Physics > Centre of Mass and Conservation of Linear Momentum > 1.0 Introduction
Centre of Mass and Conservation of Linear Momentum
1.0 Introduction
2.0 Position of centre of mass of continuous bodies
2.1 Rod
2.2 Semicircular ring
2.3 Semicircular disc
2.4 Solid sphere
2.5 Hemispherical Shell
2.6 Rectangular Plate
2.7 Square Plate
2.8 Circular Plate
2.9 Solid Cone
2.10 Hollow Cone
2.11 Questions
2.12 Centre of mass of a rigid complex bodies
3.0 Centre of mass of the remaining portion
4.0 Laws of conservation of linear momentum
5.0 Variable Mass
6.0 Impulse
7.0 Collision
8.0 Types of collision
9.0 Newton's law of restitution
10.0 Head on elastic and inelastic collision
11.0 Collision in two dimension
12.0 Oblique collision
1.1 Position of centre of mass for a system of two particles
2.2 Semicircular ring
2.3 Semicircular disc
2.4 Solid sphere
2.5 Hemispherical Shell
2.6 Rectangular Plate
2.7 Square Plate
2.8 Circular Plate
2.9 Solid Cone
2.10 Hollow Cone
2.11 Questions
2.12 Centre of mass of a rigid complex bodies
- For a system of two particles
- For a system of large number of particles
- For continuous bodies
1.1.1 Position of centre of mass for a system of two particles
Let us consider a system of two particles of masses $m_1$ & $m_2$ separated by a distance $r$.
The distance $(r)$ of centre of mass from particle is inversely proportional to the mass $(m)$ of the particle.
$$r \propto \frac{1}{m}$$ For the system, $${r_1} \propto \frac{1}{{{m_1}}}\quad \& \quad {r_2} \propto \frac{1}{{{m_2}}}$$ or $$\begin{equation} \begin{aligned} \frac{{{r_1}}}{{{r_2}}} = \frac{{{m_2}}}{{{m_1}}} \\ {m_1}{r_1} = {m_2}{r_2}\quad ...(i) \\ r = {r_1} + {r_2}\quad ...(ii) \\\end{aligned} \end{equation} $$ From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} r = {r_1} + \frac{{{m_1}{r_1}}}{{{m_2}}} \\ {r_1} = \left( {\frac{{{m_2}}}{{{m_1} + {m_2}}}} \right)r \\\end{aligned} \end{equation} $$ Similarly, $${r_2} = \left( {\frac{{{m_1}}}{{{m_1} + {m_2}}}} \right)r$$ where,
$r_1$: is the distance of centre of mass from mass $m_1$
$r_2$: is the distance of centre of mass from mass $m_2$
Note:
- Centre of mass of the system of particle may or may not lie on a particle: As the position of centre of mass in the above example is not lying on any of the particle of a system.
- Centre of mass is nearer to the particle having larger mass $${r_1} > {r_2}\quad if\quad {m_2} > {m_1}$$ or $${r_2} > {r_1}\quad if\quad {m_1} > {m_2}$$
Question 1. Find the position of centre of mass for a two-particle system having masses 2 $kg$ & 5 $kg$ separated by a distance of 7 $m$.
Solution: Let us consider the position of mass ${m_1} = 2kg$ to be $(0,0)$ in the cartesian co-ordinate system. Then the position of mass ${m_2} = 5kg$ is $(7,0)$.
Let the position of centre of mass $(COM)$ is $(5,0)$.
