[NOTES/ME-13001]-Kinetic Energy and Angular Momentum: Many Particle System

1. Introduction

In this lecture we shall discuss the kinetic energy and angular momentum of a dynamical many particle system. Transformation properties of kinetic energy and angular momentum under Galilean transformations is discussed. It is proved that that these dynamical variables can be decomposed into a sum of two parts corresponding to the motion of centre of mass and the motion relative to the centre of mass.

2.Galilean Transformations

Consider two frames $K$ and $K^\prime$. Let $\vec{x}$ and $\vec{x}^{\,\prime}$ be position vectors of a point with respect to $K$ and $K^\prime$. Let the origin of $K^\prime$ be position $\vec{a}$ w.r.t. the origin of $K$. Then \begin{equation} \vec{x} = \vec{x}^\vp + \vec{a}. \label{EQ01} \end{equation} Let $m_\Al , \alpha=1,2,...N$ denote the masses, and $\vec{x}_\Al, \vec{v}_\Al,\vec{p}_\Al$ and $\vec{x}_\Al^\vp,\vec{v}_\Al^\vp, \vec{p}_\Al^\vp$ be the position, velocity and momenta of the particle $\alpha$ as seen in the two frames $K$ and $^\prime$. If $\vec{u}$ denotes the velocity of the orign of the frame $K^\prime$ w.r.t. the frame $K$, then \begin{equation} \vec{u} = \dd[\vec{a}]{t}. \label{EQ02} \end{equation} and \begin{equation} \vec{v}_\Al = \vec{v}^\vp + \vec{a}, \qquad \vec{p} = \vec{p}^\vp_\Al + m_\Al \vec{u}. \label{EQ03} \end{equation} The total momentum of the system in the frame $K$ is \begin{equation} \vec{P} = \sum_\alpha \vec{p}_\Al^\vp + \sum_\alpha + M \vec{u}, \label{EQ04} \end{equation} where $M$ is the total mass of the system and hence \begin{equation} \vec{P}= \vec{P}^\vp + M \vec{u} \label{EQ05}. \end{equation} The total kinetic energy is then given by \begin{eqnarray} T &=& \frac{1}{2}\sum_\alpha m_\alpha \vec{v}_\Al^{\,2}\nonumber\\ &=& \frac{1}{2}\sum_\alpha m_\alpha (\vec{v}_\Al^\vp + \vec{u})^2\nonumber\\ &=& \frac{1}{2}\sum_\alpha m_\alpha (\vec{v}_\Al^{\vp\,2} + 2\vec{v}_\Al\cdot \vec{u}+ \vec{u}^{\,2} )\nonumber\\ &=& T + (\sum_\Al m_\alpha \vec{v}_\Al) \cdot \vec{u}+ \frac{1}{2}\sum_\alpha m_\Al\vec{u}^{\,2}) \label{EQ06} \end{eqnarray} Therefore we get \begin{equation} \boxed{T = T^\prime + \vec{P}^\vp\cdot \vec{u}^\vp +\frac{1}{2}M \vec{u}^2}\label{EQ07A} \end{equation} Next we obtain the transformation rule for angular momentum. \begin{eqnarray} \vec{L} &=& \sum_\alpha \vec{x}_\Al\times \vec{p}_\Al \nonumber\\ &=& \sum_\alpha (\vec{x}_\Al^\vp+ \vec{a})\times (\vec{p}_\Al^\vp + m_\alpha \vec{u}) \nonumber\\ &=& \sum_\alpha \vec{x}_\Al^\vp \times \vec{p}_\Al^\vp + \sum_\alpha m_\alpha\vec{x}_\Al^\vp \times \vec{u} + \vec{a} \times \sum_\alpha \vec{p}_\Al^\vp + (\sum_\alpha m_\alpha) \vec{a}\times \vec{u}. \label{EQ07B} \end{eqnarray} Therefore \begin{eqnarray} \boxed{\vec{L} = \vec{L}^\vp + M (\vec{X}^\vp \times \vec{u}) + \vec{a}\times \vec{P}^\vp + M \vec{a}\times \vec{u} }\,. \label{EQ08} \end{eqnarray} where $\vec{X}^\vp$ denotes the position vector of the centre of mass w.r.t. the frame $K^\prime$.

3. Special Cases

Case I: Shift of Origin with $K^\prime$ at rest w.r.t. $K$

Let the frame $K^\prime$ be at rest w.r.t the frame $K$ but the origins of the two frames do not coincide. Then \begin{equation} \vec{a}\ne 0, \qquad \vec{u}= 0 \label{EQ09}. \end{equation} Then we have \begin{eqnarray} T &=& T^\prime, \\ \vec{L} &=& \vec{L}^\vp + \vec{a} \times \vec{P}^\vp \label{EQ10}. \end{eqnarray}

Case-II: $K^\prime$ coincides with the centre of mass frame

The centre of mass frame is defined as the frame in which the momentum of the centre of mass is zero, {\it i.e.} the frame in which the centre of mass is at rest: \begin{equation} \vec{P}^\vp=0 \Rightarrow {V}_{cm}^\vp = 0. \label{EQ11} \end{equation} The transformation rules become \begin{eqnarray} T = T^\prime + \frac{1}{2}M \vec{u}^{\,2} \label{EQ12}\\ \vec{L} = \vec{L}^\vp + M \vec{a}\times \vec{u}\label{EQ13}. \end{eqnarray} Note that the origin of $K^\prime$ may or may not coincide with the position of the centre of mass. Also the axes of the two systems may or may not be parallel.

Case-III : The origin of $K^\prime$ coincides with the c.m.

If in addition to the frame $K^\prime$ moving with the velocity of the centre of mass, let us assume that the origin of $K^\prime$ coincides with the the position of the centre of mass, $\vec{a}=\vec{X}$, then we have \begin{equation} \vec{a} = \vec{X} = \frac{\sum_\alpha m_\alpha \vec{x}_\Al}{\sum_\alpha m_\alpha}, \qquad \vec{u}= \vec{V} \frac{P}{M} \label{EQ14}. \end{equation} In this case the transformation rules simplify to \begin{eqnarray} \boxed{T= T^\prime + \frac{1}{2} M \vec{V}^{\,2}} \label{EQ15}\\ \boxed{\vec{L}= \vec{L}^\vp + M \vec{X} \times \vec{V}} \label{EQ16}. \end{eqnarray} The above results will be useful in discussion of rigid body dynamics.