Chapter 6 of 14

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Haryana (BSEH)Class 11 Physics★ Unit V · part of the 17-mark block covering Units III to VI in CBSE Class 11 Physics (70 marks total); rotational dynamics is high-yield for JEE and NEET

System of Particles and Rotational Motion

Up to now every object has been a point. This chapter gives bodies size, and everything changes — a force now depends on where you apply it, mass gets replaced by a quantity that depends on the axis you chose, and a diver can spin faster by folding her arms. Two theorems most coaching notes still drill are no longer in the syllabus. CBSE Class 11 Physics Chapter 6.

⏱ 14 min readHard difficulty✓ Free · NCERT-aligned

Worksheet

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The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Locate the centre of mass, and explain why it need not lie inside the body
2Show that internal forces cannot move the centre of mass
3Compute a vector product, and get the sign and direction right
4Find torque and angular momentum, and connect them by dL/dt = torque
5State when angular momentum is conserved — zero net torque, not zero net force
6Apply both equilibrium conditions to a rigid body
7Define moment of inertia and radius of gyration, and quote values for simple shapes
8Translate any linear equation into its rotational partner

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Why this chapter matters

Give a body size and the rules change. Where you push now matters as much as how hard, mass gets replaced by a quantity that depends on which axis you picked, and a diver speeds up her spin by folding her arms without anything pushing her. Note that two theorems most coaching notes still drill are no longer in the syllabus.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Work, Energy and Power

Energy methods used in rolling-motion problems

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Class 11 Physics: Laws of Motion

Force and momentum, extended to rotation

System of Particles and Rotational Motion

1. Check this before you revise anything

Two theorems that appear in almost every coaching handout for this chapter are not in the 2026-27 syllabus and not in the textbook.

Topic	NCERT 2026-27 chapter	CBSE 2026-27
Parallel axes theorem ( $I = I_{c m} + M d^{2}$ )	Absent — the phrase does not occur	Not listed
Perpendicular axes theorem ( $I_{z} = I_{x} + I_{y}$ )	Absent as a theorem	Not listed
Rolling motion dynamics	Mentioned once, only to illustrate combined motion	Not listed
Radius of gyration	Present	Listed
Values of moment of inertia for simple shapes	Given as a table	Listed — "no derivation"

The NCERT exercises never use either theorem either, so nothing is orphaned. Do not spend revision time on them.

What CBSE does ask for is narrower than most notes suggest: centre of mass, torque, angular momentum and its conservation, equilibrium, rotational kinematics, the linear-rotational comparison, moment of inertia values (not derivations), and radius of gyration.

2. Why extended bodies need a new chapter

Every earlier chapter treated an object as a point. That works while you only care about where something goes, not how it turns.

Give a body size and two things break immediately:

Where you apply a force now matters, not just how hard. Push a door at the handle or next to the hinge and you get very different results from the same force.
The body can rotate, so a single position and velocity no longer describe it.

A rigid body — one whose particles keep fixed distances from each other — can do exactly two things: translate, rotate, or combine both. A cylinder rolling down an incline is that combination, which is the one place the chapter mentions rolling at all.

3. Centre of mass

For a system of particles, the centre of mass is the mass-weighted average position:

$R = \frac{\sum m _{i} r _{i}}{\sum m _{i}}$

The point students miss: the centre of mass need not lie inside the body. For a ring it sits at the centre, in empty space. Example 6.3 works an L-shaped lamina where it again falls outside the material. Exercise 6.1 asks this directly.

Why it earns its own chapter section

The centre of mass moves as though the entire mass were concentrated there and all external forces acted on that single point:

$M A_{c m} = F_{e x t}$

Internal forces cancel in pairs, so they cannot shift it. That is why a projectile which explodes in mid-air has fragments whose centre of mass continues along the original parabola — the explosion is internal.

It also gives the momentum result CBSE lists explicitly: with no net external force, the total momentum of the system is constant, and the centre of mass moves with constant velocity. This is why a child running about on a frictionless trolley cannot change the trolley-plus-child centre of mass velocity, which is Exercise 6.3.

4. The vector product, and why it appears here

Chapter 5 introduced the scalar product for work. This chapter needs the other one, because torque and angular momentum are both cross products.

$a \times b = ab sin θ \hat{n}$

The direction comes from the right-hand rule, and order matters:

$a \times b = - b \times a$

That sign is a common slip. The scalar product does not care about order; the vector product reverses.

5. Torque and angular momentum

Torque is the rotational analogue of force — the moment of a force about a point:

$τ = r \times F$

Only the component of the force perpendicular to $r$ turns anything. A force pointing straight at the axis produces no torque at all, however large.

Angular momentum of a particle is the moment of its momentum:

$l = r \times p$

Differentiate it and the two connect, exactly as force connects to linear momentum:

$\frac{d L}{d t} = τ_{e x t}$

The conservation trap

This is the error worth guarding against. Angular momentum is conserved when the net external torque is zero — not when the net external force is zero.

Those are different conditions. A force can be non-zero while its torque about your chosen axis is zero, because torque depends on where the force acts.

$τ_{e x t} = 0 ⟹ L = constant$

The standard illustration: a diver in mid-air. Gravity acts on her the whole time, so the net force is certainly not zero — but it acts through her centre of mass, so it exerts no torque about it. Her angular momentum is fixed. Folding into a tuck cuts her moment of inertia, so $ω$ must rise to keep $I ω$ constant, and she spins faster. Opening out slows her again.

Since $L = I ω$ is constant, a smaller $I$ forces a larger $ω$ . Nothing pushes her — the physics is bookkeeping.

6. Equilibrium of a rigid body

A rigid body needs both conditions, and checking only one is a reliable way to lose an answer.

Condition	Meaning
$\sum F_{e x t} = 0$	No translational acceleration
$\sum τ_{e x t} = 0$	No angular acceleration

A body can have zero net force and still spin up — a couple is precisely that case. Two equal and opposite forces along different lines give zero resultant force and a non-zero torque.

Example 6.7 shows that a couple's moment is the same about every point, which is why you can take moments about whichever point kills the most unknowns. That is the practical trick in Examples 6.8 and 6.9 — the bar on two knife-edges, and the ladder against a frictionless wall.

Worked: textbook example 6.8, choosing the point that kills the most unknowns

A 70 cm, 4.00 kg uniform bar rests on two knife-edges $K_{1}$ and $K_{2}$ , each 10 cm from an end. A 6.00 kg load hangs 30 cm from the $K_{1}$ end. Find the reaction at each knife-edge.

The bar's own weight acts at its centre $G$ , the midpoint (35 cm from either end), so $K_{1} G = K_{2} G = 25$ cm and the load sits 5 cm from $G$ .

Translational equilibrium: $R_{1} + R_{2} = W + W_{1} = (4.00 + 6.00) g = 98.0$ N.

Rotational equilibrium, taken about $G$ — this single choice removes the unknown weight of the bar from the moment equation entirely, since it acts exactly at $G$ :

$- R_{1} (K_{1} G) + W_{1} (P G) + R_{2} (K_{2} G) = 0 ⟹ R_{1} - R_{2} = \frac{W _{1} ( P G )}{K _{1} G} = \frac{6.00 g \times 0.05}{0.25} = 1.2 g \approx 11.76 N$

Solving the two equations together: $R_{1} \approx 54.9$ N, $R_{2} \approx 43.1$ N.

Taking moments about $G$ was the deliberate choice — about any other point, the bar's own weight would have contributed an extra term to the moment equation, needing an extra step to eliminate.

The principle of moments for a lever follows directly:

$F_{1} d_{1} = F_{2} d_{2}$

Centre of gravity is where the total gravitational torque vanishes. It coincides with the centre of mass when gravity is uniform over the body, which is every case in this chapter — but the two are defined differently and the chapter keeps them apart.

7. Moment of inertia

In rotation, mass is replaced by moment of inertia:

$I = \sum m_{i} r_{i}^{2}$

It is not a property of the body alone. Mass is fixed; moment of inertia depends on the axis you choose, because each particle contributes $m r^{2}$ about that axis. Change the axis and the number changes.

The $r^{2}$ weighting is the whole story: a particle twice as far out counts four times as much. So mass far from the axis dominates.

Radius of gyration packages this as a single distance:

$I = M k^{2} k = I / M$

It is where you could put the entire mass, as a point, and get the same moment of inertia.

The values CBSE asks for — no derivation

Body	Axis	$I$
Thin ring, radius R	perpendicular to plane, at centre	$M R^{2}$
Thin ring, radius R	diameter	$M R^{2} /2$
Thin rod, length L	perpendicular to rod, at midpoint	$M L^{2} /12$
Disc, radius R	perpendicular to disc, at centre	$M R^{2} /2$
Disc, radius R	diameter	$M R^{2} /4$
Hollow cylinder, radius R	axis of cylinder	$M R^{2}$
Solid cylinder, radius R	axis of cylinder	$M R^{2} /2$
Solid sphere, radius R	diameter	$2 M R^{2} /5$

Read the ring against the disc. Same mass, same radius, and the ring has twice the moment of inertia — because all its mass sits at $R$ , while the disc's is spread across every radius from 0 to $R$ . That single comparison is what moment of inertia means.

8. Rotational dynamics, and the map back to linear motion

For a fixed axis, every linear equation has a rotational twin. The chapter's own comparison:

Linear motion	Rotational motion about a fixed axis
Displacement $x$	Angular displacement $θ$
Velocity $v = d x / d t$	Angular velocity $ω = d θ / d t$
Acceleration $a = d v / d t$	Angular acceleration $α = d ω / d t$
Mass $M$	Moment of inertia $I$
Force $F = M a$	Torque $τ = I α$
Work $d W = F d s$	Work $W = τ d θ$
Kinetic energy $K = M v^{2} /2$	Kinetic energy $K = I ω^{2} /2$
Power $P = F v$	Power $P = τ ω$
Linear momentum $p = M v$	Angular momentum $L = I ω$

Learn the table as a translation, not as nine new formulas. Every rotational result in this chapter is a linear one with mass swapped for $I$ and the linear quantity swapped for its angular partner.

The kinematic equations translate the same way, for constant $α$ :

$ω = ω_{0} + α t θ = ω_{0} t + \frac{1}{2} α t^{2} ω^{2} = ω_{0}^{2} + 2 α θ$

Example 6.11 uses exactly these on a motor wheel going from 1200 to 3120 rpm, and Example 6.12 combines torque, angular acceleration and energy for a cord unwinding from a flywheel.

Worked: textbook example 6.11, converting rpm before touching a formula

A motor wheel's angular speed rises uniformly from 1200 rpm to 3120 rpm in 16 s. Find (i) the angular acceleration, and (ii) the number of revolutions turned in that time.

Convert both speeds to rad/s first — this is the step a rushed answer skips:

$ω_{0} = \frac{2 π \times 1200}{60} = 40 π rad/s ω = \frac{2 π \times 3120}{60} = 104 π rad/s$

$α = \frac{ω - ω _{0}}{t} = \frac{104 π - 40 π}{16} = 4 π rad/s^{2}$

For the angle turned, use $θ = ω_{0} t + \frac{1}{2} α t^{2}$ , then divide by $2 π$ to convert radians to revolutions:

$θ = (40 π) (16) + \frac{1}{2} (4 π) (16)^{2} = 640 π + 512 π = 1152 π rad ⟹ \frac{1152 π}{2 π} = 576 revolutions$

The unit conversion is the entire difficulty here — every formula used is one already in the table above.

One caution carried over from Chapter 2: these hold only while angular acceleration is constant.

9. Summary

Giving a body size breaks the point-particle picture: where a force acts now matters, and rotation needs its own description.
Centre of mass $R = \sum m_{i} r_{i} / \sum m_{i}$ need not lie inside the body, and moves as if all external force acted on it alone — internal forces cancel in pairs.
Torque $τ = r \times F$ and angular momentum $l = r \times p$ are both vector (cross) products, where order and direction matter.
$d L / d t = τ_{e x t}$ : angular momentum is conserved when net external torque is zero, not when net external force is zero.
Rigid-body equilibrium needs both $\sum F_{e x t} = 0$ and $\sum τ_{e x t} = 0$ — a couple satisfies the first while violating the second.
Moment of inertia $I = \sum m_{i} r_{i}^{2}$ depends on the axis, not on the body alone, because of the $r^{2}$ weighting; radius of gyration $k = I / M$ packages it as one distance.
A ring and a disc of equal mass and radius have different $I$ (MR² vs MR²/2) purely because of how that mass is distributed relative to the axis.
Every linear quantity has a rotational partner — mass to $I$ , force to torque, momentum to $L$ — and the constant-acceleration kinematic equations translate directly.
Parallel and perpendicular axes theorems, and rolling-motion dynamics, are not in the 2026-27 syllabus despite appearing in most coaching material.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Centre of mass

r_cm = sum(m_i r_i) / sum(m_i)

CM accelerates only due to net external force: A_cm = F_ext/M

Vector (cross) product

a x b = ab sin(theta) n-hat; a x b = -(b x a)

Direction by the right-hand rule; order matters, unlike the scalar product

Torque

tau = r x F

A force pointing straight at the axis produces zero torque

Angular momentum of a particle

l = r x p

Rotational analogue of linear momentum

Torque-angular momentum relation

tau_ext = dL/dt

Angular momentum is conserved when net external TORQUE is zero, not net force

Rigid body equilibrium

sum(F_ext) = 0 AND sum(tau_ext) = 0

Both conditions are needed; a couple satisfies only the first

Principle of moments (lever)

F1 d1 = F2 d2

Balances a lever about its pivot

Moment of inertia

I = sum(m_i r_i^2)

Ring MR^2, disc MR^2/2, solid sphere 2MR^2/5, rod (centre) ML^2/12

Radius of gyration

I = M k^2, so k = sqrt(I/M)

CBSE lists this for Ch6 — the distance at which the whole mass would give the same I

Rotational kinematics (constant alpha)

omega = omega0 + alpha t; theta = omega0 t + (1/2) alpha t^2; omega^2 = omega0^2 + 2 alpha theta

Direct rotational translation of the linear kinematic equations

Torque and angular acceleration

tau = I alpha

Rotational analogue of F = Ma

Rotational kinetic energy and power

K = (1/2) I omega^2; P = tau omega

Rotational analogues of (1/2)Mv^2 and Fv

Angular momentum of a rigid body

L = I omega

Rotational analogue of p = Mv

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Treating torque as just force

✓ Torque = r F sin(theta); a force applied at the axis (r = 0) produces zero torque.

WATCH OUT

✗ Using one fixed moment of inertia for a body

✓ Moment of inertia depends on the axis; the same body has different I about different axes.

WATCH OUT

✗ Saying angular momentum is conserved when force is zero

✓ Angular momentum is conserved when net external TORQUE is zero, not net force.

WATCH OUT

✗ Revising the parallel and perpendicular axis theorems for this chapter

✓ Neither theorem appears in the 2026-27 NCERT chapter, and CBSE does not list either one for Ch6. Coaching material still drills them heavily. CBSE asks for values of moment of inertia for simple shapes with no derivation, plus radius of gyration.

WATCH OUT

✗ Assuming the centre of mass must lie inside the material of the body

✓ It need not. A ring's centre of mass sits in the empty space at its centre; an L-shaped lamina's centre of mass can fall outside the material entirely, as the chapter's own Example 6.3 shows.

WATCH OUT

✗ Treating the vector product as commutative like the scalar (dot) product

✓ It is not: a x b = -(b x a). Reversing the order of a cross product flips its sign and direction, unlike a dot product which never depends on order.

WATCH OUT

✗ Assuming zero net force on a rigid body means it cannot start rotating

✓ A couple — two equal and opposite forces acting along different lines — has zero net force but a nonzero net torque, so the body can still spin up. Equilibrium needs BOTH conditions checked separately.

WATCH OUT

✗ Treating centre of gravity and centre of mass as different points needing separate calculation in this chapter

✓ They coincide whenever gravity is uniform over the body, which is every situation in this chapter — but they are defined differently (one from mass distribution, one from where gravitational torque vanishes), so the chapter still keeps the two concepts distinct.

WATCH OUT

✗ Assuming the moment of inertia formula for a ring or disc applies about any axis through it

✓ The standard table values are for specific stated axes (e.g. through the centre, perpendicular to the plane, or along a diameter) — quoting MR^2 for a disc about its diameter, where the correct value is MR^2/4, is a common slip.

WATCH OUT

✗ Applying v = u + at or similar linear kinematic equations directly in a rotation problem

✓ Use the rotational partner instead — omega = omega0 + alpha t — and remember it holds only while angular acceleration is constant, exactly the same restriction as in the linear case.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Centre of mass, vector algebra, torque/equilibrium, moment of inertia, and angular momentum conservation (Q6.1-6.17)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for System of Particles and Rotational Motion?

9 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

9 questions~6 min worth ~17 marks in Haryana (BSEH) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Centre of mass R = sum(m_i r_i) / sum(m_i); it need not lie inside the body.
•Internal forces cancel in pairs, so they cannot move the centre of mass.
•M A_cm = F_ext: the centre of mass moves as if all mass and all external force acted there.
•No net external force means total momentum is constant and the centre of mass moves uniformly.
•Vector product a x b = ab sin(theta) n-hat, direction by the right-hand rule.
•Order matters: a x b = -(b x a).
•Torque = r x F. A force pointing at the axis gives no torque, however large.
•Angular momentum l = r x p, and dL/dt = external torque.
•Angular momentum is conserved when net external TORQUE is zero, not net force.
•A diver tucks, I falls, so omega rises to keep I omega constant.
•Rigid-body equilibrium needs BOTH zero net force and zero net torque.
•A couple gives zero net force but non-zero torque, the same about every point.
•Principle of moments: F1 d1 = F2 d2.
•Centre of gravity coincides with centre of mass when gravity is uniform.
•I = sum(m_i r_i^2) — depends on the axis chosen, not on the body alone.
•Radius of gyration: I = M k^2, so k = sqrt(I/M).
•Ring MR^2 versus disc MR^2/2 — same mass and radius, twice the I, because the ring's mass all sits at R.
•Rotational analogues: M -> I, F -> torque, p -> L, and K = I omega^2 / 2.
•Constant angular acceleration: omega = omega_0 + alpha t; theta = omega_0 t + alpha t^2 / 2.
•Parallel and perpendicular axis theorems are NOT in the 2026-27 chapter or syllabus.

Haryana (BSEH) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit V sits inside the 17-mark block covering Units III to VI (CBSE Class 11 Physics, 70 marks)

Question type	Marks each	Typical count	What it tests
Moment of inertia / rolling	3-5	1	Axis theorems and rolling-motion energy method
Angular momentum conservation	3	1	Skater/diver problems
Centre of mass / torque	2-3	1	CM location and torque calculation

Prep strategy

Memorise standard moments of inertia
Practise both axis theorems
Use the linear-angular analogy table
Apply energy conservation to rolling bodies

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Sports technique

Divers, gymnasts, and skaters exploit angular momentum conservation to control their spins.

Machinery and flywheels

Moment of inertia governs the design of flywheels, gears, and rotating engine parts.

Astronomy

Angular momentum conservation explains Kepler's second law and the spin of collapsing stars.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Use the linear-angular analogy to recall rotational formulas

Choose the correct moment of inertia and axis

Apply energy conservation for rolling-without-slipping problems

Check whether torque (not force) is zero before conserving L

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Derive moments of inertia by integration for continuous bodies.

STRETCH

Analyse gyroscopic precession using tau = dL/dt vectorially.

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainTorque and angular acceleration of a flywheelFormula application

A flywheel of moment of inertia $2 kg m^{2}$ is at rest. A constant torque of $8$ N·m acts on it for 5 s. Find its angular velocity at the end of 5 s and the angle turned through.

Stuck? Show the approach

Find angular acceleration from tau = I alpha, then apply the rotational kinematic equations directly, exactly as you would use a = F/m and then v = u + at in the linear case.

Show the full solution

$α = \frac{τ}{I} = \frac{8}{2} = 4 rad/s^{2}$ $ω = ω_{0} + α t = 0 + (4) (5) = 20 rad/s$ $θ = ω_{0} t + \frac{1}{2} α t^{2} = 0 + \frac{1}{2} (4) (25) = 50 rad$

Answer: angular velocity 20 rad/s; angle turned 50 rad

The trap

Forgetting to convert the answer's meaning — 50 rad is not 50 revolutions. Divide by 2*pi (about 6.28) to get roughly 7.96 revolutions, if the question asks for revolutions rather than radians.

JEE MainAngular momentum of a particle in circular motionVector application

A particle of mass 0.5 kg moves in a circle of radius 2 m at a constant speed of 4 m/s. Find the magnitude of its angular momentum about the centre of the circle.

Stuck? Show the approach

For circular motion, the position vector from the centre is always perpendicular to the velocity, so the cross product r x p reduces to a simple product of magnitudes with sin(90 degrees) = 1 — no need for full vector components.

Show the full solution

$L = r p sin (90°) = r m v = (2) (0.5) (4) = 4 kg m^{2} /s$

Answer: 4 kg m^2/s

The trap

Using sin(theta) with some other angle is unnecessary complication here — for any particle in uniform circular motion, the radius vector and the velocity are always perpendicular, so the angle is always exactly 90 degrees regardless of where the particle currently sits on the circle.

JEE MainComparing rolling speeds of different shapesEnergy conservation with rotation

A solid sphere and a solid cylinder, both of the same mass and radius, are released from rest at the top of the same incline of height h and roll without slipping. Which reaches the bottom first, and what is the ratio of their speeds at the bottom?

Stuck? Show the approach

Use energy conservation for each shape separately, writing kinetic energy as translational plus rotational with the rolling condition v = R*omega, and compare the resulting speed formulas — the body with the smaller moment-of-inertia fraction wins.

Show the full solution

For any rolling body, $m g h = \frac{1}{2} m v^{2} (1 + k^{2} / R^{2})$ , so $v = \frac{2 g h}{1 + k ^{2} / R ^{2}}$ .

Sphere: $k^{2} / R^{2} = 2/5 ⟹ v_{s p h er e} = \frac{2 g h}{1.4} = \frac{10 g h}{7}$ .

Cylinder: $k^{2} / R^{2} = 1/2 ⟹ v_{cy l} = \frac{2 g h}{1.5} = \frac{4 g h}{3}$ .

$\frac{v _{s p h er e}}{v _{cy l}} = \frac{10/7}{4/3} = \frac{30}{28} = \frac{15}{14} \approx 1.036$

Answer: the solid sphere reaches the bottom first; v_sphere / v_cyl is approximately 1.036

The trap

Assuming mass or radius decides the race is the standard error — for rolling without slipping, both cancel out completely, and only the dimensionless ratio k^2/R^2 (equivalently the fraction of mass near versus far from the axis) determines who wins.

JEE AdvancedRod struck by an impulsive force — combined translation and rotationMulti-part mechanics problem

A uniform rod of mass M and length L lies at rest on a frictionless horizontal surface. An impulsive force J is applied perpendicular to the rod at a distance L/4 from its centre. Find (a) the speed of the centre of mass immediately after the impulse, and (b) the angular speed of the rod immediately after the impulse, about its centre of mass.

Stuck? Show the approach

Treat the impulse exactly like a very large force acting for a very short time: it changes linear momentum by J directly (regardless of where it is applied) and changes angular momentum about the centre of mass by J times the perpendicular distance from the centre of mass to the line of the force.

Show the full solution

(a) Linear impulse-momentum theorem, independent of where the force is applied: $J = M v_{c m} ⟹ v_{c m} = \frac{J}{M}$ .

(b) Angular impulse about the centre of mass: $J (\frac{L}{4}) = I_{c m} ω$ , where $I_{c m} = \frac{M L ^{2}}{12}$ for a rod about its centre.

$ω = \frac{J ( L /4 )}{M L ^{2} /12} = \frac{3 J}{M L}$

Answer: v_cm = J/M; omega = 3J/(ML)

The trap

Trying to find a single 'point of application' velocity and treating the whole rod as moving at that one speed ignores that the rod undergoes BOTH translation of its centre of mass AND independent rotation about that centre of mass simultaneously — the two must be found separately, using linear and angular impulse respectively.

JEE AdvancedTwo-body collision with a rod pivoted at one endAngular momentum conservation with a trap

A uniform rod of mass M and length L is pivoted at one end and hangs vertically at rest. A small ball of mass m moving horizontally with speed $v_{0}$ strikes the rod at its free lower end and sticks to it. Find the angular speed of the rod-ball system immediately after the collision.

Stuck? Show the approach

Linear momentum is NOT conserved here, because the pivot exerts an external impulsive force on the system during the collision. Angular momentum about the pivot, however, is conserved, because the pivot's force acts exactly at the pivot and so contributes zero torque about that point.

Show the full solution

Angular momentum about the pivot, just before collision: $L_{i} = m v_{0} L$ (only the ball contributes, moving perpendicular to the rod at distance L).

Moment of inertia of the combined system about the pivot, just after: $I = \frac{M L ^{2}}{3} + m L^{2}$ (rod about its end, plus the ball as a point mass at distance L).

Conservation of angular momentum: $m v_{0} L = (\frac{M L ^{2}}{3} + m L^{2}) ω$

$ω = \frac{m v _{0}}{L ( \frac{M}{3} + m )} = \frac{3 m v _{0}}{L ( M + 3 m )}$

Answer: omega = 3 m v0 / [L(M + 3m)]

The trap

Reaching for linear momentum conservation out of habit is the standard error — the hinge/pivot exerts an unknown external force during the impact, which rules out linear momentum conservation entirely, even though angular momentum about that same pivot point remains perfectly conserved since the pivot force itself has zero moment arm about the pivot.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Rotational Motion)Very High

NEET PhysicsHigh

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With no external torque about her spin axis, her angular momentum L = I omega is conserved. Pulling her arms in moves mass closer to the axis, decreasing her moment of inertia I. To keep L constant, her angular velocity omega must increase, so she spins faster. Stretching the arms out reverses the effect.

A hollow cylinder has a larger moment of inertia (MR^2) than a solid cylinder (MR^2/2) of the same mass and radius, because its mass is concentrated farther from the axis. With the same gravitational potential energy available, the hollow cylinder puts a greater fraction into rotational kinetic energy, leaving less for translation, so it reaches the bottom with a smaller speed and later.

No, and this catches a lot of students out because almost every coaching handout still drills them. The 2026-27 NCERT chapter never states either theorem, and the CBSE 2026-27 syllabus does not list them under this chapter either. What CBSE does ask for is the VALUES of moment of inertia for simple shapes like rings, discs, rods, cylinders and spheres about stated axes, learned directly rather than derived from a more general axis theorem, plus the radius of gyration.

No, provided no new external force is introduced by the explosion itself. The forces that blow the object apart are internal to the system (the fragments push on each other, in equal and opposite pairs by Newton's third law), and internal forces can never move the centre of mass. So the centre of mass of all the fragments together continues along exactly the same trajectory the original object was already following -- typically the same parabola under gravity -- even as the individual fragments scatter in every direction.

A couple is two equal and opposite forces acting along different, parallel lines rather than along the same line. Because the forces are equal and opposite, they add to zero and produce no net translational acceleration whatsoever -- but because they act along different lines, their torques about any point do not cancel; instead they add up. The result is a body that experiences pure rotation with no tendency to translate at all, which is exactly why equilibrium requires checking force and torque as two separate, independent conditions.

They carry the same information, just packaged differently. The moment of inertia I is a number with unusual units (kg m^2) that is hard to build intuition for directly. The radius of gyration k, defined by I = Mk^2, converts that same information into a single distance: it is the radius at which you could concentrate the entire mass of the body, as an idealised point or thin shell, and get exactly the same moment of inertia. A larger k means the same mass is, in effect, distributed further from the axis on average.

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Related chapters

Students who read System of Particles and Rotational Motion usually read these next.

Units and Measurements

The seven SI base units are no longer defined by objects — since 2018 every one is pinned to a constant of nature. This chapter covers that, plus significant figures and dimensional analysis, worked through all five of the NCERT chapter's own solved examples. CBSE Class 11 Physics Chapter 1.

Motion in a Straight Line

Drop a ruler through a friend's fingers, measure how far it fell before they caught it, and you have measured their reaction time. This chapter builds the tools for exactly that — velocity at a single instant, found by shrinking a time window until the number stops moving; Galileo settling whether acceleration is measured against distance or against time; and three equations derived twice over, once from the area under a graph and once by integration. It never asks what causes motion. That is the next chapter. CBSE Class 11 Physics Chapter 2.

Motion in a Plane

Along a straight line, a plus and a minus sign were enough. In a plane they are not, so this chapter builds vector algebra from scratch and then spends it — showing that a projectile's horizontal and vertical motions are completely independent, and that moving in a circle at constant speed is still accelerated motion. It also carries a syllabus gap worth knowing before you revise. CBSE Class 11 Physics Chapter 3.

Laws of Motion

For two thousand years the obvious answer was that a moving body needs something to keep it moving. It took Galileo to see that the obvious answer was wrong, and Newton to replace it with three laws. This chapter is where kinematics stops describing motion and starts explaining it — through momentum, impulse, friction, and the forces that hold a car on a curve. CBSE Class 11 Physics Chapter 4.

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