Class 11 Physics: System of Particles and Rotational Motion | Physics Q&A

CBSE Class 11 › Unit V: Rigid Body Motion

System of Particles and Rotational Motion

NCERT Chapter 6 • Centre of Mass, Torque, Moment of Inertia & Rolling

NCERT 2025–26 Unit V • ~10 Marks JEE Main • 3-4 Questions

Spinning gyroscope illustrating angular momentum and torque. — Rotation: Where forces turn into torque and mass becomes inertia.

1. Centre of Mass (CM)

The Centre of Mass of a system of particles is the point that moves as if all the mass of the system were concentrated at that point and all external forces were applied at that point.

Location of Centre of Mass for a two-particle system. — The Centre of Mass is the balance point where the system’s entire mass is concentrated.

1.1 Formula for CM

For a system of $n$ particles with masses $m_i$ at positions $\vec{r}_i$ :

\vec{R}_{CM} = \dfrac{\sum m_i \vec{r}_i}{\sum m_i} = \dfrac{1}{M} \sum m_i \vec{r}_i

For a two-particle system (on x-axis):

X = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}

1.2 Motion of the Centre of Mass

The velocity of the CM is given by:

\vec{V}_{CM} = \dfrac{\sum m_i \vec{v}_i}{M} = \dfrac{\vec{P}_{total}}{M}

Differentiating again, we get Newton’s Second Law for a system:

M \vec{A}_{CM} = \vec{F}_{ext}

This means the CM moves as if all the mass were concentrated at that point and all external forces acted there.

Key Example:

In an exploding projectile, internal forces cause fragments to fly apart, but the CM continues on the original parabolic trajectory because $\vec{F}_{ext} = m\vec{g}$ remains unchanged.

1.3 Conservation of Linear Momentum

If the net external force on a system is zero ( $\vec{F}_{ext} = 0$ ), then:

\vec{P}_{total} = \text{Constant}

This principle explains recoil of guns, motion of binary stars, and radioactive decay.

Centre of Gravity (CG) vs Centre of Mass (CM):

The CG is the point where the total gravitational torque is zero. For small bodies in a uniform gravitational field, CG coincides with CM. For large bodies (e.g., mountains), they may differ.

Problem 1: Two bodies of masses 1 kg and 2 kg are located at (1, 2) and (-1, 3) respectively. Find the coordinates of the Centre of Mass.

Show Answer

$X_{cm} = \dfrac{1(1) + 2(-1)}{1+2} = \dfrac{1-2}{3} = -1/3$ .
$Y_{cm} = \dfrac{1(2) + 2(3)}{1+2} = \dfrac{2+6}{3} = 8/3$ .
CM is at $(-1/3, 8/3)$ .

2. Vector Product (Cross Product)

To describe rotation, we use the vector product. For two vectors $\vec{A}$ and $\vec{B}$ :

\vec{A} \times \vec{B} = AB \sin\theta \, \hat{n}

Where $\hat{n}$ is a unit vector perpendicular to both $\vec{A}$ and $\vec{B}$ given by the Right-Hand Rule.

Properties:

1. Not Commutative: $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$
2. Parallel vectors: $\vec{A} \times \vec{A} = \vec{0}$

3. Rotational Kinematics

The equations for rotational motion with constant angular acceleration ( $\alpha$ ) are analogous to linear motion.

Linear ( $a$ )	Rotational ( $\alpha$ )
$v = v_0 + at$	$\omega = \omega_0 + \alpha t$
$x = v_0t + \dfrac{1}{2}at^2$	$\theta = \omega_0t + \dfrac{1}{2}\alpha t^2$
$v^2 = v_0^2 + 2ax$	$\omega^2 = \omega_0^2 + 2\alpha\theta$

Relation between Linear and Angular Variables

v = r\omega \quad \text{and} \quad a_t = r\alpha

4. Torque and Angular Momentum

4.1 Torque ( $\vec{\tau}$ )

Torque is the “turning effect” of a force. It is the moment of force.

Diagram showing the cross product relationship between Radius, Force, and Torque. — Torque ( $\vec{\tau} = \vec{r} \times \vec{F}$ ) is the turning effect of a force.

\vec{\tau} = \vec{r} \times \vec{F} = rF \sin\theta

4.2 Angular Momentum ( $\vec{l}$ )

Angular momentum is the rotational equivalent of linear momentum.

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

For a rigid body rotating about a fixed axis, $\vec{L} = I\vec{\omega}$ .

4.3 Relation between Torque and Angular Momentum

Derivation: $\vec{\tau} = \frac{d\vec{l}}{dt}$ 3 Marks

Step 1: Differentiate $\vec{l}$
$\vec{l} = \vec{r} \times \vec{p}$ .
$\dfrac{d\vec{l}}{dt} = \dfrac{d}{dt}(\vec{r} \times \vec{p})$ .

Step 2: Product Rule
$\dfrac{d\vec{l}}{dt} = (\dfrac{d\vec{r}}{dt} \times \vec{p}) + (\vec{r} \times \dfrac{d\vec{p}}{dt})$ .

Step 3: Simplify
$\dfrac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v}$ . So, $\vec{v} \times m\vec{v} = \vec{0}$ (Parallel).
$\dfrac{d\vec{p}}{dt} = \vec{F}$ (Newton’s 2nd Law).

Step 4: Final Relation
$\dfrac{d\vec{l}}{dt} = \vec{0} + (\vec{r} \times \vec{F}) = \vec{\tau}$ .

4.4 Conservation of Angular Momentum

If the net external torque on a system is zero ( $\vec{\tau}_{ext} = 0$ ), then $\dfrac{d\vec{L}}{dt} = 0$ , which implies $\vec{L} = \text{Constant}$ .

Example:

A spinning ice skater brings her arms in (decreasing Moment of Inertia $I$ ). To conserve $L = I\omega$ , her angular speed $\omega$ increases.

5. Equilibrium of a Rigid Body

For a rigid body to be in mechanical equilibrium, two conditions must be satisfied:

Translational Equilibrium: Net external force is zero ( $\sum \vec{F}_{ext} = 0$ ).
Rotational Equilibrium: Net external torque is zero ( $\sum \vec{\tau}_{ext} = 0$ ).

5.1 Principle of Moments

For a lever in equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments:

F_1 d_1 = F_2 d_2

This is the basis of balances and see-saws. The mechanical advantage is $MA = \dfrac{d_2}{d_1}$ .

6. Moment of Inertia ( $I$ )

Moment of Inertia is the rotational analogue of mass. It measures the resistance of a body to change its rotational motion.

I = \sum m_i r_i^2

For continuous bodies: $I = \int r^2 dm$

6.1 Theorems of Moment of Inertia

Perpendicular Axes Theorem: (Planar bodies only)
$I_z = I_x + I_y$ .
Example: For a disc, $I_z = \frac{1}{2}MR^2$ , so $I_x = I_y = \frac{1}{4}MR^2$ .

Parallel Axes Theorem:
$I = I_{CM} + Md^2$ , where $d$ is distance between axes.
Example: For a rod about one end: $I = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = \frac{1}{3}ML^2$ .

Radius of Gyration ( $k$ )

The distance from the axis where the entire mass $M$ can be assumed to be concentrated to give the same Inertia.

I = Mk^2 \quad \Rightarrow \quad k = \sqrt{\dfrac{I}{M}}

Common Moments of Inertia

Body	Axis	I
Thin Ring	Center (Perpendicular)	$MR^2$
Disc	Center (Perpendicular)	$MR^2/2$
Solid Sphere	Diameter	$2MR^2/5$
Hollow Sphere	Diameter	$2MR^2/3$
Rod (Length L)	Center (Perpendicular)	$ML^2/12$

Dynamics of Rotation

Just as $F = ma$ , for rotation we have:

\tau = I\alpha

Work and Power in Rotation

Work done by torque: $dW = \tau \, d\theta$ .

W = \int_{\theta_1}^{\theta_2} \tau \, d\theta \quad \text{and} \quad P = \tau \omega

Rotational Work-Energy Theorem: $W = \dfrac{1}{2}I\omega_f^2 - \dfrac{1}{2}I\omega_i^2$ .

Problem 2: A cord is wound around a solid cylinder of mass 50 kg and radius 0.5 m. A force of 25 N is applied to pull the cord. Find angular acceleration.

Show Answer

Torque $\tau = F \times R = 25 \times 0.5 = 12.5$ Nm.
Inertia (Solid Cylinder) $I = \dfrac{1}{2}MR^2 = 0.5(50)(0.5)^2 = 6.25$ kg m².
$\alpha = \tau / I = 12.5 / 6.25 = 2$ rad/s².

7. Rolling Motion

Rolling without slipping is a combination of Translation and Rotation.

v_{cm} = R\omega

Velocity distribution of a wheel rolling without slipping. — In pure rolling, the point of contact is momentarily at rest.

Kinetic Energy of Rolling

Total KE = Translational KE + Rotational KE

K = \dfrac{1}{2}mv_{cm}^2 + \dfrac{1}{2}I\omega^2

K = \dfrac{1}{2}mv^2 \left( 1 + \dfrac{k^2}{R^2} \right)

7.1 Rolling Down an Incline

For a body rolling down an incline of angle $\theta$ , the linear acceleration is:

a = \dfrac{g \sin\theta}{1 + \dfrac{I}{mR^2}}

This shows that for the same $m$ and $R$ , the body with smaller $I$ (e.g., solid sphere) accelerates faster.

7.2 Role of Friction in Rolling

Friction provides the torque for rotation. Its direction depends on the situation:

Accelerating car: Friction on driving wheels acts forward.
Braking car: Friction acts backward.
Pure rolling at constant speed: Friction is zero.

The condition for rolling without slipping is $f_s \leq \mu_s N$ .

Problem 3 (JEE): A solid sphere and a hollow sphere of same mass and radius roll down an incline. Which reaches the bottom first?

Show Answer

Acceleration $a = \dfrac{g \sin\theta}{1 + k^2/R^2}$ .
Solid Sphere: $k^2/R^2 = 2/5 = 0.4$ .
Hollow Sphere: $k^2/R^2 = 2/3 \approx 0.67$ .
Since Solid Sphere has smaller $k^2/R^2$ , it has larger acceleration and reaches first.

← Previous Chapter Next Chapter: Gravitation →

1. Centre of Mass (CM)

1.1 Formula for CM

1.2 Motion of the Centre of Mass

1.3 Conservation of Linear Momentum

2. Vector Product (Cross Product)

3. Rotational Kinematics

Relation between Linear and Angular Variables

4. Torque and Angular Momentum

4.1 Torque ()

4.2 Angular Momentum ()

4.3 Relation between Torque and Angular Momentum

4.4 Conservation of Angular Momentum

5. Equilibrium of a Rigid Body

5.1 Principle of Moments

6. Moment of Inertia ()

6.1 Theorems of Moment of Inertia

Radius of Gyration ()

Common Moments of Inertia

Dynamics of Rotation

Work and Power in Rotation

7. Rolling Motion

Kinetic Energy of Rolling

7.1 Rolling Down an Incline

7.2 Role of Friction in Rolling

4.1 Torque ( $\vec{\tau}$ )

4.2 Angular Momentum ( $\vec{l}$ )

6. Moment of Inertia ( $I$ )

Radius of Gyration ( $k$ )