Dynamics of a Particle: Free Motion

The differential equations for the motion of a particle under any forces when we use rectangular coordinates are known to be

$\begin{equation*} \left. \begin{aligned} m\ddot{x}&=X,\ m\ddot{y}&=Y,\ m\ddot{z}&=Z. \end{aligned}\right\} \tag{1} \end{equation*}$

$X,Y,$ and $Z$ , the components of the actual forces on the particle resolved parallel to the fixed rectangular axes, or rather their equivalents $m\ddot{x},m\ddot{y},m\ddot{z}$ , are called the effective forces on the particle. They are of course a set of forces mechanically equivalent to the actual forces acting on the particle.

The equations of motion of the particle in terms of any other system of coordinates are easily obtained.

Let $q_1,q_2,q_3$ be the coordinates in question. The appropriate formulas for transformation of coordinates express $x,y,$ and $z$ in terms of $q_1,q_2,$ and $q_3$ :

$\displaystyle x=f_1(q_1,q_2,q_3),\qquad y=f_2(q_1,q_2,q_3),\qquad z=f_3(q_1,q_2,q_3).$

For the component velocity $\dot{x}$ we have

$\displaystyle \dot{x} = \frac{\partial x}{\partial q_1}\dot{q}_1 +\frac{\partial x}{\partial q_2}\dot{q}_2 +\frac{\partial x}{\partial q_3}\dot{q}_3,$

and $\dot{x},\dot{y},\dot{z}$ are explicit functions of $q_1,q_2,q_3,\dot q_1,\dot q_2,\dot q_3$ , linear and homogeneous in terms of $\dot q_1,\dot q_2,\dot q_3$ .¹

We may note in passing that it follows from this fact that $\dot{x}^2,\dot{y}^2,$ and $\dot{z}^2$ are homogeneous quadratic functions of $\dot q_1,\dot q_2,$ and $\dot q_3$ .

Obviously

$\displaystyle \frac{\partial\dot{x}}{\partial\dot q_1} = \frac{\partial x}{\partial q_1}, \tag{2}$

and since

$\displaystyle \frac{d}{dt}\frac{\partial x}{\partial q_1} = \frac{\partial^2x}{... ..._2\partial q_1}\dot q_2 +\frac{\partial^2x}{\partial q_3\partial q_1}\dot q_3,$

and

$\displaystyle \frac{\partial\dot{x}}{\partial q_1} = \frac{\partial^2x}{\partia... ..._1\partial q_2}\dot q_2 +\frac{\partial^2x}{\partial q_1\partial q_3}\dot q_3,$

we have

$\displaystyle \frac{d}{dt}\frac{\partial x}{\partial q_1} = \frac{\partial\dot{x}}{\partial q_1}. \tag{3}$

Let us find now an expression for the work $\delta_{q_1}W$ done by the effective forces when the coordinate $q_1$ is changed by an infinitesimal amount $\delta q_1$ without changing $q_2$ or $q_3$ . If $\delta x,\delta y,\delta z$ are the changes thus produced in $x,y,z$ , obviously

$\displaystyle \delta_{q_1}W = m\left[\ddot{x}\,\delta x+\ddot{y}\,\delta y+\ddot{z}\,\delta z\right].$

If expressed in rectangular coordinates, we need, however, to express $\delta_{q_1}W$ in terms of our coordinates $q_1,q_2,q_3$ :

$\displaystyle \delta_{q_1}W = m\left[ \ddot{x}\frac{\partial x}{\partial q_1} +... ...l y}{\partial q_1} +\ddot{z}\frac{\partial z}{\partial q_1} \right]\delta q_1.$

Now

$\displaystyle \ddot{x}\frac{\partial x}{\partial q_1} = \frac{d}{dt}\left( \dot... ...x}{\partial q_1} \right) - \dot{x}\frac{d}{dt}\frac{\partial x}{\partial q_1},$

but by (2) and (3),

$\displaystyle \frac{\partial x}{\partial q_1} = \frac{\partial\dot{x}}{\partial... ...{d}{dt}\frac{\partial x}{\partial q_1} = \frac{\partial\dot{x}}{\partial q_1}.$

Hence

$\displaystyle \ddot{x}\frac{\partial x}{\partial q_1} = \frac{d}{dt}\left( \dot... ...2}{2}\right) - \frac{\partial}{\partial q_1} \left(\frac{\dot{x}^2}{2}\right).$

Therefore

$\displaystyle \delta_{q_1}W = \left[ \frac{d}{dt}\frac{\partial T}{\partial\dot q_1} - \frac{\partial T}{\partial q_1} \right]\delta q_1, \tag{4}$

where

$\displaystyle T=\frac{m}{2}\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)$

is the kinetic energy of the particle.

To get our differential equation we have only to write the second member of (4) equal to the work done by the actual forces when $q_1$ is changed by $\delta q_1$ .

If we represent the work in question by $Q_1\delta q_1$ , our equation is

$\displaystyle \frac{d}{dt}\frac{\partial T}{\partial\dot q_1} - \frac{\partial T}{\partial q_1} = Q_1, \tag{5}$

and of course we get such an equation for every coordinate.

It must be noted that usually equation (5) will contain $q_2$ and $q_3$ and their time derivatives as well as $q_1$ , and therefore cannot be solved without the aid of the other equations of the set.

In any concrete problem, $T$ must be expressed in terms of $q_1,q_2,q_3$ and their time derivatives before we can form the expression for the work done by the effective forces. The work done by the actual forces must be obtained from direct examination of the problem.

Source

William Elwood Byerly, An Introduction to the Use of Generalized Coördinates in mechanics and Physics, Ginn and Company, 1916. Chapter I, “Introduction.”

The 1916 source work is in the public domain in the United States.

Footnotes

....¹: For time derivatives we shall use the Newtonian fluxion notation, so that we shall write $\dot{x}$ for , $\ddot{x}$ for .

"dynamics of a particle: free_motion" is owned by bloftin.
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