Systems of Particles and Rotational Motion Class 11 Notes: CBSE Physics Ch 6

These Systems of Particles and Rotational Motion Class 11 notes condense CBSE Physics Chapter 6 into a revision-ready page: definitions, formulas, SI units, worked examples, and exam patterns straight from the NCERT textbook. Use them after your first read of the chapter — they are built for one focused revision sitting, not for first learning.

The chapter’s core idea: real bodies are not point masses, and any extended body can be treated as a system of particles. Its motion splits into translation of the centre of mass plus rotation about that point. New quantities — torque, angular momentum, moment of inertia — are the rotational twins of force, momentum, and mass.

The sections below follow exactly that build-up.

Centre of Mass — Where the Whole System Balances

The centre of mass (CM) of a system of particles is its mass-weighted average position — the balancing point of the whole system (NCERT, p. 96).

For two particles on the x-axis, the CM lies at:

\[ X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]

• Equal masses: the CM is exactly the midpoint between the two particles.
• n particles in space (Eqs. 6.4a–c): \( \mathbf{R} = \frac{\sum m_i \mathbf{r}_i}{M} \), so \( X = \frac{\sum m_i x_i}{M} \), \( Y = \frac{\sum m_i y_i}{M} \), \( Z = \frac{\sum m_i z_i}{M} \).
• Continuous bodies: sums become integrals — \( \mathbf{R} = \frac{1}{M}\int \mathbf{r}\,dm \) (Eq. 6.5b), where \( dm \) is each tiny mass element.

Symmetry shortcut: the CM of a homogeneous (uniform-density) rod, ring, disc, or solid sphere lies at its geometric centre. Reflection symmetry is the reason: every mass element \( dm \) at position \( +x \) has a twin of equal mass at \( -x \), so \( \int x\,dm = 0 \) (NCERT, p. 97).

Worked example — three particles at the vertices of an equilateral triangle. Masses \( 200\ \text{g} \), \( 300\ \text{g} \), and \( 500\ \text{g} \) sit at vertices \( O(0,0) \), \( A(0.5,0) \), and \( B(0.25,\ 0.25\sqrt{3}) \) of an equilateral triangle of side 0.5 m. Find the CM of the system.

Motion of Centre of Mass and Linear Momentum of a System

Differentiate \( M\mathbf{R} = \sum m_i \mathbf{r}_i \) twice with respect to time and apply Newton’s second law to each particle. Internal forces cancel in pairs by Newton’s third law — see the Laws of Motion notes — leaving only the external forces:

\[ M\mathbf{A} = \mathbf{F}_{\text{ext}} \quad \text{(Eq. 6.11, NCERT, p. 100)} \]

Meaning: the centre of mass moves as if all the system’s mass were concentrated at it and all external forces acted there. To predict the CM’s motion, you never need the internal forces.

The classic picture is Fig. 6.12 of the textbook: a projectile explodes mid-air into fragments. The explosion forces are internal, so the CM of the fragments continues exactly along the original parabola, as if no explosion had happened.

Total linear momentum: \( \mathbf{P} = \sum m_i\mathbf{v}_i = M\mathbf{V} \) — the vector sum of momenta equals total mass times CM velocity (Eq. 6.15, NCERT, p. 101). Differentiating gives:

\[ \frac{d\mathbf{P}}{dt} = \mathbf{F}_{\text{ext}} \]

Conservation: if \( \mathbf{F}_{\text{ext}} = 0 \), then \( \mathbf{P} \) is constant and the CM moves like a free particle. This is how radioactive decay is understood: radium splits into radon plus an alpha particle by internal forces, so the CM of the products keeps the original path (Fig. 6.13). In the CM frame, the two products fly back-to-back.

Worked example — two ice skaters push apart. A 60 kg skater and a 40 kg skater, at rest on frictionless ice, push each other. The 60 kg skater moves off at 2.0 m/s. Find the velocity of the 40 kg skater.

Vector Product of Two Vectors — the Tool for Rotation

Torque and angular momentum are both vector products, so master the cross product first (NCERT, pp. 102–103). For vectors \( \mathbf{a} \) and \( \mathbf{b} \) with magnitudes \( a \), \( b \) and included angle \( \theta \):

• Magnitude: \( |\mathbf{a}\times\mathbf{b}| = ab\sin\theta \).
• Direction: perpendicular to the plane of \( \mathbf{a} \) and \( \mathbf{b} \), fixed by the right-hand screw rule — turn the screw from \( \mathbf{a} \) to \( \mathbf{b} \); the tip advances along \( \mathbf{a}\times\mathbf{b} \).

Properties to quote in answers:

• Not commutative: \( \mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a}) \).
• Distributive over addition: \( \mathbf{a}\times(\mathbf{b}+\mathbf{c}) = \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c} \).
• \( \mathbf{a}\times\mathbf{a} = \mathbf{0} \), the null vector.
• Unit vectors are cyclic: \( \hat{\mathbf{i}}\times\hat{\mathbf{j}} = \hat{\mathbf{k}} \), \( \hat{\mathbf{j}}\times\hat{\mathbf{k}} = \hat{\mathbf{i}} \), \( \hat{\mathbf{k}}\times\hat{\mathbf{i}} = \hat{\mathbf{j}} \); any anti-cyclic order carries a minus sign.

In component form, compute \( \mathbf{a}\times\mathbf{b} \) as a determinant:

\[ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \]

You first met the scalar product in the Work, Energy and Power notes; the cross product is its vector-valued cousin.

Worked example — scalar and vector products. Let \( \mathbf{a} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}} – \hat{\mathbf{k}} \) and \( \mathbf{b} = \hat{\mathbf{i}} – 2\hat{\mathbf{j}} + 4\hat{\mathbf{k}} \). Find \( \mathbf{a}\cdot\mathbf{b} \) and \( \mathbf{a}\times\mathbf{b} \).

Angular Velocity, Angular Acceleration and Relation with Linear Velocity

For rotation about a fixed axis, every particle of the rigid body moves in a circle lying in a plane perpendicular to the axis, with its centre on the axis; particles on the axis itself stay fixed (NCERT, p. 93).

Angular velocity \( \boldsymbol{\omega} \) is the same for all particles of the body at any instant. It is a vector along the axis of rotation, directed by the right-hand screw rule. The linear velocity of a particle at position vector \( \mathbf{r} \) is (Eq. 6.20, NCERT, p. 105):

\[ \mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}, \qquad |\mathbf{v}| = v = \omega r \]

• \( v \) is always tangential — along the tangent to the particle’s circle, perpendicular to both the axis and the radius.
• Particles on the axis (\( r = 0 \)) have zero speed — that is what “fixed axis” means.

Angular acceleration: \( \alpha = d\omega/dt \) (Eq. 6.21). For uniform \( \alpha \), the kinematics mirror linear motion exactly (NCERT, p. 117):

\[ \omega = \omega_0 + \alpha t, \qquad \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2, \qquad \omega^2 = \omega_0^2 + 2\alpha(\theta-\theta_0) \]

WHY the same algebra works: each rotational variable is defined from its linear twin by the same rate — \( \theta \) from \( x \), \( \omega \) from \( v \), \( \alpha \) from \( a \) — so every linear formula has a direct rotational copy.

Worked example — motor wheel. A motor wheel speeds up uniformly from 600 rpm to 1800 rpm in 10 s. Find (a) the angular acceleration and (b) the number of revolutions made in this time.

Torque and Angular Momentum — Definitions and Relation

Torque (moment of force) is the rotational analogue of force: \( \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} \) (Eq. 6.23, NCERT, p. 106). Its magnitude takes three useful forms:

\[ \tau = rF\sin\theta = r_\perp F = rF_\perp \]

• \( r_\perp = r\sin\theta \) is the lever arm — the perpendicular distance of the force’s line of action from the axis.
• \( F_\perp \) is the component of force perpendicular to \( \mathbf{r} \).
• SI unit \( \text{N m} \); dimensions \( ML^2T^{-2} \). Same dimensions as work, but torque is a vector and work is a scalar.
• \( \tau = 0 \) when the force’s line of action passes through the axis (\( \theta = 0^\circ \) or \( 180^\circ \)).

The door example makes this physical: a force on the hinge line rotates nothing, while the same force at the outer edge (maximum lever arm) rotates the door easily. Rotation depends on where and how a force is applied, not just on its size.

Angular momentum of a particle: \( \mathbf{l} = \mathbf{r}\times\mathbf{p} = m(\mathbf{r}\times\mathbf{v}) \) (Eq. 6.25a, NCERT, p. 107). Magnitude \( l = rp\sin\theta \); SI unit \( \text{kg m}^2/\text{s} \). Differentiating the product gives the central link:

\[ \frac{d\mathbf{l}}{dt} = \boldsymbol{\tau}, \qquad \text{and for a system: } \frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{\text{ext}} \]

(Eqs. 6.27, 6.28b, NCERT, p. 108). Internal torques drop out only when the internal forces are central — directed along the line joining each pair of particles.

Worked example — torque about the origin. A force \( \mathbf{F} = 2\hat{\mathbf{i}} – \hat{\mathbf{j}} + 3\hat{\mathbf{k}} \) acts at the point \( \mathbf{r} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} – \hat{\mathbf{k}} \). Find the torque about the origin.

Equilibrium of Rigid Bodies — Translational and Rotational

A rigid body is in mechanical equilibrium when it has neither linear nor angular acceleration (NCERT, p. 110):

• Translational: \( \sum \mathbf{F}_i = \mathbf{0} \) — no net force.
• Rotational: \( \sum \boldsymbol{\tau}_i = \mathbf{0} \) — no net torque.

These two vector equations are equivalent to six scalar conditions. If all forces are coplanar, only three conditions are needed: two for forces, one for torques. A body can be in partial equilibrium — balanced translationally but spinning, or vice versa.

Couple: two equal, opposite, non-collinear parallel forces. The net force is zero but the moments act in the same sense, so a couple produces pure rotation without translation — exactly how two fingers turn a bottle lid (NCERT, p. 111).

The moment of a couple equals the magnitude of one force times the perpendicular separation of the forces, and is independent of the point about which moments are taken.

Principle of moments (lever, NCERT, p. 112):

\[ \text{load} \times \text{load arm} = \text{effort} \times \text{effort arm}, \qquad \text{M.A.} = \frac{\text{effort arm}}{\text{load arm}} \]

A lever with a longer effort arm multiplies your effort — a small effort lifts a heavy load. A see-saw and a beam balance are both levers.

Centre of gravity (CG) is the point at which the total gravitational torque on the body is zero — the spot where a cardboard balances on a pencil tip (NCERT, p. 113). The CG coincides with the CM when \( \mathbf{g} \) is uniform over the body; the two are conceptually different, and the CM has nothing to do with gravity.

You will meet this again in the Gravitation notes.

Worked example — reactions on a bar. A uniform 3 kg metal bar, 80 cm long, rests on two knife-edges placed 10 cm from each end. A 5 kg load hangs 25 cm from the left end. Find the reactions at the knife-edges.

Worked example — ladder against a frictionless wall. A 4 m ladder of mass 15 kg leans against a smooth vertical wall, its foot 1.5 m from the wall. Find the reaction forces of the wall and the floor.

Moment of Inertia and Rotational Kinetic Energy

A particle at distance \( r_i \) from the axis moves with speed \( v_i = r_i\omega \), so its kinetic energy is \( \frac{1}{2}m_i r_i^2\omega^2 \). Summing over all particles and defining the moment of inertia (NCERT, p. 115):

\[ I = \sum_i m_i r_i^2, \qquad K = \frac{1}{2}I\omega^2 \]

• SI unit \( \text{kg m}^2 \); dimensions \( ML^2 \).
• I is the rotational analogue of mass — a measure of rotational inertia that quantifies how mass is distributed about the axis.
• Unlike mass, I is not a fixed property of a body: it depends on the chosen axis and on the mass distribution.

Radius of gyration \( k \), defined by \( I = Mk^2 \), is the distance from the axis at which the whole mass M could be concentrated to give the same I (NCERT, p. 116). For a rod about its centre, \( k = L/\sqrt{12} \); for a disc about a diameter, \( k = R/2 \).

Table 6.1 — moments of inertia of standard bodies (NCERT, p. 116). Memorise the axis with each value:

Memory device: a ring is “its own everything” — every bit of mass sits at distance R, so \( I = MR^2 \). A disc spreads its mass over the face, so about its own axis \( I = MR^2/2 \). The solid sphere’s \( 2MR^2/5 \) is the odd one out — recall it by repetition.

For thin flat bodies, the value about a diameter is half the value about the central perpendicular axis.

Real-life use — the flywheel: engines fit a heavy disc of large I on the crankshaft because rotational inertia resists sudden speed changes. The flywheel smooths jerky motion and keeps the ride even (NCERT, p. 116).

Micro-example — inertia of a flywheel. A flywheel disc has mass 30 kg and radius 25 cm. About its axle:

\[ I = \frac{MR^2}{2} = \frac{30(0.25)^2}{2} = 0.9375\ \text{kg m}^2 \]

Dynamics of Rotation — Work, Power and Newton’s Second Law for Rotation

Rotational motion about a fixed axis mirrors linear motion quantity by quantity (Table 6.2, NCERT, p. 119):

For rotation about a fixed axis, only forces lying in planes perpendicular to the axis, and only torque components along the axis, need to be considered (NCERT, p. 119). A force acting through an arc \( ds = r\,d\theta \) does work \( dW = F\,ds = Fr\,d\theta = \tau\,d\theta \). Summing over forces:

\[ dW = \tau\,d\theta, \qquad P = \frac{dW}{dt} = \tau\omega \]

In a perfectly rigid body, work by external torques is not dissipated; it becomes rotational kinetic energy. Equating power \( P = \tau\omega \) with the rate of change of kinetic energy \( \frac{d}{dt}(\frac{1}{2}I\omega^2) = I\omega\alpha \) gives the rotational form of Newton’s second law (Eq. 6.41, NCERT, p. 120):

\[ \tau = I\alpha \]

WHY it makes sense: torque plays the role of force and I the role of mass — the same applied torque produces less angular acceleration when the body is harder to rotate (larger I).

Worked example — cord pulled on a flywheel. A cord is wound on the rim of a flywheel of mass 15 kg and radius 30 cm, mounted on frictionless bearings. A steady pull of 40 N unwinds 2 m of cord from rest. Find (a) the angular acceleration, (b) the work done by the pull, (c) the kinetic energy gained.

Angular Momentum About a Fixed Axis and Its Conservation

For a particle of a rotating body, the component of angular momentum along the fixed axis is \( l_z = mr_\perp^2\omega \). Summing over the whole body (NCERT, p. 122):

\[ \mathbf{L}_z = I\omega\,\hat{\mathbf{k}} \]

For a body symmetric about the rotation axis, \( \mathbf{L} = \mathbf{L}_z = I\boldsymbol{\omega} \): the angular momentum vector lies exactly along the axis. For asymmetric bodies, L also has a component perpendicular to the axis.

Conservation of angular momentum: when the total external torque is zero, \[ \frac{d\mathbf{L}}{dt} = 0 \quad \Rightarrow \quad L_z = I\omega = \text{constant} \quad \text{(Eq. 6.44, NCERT, p. 123)} \]

The swivel-chair experiment (Fig. 6.32 of the textbook) is the standard demonstration: sit on a rotating chair with your arms out, then pull them in. Your moment of inertia drops and your angular speed jumps — because \( I\omega \) stays constant when no external torque acts about the chair’s axis.

Performers exploit the same principle: a diver tucks to spin faster during a somersault and opens out to slow down before entering the water; skaters and classical dancers pull their arms in to speed up a pirouette (NCERT, pp. 122–123).

Worked example — diver tucking. A diver leaves the board spinning at 2 rad/s with moment of inertia 8 kg m². Tucking reduces the moment of inertia to 2 kg m². Find the new spin rate.

Common Mistakes and Exam Pointers for Rotational Motion

Common mistakes in systems of particles and rotational motion

One subtlety examiners like: internal forces always cancel in the force sum, but internal torques cancel only when those forces are central — directed along the line joining the particles. Quote this condition when you write \( d\mathbf{L}/dt = \boldsymbol{\tau}_{\text{ext}} \).

Exam pointers — what earns the marks

• Define each term before using it: torque, angular momentum, moment of inertia — with SI units and dimensions (N m, \( ML^2T^{-2} \); \( \text{kg m}^2\text{s}^{-1} \), \( ML^2T^{-1} \); kg m², \( ML^2 \)).
• State the conservation condition explicitly: “external torque is zero, hence \( L = I\omega \) is constant”, then substitute numbers.
• Write the formula line (\( \tau = I\alpha \), \( L = I\omega \), \( K = \frac{1}{2}I\omega^2 \)) before plugging in values — substitution alone can lose the formula mark.
• Table 6.2 (translation–rotation analogy) is frequently tested side by side: learn the pairs \( x-\theta \), \( v-\omega \), \( m-I \), \( F-\tau \), \( p-L \).
• In equilibrium problems, draw the body, mark every force and its lever arm about one chosen point, then impose \( \sum \mathbf{F} = 0 \) and \( \sum \boldsymbol{\tau} = 0 \) separately.

Quick Revision Recap — One-Minute Summary

The whole chapter on one screen — formulas, units, and conditions:

Mnemonics to lock it in:

• “Ring is its own everything” — ring about its own axis: \( I = MR^2 \).
• “Flat things go half” — disc about its own axis and ring about a diameter: \( I = MR^2/2 \).
• “Rod is unlucky twelve” — \( I = ML^2/12 \) about its centre.
• “No external torque, no change in I omega” — the conservation pair.

Before the exam, rework one torque/angular-momentum problem and one equilibrium problem from scratch — together they cover most of the chapter’s numerical marks. All formulas here match the official NCERT Class 11 Physics Part I textbook (Chapter 6); open the PDF to verify any expression.

For the rest of the syllabus, browse the Class 11 Physics notes, the wider Class 11 notes hub, or the full CBSE notes library.

Frequently Asked Questions on Systems of Particles and Rotational Motion

Why does a diver or figure skater spin faster when they pull their arms and legs in?

External torque about the spin axis is negligible, so angular momentum \( L = I\omega \) is conserved. Pulling the limbs in brings mass closer to the axis, decreasing I; since the product stays constant, \( \omega \) must increase.

Is torque zero when a force acts through the centre of mass?

Yes, for rotation about the CM. \( \tau = rF\sin\theta \), and when the line of action passes through the CM the lever arm is zero (\( \theta = 0^\circ \) or \( 180^\circ \)), so \( \tau = 0 \). The body then gains translational acceleration, not angular acceleration.

Under what condition is angular momentum conserved for a rigid body?

When the net external torque is zero: \( \boldsymbol{\tau}_{\text{ext}} = \mathbf{0} \), which makes \( d\mathbf{L}/dt = 0 \) and \( L_z = I\omega \) constant (Eq. 6.44). The condition is zero torque, not zero force.

What is the difference between centre of mass and centre of gravity?

The CM is a mass-distribution property: \( \mathbf{R} = \sum m_i\mathbf{r}_i/M \). The CG is the point where total gravitational torque is zero. They coincide when \( \mathbf{g} \) is uniform over the body, and separate only if \( \mathbf{g} \) varies across a very extended body.

Can the moment of inertia of a body ever be zero?

Only when every particle lies on the axis, so all \( r_i = 0 \) — for example, a thin rod spinning about its own length. For any real rotation about an axis at a distance from the mass, \( I = \sum m_i r_i^2 \) is strictly positive.

How do I remember the moment of inertia formulas from Table 6.1?

Use the pattern inside them: every formula is M times a length squared. The ring is “its own everything” (\( MR^2 \)); flat bodies about their central axis are half that (\( MR^2/2 \)); the rod about its centre is the lone twelfth (\( ML^2/12 \)); the solid sphere is the lone fifth (\( 2MR^2/5 \)). Before the exam, write the whole table once from memory.

Reference: NCERT Class 11 Physics textbook, chapter Systems of Particles and Rotational Motion.