What is System of Particles and Rotational Motion Class 11: Complete Guide

Introduction to System of Particles in Class 11 Physics

A system of particles is a collection of many particles that may interact with each other. Instead of studying each particle separately, physics analyses the system as a whole to simplify complex problems. This approach is essential in Class 11 NCERT Physics to understand real-world phenomena like motion of planets, molecules, or connected bodies.

Key points:

• Each particle has mass and position.
• The system's total mass is the sum of individual masses.
• The centre of mass represents the average position weighted by mass.
• External and internal forces act on the system.

Studying the system of particles helps in deriving important quantities like total momentum, kinetic energy, and understanding how forces affect the entire system rather than just individual parts.

Understanding Centre of Mass and Its Importance

The centre of mass (COM) is a crucial concept in the system of particles. It is the point where the total mass of the system can be considered to be concentrated for translational motion analysis.

Mathematically, for particles with masses $m_i$ at positions $\vec{r_i}$,

$$ \vec{R} = \frac{\sum m_i \vec{r_i}}{\sum m_i} $$

where $\vec{R}$ is the position vector of the centre of mass.

Why is COM important?

• Simplifies motion analysis of complex systems.
• External forces cause acceleration of the centre of mass.
• Internal forces cancel out and do not affect COM motion.

Example: When a person jumps on a trampoline, the trampoline moves according to the combined motion of the person and the trampoline’s centre of mass.

Basics of Rotational Motion in Class 11 Physics

Rotational motion occurs when a body spins about a fixed axis. This chapter explains how angular quantities describe such motion, analogous to linear motion.

Key terms:

• Angular displacement ($\theta$): The angle rotated in radians.
• Angular velocity ($\omega$): Rate of change of angular displacement.
• Angular acceleration ($\alpha$): Rate of change of angular velocity.

Formulas linking angular and linear quantities:

• $v = r\omega$ (linear speed at radius $r$)
• $a_t = r\alpha$ (tangential acceleration)

Understanding these helps in solving problems involving rotating wheels, discs, or planets.

Moment of Inertia: Rotational Equivalent of Mass

The moment of inertia (I) measures how difficult it is to change an object's rotational motion about an axis. It depends on the mass distribution relative to the axis.

Formula:

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

where $m_i$ is the mass of the $i^{th}$ particle and $r_i$ is its distance from the axis.

Comparison with mass:

Different shapes have different moments of inertia. For example:

• Solid sphere: $I = \frac{2}{5}MR^2$
• Thin rod about centre: $I = \frac{1}{12}ML^2$

Worked example:

Calculate $I$ for two masses 2 kg and 3 kg placed 1 m and 2 m from axis:

$$ I = 2 \times 1^2 + 3 \times 2^2 = 2 + 12 = 14 \text{ kg m}^2 $$

Torque and Its Role in Rotational Motion

Torque ($\tau$) is the rotational equivalent of force. It causes angular acceleration in a body.

Definition:

$$ \tau = r F \sin \theta $$

where $r$ is the lever arm (distance from axis), $F$ is the applied force, and $\theta$ is the angle between force and lever arm.

Key points:

• Torque direction is given by the right-hand rule.
• Larger torque means greater tendency to rotate.
• Torque and angular acceleration are related by:

$$ \tau = I \alpha $$

Example:

A force of 10 N is applied perpendicular to a wrench 0.3 m long. Torque:

$$ \tau = 0.3 \times 10 = 3 \text{ Nm} $$

Angular Momentum and Its Conservation

Angular momentum ($L$) is the rotational analogue of linear momentum. It is defined as:

$$ L = I \omega $$

where $I$ is moment of inertia and $\omega$ is angular velocity.

Conservation principle:

• In absence of external torque, total angular momentum remains constant.
• This explains phenomena like figure skaters spinning faster by pulling arms in.

Worked example:

A rotating disc of $I = 5$ kg m$^2$ spins at $\omega = 10$ rad/s. If a 2 kg mass is dropped at radius 0.5 m and sticks, find new angular velocity.

Initial $L_i = 5 \times 10 = 50$ kg m$^2$/s

New $I = 5 + 2 \times 0.5^2 = 5 + 0.5 = 5.5$

Using $L_i = L_f$:

$$ 50 = 5.5 \times \omega_f \Rightarrow \omega_f = \frac{50}{5.5} \approx 9.09 \text{ rad/s} $$