CBSE Class 11 Physics Revision Notes Chapter 6 System of Particles and Rotational Motion

System of Particles and Rotational Motion explains how extended bodies translate, rotate and respond to external force and torque. For CBSE Class 11 Physics, the chapter connects centre of mass, momentum, angular motion, equilibrium and rotational dynamics.

A real body has a finite size and contains a large number of particles. Its motion may involve translation, rotation or both. The particle model alone cannot fully describe such motion.

These CBSE Class 11 Physics Revision Notes Chapter 6 follow the 2026–27 chapter sequence. They cover the centre of mass, vector product, torque, angular momentum, moment of inertia, equilibrium and fixed-axis rotation.

Key Takeaways

• Centre of mass: Its motion depends only on the net external force acting on the system.
• Torque: The turning effect of a force is τ = r × F.
• Moment of inertia: It measures resistance to angular acceleration about a chosen axis.
• Angular momentum: It remains constant when the net external torque is zero.

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Access Class 11 Physics Chapter 6 System of Particles and Rotational Motion Notes in 30 Minutes

These Class 11 System of Particles and Rotational Motion Revision Notes organise the chapter into linked translational and rotational concepts.

The sequence also makes these Class 11 Physics Chapter 6 Notes useful for formula-based revision.

Motion of a Rigid Body

A rigid body is an ideal body whose shape does not change. The distance between any two particles of the body remains constant.

No real body is perfectly rigid. However, deformation is negligible in many situations, so bodies may be treated as rigid.

Pure Translational Motion

In pure translational motion, every particle of the body has the same velocity at a given instant.

A block sliding down an inclined plane without rotating shows pure translation.

The orientation of the body remains unchanged during pure translation.

Pure Rotational Motion

In pure rotational motion about a fixed axis, every particle moves in a circle.

Each circle:

• Lies in a plane perpendicular to the axis
• Has its centre on the axis
• Has a radius equal to the particle’s perpendicular distance from the axis

A ceiling fan and a potter’s wheel show rotation about a fixed axis.

Particles lying on the axis remain stationary.

Translation Combined with Rotation

Rolling motion combines translation and rotation.

A cylinder rolling down an inclined plane translates through the motion of its centre of mass. It also rotates about an axis through the centre of mass.

The particles of the body have different velocities during combined motion.

Centre of Mass of a System of Particles

The centre of mass is the mass-weighted mean position of all particles in a system.

It describes the translational motion of the system as a whole.

Centre of Mass of Two Particles

For masses m1 and m2 at positions x1 and x2:

X = (m1x1 + m2x2)/(m1 + m2)

If m1 = m2:

X = (x1 + x2)/2

The centre of mass lies midway between two particles of equal mass.

For unequal masses, it lies closer to the heavier particle.

Centre of Mass of Many Particles

For n particles along the x-axis:

X = Σmixi/Σmi

Since total mass M = Σmi:

X = Σmixi/M

In three dimensions:

X = Σmixi/M

Y = Σmiyi/M

Z = Σmizi/M

In vector form:

R = Σmiri/M

Here:

• R = position vector of the centre of mass
• ri = position vector of the ith particle
• M = total mass of the system

Centre of Mass of a Continuous Body

For a continuous mass distribution:

R = (1/M)∫r dm

The coordinate forms are:

X = (1/M)∫x dm

Y = (1/M)∫y dm

Z = (1/M)∫z dm

The summation over particles becomes integration over mass elements.

Centre of Mass of Symmetrical Bodies

For a homogeneous body with suitable symmetry, the centre of mass lies at the geometric centre.

Examples include:

• Uniform rod
• Ring
• Disc
• Sphere
• Uniform rectangular plate

The centre of mass may lie outside the material of a body. For example, the centre of mass of a ring lies at its empty centre.

Motion and Linear Momentum of the Centre of Mass

The motion of centre of mass helps separate the translation of an extended body from its internal or rotational motion.

Velocity and Acceleration of the Centre of Mass

For a system of particles:

MR = Σmiri

Differentiating with respect to time:

MV = Σmivi

Therefore:

V = Σmivi/M

Differentiating again:

MA = Σmiai

Here:

• V = velocity of the centre of mass
• A = acceleration of the centre of mass

Effect of External Forces

Internal forces occur in equal and opposite pairs. Their total contribution to the motion of the centre of mass is zero.

Therefore:

Fext = MA

The centre of mass moves as if:

• The entire mass were concentrated at the centre of mass.
• The total external force acted at that point.

An explosion changes the motion of individual fragments. However, the centre of mass continues under the same external force.

Linear Momentum of a System

The linear momentum of a system of particles is the vector sum of the momenta of all particles.

P = Σmivi

Since:

MV = Σmivi

Therefore:

P = MV

The total momentum equals the total mass multiplied by the velocity of the centre of mass.

Differentiating:

Fext = dP/dt

If Fext = 0:

P = constant

The total linear momentum remains conserved, and the centre of mass moves with constant velocity.

Vector Product in Rotational Motion

The vector product or cross product produces a vector perpendicular to the plane containing the original vectors.

It is used to define torque and angular momentum.

Magnitude and Direction of Cross Product

For vectors A and B:

A × B = C

The magnitude is:

C = AB sin θ

Here, θ is the smaller angle between A and B.

The direction follows the right-hand rule.

Curl the fingers of the right hand from A towards B. The thumb gives the direction of A × B.

The vector product is not commutative:

A × B = −(B × A)

It follows the distributive law:

A × (B + C) = A × B + A × C

Cross Products of Unit Vectors

The cyclic products are:

i × j = k

j × k = i

k × i = j

Reversing the order changes the sign:

j × i = −k

k × j = −i

i × k = −j

Also:

i × i = j × j = k × k = 0

Angular Velocity and Linear Velocity

Every particle in a rigid body rotating about a fixed axis has the same angular displacement, angular velocity and angular acceleration.

Angular Displacement, Velocity and Acceleration

Angular displacement is represented by θ.

The average angular velocity is:

ωavg = Δθ/Δt

Instantaneous angular velocity is:

ω = dθ/dt

Angular acceleration is:

α = dω/dt

The SI units are:

• Angular displacement: radian
• Angular velocity: rad s⁻¹
• Angular acceleration: rad s⁻²

The direction of angular velocity follows the right-hand rule.

Relation Between Linear and Angular Velocity

For a particle at position r from the axis:

v = ω × r

The magnitude is:

v = ωr sin θ

For a particle moving in a circle perpendicular to the axis:

v = rω

The direction of linear velocity is tangential to the circular path.

The relation between linear velocity and angular velocity shows that particles farther from the axis move faster for the same ω.

Tangential acceleration is:

at = rα

Radial or centripetal acceleration is:

ar = rω²

Torque and Angular Momentum

Force changes linear momentum. Similarly, torque changes angular momentum.

Moment of Force or Torque

The torque of a force F about a point is:

τ = r × F

Its magnitude is:

τ = rF sin θ

It may also be written as:

τ = Fd

Here, d is the perpendicular distance between the axis and the line of action of force.

The SI unit of torque is N m.

Torque depends on:

• Magnitude of force
• Distance from the axis
• Angle between r and F

A force passing through the axis produces zero torque.

Angular Momentum of a Particle

The angular momentum of a particle about a point is:

L = r × p

Since p = mv:

L = r × mv

The magnitude is:

L = mvr sin θ

For circular motion with r perpendicular to v:

L = mvr

The direction follows the right-hand rule.

Torque-Angular Momentum Relation

Torque equals the rate of change of angular momentum.

τ = dL/dt

For a system of particles:

τext = dL/dt

Internal torques cancel when the forces satisfy Newton’s third law along the line joining the particles.

Conservation of Angular Momentum

If the net external torque is zero:

τext = 0

Therefore:

dL/dt = 0

Hence:

L = constant

This is the conservation of angular momentum.

For rotation about a fixed axis:

L = Iω

Therefore:

I1ω1 = I2ω2

A skater spins faster by drawing the arms inward because the moment of inertia decreases.

Equilibrium of a Rigid Body

The equilibrium of a rigid body requires both force balance and torque balance.

Translational Equilibrium

For translational equilibrium:

ΣF = 0

This means:

ΣFx = 0

ΣFy = 0

ΣFz = 0

The centre of mass has zero acceleration.

Rotational Equilibrium

For rotational equilibrium:

Στ = 0

The angular acceleration is zero.

A body may satisfy translational equilibrium but still rotate if the net torque is non-zero.

It may also have zero net torque but accelerate linearly if the net force is non-zero.

Complete equilibrium requires:

ΣF = 0

Στ = 0

Principle of Moments

For a body in rotational equilibrium:

Sum of clockwise moments = Sum of anticlockwise moments

For two forces:

F1d1 = F2d2

This principle is used in balances, levers and seesaws.

Centre of Gravity

The centre of gravity is the point through which the resultant gravitational force acts.

In a uniform gravitational field, the centre of gravity coincides with the centre of mass.

The centre of gravity may lie outside the body, as in a ring.

A body remains more stable when:

• Its centre of gravity is lower.
• Its base is wider.
• The vertical line through the centre of gravity falls within the base.

Moment of Inertia and Radius of Gyration

Mass measures resistance to linear acceleration. The moment of inertia measures resistance to angular acceleration.

Definition of Moment of Inertia

For particles of masses mi at perpendicular distances ri from an axis:

I = Σmiri²

For a continuous body:

I = ∫r² dm

The SI unit is kg m².

The dimensions are:

[ML²]

The moment of inertia depends on the chosen axis. The same body can have different moments of inertia about different axes.

Radius of Gyration

The radius of gyration k is defined by:

I = Mk²

Therefore:

k = √(I/M)

It is the distance from the axis at which the entire mass may be considered concentrated without changing the moment of inertia.

Factors Affecting Moment of Inertia

Moment of inertia depends on:

• Total mass of the body
• Distribution of mass
• Shape and size of the body
• Position and direction of the axis

More mass located farther from the axis produces a larger moment of inertia.

Moments of Inertia of Standard Bodies

Parallel and Perpendicular Axis Theorems

Axis theorems help calculate the moment of inertia about a new axis from a known value.

Parallel Axis Theorem

The parallel axis theorem states:

I = ICM + Md²

Here:

• ICM = moment of inertia about a parallel axis through the centre of mass
• M = total mass
• d = perpendicular distance between the axes

This theorem applies to any rigid body.

For a rod about one end:

Iend = ICM + M(L/2)²

Iend = 1/12 ML² + 1/4 ML²

Iend = 1/3 ML²

Perpendicular Axis Theorem

The perpendicular axis theorem applies to a plane lamina.

If x and y are perpendicular axes in its plane and z is perpendicular to the plane:

Iz = Ix + Iy

All three axes must pass through the same point.

For a circular disc:

Ix = Iy

Since Iz = 1/2 MR²:

Ix = Iy = 1/4 MR²

Kinematics of Rotational Motion About a Fixed Axis

The equations of rotational motion resemble the equations of linear motion.

For constant angular acceleration:

ω = ω0 + αt

θ = ω0t + 1/2 αt²

ω² = ω0² + 2αθ

θ = [(ω0 + ω)/2]t

These equations apply when α is constant.

The Rotational Motion Class 11 Notes use θ, ω and α in the same way that linear motion uses x, v and a.

Dynamics of Rotational Motion About a Fixed Axis

Rotational dynamics connects torque with angular acceleration and rotational inertia.

Torque and Angular Acceleration

For a rigid body rotating about a fixed axis:

τ = Iα

Here:

• τ = net external torque
• I = moment of inertia
• α = angular acceleration

This is the rotational form of F = ma.

A larger moment of inertia requires greater torque for the same angular acceleration.

Rotational Kinetic Energy

A rotating body possesses rotational kinetic energy.

Krot = 1/2 Iω²

Each particle contributes 1/2 mivi² to the total energy.

Since vi = riω:

Krot = 1/2 Σmiri²ω²

Therefore:

Krot = 1/2 Iω²

Work and Power in Rotational Motion

For a small angular displacement dθ:

dW = τ dθ

Therefore:

W = ∫τ dθ

For constant torque:

W = τθ

Power is:

P = dW/dt

Therefore:

P = τω

These are the rotational forms of W = Fs and P = Fv.

Linear and Rotational Motion Comparison

System of Particles and Rotational Motion Formula Sheet

These System of Particles and Rotational Motion Class 11 Notes bring the main relations together for quick revision.

These System of Particles and Rotational Motion Notes connect translational motion with their rotational equivalents instead of presenting formulas separately.

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