre of mass of a uniform ring located?

2 Mark Questions

  1. State the rule for locating the centre of mass of two particles of unequal masses.
  2. Show that internal forces cannot change the motion of the centre of mass.
  3. Two 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

  1. Derive 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.
  2. Three 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.

τ⃗=r⃗×F⃗,∣τ∣=rFsin⁡θ\vec{\tau} = \vec{r} \times \vec{F}, \qquad |\tau| = rF\sin\theta
Torque about a point

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

To solve an equilibrium problem, pick a pivot where an unknown force acts — its torque is then zero, leaving fewer unknowns in Στ = 0.

MCQ Questions

1The SI unit of torque is —

2Torque is maximum when the force makes an angle of —

3A torque is zero when the line of action of the force —

4A couple consists of two equal and opposite forces. It produces —

1 Mark Questions

  1. Define torque. Write its SI unit.
  2. State the conditions for translational equilibrium of a rigid body.
  3. What is a couple?

2 Mark Questions

  1. A 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.
  2. Explain why the effectiveness of a wrench is increased by making its handle longer.
  3. State the conditions for equilibrium of a rigid body under the action of several forces.

3 Mark Questions

  1. A 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.
  2. Derive 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.

L⃗=r⃗×p⃗,L=Iω,τ⃗=dL⃗dt\vec{L} = \vec{r} \times \vec{p}, \qquad L = I\omega, \qquad \vec{\tau} = \frac{d\vec{L}}{dt}
Angular momentum and its relation to torque

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

A figure skater pulling in arms from I₁ to I₂ (smaller) speeds up from ω₁ to ω₂ such that I₁ω₁ = I₂ω₂. Angular momentum stays constant; rotational kinetic energy and speed increase.

MCQ Questions

1Angular momentum is conserved when the net external —

2For a rigid body rotating about a fixed axis, angular momentum L = —

3When an ice skater pulls her arms in while spinning, her angular speed —

4The relation between torque and angular momentum is —

1 Mark Questions

  1. Define angular momentum of a particle about a point.
  2. Write the relation between torque and angular momentum.
  3. State the law of conservation of angular momentum.

2 Mark Questions

  1. Show that τ = dL/dt follows from Newton's second law for a particle.
  2. A diver curls into a tight ball during a dive then straightens out on entry. Explain the change in angular speed using angular momentum conservation.
  3. State the law of conservation of angular momentum and give two examples where it applies.

3 Mark Questions

  1. Prove 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.
  2. A 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.

I=∑miri2,τ=IαI = \sum m_i r_i^2, \qquad \tau = I\alpha
Moment of inertia and the rotational analogue of Newton's second law

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.

I=ICM+Md2(parallel axes),Iz=Ix+Iy(perpendicular axes)I = I_{\text{CM}} + Md^2 \quad \text{(parallel axes)}, \qquad I_z = I_x + I_y \quad \text{(perpendicular axes)}
Two important theorems

Remember the thin disc

For a uniform thin disc of radius R: about the centre (⊥ plane) I = ½MR²; about a diameter I = ¼MR² (by the perpendicular-axes theorem, two diameters sum to ½MR²). These appear constantly in JEE and NEET.

MCQ Questions

1The moment of inertia of a body depends on —

2The moment of inertia of a uniform ring of mass M and radius R about an axis through its centre and perpendicular to its plane is —

3The parallel axes theorem is written as —

4The radius of gyration of a body is related to its moment of inertia by —

1 Mark Questions

  1. Define moment of inertia. Write its SI unit.
  2. State the theorem of parallel axes.
  3. Define the radius of gyration.

2 Mark Questions

  1. State and prove the theorem of perpendicular axes for a plane lamina.
  2. A ring and a disc of the same mass and radius rotate about their central axes. Compare their moments of inertia.
  3. Write 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

  1. State 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.
  2. Find 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.

KErot=12Iω2,KEtotal=12Mv2+12Iω2\text{KE}_{\text{rot}} = \frac{1}{2}I\omega^2, \qquad \text{KE}_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2
Rotational and rolling kinetic energy

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

Rolling down a given incline, a solid sphere (I = 2/5 MR²) reaches the bottom before a cylinder and before a ring, because a smaller fraction of the available energy goes into rotation — so more is available for speeding up.

MCQ Questions

1The rotational kinetic energy of a body rotating with angular speed ω is —

2For rolling without slipping, the relation between linear speed v and angular speed ω is —

3A ring, a disc and a solid sphere of the same mass and radius roll down a slope. The one that reaches the bottom first is —

4The work done by a torque τ rotating a body through an angle θ is —

1 Mark Questions

  1. Write the expression for the rotational kinetic energy of a rigid body.
  2. State the rolling condition for a body moving without slipping.
  3. Give the expression for the total kinetic energy of a rolling body.

2 Mark Questions

  1. A flywheel of moment of inertia 2 kg m² rotates at 300 rpm. Find its rotational kinetic energy.
  2. Explain why a solid sphere rolls down an incline faster than a ring of the same mass and radius.
  3. Derive the total kinetic energy of a rolling body as ½Mv² + ½Iω².

3 Mark Questions

  1. A solid sphere rolls down an inclined plane of height h starting from rest. Obtain the velocity of its centre of mass at the bottom.
  2. State and prove that the total kinetic energy of a rolling body equals ½Mv²(1 + k²/R²), where k is its radius of gyration.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Centre of mass

R⃗=∑mir⃗i∑mi\vec{R} = \frac{\sum m_i\vec{r}_i}{\sum m_i}

Torque

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

Angular momentum

L⃗=r⃗×p⃗=Iω\vec{L} = \vec{r} \times \vec{p} = I\omega

Torque–angular momentum

τ⃗=dL⃗dt\vec{\tau} = \frac{d\vec{L}}{dt}

Moment of inertia

I=∑miri2,τ=IαI = \sum m_i r_i^2, \quad \tau = I\alpha

Parallel axes

I=ICM+Md2I = I_{\text{CM}} + Md^2

Perpendicular axes

Iz=Ix+IyI_z = I_x + I_y

Rotational KE

KE=12Iω2\text{KE} = \frac{1}{2}I\omega^2

Rolling

v=rω,KE=12Mv2+12Iω2v = r\omega, \quad \text{KE} = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • 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.

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

Centre of mass of two particles

Two masses of 2 kg and 3 kg are placed at the origin and at (2, 0) m respectively. Find the centre of mass.

  1. x_CM = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2×0 + 3×2)/5 = 6/5 = 1.2 m.
  2. y_CM = 0. The centre of mass is at (1.2, 0) m, closer to the 3 kg mass.

Answer

(1.2, 0) m

2

Torque about a pivot

A force of 12 N is applied at the end of a rod of length 0.4 m making an angle of 30° with the rod. Find the torque about the pivot at the other end.

  1. τ = rF sinθ = 0.4 × 12 × sin30° = 0.4 × 12 × 0.5 = 2.4 N m.
  2. The turning effect is 2.4 N m about the pivot.

Answer

2.4 N m

3

Rolling sphere down an incline

A solid sphere of radius R rolls down an incline of height h from rest. Find the speed of its centre of mass at the bottom.

  1. Energy conservation: mgh = ½mv² + ½Iω², with I = 2/5 mR² and v = Rω.
  2. mgh = ½mv² + ½(2/5 mR²)(v²/R²) = ½mv² + (1/5)mv² = (7/10)mv².
  3. v² = (10/7)gh, so v = √(10gh/7).

Answer

v = √(10gh/7)

FAQ

Common 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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