Chapter 3: Laws of Conservation in Physics

Chapter 3: Laws of Conservation

3.1 Basic Concepts

This section introduces the concept of a rigid body and the foundational idea of the center of mass in systems of particles. These concepts are essential for understanding the laws of conservation in physics.

• Rigid Body: A rigid body is a system of particles in which the distance between any two points remains unchanged during motion.

• Center of Mass: The center of mass (G) is a point that satisfies the balance condition of the system, representing the average position of the mass distribution.

Center of Mass

The center of mass of a system of particles is calculated using the weighted average of their positions. It is crucial for analyzing the motion and balance of physical systems.

• Definition: For a system of n particles with masses $m_1, m_2, ..., m_n$ and position vectors $extbf{M}_1, extbf{M}_2, ..., extbf{M}_n$, the center of mass G satisfies: $m_1 extbf{M}_1 extbf{G} + m_2 extbf{M}_2 extbf{G} + ... + m_n extbf{M}_n extbf{G} = 0$ or $oxed{ extbf{R} = rac{ extstyle extsum_{i=1}^n m_i extbf{r}_i}{ extstyle extsum_{i=1}^n m_i}}$

• Coordinates: The coordinates of the center of mass $(X, Y, Z)$ are given by: $X = rac{ extstyle extsum_{i=1}^n m_i x_i}{ extstyle extsum_{i=1}^n m_i}$ $Y = rac{ extstyle extsum_{i=1}^n m_i y_i}{ extstyle extsum_{i=1}^n m_i}$ $Z = rac{ extstyle extsum_{i=1}^n m_i z_i}{ extstyle extsum_{i=1}^n m_i}$

• Example (Bar with Attached Masses): For a uniform bar AB (length 1 m, mass 100 g) with masses 20 g at 20 cm and 40 g at 40 cm from A: $m_1(x-20) + m_2(x-40) + m_3(x-50) = 0$ Solution: $x = 43.75$ cm

• Example (Masses at Vertices of a Square): Four masses (10 g, 20 g, 30 g, 40 g) at the vertices of a square (20 cm side): $X_G = rac{40 imes 20 + 30 imes 20}{100} = 14$ cm $Y_G = rac{20 imes 20 + 30 imes 20}{100} = 10$ cm

Momentum and the Theorem of Momentum

This section covers the concept of momentum and its conservation in isolated systems, including the velocity of the center of mass and the total momentum of a system.

• Velocity of the Center of Mass: $rac{d extbf{R}}{dt} = rac{ extstyle extsum_i m_i rac{d extbf{r}_i}{dt}}{ extstyle extsum_i m_i}$ $extbf{V}_G = rac{ extstyle extsum_i m_i extbf{v}_i}{ extstyle extsum_i m_i}$

• Total Momentum: $extbf{P} = extstyle extsum_i m_i extbf{v}_i$ The total momentum of the system equals the momentum of a single particle located at the center of mass, with mass equal to the total mass.

Key Points and Applications

• Rigid Body: Maintains constant distances between all points during motion.

• Center of Mass: Used to simplify analysis of motion and collisions in systems of particles.

• Momentum: Fundamental for understanding collisions and the motion of systems.

• Examples: Calculating the center of mass for bars and squares with attached masses demonstrates practical applications in mechanics.

Additional info: These notes cover the introductory concepts and mathematical foundations for the laws of conservation in physics, including center of mass and momentum, which are essential for further study of collisions, energy, and angular momentum.