PhysicsCBSE Class 11 · Chapter 6

Systems of Particles and Rotational Motion Class 11: Notes and Numericals

Updated 14 October 2026 · Fact-checked

Systems of Particles and Rotational Motion is Chapter 6 of CBSE Class 11 Physics. It covers the centre of mass and its motion, torque and angular momentum, conservation of angular momentum, equilibrium of rigid bodies, moment of inertia and radius of gyration (values only), and the kinematics and dynamics of rotation about a fixed axis.

What Systems of Particles and Rotational Motion is about

So far every object was treated as a point. Real objects have size and can spin. This chapter treats an object as a system of particles, or as a rigid body whose particles keep fixed distances. The first idea is the centre of mass: a single point that moves as if all the mass were there and all the external forces acted on it. A shell that explodes in mid-air has its centre of mass still following the original parabola.

The second idea is rotation about a fixed axis. Every linear quantity has a rotational partner: angle for displacement, angular velocity ω for velocity, torque $τ = r \times F$ for force, moment of inertia $I = \sum m r^{2}$ for mass, and angular momentum $L$ for momentum. Newton's second law becomes $τ = I α$ , and when no external torque acts, $L = I ω$ stays constant. That is why a skater spins faster when she pulls her arms in.

You also learn the conditions for a rigid body to be in equilibrium (no net force and no net torque) and the principle of moments. For 2026-27, moments of inertia are needed as values only (no derivation), and the theorems of parallel and perpendicular axes and rolling motion have been removed. This chapter is the base for rotational problems in JEE and NEET and for torque on dipoles in Class 12.

Weightage: This chapter is part of Unit V, Motion of System of Particles and Rigid Body. The syllabus groups Unit V with Unit IV, Work, Energy and Power and Unit VI, Gravitation: together they carry 17 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit V Motion of System of Particles and Rigid Body

Key concepts

Rigid bodies and the centre of mass

NCERT section 6.1 and 6.2

A rigid body is one whose shape and size do not change: the distances between its particles stay fixed. It can translate (every particle has the same velocity) or rotate about a fixed axis (every particle moves in a circle centred on the axis), or do a combination.

The centre of mass (CM) of a system of particles is the point whose position is the mass-weighted average of their positions:
$X = \frac{\sum m _{i} x _{i}}{M}$ , $Y = \frac{\sum m _{i} y _{i}}{M}$ , $Z = \frac{\sum m _{i} z _{i}}{M}$ , or $R = \frac{\sum m _{i} r _{i}}{M}$ .

Two particles: the CM lies on the line joining them, nearer the heavier one; with $m_{1}$ at the origin and $m_{2}$ at distance d, $X = \frac{m _{2} d}{m _{1} + m _{2}}$ . For equal masses it is the midpoint.

Worked example

Particles of 2 kg and 3 kg are at x = 0 and x = 5 m. Find their centre of mass. [1 mark]

Try it first, then show the solutionSolution

$X = \frac{2 \times 0 + 3 \times 5}{2 + 3} = \frac{15}{5}$
X = 3 m (nearer the heavier particle)

Answer: At x = 3 m

Centre of mass of a rigid body and of a uniform rod

NCERT section 6.2

For a continuous body, replace the sum by an integral: $X = \frac{1}{M} \int x d m$ .

Uniform rod of length L and mass M along the x-axis from 0 to L: mass per unit length $λ = \frac{M}{L}$ , so $d m = λ d x$ and
$X = \frac{1}{M} \int_{0}^{L} x λ d x = \frac{λ}{M} \cdot \frac{L ^{2}}{2} = \frac{L}{2}$ : the CM is at the midpoint.

For symmetric uniform bodies the CM is at the geometric centre: rod, ring, disc, sphere, cube. The CM need not lie inside the material: for a ring it is at the empty centre.

For several particles in a plane, find X and Y separately.

Worked example

Masses of 1 kg, 2 kg and 3 kg are placed at (1, 2), (3, 0) and (−1, 4) metres. Find their centre of mass. [2 marks]

Try it first, then show the solutionSolution

$X = \frac{1 ( 1 ) + 2 ( 3 ) + 3 ( - 1 )}{6} = \frac{4}{6} \approx 0.67 m$
$Y = \frac{1 ( 2 ) + 2 ( 0 ) + 3 ( 4 )}{6} = \frac{14}{6} \approx 2.33 m$

Answer: About (0.67 m, 2.33 m)

Motion of the centre of mass and momentum of a system

NCERT section 6.3 and 6.4

Differentiating $M R = \sum m_{i} r_{i}$ gives:

$M V = \sum m_{i} v_{i} = P$ : the total momentum equals the total mass times the velocity of the CM.
$M A = F_{e x t}$ : the CM moves as if all the mass were concentrated there and all external forces acted on it. Internal forces cancel in pairs (third law).

Conservation: if $F_{e x t} = 0$ , then $P$ is constant and the CM moves with constant velocity (or stays at rest), whatever the internal forces do.

Examples: a shell bursting in flight; its fragments scatter but the CM continues along the same parabola. A radioactive nucleus at rest decays; the products fly apart in opposite directions so the CM stays put.

Worked example

A 1 kg particle moves at $2 \hat{i}$ m s $^{- 1}$ and a 3 kg particle at $- 2 \hat{i}$ m s $^{- 1}$ . Find the velocity of their centre of mass. [1 mark]

Try it first, then show the solutionSolution

$V = \frac{1 ( 2 ) + 3 ( - 2 )}{4} = \frac{- 4}{4}$
$V = - 1 \hat{i}$ m s $^{- 1}$

Answer: 1 m s $^{- 1}$ in the −x direction

Angular velocity, angular acceleration and rotational kinematics

NCERT section 6.6 and 6.10

In rotation about a fixed axis, every particle turns through the same angle θ in the same time.

Angular velocity $ω = \frac{d θ}{d t}$ ; linear speed of a particle at distance r from the axis: $v = ω r$ (in vector form $v = ω \times r$ ). $ω$ points along the axis by the right-hand rule.
Angular acceleration $α = \frac{d ω}{d t}$ .

For constant α, the equations mirror linear motion:
$ω = ω_{0} + α t$ , $θ = ω_{0} t + \frac{1}{2} α t^{2}$ , $ω^{2} = ω_{0}^{2} + 2 α θ$ .

Convert rpm to rad s $^{- 1}$ by multiplying by $\frac{2 π}{60}$ ; number of revolutions $= \frac{θ}{2 π}$ .

Worked example

A wheel starts from rest with a constant angular acceleration of 2 rad s $^{- 2}$ . Find its angular velocity after 5 s and the number of revolutions it makes in that time. [2 marks]

Try it first, then show the solutionSolution

$ω = 0 + 2 \times 5 = 10 r a d s^{- 1}$
$θ = \frac{1}{2} \times 2 \times 25 = 25 r a d$ ; revolutions $= \frac{25}{2 π} \approx 3.98$

Answer: 10 rad s $^{- 1}$ ; about 4 revolutions

Torque and angular momentum

NCERT section 6.5 and 6.7

Torque (moment of a force) about a point O: $τ = r \times F$ , with magnitude $τ = r F sin θ = r_{⊥} F$ (force × perpendicular distance of its line of action). SI unit N m; it is the turning effect of a force. A door is easiest to open by pushing far from the hinges, at right angles.

Angular momentum of a particle about O: $l = r \times p$ , magnitude $l = r p sin θ = r_{⊥} p$ . Unit kg m $^{2}$ s $^{- 1}$ (= J s).

Key relation: $\frac{d l}{d t} = τ$ . (Proof: $\frac{d}{d t} (r \times p) = v \times m v + r \times F = τ$ , since $v \times v = 0$ .) For a system, $\frac{d L}{d t} = τ_{e x t}$ : internal torques cancel.

A particle moving in a straight line can still have angular momentum about a point not on that line: $l = m v d$ , where d is the perpendicular distance.

Worked example

A force $F = 2 \hat{i} + 3 \hat{j}$ N acts at the point $r = \hat{i} + 2 \hat{j}$ m. Find the torque about the origin. [2 marks]

Try it first, then show the solutionSolution

$τ = r \times F$ ; for vectors in the x-y plane only a $\hat{k}$ part survives
$τ_{z} = x F_{y} - y F_{x} = 1 (3) - 2 (2) = - 1$ , so $τ = - \hat{k}$ N m

Answer: $- \hat{k}$ N m (1 N m, clockwise seen from +z)

Equilibrium of a rigid body, principle of moments and centre of gravity

NCERT section 6.8

A rigid body is in mechanical equilibrium when:
1. Translational equilibrium: $\sum F_{e x t} = 0$ (no linear acceleration);
2. Rotational equilibrium: $\sum τ_{e x t} = 0$ about any point (no angular acceleration).

A couple is a pair of equal and opposite forces with different lines of action: net force zero but net torque non-zero (turning a tap or a steering wheel). Its moment is the same about every point.

Principle of moments (lever): for a lever in equilibrium, load × load arm = effort × effort arm. Mechanical advantage $= \frac{load}{effort} = \frac{effort arm}{load arm}$ .

Centre of gravity (CG): the point about which the total gravitational torque is zero. In a uniform gravitational field it coincides with the centre of mass (they differ only for very large bodies).

Worked example

A uniform metre rule is balanced at its 50 cm mark. A 20 g mass hangs at the 10 cm mark. Where must a 40 g mass be hung to keep it balanced? [2 marks]

Try it first, then show the solutionSolution

Anticlockwise moment $= 20 \times (50 - 10) = 800$ g cm
$40 \times d = 800 \Rightarrow d = 20$ cm on the other side of the pivot

Answer: At the 70 cm mark

Moment of inertia and radius of gyration (values only)

NCERT section 6.9

Kinetic energy of a rigid body rotating at ω about a fixed axis:
$K = \sum \frac{1}{2} m_{i} (r_{i} ω)^{2} = \frac{1}{2} (\sum m_{i} r_{i}^{2}) ω^{2} = \frac{1}{2} I ω^{2}$

The moment of inertia $I = \sum m_{i} r_{i}^{2}$ (kg m $^{2}$ ) measures rotational inertia: it plays the role of mass in rotation. It depends on the mass, how the mass is spread about the axis, and the choice of axis. Flywheels have most of their mass at the rim to give a large I.

Radius of gyration k: $I = M k^{2}$ , the distance from the axis at which the whole mass could be put to give the same I.

Values to remember (no derivation needed):

Body	Axis	I
Thin ring, radius R	through centre, perpendicular to plane	$M R^{2}$
Thin ring	a diameter	$\frac{1}{2} M R^{2}$
Thin rod, length L	through centre, perpendicular to rod	$\frac{M L ^{2}}{12}$
Disc, radius R	through centre, perpendicular to plane	$\frac{1}{2} M R^{2}$
Disc	a diameter	$\frac{1}{4} M R^{2}$
Hollow cylinder	its axis	$M R^{2}$
Solid cylinder	its axis	$\frac{1}{2} M R^{2}$
Solid sphere	a diameter	$\frac{2}{5} M R^{2}$

Worked example

Find the radius of gyration of a uniform disc of radius R about an axis through its centre perpendicular to its plane. [1 mark]

Try it first, then show the solutionSolution

$M k^{2} = \frac{1}{2} M R^{2}$
$k = \frac{R}{2} \approx 0.71 R$

Answer: $k = R / 2$

Dynamics of rotation about a fixed axis

NCERT section 6.11 and 6.12

For rotation about a fixed axis:

Second law for rotation: $τ = I α$ .
Angular momentum: $L = I ω$ (along the axis), and $τ = \frac{d L}{d t}$ .
Work by a torque: $W = τ θ$ ; power $P = τ ω$ .
Kinetic energy: $K = \frac{1}{2} I ω^{2}$ ; work-energy theorem: $W = Δ (\frac{1}{2} I ω^{2})$ .

Linear and rotational motion compared:

Linear	Rotational (fixed axis)
displacement x	angle θ
velocity v	angular velocity ω
acceleration a	angular acceleration α
mass m	moment of inertia I
force F = ma	torque τ = Iα
momentum p = mv	angular momentum L = Iω
work Fs	work τθ
KE ½mv²	KE ½Iω²
power Fv	power τω

Worked example

A light cord wound round a flywheel of radius 0.2 m and moment of inertia 0.4 kg m $^{2}$ is pulled with a steady force of 20 N. Find the angular acceleration of the flywheel. [2 marks]

Try it first, then show the solutionSolution

$τ = F R = 20 \times 0.2 = 4 N m$
$α = \frac{τ}{I} = \frac{4}{0.4} = 10 r a d s^{- 2}$

Answer: 10 rad s $^{- 2}$

Conservation of angular momentum

NCERT section 6.12.1

If the net external torque on a system is zero, its total angular momentum is constant:
$τ_{e x t} = 0 \Rightarrow L = I ω = constant$ , so $I_{1} ω_{1} = I_{2} ω_{2}$ .

Applications:

A skater or a ballet dancer spins faster on pulling in the arms (I falls, ω rises).
A diver curls up to somersault quickly, then straightens to slow down before entering the water.
A person on a rotating stool with dumbbells speeds up on bringing them in.

Kinetic energy is not conserved in these cases: $K = \frac{L ^{2}}{2 I}$ increases when I decreases, the extra energy coming from the work done by the muscles.

Worked example

A skater spins at 3 rad s $^{- 1}$ with moment of inertia 4 kg m $^{2}$ . She pulls her arms in, reducing I to 1.6 kg m $^{2}$ . Find her new angular speed. [2 marks]

Try it first, then show the solutionSolution

$I_{1} ω_{1} = I_{2} ω_{2} \Rightarrow 4 \times 3 = 1.6 ω_{2}$
$ω_{2} = 7.5 r a d s^{- 1}$

Answer: 7.5 rad s $^{- 1}$

Physics in real life: Systems of Particles and Rotational Motion

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Torque and lever arm

Suppose a door needs a torque of 6 N m about its hinges to start opening.

What force, applied at right angles to the door, is needed if you push (a) at the handle, 0.8 m from the hinges, (b) at 0.2 m from the hinges?

Try it first, then show the solutionSolution

$τ = F r_{⊥} \Rightarrow F = \frac{τ}{r}$
(a) $F = \frac{6}{0.8} = 7.5 N$
(b) $F = \frac{6}{0.2} = 30 N$ , four times as much

Answer: (a) 7.5 N (b) 30 N

Memory hook: Longer arm, smaller push: that is why handles are far from hinges.

Conservation of angular momentum

Imagine the Earth suddenly shrank to half its present radius without losing any mass or getting any external torque.

Suppose the Earth is a uniform sphere. How long would a day become?

Try it first, then show the solutionSolution

$I = \frac{2}{5} M R^{2}$ ; halving R makes I one quarter
$I ω$ constant, so ω becomes 4 times
Period $T \propto \frac{1}{ω}$ : the day becomes $\frac{24}{4} = 6$ hours

Answer: 6 hours

Memory hook: Shrink the spinner, speed the spin.

Rotational kinetic energy

Suppose a bus uses a flywheel, a uniform disc of mass 50 kg and radius 0.5 m, to store energy while braking, and it spins up to 300 rad s $^{- 1}$ .

How much energy does the flywheel store?

Try it first, then show the solutionSolution

$I = \frac{1}{2} M R^{2} = \frac{1}{2} \times 50 \times 0.25 = 6.25 k g m^{2}$
$K = \frac{1}{2} I ω^{2} = \frac{1}{2} \times 6.25 \times 90000 \approx 2.8 \times 1 0^{5} J$

Answer: About 0.28 MJ

Memory hook: Energy grows with ω squared: spin twice as fast, store four times as much.

Topics in this chapter

Systems of Particles and Rotational Motion formula sheet

Centre of mass: $R = \frac{\sum m _{i} r _{i}}{M}, X = \frac{1}{M} \int x d m$
Motion of the centre of mass: $P = M V, M A = F_{e x t}$
Rotational kinematics (constant α): $ω = ω_{0} + α t, θ = ω_{0} t + \frac{1}{2} α t^{2}, ω^{2} = ω_{0}^{2} + 2 α θ$
Torque and angular momentum: $τ = r \times F, l = r \times p, \frac{d L}{d t} = τ_{e x t}$
Moment of inertia and radius of gyration: $I = \sum m_{i} r_{i}^{2} = M k^{2}$
Standard moments of inertia: $I_{r in g} = M R^{2}, I_{d i sc} = \frac{1}{2} M R^{2}, I_{r o d} = \frac{M L ^{2}}{12}, I_{s p h er e} = \frac{2}{5} M R^{2}$; Ring and disc about the perpendicular axis through the centre; rod about the perpendicular axis through its middle; solid sphere about a diameter.
Rotational dynamics: $τ = I α, L = I ω, K = \frac{1}{2} I ω^{2}, W = τ θ, P = τ ω$
Conservation of angular momentum: $I_{1} ω_{1} = I_{2} ω_{2} (τ_{e x t} = 0)$
Equilibrium of a rigid body: $\sum F = 0, \sum τ = 0$

How the school exam asks Systems of Particles and Rotational Motion

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Centre of mass	CM of two or three particles, of a rod, of symmetric bodies; velocity of CM; what happens to the CM in an explosion.	Mass-weighted average, coordinate by coordinate; external force alone decides the CM's motion.
Torque and angular momentum	Torque from $r \times F$ ; angular momentum of a particle moving in a line; proof of $d l / d t = τ$ .	Cross product in components; $l = m v r_{⊥}$ .
Equilibrium	Beams on supports, metre rules with weights, levers, ladders; conditions for equilibrium; couples.	Take moments about the point that removes the most unknowns; then use $\sum F = 0$ .
Moment of inertia	Definition, radius of gyration, factors affecting I, standard values; I of point masses on a light rod.	$I = \sum m r^{2}$ for particles; use the table for standard bodies (no derivation).
Rotational dynamics	Cord on a flywheel, mass hanging from a pulley; rotational KE, work and power; analogy table.	$τ = I α$ for the wheel and $F = ma$ for hanging masses, linked by $a = α R$ .
Conservation of angular momentum	Skaters, divers, turntables; two discs coupled together; change in KE.	$I_{1} ω_{1} = I_{2} ω_{2}$ ; then compare $\frac{1}{2} I ω^{2}$ before and after.

Tricks to remember Systems of Particles and Rotational Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

Swap the letters
m → I, v → ω, a → α, F → τ, p → L: every linear formula becomes its rotational twin.
Use it when: Writing rotational equations quickly.
Ring 1, disc ½, sphere 2/5, rod 1/12
Mass far from the axis means more I: ring ( $M R^{2}$ ) > disc ( $\frac{1}{2} M R^{2}$ ) > solid sphere ( $\frac{2}{5} M R^{2}$ ). Rod about its middle: $\frac{1}{12} M L^{2}$ .
Use it when: Standard MI values.
Two conditions for balance
No push (ΣF = 0) and no twist (Στ = 0).
Use it when: Equilibrium questions.
Skater rule
Arms in: I down, ω up, L same, KE up.
Use it when: Conservation of angular momentum MCQs.
CM ignores internal forces
Explosions, springs, people walking on a boat: only external forces move the CM.
Use it when: CM motion questions.

Systems of Particles and Rotational Motion: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Moment of inertia of a disc · 1 mark · MCQExam level

The moment of inertia of a uniform disc of mass M and radius R about an axis through its centre perpendicular to its plane is:
(a) $M R^{2}$
(b) $\frac{1}{2} M R^{2}$
(c) $\frac{2}{5} M R^{2}$
(d) $\frac{1}{4} M R^{2}$

Try it first, then show the solutionSolution

Standard value for a disc about its perpendicular central axis.

Answer: (b) $\frac{1}{2} M R^{2}$

Example 2: Diving somersaults · 1 mark · Assertion-reasonExam level

Assertion (A): A diver curls up her body to turn more somersaults in the air.
Reason (R): With no external torque, angular momentum is conserved, so reducing I increases ω.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Gravity acts at the CM and gives no torque about it; $I ω$ stays constant, so curling up (smaller I) spins her faster.

Answer: (a)

Example 3: Two masses on a light rod · 2 marks · Very short answerHarder exam level

Particles of 1 kg and 3 kg are joined by a light rod 1 m long. Find the moment of inertia of the system about an axis perpendicular to the rod through the centre of mass. [2 marks]

Try it first, then show the solutionSolution

CM from the 1 kg particle: $\frac{3 \times 1}{4} = 0.75 m$ ; so 0.25 m from the 3 kg particle
$I = 1 (0.75)^{2} + 3 (0.25)^{2} = 0.5625 + 0.1875 = 0.75 k g m^{2}$

Answer: 0.75 kg m $^{2}$

Example 4: Beam with a load · 3 marks · Short answerHarder exam level

A uniform plank 6 m long and weighing 300 N rests on two supports, one at each end. A box of weight 600 N is placed 2 m from the left end. Find the force on each support. [3 marks]

Try it first, then show the solutionSolution

Moments about the left support: $R_{R} \times 6 = 300 \times 3 + 600 \times 2 = 2100$
$R_{R} = 350 N$
$R_{L} = 900 - 350 = 550 N$

Answer: Left 550 N, right 350 N

Example 5: Wheel speeding up · 2 marks · Very short answerHarder exam level

A wheel's speed rises uniformly from 600 rpm to 1200 rpm in 10 s. Find its angular acceleration and the number of revolutions made in this time. [2 marks]

Try it first, then show the solutionSolution

$ω_{0} = 20 π$ , $ω = 40 π$ rad s $^{- 1}$ ; $α = \frac{20 π}{10} = 2 π \approx 6.28 r a d s^{- 2}$
Average speed 900 rpm for $\frac{10}{60}$ min: 150 revolutions

Answer: About 6.28 rad s $^{- 2}$ ; 150 revolutions

Example 6: Torque, work and kinetic energy of a flywheel · 5 marks · Long answerHarder exam level

(a) Show that the work done by a constant torque τ in turning a body through angle θ is τθ, and that the power is τω. (b) A flywheel of moment of inertia 0.5 kg m $^{2}$ starts from rest under a constant torque of 2 N m. Find its angular acceleration and its angular speed after 10 s. (c) Find the angle turned and the work done in that time, and check that the work equals the gain in kinetic energy. [5 marks]

Try it first, then show the solutionSolution

(a) A force F at radius r moving through arc $d s = r d θ$ does $d W = F r d θ = τ d θ$ ; for constant τ, $W = τ θ$ , and $P = \frac{d W}{d t} = τ ω$
(b) $α = \frac{τ}{I} = \frac{2}{0.5} = 4 r a d s^{- 2}$ ; $ω = 4 \times 10 = 40 r a d s^{- 1}$
(c) $θ = \frac{1}{2} \times 4 \times 100 = 200 r a d$
$W = τ θ = 2 \times 200 = 400 J$
$K = \frac{1}{2} \times 0.5 \times 4 0^{2} = 400 J$ ✓

Answer: 4 rad s $^{- 2}$ ; 40 rad s $^{- 1}$ ; 200 rad; 400 J = gain in KE

Example 7: Skater's kinetic energy · 3 marks · Short answerHOTS / competency

A skater with moment of inertia 4 kg m $^{2}$ spins at 3 rad s $^{- 1}$ , then pulls her arms in so that I becomes 1.6 kg m $^{2}$ . Find her new angular speed and her kinetic energy before and after. Where does the extra energy come from? [3 marks]

Try it first, then show the solutionSolution

$ω_{2} = \frac{4 \times 3}{1.6} = 7.5 r a d s^{- 1}$
$K_{1} = \frac{1}{2} \times 4 \times 9 = 18 J$ ; $K_{2} = \frac{1}{2} \times 1.6 \times 56.25 = 45 J$
The extra 27 J comes from the work her muscles do in pulling her arms inwards.

Answer: 7.5 rad s $^{- 1}$ ; 18 J → 45 J; from muscular work

Example 8: Ring dropped on a spinning disc · 3 marks · Short answerHOTS / competency

A uniform disc of mass M and radius R spins freely at ω about its vertical axis. A ring of the same mass and radius is gently placed on it, coaxially, and sticks. Find the new angular speed and the fraction of kinetic energy lost. [3 marks]

Try it first, then show the solutionSolution

$I_{1} = \frac{1}{2} M R^{2}$ ; $I_{2} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$
$I_{1} ω = I_{2} ω^{'} \Rightarrow ω^{'} = \frac{ω}{3}$
$\frac{K _{2}}{K _{1}} = \frac{I _{2} ω ^{'2}}{I _{1} ω ^{2}} = 3 \times \frac{1}{9} = \frac{1}{3}$ , so $\frac{2}{3}$ of the KE is lost (as heat in the sticking)

Answer: ω/3; two-thirds of the KE is lost

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Spinning on an ice rink

Suppose a figure skater spins on ice with arms stretched out, her moment of inertia then being 3.6 kg m $^{2}$ , at 2 revolutions per second. Friction from the ice is negligible.

(i) · 1 mark
What is her angular momentum (in kg m $^{2}$ s $^{- 1}$ )?
Answer
Answer: About 45.2
$I ω = 3.6 \times 4 π$
(ii) · 1 mark
She pulls her arms in and I drops to 1.2 kg m $^{2}$ . How many revolutions per second does she now make?
Answer
Answer: 6
I falls to one third, so frequency triples.
(iii) · 1 mark
Which quantity stays constant: angular momentum or kinetic energy?
Answer
Answer: Angular momentum
No external torque about the vertical axis.
(iv) · 1 mark
By what factor does her kinetic energy change?
Answer
Answer: It becomes 3 times
$K = \frac{L ^{2}}{2 I}$ ; I falls to a third.

A seesaw in the park

Suppose a uniform seesaw 4 m long, pivoted at its centre, has a child of weight 300 N sitting 1.6 m from the pivot on the left.

(i) · 1 mark
What torque does the child produce about the pivot?
Answer
Answer: 480 N m
$300 \times 1.6$
(ii) · 1 mark
Where must a 400 N child sit on the right to balance it?
Answer
Answer: 1.2 m from the pivot
$400 \times d = 480$
(iii) · 1 mark
What force does the pivot then exert (seesaw weight 200 N)?
Answer
Answer: 900 N upwards
Translational equilibrium: 300 + 400 + 200
(iv) · 1 mark
Why does the seesaw's own weight not affect the balance position?
Answer
Answer: It acts at the pivot
Its CG is at the centre, so its torque about the pivot is zero.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Define radius of gyration. Find it for a thin ring of radius R about an axis through its centre perpendicular to its plane. [2 marks] (2 marks)

Step	Marks
k is the distance from the axis at which the whole mass, if concentrated, would give the same moment of inertia: $I = M k^{2}$	1
Ring: $I = M R^{2} = M k^{2} \Rightarrow k = R$	1

Show that the rate of change of angular momentum of a particle equals the torque acting on it. [3 marks] (3 marks)

Step	Marks
$l = r \times p$ ; differentiate: $\frac{d l}{d t} = \frac{d r}{d t} \times p + r \times \frac{d p}{d t}$	1
$\frac{d r}{d t} \times p = v \times m v = 0$ (parallel vectors)	1
$\frac{d p}{d t} = F$ , so $\frac{d l}{d t} = r \times F = τ$	1

A light string is wound round a uniform disc of mass 2 kg and radius 0.1 m, free to turn about its fixed horizontal axis. A 1 kg mass hangs from the free end. (a) Write the equations of motion for the mass and the disc. (b) Find the acceleration of the mass, the tension and the angular acceleration of the disc. (c) Find the kinetic energy of the disc after the mass has fallen for 1 s from rest. (g = 10 m s $^{- 2}$ ) [5 marks] (5 marks)

Step	Marks
(a) Mass: $m g - T = ma$ . Disc: $T R = I α$ with $I = \frac{1}{2} M R^{2}$ and $a = α R$ , so $T = \frac{1}{2} M a$	1
(b) $m g = (m + \frac{M}{2}) a \Rightarrow a = \frac{10}{2} = 5 m s^{- 2}$	1
$T = \frac{1}{2} \times 2 \times 5 = 5 N$ ; $α = \frac{a}{R} = 50 r a d s^{- 2}$	1
(c) After 1 s: $ω = 50 r a d s^{- 1}$ ; $I = \frac{1}{2} \times 2 \times 0.01 = 0.01 k g m^{2}$	1
$K = \frac{1}{2} \times 0.01 \times 2500 = 12.5 J$	1

Where students lose marks in Systems of Particles and Rotational Motion

The slips examiners see most often, and what to write instead.

Mistake: Using the length of $r$ instead of the perpendicular distance
Forgetting $sin θ$ .
Do this: $τ = r F sin θ = r_{⊥} F$ ; a force along the line through the axis has no torque.
Usually costs: 1-2 marks.
Mistake: Deriving moments of inertia at length
Older books did.
Do this: The 2026-27 syllabus asks for values only, without derivation. Learn the table and use it.
Usually costs: Time.
Mistake: Using the parallel or perpendicular axis theorem
Old notes and coaching material include them.
Do this: Both theorems were removed for 2026-27. Questions will give I directly or use the standard values.
Usually costs: Time, and possibly marks if the method is not accepted.
Mistake: Assuming kinetic energy is conserved when a skater pulls in her arms
Mixing up conservation laws.
Do this: Only angular momentum is conserved; KE = $L^{2} /2 I$ increases as I decreases.
Usually costs: 1 mark.
Mistake: Forgetting that the tension in a pulley problem differs on each side or from mg
Treating the hanging mass as in equilibrium.
Wrong working
T = mg for a mass falling from a pulley
Right working
mg − T = ma and TR = Iα with a = αR
Do this: Write separate equations for the mass and the wheel.
Usually costs: Whole numerical.

NCERT exercise guide: Systems of Particles and Rotational Motion

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 6.1-6.3 (centre of mass)

What it tests: CM of symmetric bodies, of a diatomic molecule, and the CM velocity of a trolley-child system.

How to approach it: $X = \frac{m _{2} d}{m _{1} + m _{2}}$ for two particles; internal motion does not change the CM velocity.

Same kind as Exercise 6.2

In a carbon monoxide molecule the C and O nuclei are $1.13 \times 1 0^{- 10}$ m apart. The O atom is about 16/12 times as heavy as the C atom. Find the CM from the C nucleus.

Show the solutionSolution

$X = \frac{m _{O} d}{m _{C} + m _{O}} = \frac{16 \times 1.13 \times 1 0 ^{- 10}}{12 + 16}$
$X \approx 0.646 \times 1 0^{- 10} m$

Answer: About $6.5 \times 1 0^{- 11}$ m from C, towards O

Exercises 6.4-6.9 (vectors, angular momentum and equilibrium)

What it tests: Areas and volumes from cross products, components of angular momentum, a bar hung by strings, and the load on the axles of a car.

How to approach it: Cross products in components; for equilibrium take moments about a convenient point and balance forces.

Same kind as Exercise 6.9

A uniform beam 4 m long weighing 200 N rests on supports at its ends. A man weighing 300 N stands 1 m from the left end. Find the reaction at each support.

Show the solutionSolution

Moments about the left end: $R_{B} \times 4 = 200 \times 2 + 300 \times 1 = 700 \Rightarrow R_{B} = 175 N$
Forces: $R_{A} + R_{B} = 500 \Rightarrow R_{A} = 325 N$

Answer: Left 325 N, right 175 N

Exercises 6.10-6.17 (rotational dynamics)

What it tests: Equal torques on different bodies, rotational KE, turntables and angular momentum, a rope on a hollow cylinder, engine torque, and a metre stick with coins.

How to approach it: $α = τ / I$ ; $K = \frac{1}{2} I ω^{2}$ ; $L = I ω$ conserved when external torque is zero; $P = τ ω$ .

Same kind as Exercise 6.11

A solid sphere of mass 5 kg and radius 0.2 m spins about a diameter at 30 rad s $^{- 1}$ . Find its rotational kinetic energy and angular momentum.

Show the solutionSolution

$I = \frac{2}{5} \times 5 \times 0.04 = 0.08 k g m^{2}$
$K = \frac{1}{2} \times 0.08 \times 900 = 36 J$ ; $L = 0.08 \times 30 = 2.4 k g m^{2} s^{- 1}$

Answer: 36 J; 2.4 kg m $^{2}$ s $^{- 1}$

Practice questions on Systems of Particles and Rotational Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The rotational analogue of mass is:
(a) torque
(b) angular momentum
(c) moment of inertia
(d) angular velocity
Hint
τ = Iα vs F = ma.
Answer
(c) moment of inertia
Q2 · 1 mark · MCQExam level
The SI unit of angular momentum is:
(a) kg m s $^{- 1}$
(b) kg m $^{2}$ s $^{- 1}$
(c) N m
(d) kg m $^{2}$ s $^{- 2}$
Hint
L = Iω.
Answer
(b) kg m $^{2}$ s $^{- 1}$ (same as J s)
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The centre of mass of a body may lie where there is no mass.
Reason (R): The centre of mass of a uniform ring is at its centre.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Ring.
Answer
(a): the ring's centre is empty space, yet it is the CM.
Q4 · 1 mark · Very short answerExam level
State two factors on which the moment of inertia of a body depends.
Hint
$I = \sum m r^{2}$ .
Answer
The mass of the body, the way the mass is distributed about the axis, and the position and direction of the axis (any two).
Q5 · 2 marks · Very short answerHarder exam level
Find the torque of the force $F = \hat{i} - \hat{j} + \hat{k}$ N acting at the point $r = 2 \hat{i} + \hat{j} - \hat{k}$ m, about the origin.
Hint
Determinant with $\hat{i}, \hat{j}, \hat{k}$ on top.
Answer
$τ = r \times F = (1 \cdot 1 - (- 1) (- 1)) \hat{i} + ((- 1) (1) - 2 \cdot 1) \hat{j} + (2 (- 1) - 1 \cdot 1) \hat{k} = 0 \hat{i} - 3 \hat{j} - 3 \hat{k}$ N m.
Q6 · 2 marks · Very short answerHarder exam level
A 2 kg particle moves at 3 m s $^{- 1}$ along the line y = 4 m, parallel to the x-axis. Find its angular momentum about the origin. Does it change as the particle moves?
Hint
$l = m v r_{⊥}$ .
Answer
$l = 2 \times 3 \times 4 = 24$ kg m $^{2}$ s $^{- 1}$ . It stays constant, because no force (hence no torque) acts.
Q7 · 2 marks · Very short answerHarder exam level
A uniform metre rule of mass 100 g balances on a knife edge at the 40 cm mark when a mass m is hung at the 10 cm mark. Find m.
Hint
The rule's weight acts at 50 cm.
Answer
$m \times (40 - 10) = 100 \times (50 - 40) \Rightarrow m = \frac{1000}{30} \approx 33.3$ g.
Q8 · 2 marks · Very short answerHarder exam level
Write the rotational analogues of: (a) $F = ma$ (b) $p = m v$ (c) $W = F s$ (d) $P = F v$ (e) $K = \frac{1}{2} m v^{2}$ .
Hint
Swap the letters.
Answer
(a) $τ = I α$ (b) $L = I ω$ (c) $W = τ θ$ (d) $P = τ ω$ (e) $K = \frac{1}{2} I ω^{2}$ .
Q9 · 3 marks · Short answerHarder exam level
A disc of moment of inertia 0.2 kg m $^{2}$ rotating at 50 rad s $^{- 1}$ is brought to rest in 10 s by a constant frictional torque. Find the torque and the number of revolutions made while stopping.
Hint
α first.
Answer
$α = \frac{0 - 50}{10} = - 5$ rad s $^{- 2}$ ; torque $= 0.2 \times 5 = 1$ N m. $θ = \frac{50 + 0}{2} \times 10 = 250$ rad, about $\frac{250}{2 π} \approx 39.8$ revolutions.
Q10 · 3 marks · Short answerHOTS / competency
A shell is fired from a gun and explodes at the top of its path into two equal pieces. One piece falls straight down from the point of explosion. Where does the other piece land, compared with where the unexploded shell would have landed (range R)? Explain.
Hint
The CM keeps following the parabola.
Answer
The CM still lands at R. One piece lands at R/2 (directly below the top), so for the CM of two equal masses to be at R, the other must land at $2 R - \frac{R}{2} = \frac{3 R}{2}$ from the gun.
Q11 · 3 marks · Short answerHOTS / competency
A man stands at the centre of a turntable that rotates freely, holding his arms out with dumbbells. Explain what happens to the angular speed and the kinetic energy when he pulls the dumbbells in, and why the turntable's bearings must be nearly frictionless for this to work.
Hint
External torque must be zero.
Answer
Pulling in reduces I, so ω increases ( $I ω$ constant) and KE ( $L^{2} /2 I$ ) increases, using the work done by his arms. If the bearings had friction there would be an external torque, L would not be conserved and he would slow down.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Where is the centre of mass of a uniform rod?Show answer
At its midpoint.
2.State the condition for the CM of a system to move with constant velocity.Show answer
The net external force on the system must be zero.
3.Define torque.Show answer
The turning effect of a force: $τ = r \times F$ , magnitude $r F sin θ$ .
4.What is the relation between torque and angular momentum?Show answer
$τ = d L / d t$ .
5.Define radius of gyration.Show answer
The distance k from the axis such that $I = M k^{2}$ .
6.Write the moment of inertia of a solid sphere about a diameter.Show answer
$\frac{2}{5} M R^{2}$ .
7.State the two conditions for equilibrium of a rigid body.Show answer
Net external force zero and net external torque zero.
8.State the law of conservation of angular momentum.Show answer
If the net external torque on a system is zero, its total angular momentum stays constant.

Best order to study it

Learn rigid bodies, types of motion and the centre of mass of particles.
Find the CM of a uniform rod by integration and learn the CM of symmetric bodies.
Learn the motion of the CM and the momentum of a system.
Revise the vector product, then learn torque and angular momentum and prove $d l / d t = τ$ .
Study equilibrium of rigid bodies, couples, the principle of moments and the centre of gravity.
Learn moment of inertia, radius of gyration and the table of values.
Learn rotational kinematics and dynamics with the linear-rotational comparison.
Finish with conservation of angular momentum and its examples.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find the CM of two or three particles in a plane?
Can you show the CM of a uniform rod is at its middle?
Can you explain what happens to the CM of a shell that explodes in flight?
Can you compute a torque as a cross product?
Can you prove $d l / d t = τ$ for a particle?
Can you solve a beam or metre-rule balance problem with moments?
Can you state I for a ring, disc, rod and sphere?
Can you write the rotational analogue of every linear formula?
Can you solve a mass-on-a-pulley problem using $τ = I α$ ?
Can you explain a skater's spin using conservation of angular momentum?

Systems of Particles and Rotational Motion: frequently asked questions

Are the parallel and perpendicular axes theorems in the 2026-27 syllabus?

No. Both theorems were removed in the NCERT rationalisation, and rolling motion was removed too. Moments of inertia of simple bodies are needed as values only, with no derivation.

Is the vector product taught in this chapter?

Yes. NCERT section 6.5 introduces the vector product because torque and angular momentum are defined with it. The CBSE syllabus also lists vector products under Motion in a Plane.

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

The centre of mass depends only on the distribution of mass. The centre of gravity is where the total gravitational torque is zero. In a uniform field (any everyday object) they are the same point.

Can a body be in translational equilibrium but not rotational equilibrium?

Yes. A couple (two equal, opposite, parallel forces with different lines of action) gives zero net force but a net torque, so the body starts rotating.

Why is this chapter important later?

Torque and angular momentum appear in Class 12 for the torque on electric and magnetic dipoles (\vec\tau = \vec p\times\vec E) and in Bohr's quantisation of angular momentum. JEE and NEET set many questions on rotation.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part I, Textbook for Class XI, Chapter 6 (Reprint 2026-27)