Brief Summary
This is a summary lecture on Center of Mass and System of Particles, focusing on key concepts and problem-solving techniques. It covers topics from mass moment and the mathematical definition of the center of mass to continuous mass distributions, additive and negative systems, and collisions. The lecture emphasizes understanding the underlying principles and applying them to various scenarios, including boat problems, recoil of a gun, explosions, and collisions.
- Mass Moment and Center of Mass Definition
- Continuous Mass Distribution and Integration Techniques
- Additive and Negative Systems for Complex Shapes
- Boat Problems and External Forces
- Collisions and Coefficient of Restitution
Introduction
The lecture is a summary of the Center of Mass and System of Particles chapter, intended for revision rather than initial learning. It highlights important concepts and problem-solving techniques discussed in the Arjuna JEE 2025 batch. The lecture includes theoretical points and selected questions for review, with an emphasis on marking and revising additional important questions, especially from the advanced PYQs in lecture 14.
Mass Moment and Center of Mass Definition
The lecture starts with the definition of mass moment, which is the product of the mass of a particle and its position vector with respect to a point. The center of mass is defined as a point in space where the mass moment is zero. It's important to note that the center of mass doesn't necessarily lie within the body. For regular bodies like rectangles or rings, the center of mass is at the geometric center.
Center of Mass of Discrete Particles
For discrete particles, the position vector of the center of mass is calculated using the formula: Rcm = (m1r1 + m2r2 + m3r3) / (m1 + m2 + m3). The x, y, and z coordinates of the center of mass can be found by applying the formula to each coordinate separately. For two-point particles, the distance of the center of mass from a particle is given by m2d / (m1 + m2), where d is the distance between the particles.
Continuous Mass Distribution
For continuous mass distributions, integration is used instead of summation. The formula becomes Xcm = ∫x dm / ∫dm, where dm is the mass of an element. An element is a repeating structure that makes up the system. Three types of integration are discussed: linear, aerial, and angular.
Linear, Areal, and Angular Integration
Linear integration involves mass per unit length (λ = m/l), aerial integration involves mass per unit area (σ = m/A), and angular integration is used for objects like rings. For a rod, the element is a small strip of length dx. For a disc, the element is a ring of radius r and thickness dr. The mass of the element (dm) is calculated differently for each type of integration.
Center of Mass of Common Shapes
The lecture provides the center of mass locations for common shapes: uniform rod (l/2), disc (geometric center), ring (geometric center), sphere (geometric center), semi-circular ring (2r/π), semi-circular disc (4r/3π), hollow cone (h/3), solid cone (h/4), cylinder (geometric center), cuboid (l/2, b/2, h/2), solid hemisphere (3r/8), and hollow hemisphere (r/2). A general formula for the center of mass of an angular ring is given as r sin(θ) / θ, where 2θ is the total angle.
Additive and Negative Systems
Additive systems involve combining shapes, while negative systems involve removing a shape from another. Problems can be based on mass, area, or volume. For additive systems, the formula is Vcm = (v1x1 + v2x2) / (v1 + v2). For negative systems, the formula is Acm = (a1x1 - a2x2) / (a1 - a2).
Shift of Center of Mass
The shift of the center of mass is calculated using the formula: Δrcm = (m1Δr1 + m2Δr2) / (m1 + m2). This concept is applied to problems where objects are displaced, and the goal is to find the new position of the center of mass. The lecture also covers velocity and acceleration of the center of mass, emphasizing the importance of using vector notation.
Boat Problems
Boat problems involve a person moving on a boat or plank. The key concept is that if there is no external force along the x-axis, the center of mass of the system (person + boat) does not shift. The shift of the boat can be calculated by setting the shift of the center of mass to zero. The lecture also discusses a generalized formula for boat problems and provides examples of different scenarios.
Recoil of a Gun
The recoil of a gun is explained in terms of conservation of momentum. The initial momentum of the gun and bullet is zero. When the gun is fired, the bullet moves forward, and the gun recoils backward. The lecture discusses two cases: when the bullet's velocity is given with respect to the ground and when it's given with respect to the gun (muzzle velocity). The recoil velocity is calculated using the formula: v = -m/M * u, where m is the mass of the bullet, M is the mass of the gun, and u is the velocity of the bullet.
Explosions
Explosions are discussed in terms of conservation of momentum and energy. If a bomb is at rest, the initial momentum is zero, and the final momentum is also zero. The lecture introduces the concept of reduced mass and discusses how the kinetic energy is distributed among the fragments.
Bullet and Pendulum Problems
Two cases are considered: when the bullet sticks to the pendulum and when the bullet passes through the pendulum. In both cases, momentum is conserved. The lecture provides formulas for calculating the angle of rotation and the height reached by the pendulum.
Collisions: Introduction and Types
The lecture introduces the concept of collisions and discusses different types of collisions: head-on and oblique. The impact factor is defined as the perpendicular distance between the lines of velocity. The lecture also discusses different types of materials: elastic, inelastic, and perfectly inelastic.
Collisions: Momentum and Energy Conservation
In any collision, the momentum of the system is conserved along the line of impact. The lecture discusses the coefficient of restitution (e), which is defined as the ratio of the velocity of separation to the velocity of approach. For perfectly elastic collisions, e = 1, and for perfectly inelastic collisions, e = 0.
Coefficient of Restitution
The coefficient of restitution (e) is defined as the ratio of the velocity of separation to the velocity of approach. The formula is e = (v2 - v1) / (u1 - u2). For perfectly elastic collisions, e = 1, and for perfectly inelastic collisions, e = 0. The lecture emphasizes that the velocities used in the formula must be along the line of impact.
General Collision Equations and Methods
The lecture presents two methods for solving collision problems: a formula method and a method based on writing equations for momentum and the coefficient of restitution. The formula method involves using the formulas for v1 and v2, while the equation method involves writing equations for momentum and the coefficient of restitution and solving them simultaneously.
Collisions with Heavy Bodies
The lecture discusses collisions with heavy bodies, such as a train colliding with a bicycle. In this case, the velocity of the heavy body is not affected, and the velocity of the light body is given by v1 = 2u2 - u1. The lecture also discusses the case where a light body collides with a heavy body that is at rest.
Ball Bouncing on the Ground
The lecture discusses the case of a ball bouncing on the ground. The velocity of the ball after each collision is given by v = e * u, where e is the coefficient of restitution and u is the velocity before the collision. The lecture provides formulas for calculating the total time, total distance, and total change in momentum.
Oblique Collisions
The lecture discusses oblique collisions, where the particles collide at an angle. Two types of problems are considered: when the sizes of the particles are given and when they are not. When the sizes are given, the line of impact can be drawn, and the momentum of the system can be conserved along the line of impact. When the sizes are not given, the momentum can be conserved along the x and y axes.
Additional Problems and Examples
The lecture concludes with a discussion of additional problems and examples, including finding the center of mass of a rod with variable density, the semi-circular ring, jumping off a cart, a bomb exploding into three particles, and a projectile exploding at the top of its trajectory.
