Academia Aeternum
तमसो मा ज्योतिर्गमय
Class XI Physics · Chapter 6
System of Particles
& Rotational Motion
NCERT | Class 11 | Physics
From the pirouette of a skater to the spin of a neutron star — rotational mechanics governs every turning, rolling, and spinning object in the universe. Master the language of angular momentum.
\( \vec{\tau} = I\,\vec{\alpha} \)
Explore
10+
Key Concepts
2
Types of Motion
33
NCERT Exercises
τ=Iα
Rotational Newton II
JEE
Very High Weightage
Conceptual Framework
Core Topics at a Glance
Centre of Mass
The point where the entire mass of a system can be assumed to be concentrated for translational analysis. For uniform bodies it coincides with the geometric centre.
\(x_{cm} = \frac{\sum m_i x_i}{\sum m_i}\)
Motion of Centre of Mass
The centre of mass of a system moves as if all external forces act on a single particle of mass equal to the total mass of the system.
\(M\vec{a}_{cm} = \vec{F}_{\text{ext}}\)
Angular Velocity & Acceleration
Angular velocity ω is the rate of change of angular displacement. Angular acceleration α = dω/dt. They are axial vectors obeying the right-hand screw rule.
\(\vec{v} = \vec{\omega}\times\vec{r}\)
Torque
Torque is the rotational analogue of force. It is the cross product of the position vector and the force. A net torque causes angular acceleration.
\(\vec{\tau} = \vec{r}\times\vec{F} = I\vec{\alpha}\)
Moment of Inertia
The rotational analogue of mass. It depends on both the total mass and its distribution about the rotation axis. Determined by the parallel and perpendicular axis theorems.
\(I = \sum m_i r_i^2\)
Angular Momentum
Angular momentum L = Iω (for rigid body) or L = r × p (for a particle). It is conserved when net external torque is zero — the most important conservation law in rotation.
\(\vec{L} = I\vec{\omega} = \vec{r}\times\vec{p}\)
Conservation of Angular Momentum
When net external torque = 0, angular momentum is conserved. This explains why a figure skater spins faster when arms are pulled in (I decreases → ω increases).
\(I_1\omega_1 = I_2\omega_2\)
Equilibrium of Rigid Body
A rigid body is in mechanical equilibrium when: (i) net external force = 0 (translational equilibrium) and (ii) net external torque = 0 (rotational equilibrium).
\(\sum F = 0\;\text{and}\;\sum\tau = 0\)
Rolling Motion
Rolling without slipping combines translation of the centre of mass and rotation about it. The contact point has zero instantaneous velocity.
\(v_{cm} = R\omega,\; KE = \tfrac{1}{2}mv^2(1+k^2/R^2)\)
Theorems of MI
Parallel Axis Theorem: I = I_cm + Md². Perpendicular Axis Theorem (laminae): I_z = I_x + I_y. These allow MI calculation for any axis from a known standard axis.
\(I = I_{cm} + Md^2\)
Most Powerful Study Tool
Translation ↔ Rotation Analogy Table
Every translational quantity has a perfect rotational counterpart. Mastering this table lets you derive any rotational formula instantly — and is directly tested in JEE and NEET.
| 🔵 Translational (Linear) | 🔴 Rotational (Angular) |
|---|---|
| Mass (m) | Moment of Inertia (I) |
| Force (F) | Torque (τ) |
| Linear velocity (v) | Angular velocity (ω) |
| Linear acceleration (a) | Angular acceleration (α) |
| Linear momentum (p = mv) | Angular momentum (L = Iω) |
| KE = ½mv² | KE = ½Iω² |
| F = ma | τ = Iα |
| W = F·d | W = τ·θ |
| Power = Fv | Power = τω |
Quick Reference
Key Formulae
| Quantity | Formula | Remarks |
|---|---|---|
| Centre of Mass | \(\vec{r}_{cm}=\dfrac{\sum m_i\vec{r}_i}{M}\) | M = total mass |
| Torque | \(\vec{\tau}=\vec{r}\times\vec{F}=I\vec{\alpha}\) | Rotational analogue of F=ma |
| Moment of Inertia | \(I=\sum m_i r_i^2 = \int r^2\,dm\) | Depends on axis of rotation |
| Angular Momentum | \(\vec{L}=I\vec{\omega}=\vec{r}\times\vec{p}\) | dL/dt = τ |
| Rotational KE | \(KE_{\text{rot}}=\tfrac{1}{2}I\omega^2\) | Analogue of ½mv² |
| Parallel Axis Theorem | \(I=I_{cm}+Md^2\) | d = distance between axes |
| Perpendicular Axis (Lamina) | \(I_z=I_x+I_y\) | Only for planar bodies |
| MI — Solid Sphere | \(I=\tfrac{2}{5}MR^2\) | About diameter |
| MI — Hollow Sphere | \(I=\tfrac{2}{3}MR^2\) | About diameter |
| MI — Solid Cylinder / Disc | \(I=\tfrac{1}{2}MR^2\) | About central axis |
| MI — Ring / Hollow Cylinder | \(I=MR^2\) | About central axis |
| MI — Rod (centre) | \(I=\tfrac{1}{12}ML^2\) | About perpendicular bisector |
| Rolling KE | \(KE=\tfrac{1}{2}Mv_{cm}^2\!\left(1+\dfrac{k^2}{R^2}\right)\) | k = radius of gyration |
| Conservation of L | \(I_1\omega_1=I_2\omega_2\) | When τ_ext = 0 |
Study Material
All Resources for This Chapter
Chapter Notes
Comprehensive Notes
Complete theory for System of Particles and Rotational Motion — Centre of mass, MI derivations, rolling motion, and all NCERT derivations step-by-step.
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MCQ Practice
MCQ Practice Bank
50+ MCQs on MI of standard bodies, torque numericals, rolling motion, angular momentum conservation — with full solution hints.
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True / False Quiz
True–False Quiz
Targeted True/False statements to sharpen conceptual clarity on torque, angular momentum, rolling, and equilibrium of rigid bodies.
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NCERT Exercises
Exercise Solutions
Step-by-step solutions to all 33 NCERT back-exercise questions — worked in full with highlighted final answers for Board preparation.
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Previous Year Qs
PYQ Bank
Curated PYQs from JEE Main, JEE Advanced, NEET, and CBSE Boards — sorted topic-wise so you focus on the highest-frequency patterns.
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Exam-Ready Insights
Important Points to Remember
01
Centre of mass ≠ centre of gravity in a non-uniform gravitational field, but they coincide near Earth's surface.
02
The velocity of centre of mass of an isolated system remains constant — a direct consequence of Newton's third law.
03
Torque depends on the choice of origin (pivot). Always state the axis about which torque is calculated.
04
Moment of inertia is not a fixed property — it changes with axis. I is minimum about the axis through the centre of mass.
05
τ = Iα is the rotational analogue of F = ma. Here I plays the role of mass and α the role of linear acceleration.
06
Angular momentum is conserved when net torque = 0, not net force = 0. A planet's elliptical orbit obeys this.
07
In rolling without slipping, friction does no work — it only provides the torque needed for rolling. v_cm = Rω.
08
A hollow cylinder has greater I than a solid cylinder of the same mass and radius — mass is farther from the axis.
09
For a rigid body in equilibrium, torques about ANY point sum to zero — choose the pivot at an unknown force to simplify.
10
The radius of gyration k is defined by I = Mk² — it is the RMS distance of mass elements from the rotation axis.
Competitive Exams
Exam Corner
System of Particles and Rotational Motion is one of the highest-weightage chapters in JEE and NEET. These are the most frequently tested topics across all exams:
JEE
MI of standard bodies — derivation & numerical
All
Torque and angular acceleration — τ = Iα
JEE
Parallel & perpendicular axis theorems
All
Conservation of angular momentum — problems
NEET
Rolling motion — KE, velocity, acceleration
Board
Equilibrium of rigid bodies — beam & ladder
JEE
Centre of mass of composite bodies
NEET
Analogy table: translation ↔ rotation
All
Angular momentum of a planet (Kepler II)
JEE
Toppling vs sliding — comparing friction
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