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NCERT Class 11 Physics Chapter 7 notes System of Particles and Rotational Motion - Download PDF

NCERT Class 11 Physics Chapter 7 notes System of Particles and Rotational Motion - Download PDF

Edited By Vishal kumar | Updated on Mar 19, 2024 11:55 AM IST

Hello, Class 11 Physics enthusiasts! We're excited to provide you with comprehensive System of Particles and Rotational Motion class 11 notes. These notes, created by our experts, are intended to provide a thorough understanding of each concept, taking into account both theoretical knowledge and practical applications.

Physics is more than just formulas; it is about comprehending how these concepts apply in the real world. Our class 11 physics chapter 7 notes not only explain the theories, but also explore the practical applications of each concept. In today's exam environment, where application-based questions are common, these CBSE class 11 physics ch 7 notes will not only help you prepare for exams, but will also foster a genuine interest in the subject.

Don't miss out on this opportunity to improve your understanding and excel in your exams. Download the physics class 11 chapter 7 notes pdf now and take a journey of discovery through Chapter 7.

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NCERT Class 11 Physics Chapter 7 Notes

Rotational Motion

Rotational motion describes the movement of an object as it spins or revolves around an axis. Consider a spinning top, a rotating wheel, or the Earth turning on its axis; these are all examples of rotational motion.

In this type of motion, each point in the object follows a circular path around a fixed line known as the axis. The motion is a rotation rather than a linear translation.

Centre of Mass

The centre of mass is an imaginary point in a system where all of the system's mass is concentrated. When two bodies are joined, their centre of mass is along the line that connects them.


(Here m1 & m2 are two bodies in which m1 is at a distance x1, & m2 at a distance x2 from the origin O).

  • The coordinates of a system's centre of mass (X, Y, Z) are generally given by


  • Considering a thin rod of length l, the origin is at the geometric centre of the rod. and the X-axis is along the length of the rod, and for every element dm of the rod at x, there is an element of the same mass dm located at –x.


  • The total contribution of each such integral and the integral x dm is zero. Therefore, the COM coincides with its geometric centre.

  • The centre of mass coincides with the geometric centre and with the same geometry is applicable to homogenous rings, discs, spheres, or even thick rods of circular or rectangular cross-sections;


Motion of COM

The centre of mass of a system of particles moves in a manner such that the entire mass of the system is considered to be concentrated at the centre, and all external forces are treated as if they act solely on that point.


Differentiating w.r.t. time, we get,



Again, differentiating w.r.t. time, we get,


Ma = Fext

  • Instead of considering extended bodies as single particles, we can consider them as systems of particles.

The translational component of their motion could be attained by taking the mass of the whole system and all the external forces on the system to be acting at the centre of mass.

Linear Momentum of System of Particles

The linear momentum of a system of particles is the product of the system's total mass and the velocity of its centre of mass. This relationship is expressed as follows:

Linear Momentum=Total Mass×Velocity of Center of Mass

P = p1 + p2 + …. + pn

= m1v1 + m2v2 + …. + mnvn

P = MV;

Where P = Momentum particles and V = Velocity of COM

Newton’s Second Law in reference to the system of particles is represented by the following expression:

dP/dt = Fext

  • In case when the total external force acting on a system of particles is zero (Fext = 0), then, in that case, the total linear momentum of the system is constant and can be represented as:

(dP/dt = 0; P = constant).

  • The Centre of mass having velocity also remains constant (Since P = mv = Constant).

  • If the total external force on a body is zero, then the internal forces cause complex trajectories of individual particles in spite of the COM moving with a constant velocity.

Angular Velocity & Linear Velocity

Each particle of a rotating body is considered to move in a circle. We define the angular displacement of a given particle about its centre in unit time as angular velocity.


  • Average angular velocity =∆θ/∆t
  • Instantaneous angular velocity, ω=dθ/dt
  • In a pure rotation motion, all the parts of the moving body having the same angular velocity.
  • It is a vector quantity
  • v = ω x r; where v is the linear velocity of a particle and r is the radius of the circle
  • Angular acceleration is outlined as the rate of amendment of angular speed with relevance time.α = dω / dt

Torque & Angular Momentum

The rotational analogue of force is a moment of force which is also known as Torque).


Torque is a vector quantity.

Equilibrium of Rigid Body

  • The translational state of the motion of the rigid body is due to the changes in force, i.e. its total linear momentum changes.

  • The rotational state of motion of the rigid body is due to the changes in torque i.e. the total angular momentum of the body changes.

Unless stated otherwise, only external forces and torques should be dealt.

A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum are constant with time. This means

Total force zero = Translational Equilibrium

Total torque zero = Rotational Equilibrium


A couple or torque can be defined as a pair of equal and opposite forces with different lines of action is termed as couple or torque.

Without translation, the couple can produce the rotation.

For example, when you open the lid of a jar, you put a couple on it.

Centre of Gravity

The centre of gravity of a body is the point on the body where the total gravitational torque on the body is zero.

In uniform gravity or gravity-free space, the centre of gravity of the body coincides with the centre of mass.

The condition for the centre of gravity and centre of mass will not coincide is that if the value of g will vary from part to part of the body.

Moment of Inertia

Moment of inertia (I) is the analogue of the mass in rotational motion.

A change in rotational motion is resisted by the moment of inertia about a given axis of rotation. It can be regarded as a measure of rotational inertia of the body.

It is a measure of how the different parts of the body are distributed at different distances from the axis.

The moment of inertia of a rigid body depends on the mass of the body, its form and size, and therefore the distribution of mass concerning the axis of rotation.


The radius of gyration of a body concerning association in axis is outlined as the distance from the axis of a mass whose mass is up to the mass of the total body and whose moment of inertia is up to the instant of inertia of the body concerning the axis.

I = M k2, where k is the radius of gyration.

The Theorem of Perpendicular Axis

Perpendicular Axis Theorem: The moment of inertia of a two-dimensional body (lamina) regarding an axis perpendicular to its plane is the sum of its moments of inertia regarding two perpendicular axes synchronous with perpendicular axis and lying within the plane of the body.


It is applicable only to planar bodies.

The Theorem of Parallel Axis

Parallel Axis Theorem: The moment of inertia of a body about an axis parallel to an axis passing through the center of mass is equal to the sum of the moment of inertia of body about an axis passing through center of mass and product of mass & square of the distance between two axes.This theorem is applicable to a body of any form.



Kinematics of Rotational Motion for a Fixed Axis

We can derive equations of motion similar to translational motion.

Translational motion




Rotational Motion




Dynamics of Rotational Motion for a Fixed Axis

The components of torques are considered along the direction of the fixed axis and this is so because the component of the torque is perpendicular to the axis of rotation and this will tend to turn the axis from its position.

The forces that lie in planes perpendicular to the axis are only considered. Forces parallel to the axis give torques perpendicular to the axis.

Also, the components of the position vectors which are perpendicular to the axis are only considered. Components of position vectors on the axis can lead to torques perpendicular to the axis.


Rolling Motion

Rolling motion may be defined as a combination of rotation and translation motion.

Particles of a rolling body have two velocities.

  1. Transnational, which is the velocity of COM.

  2. The linear velocity of rotational motion.


In the figure, we observe that every point has two velocities.

One Velocity is found in the direction COM and the other velocity is found to be perpendicular to the line joining centre and the point.

At Point Po have velocities and both are in opposite direction to each other.

For no-slipping condition: Velocity is found has zero value, so

Vcom = ω R

There is an instantaneous axis of rotation that can be seen through PO and parallel to ‘w’ .

At Point, P1 Velocities are added up.

Kinetic Energy of Rolling Motion

K.E.rolling = K.E.translation + K.E.rotation

Kinetic energy of a rolling body


where k is radius of gyration of body

Significance of NCERT Class 11 Physics Chapter 7 Notes

Convenience: The class 11 physics chapter 7 notes provide a concise overview of the System of Particles and Rotational Motion, allowing for a better understanding of the chapter's concepts.

Precision: These System of Particles and Rotational Motion class 11 notes highlight key points, allowing for a more precise understanding of the material, which is critical for performing well on CBSE board exams.

Comprehensive Coverage: They cover all of the chapter's essential highlights, ensuring that no important details are overlooked while studying.

Guidance: The notes point students in the right direction, clarifying complex topics and keeping them focused on key concepts.

Scoring Aid: Following these System of Particles and Rotational Motion notes class 11 improves your chances of getting a good score on the CBSE board exam by ensuring thorough preparation.

Flexibility: These CBSE class 11 physics ch 7 notes are available in both online and offline modes, allowing students to study whenever and wherever it is convenient for them.

Availability: Once downloaded, users can access the physics class 11 chapter 7 notes pdf at their leisure, making them an excellent resource for offline study sessions.

Complementing Learning: These ch 7 physics class 11 notes supplement classroom instruction, allowing for better retention and understanding of the subject matter.

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Frequently Asked Question (FAQs)

1. For small objects that the centre of gravity earth coincides with the mass but it is not found in the case of extended objects. What is the qualitative approach to the words ‘small’ and ‘extended mean in this regard?

Ans- The centre of gravity is the geometric centre whereas the centre of mass is the mass of the point where the entire mass of the body is considered. The object is said to be small when the vertical height of the object is close to the earth’s surface. If the distance between the object and the surface of the earth is large then the object is an extended object.

Buildings and ponds are considered to be small objects while a mountain and a lake are considered to be extended objects.

2. What are the main topics covered in the System of Particles and Rotational motion Class 11 notes.

Every topic is important in the chapter no particular topic is, therefore, marked to be an important one hence proper revision of all the topics through NCERT notes for Class 11 Physics chapter 7 needs to be done to secure good marks in CBSE Board examinations.  

3. For which of the following does the centre of mass lie outside the body according to the Class 11 Physics chapter 7 notes Class 11th physics chapter 7 notes and class 11 System of Particles and Rotational motion?

a) a pencil

b) a shotput

c) a dice

d) a bangle


The correct answer is d) a bangle

4. What is the value of gravitational torque on the body which we discussed in the NCERT Class 11 Physics chapter 7 notes.

NCERT Class 11 Physics chapter 7 notes the centre of gravity of a body is the point on the body where the total gravitational torque on the body is zero.

5. Is System of Particles and Rotational Motion notes class 11 important for JEE?

Yes, understanding System of Particles and Rotational Motion notes from Class 11 Physics is crucial for JEE preparation as it forms the foundation for advanced topics in mechanics and rotational dynamics tested in the exam.


Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg


An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)


Option 2)

\; K\;

Option 3)


Option 4)


In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)


Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)


Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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