Equilibrium of a Rigid Body Class 11 Notes

Physics System of Particles and Rotational Motion 2025

Syed Aquib Ur Rahman
Updated on Jul 28, 2025 08:52 IST

By Syed Aquib Ur Rahman, Assistant Manager

As you advance as a student of physics, you begin to analyse real-world and macroscopic systems. Earlier in Class 11, you had studied the motion of point mass particles with no size, which made calculations about how these particles moved much easier. 

In Chapter 6, you transition into studying bigger systems that have mass and how they move. Even at this stage, for simplifying concepts such as the centre of mass and its motion, we consider the ideal rigid bodies that don’t deform. 

For this article, we will focus on another idealised state of these rigid bodies when torque and angular momentum in sum are zero. You will learn how the net effect of the cross product of vectors is considered balanced. 

Equilibrium of Rigid Body Class 11
Table of content
  • Equilibrium of Rigid Bodies Simplified for Class 11
  • Conditions for Equilibrium
  • Types of Equilibrium of Rigid Bodies Based on Stability
  • How to Solve Problems for Equilibrium in a Rigid Body
  • Class 11 Physics Revision Notes
  • Detailed NCERT Solutions for Physics Class 11
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Equilibrium of Rigid Bodies Simplified for Class 11

Equilibrium of a rigid body explains how it maintains its state by balancing forces to prevent motion and torques to prevent rotation. 

A single particle needs only force balance. But a rigid body demands both translational and rotational stability. This dual requirement for the rigid body to be in equilibrium is important to understand for two reasons. 

  1. We can better explain systems like seesaws or suspended rods. 
  2. We can better define how the placement of forces matters, just as their magnitude.

For school exams and even for approaching JEE Main prep, remember that 

  • Translational equilibrium prevents linear acceleration.
  • Rotational equilibrium prevents angular acceleration.
  • Equilibrium can be static (at rest) or dynamic (constant velocity)

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Conditions for Equilibrium

Section 6.8 of the Systems of Particles and Rotational Motion Chapter explains that 

A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration.”

The explanation for these conditions are 

  • Translational Equilibrium: The net external force is zero: F e x t = 0

This ensures no acceleration in any direction.

  • Rotational Equilibrium: The net external torque about any point is zero: τ e x t = 0

This prevents rotation.

So, if you recall, torque is defined as

τ = r × F

Here r is the position vector from the pivot to the force's point of application.

This is because zero torque about one point indicates zero torque about all points.

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Types of Equilibrium of Rigid Bodies Based on Stability

Equilibrium stability depends on how a body responds to small displacements. This is determined by the centre of mass (COM) behaviour. 

The types are

  • Stable Equilibrium of Rigid Body: A slight displacement raises the COM, and the body returns to its original position. Take the example of a cone on its base. 
  • Unstable Equilibrium of Rigid Body: A slight displacement lowers the COM, and the body moves further away. Consider a pencil balanced on its tip.
  • Neutral Equilibrium of Rigid Body: A displacement neither raises nor lowers the COM, and the body stays in the new position. For instance, a cylinder on its side.
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How to Solve Problems for Equilibrium in a Rigid Body

If you are preparing for exams, use these steps to solve equilibrium in a rigid body. 

  • Draw a free-body diagram (FBD).
  • Don't forget to mark all forces and their points of application. 
  • Apply F = 0 in components ( F x = 0 , F y = 0 ).
  • Select a pivot point so that you know where unknown forces usually act.
  • Apply τ = 0 as the rigid body is in equilibrium. This helps when you use a consistent sign convention for torques.
  • Solve the resulting equations.

Try yourself on the example problem below. 

Example Problem: A uniform rod of mass 3 kg and length 2 m is hinged at one end and supported by a string at the other, making 60 with the horizontal. Find the tension.

 

Solution: Free-body diagram: Weight m g = 3 × 9.8 = 29.4 N  at the centre, hinge force, and tension T . Translational equilibrium: F y = 0 : T s i n 60 + F hinge   , y - 29.4 = 0

Rotational equilibrium about the hinge: τ = 0 : T c o s 60 × 2 - 29.4 × 1 = 0 T × 0.5 × 2 = 29.4 T = 29.4 N

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Class 11 Physics Revision Notes

Find well-explained notes on all Physics chapters for Class 11. 

Units and Measurements Class 11 Notes Mechanical Properties of Solids Class 11 Notes
Motion in a Straight Line Class 11 Notes Mechanical Properties of Fluids Class 11 Notes
NCERT Class 11 Notes for Motion in a Plane Thermal Properties of Matter Class 11 Notes
Laws of Motion Class 11 Notes Thermodynamics Class 11 Notes
Work, Energy, and Power Class 11 Notes Kinetic Theory of Gas Class 11 Notes
System of Particles and Rotational Motion Class 11 Notes Oscillations Class 11 Notes
Gravitation Class 11 Notes Waves Class 11 Notes

Get an overview of what you have to cover in Physics in Class 11, along with Maths and Chemistry. 

NCERT Class 11 Notes for PCM
NCERT Class 11 Physics Notes

 

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Physics System of Particles and Rotational Motion Exam

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