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ICSE • Class XI • Physics • Ch 5
Estimated Time: 45 Mins
Study Progress: In Progress

Motion of System of Particles and Rigid Body

Comprehensive ISC Class 11 Physics chapter on center of mass, motion of a system of particles, rotational kinematics, torque, angular momentum, moment of inertia, rolling motion, and rigid-body dynamics with board-level derivations and numericals.

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Why This Chapter Matters

This chapter is central to understanding the physical world and forms the foundation for higher-level physics, engineering, and scientific reasoning.

Before You Begin (Prerequisites)

  • Basic algebra and units
  • Graph reading and interpretation
  • Familiarity with physical quantities and measurements

What You Will Learn (Core Objectives)

  • Explain the key concepts of the chapter clearly.
  • Apply formulas accurately in numericals and derivations.
  • Interpret physical phenomena using scientific reasoning.
  • Differentiate between similar concepts and avoid common mistakes.

Chapter Roadmap & Progression

1 1. System of Particles and Center o...
2 2. Motion of Center of Mass
3 3. Torque and Angular Momentum
4 4. Moment of Inertia and Rotational...
5 5. Rolling without Slipping
6 6. Conservation of Angular Momentum

Complete Concept Guide (100% Curriculum Coverage)

1. System of Particles and Center of Mass

Centre of Mass
Definition

The centre of mass of a system of particles is that point at which the entire mass of the system may be imagined to be concentrated. For a system of $n$ particles:

$$\vec{R}_{cm} = \frac{\sum_{i=1}^{n} m_i \vec{r}_i}{\sum_{i=1}^{n} m_i} = \frac{\sum m_i \vec{r}_i}{M}$$

In one dimension,

$$x_{cm} = \frac{\sum m_i x_i}{M}, \quad y_{cm} = \frac{\sum m_i y_i}{M}$$

For continuous bodies, the sum is replaced by an integral: $\vec{R}_{cm} = \frac{1}{M} \int \vec{r} \, dm$.

Important: the centre of mass need not lie inside the material of the body; for a ring it lies at the centre, which is empty space.

2. Motion of Center of Mass

CM Motion
External Forces Only

Internal forces between particles cancel in pairs, hence only external forces influence the motion of the centre of mass.

$$\vec{F}_{ext} = M\vec{a}_{cm}$$

The translational motion of the body is therefore determined by the resultant external force. It is similar to treating the whole body as a point mass located at its centre of mass.

For translational motion with no rotation, the body behaves exactly as a particle of mass $M$ located at the centre of mass.

3. Torque and Angular Momentum

Rotational Dynamics
Torque

Torque is the turning effect of force about an axis:

$$\vec{\tau} = \vec{r} \times \vec{F}$$

It is the rotational analogue of force. In scalar form for planar motion, $\tau = rF\sin\theta$.

Angular Momentum

For a particle, angular momentum about a point is

$$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$$

For a rigid body rotating about a fixed axis,

$$L = I\omega$$

The relation between torque and angular momentum is

$$\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$$

4. Moment of Inertia and Rotational Kinetic Energy

Moment of Inertia
Definition

Moment of inertia is the rotational analogue of mass:

$$I = \sum m_i r_i^2$$

It measures how difficult it is to change the rotational motion of a body about an axis.

For a continuous body,

$$I = \int r^2 \, dm$$

Rotational kinetic energy is

$$K_{rot} = \frac{1}{2}I\omega^2$$

Hence the total kinetic energy of a rigid body in translation and rotation is

$$K = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I\omega^2$$

5. Rolling without Slipping

Rolling Motion
Pure Rolling Condition

For a body rolling without slipping on a plane surface, the point of contact is instantaneously at rest relative to the surface.

Therefore,

$$v_{cm} = R\omega$$

and, differentiating,

$$a_{cm} = R\alpha$$

This condition is central in all problems involving cylinders, spheres, wheels, and ring rolling down inclines.

6. Conservation of Angular Momentum

Conservation Law
When is angular momentum conserved?

If the net external torque on a system is zero, then

$$\vec{\tau}_{ext} = \frac{d\vec{L}}{dt} = 0$$

Hence,

$$\vec{L} = \text{constant}$$

This law is extremely useful for problems involving ice skaters, spinning discs, and collisions of rotating bodies. The principle is especially important in twisting or folding motion changes where the moment of inertia changes but angular momentum remains constant.

Visual Learning & Conceptual Map

ISC Physics: System of Particles & Rigid Body 1. Centre of Mass R = Σ mᵢ rᵢ / M For discrete masses x_cm = Σ mᵢxᵢ / M CM may be outside body Internal forces cancel Motion of CM follows total external force 2. General Motion F_ext = M a_cm Total external force decides translational acceleration K = ½ M v_cm² + ½ I ω² Translation + rotation Pure rolling Energy split between CM and spin 3. Torque & Angular Momentum τ = r × F Rotational effect of force L = r × p Angular momentum τ_ext = dL/dt Angular momentum is conserved when τ_ext=0 4. Moment of Inertia I = Σ mᵢ rᵢ² Analog of mass in rotation K_rot = ½ I ω² Rotational kinetic energy I depends on axis Different shapes, different inertia High-Yield Formulae • CM: R = Σ mᵢrᵢ / M • Torque: τ = r × F ; τ_ext = dL/dt ; Angular momentum: L = Iω • Rotational KE: K_rot = ½ Iω² ; For pure rolling: v_cm = Rω and a_cm = Rα

Chapter Summary & 10 Key Takeaways

Takeaway 1
The centre of mass is the point where the entire mass of a system may be considered to be concentrated; for a system of particles, $ ec{R}_{cm} = rac{\sum m_i ec{r}_i}{M}$, with $M = \sum m_i$.
Takeaway 2
The motion of the centre of mass is governed by the external forces only: $ ec{F}_{ext} = M ec{a}_{cm}$, while internal forces cancel in pairs by Newton's third law.
Takeaway 3
The total kinetic energy of a rigid body can be separated into translational and rotational parts: $K = rac{1}{2}Mv_{cm}^2 + rac{1}{2}I\omega^2$.
Takeaway 4
Torque is the rotational analogue of force: $ ec{ au} = ec{r} imes ec{F}$ and its time rate of change equals the rate of change of angular momentum: $ ec{ au}_{ext} = rac{d ec{L}}{dt}$.
Takeaway 5
Angular momentum is $ ec{L} = ec{r} imes ec{p}$ for a particle and $ ec{L} = I ec{\omega}$ for a rigid body about a fixed axis.
Takeaway 6
Moment of inertia $I = \sum m_i r_i^2$ measures the resistance offered by a body to angular acceleration; it depends on mass distribution and axis of rotation.
Takeaway 7
Pure rolling is a special combined motion where the point of contact is instantaneously at rest: $v_{cm} = R\omega$ and $a_{cm} = Rlpha$ for rolling without slipping.
Takeaway 8
In the absence of net external torque, angular momentum remains conserved: $ ec{L}_{initial} = ec{L}_{final}$.
Takeaway 9
The rotational analogue of Newton's second law is $ au = Ilpha$ for rigid bodies rotating about a fixed axis.
Takeaway 10
The mass distribution relative to the axis determines the value of $I$, so a ring, disc, and sphere of same mass have different rotational inertias about the same diameter or central axis.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Define centre of mass. Why does the centre of mass of a uniform ring lie at its geometric centre though no mass is present there?
Reveal Answer & Explanation
Answer: The centre of mass is the point where the entire mass of the system is effectively concentrated. For a uniform ring, the mass distribution is symmetric and the mean position of all mass elements is at the geometric centre; hence the centre of mass lies there even though the ring has no material at that point.
2
State and explain the theorem of motion of centre of mass.
Reveal Answer & Explanation
Answer: The resultant external force acting on a system equals the total mass times the acceleration of the centre of mass: $ ec{F}_{ext} = M ec{a}_{cm}$. This follows because internal forces cancel in pairs, leaving only external force to change the translational motion of the system.
3
How is angular momentum related to torque? Give the physical significance.
Reveal Answer & Explanation
Answer: The external torque equals the rate of change of angular momentum: $ au_{ext} = dL/dt$. This means torque is the agent that changes angular momentum, just as force changes linear momentum.
4
Differentiate between translational kinetic energy and rotational kinetic energy. Write the total kinetic energy of a rolling body.
Reveal Answer & Explanation
Answer: Translational kinetic energy is $ rac12 Mv_{cm}^2$ and rotational kinetic energy is $ rac12 I\omega^2$. For a rolling body, total kinetic energy is $K = rac12 Mv_{cm}^2 + rac12 I\omega^2$.
5
What is pure rolling? Derive the condition for rolling without slipping.
Reveal Answer & Explanation
Answer: Pure rolling means the point of contact relative to the surface is instantaneously at rest. Thus $v_{cm} = R\omega$ and $a_{cm} = Rlpha$ for rolling without slipping.
6
Define moment of inertia and explain factors on which it depends.
Reveal Answer & Explanation
Answer: Moment of inertia measures resistance to angular acceleration: $I = \sum m_i r_i^2$. It depends on the mass, its distribution relative to the axis, and the axis chosen.
7
Why is the angular momentum of a body conserved in the absence of external torque?
Reveal Answer & Explanation
Answer: Because $ au_{ext} = dL/dt$. If $ au_{ext}=0$, then $dL/dt=0$ so angular momentum remains constant.
8
State the significance of the law of conservation of angular momentum in daily life or sports.
Reveal Answer & Explanation
Answer: It explains why a spinning skater speeds up when arms are folded inward and slows down when arms are extended. The same principle is used in diving, gymnastics, and acrobatics to control spin.
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