Further Constrained-Motion Examples

The constraint may not be so simple as that imposed by compelling the moving particle to remain on a given surface or on a given curve.

(a) The Tractrix Problem

Take, for example, the tractrix problem, when the particle moves on a smooth horizontal plane.

Let a particle of mass $m$ , attached to a string of length $a$ , rest on a smooth horizontal plane. The string lies straight on the plane at the start, and then the end not attached to the particle is drawn with uniform velocity along a straight line perpendicular to the initial position of the string and lying in the plane.

Let us take as our coordinates $x$ , the distance traveled by that end of the string which is not attached to the particle, and $\theta$ , the angle made by the string with its initial position. Let $R$ be the tension of the string and $n$ the velocity with which the end of the string is drawn along. Let $X,Y$ be the rectangular coordinates of the particle, referred to the fixed line and to the initial position of the string as axes.

Regenerated diagram for Byerly, Chapter I, Art. 6(a): tractrix setup.

$\displaystyle X=x-a\sin\theta, \qquad Y=a\cos\theta,$

$\displaystyle \dot X=\dot x-a\cos\theta\,\dot\theta, \qquad \dot Y=-a\sin\theta\,\dot\theta.$

Hence

$\displaystyle T=\frac{m}{2}\left(\dot X^2+\dot Y^2\right) = \frac{m}{2} \left[ \dot x^2+a^2\dot\theta^2-2a\cos\theta\,\dot x\dot\theta \right].$

$\displaystyle \frac{\partial T}{\partial\dot x} = m\left(\dot x-a\cos\theta\,\dot\theta\right),$

$\displaystyle \frac{\partial T}{\partial\dot\theta} = m\left(a^2\dot\theta-a\cos\theta\,\dot x\right),$

$\displaystyle \frac{\partial T}{\partial\theta} = ma\sin\theta\,\dot x\dot\theta.$

The equations are

$\displaystyle m\frac{d}{dt} \left( \dot x-a\cos\theta\,\dot\theta \right)\delta x = R\sin\theta\,\delta x,$

and

$\displaystyle m\left[ \frac{d}{dt} \left( a^2\dot\theta-a\cos\theta\,\dot x \right) -a\sin\theta\,\dot x\dot\theta \right]\delta\theta = 0.$

Adding the condition

$\displaystyle x=nt,$

and reducing,

$\displaystyle -ma\left(\cos\theta\,\ddot\theta-\sin\theta\,\dot\theta^2\right) = R\sin\theta,$

$\displaystyle ma^2\ddot\theta=0.$

Therefore

$\displaystyle \ddot\theta=0, \qquad R=ma\dot\theta^2.$

Integrating,

$\displaystyle \dot\theta=C=\frac{n}{a}, \qquad R=\frac{mn^2}{a}.$

The particle revolves with uniform angular velocity about the moving center, and the pull on the string is constant.

(b) A Particle in a Rotating Horizontal Tube

A particle is at rest in a smooth horizontal tube. The tube is then made to revolve in a horizontal plane with uniform angular velocity $\omega$ . Find the motion of the particle.

Suggestion. Take the polar coordinates $r,\phi$ of the particle as our coordinates, and let $R$ be the pressure of the particle on the tube.

$\displaystyle T=\frac{m}{2}\left(\dot r^2+r^2\dot\phi^2\right),$

$\displaystyle \frac{\partial T}{\partial\dot r}=m\dot r, \qquad \frac{\partial ... ...tial r}=mr\dot\phi^2, \qquad \frac{\partial T}{\partial\dot\phi}=mr^2\dot\phi.$

Thus

$\displaystyle m(\ddot r-r\dot\phi^2)\delta r=0,$

$\displaystyle m\frac{d}{dt}(r^2\dot\phi)\delta\phi=Rr\,\delta\phi.$

Adding the condition

$\displaystyle \phi=\omega t,$

and reducing,

$\displaystyle \ddot r-\omega^2r=0, \qquad 2m\omega r\dot r=Rr.$

Solving,

$\displaystyle r=A\cosh\omega t+B\sinh\omega t.$

Since

$\displaystyle r=a,\qquad \dot r=0$

at the start,

$\displaystyle r=a\cosh\omega t=a\cosh\phi,$

and

$\displaystyle R=2ma\omega^2\sinh\omega t = 2ma\omega^2\sinh\phi.$

If we are interested only in the motions and not in the reactions, problems (a) and (b) can be solved more simply. If in each we were to use one less coordinate— $\theta$ only in (a), and $r$ only in (b)—rectangular coordinates $X,Y$ for the particle could be obtained whenever the time was given, and therefore could be expressed explicitly in terms of $\theta$ or $r$ and $t$ . A careful examination of Art. 2 will show that the reasoning is extended easily to such a case, and that the work done by the effective forces when $q_1$ only is changed is still

$\displaystyle \left[ \frac{d}{dt}\frac{\partial T}{\partial\dot q_1} - \frac{\partial T}{\partial q_1} \right]\delta q_1.$

It is to be noted, however, that when the rectangular coordinates are functions of $t$ as well as of $q_1,q_2,$ etc., the energy $T$ is no longer a homogeneous quadratic in $\dot q_1,\dot q_2,$ etc.

For (a),

$\displaystyle X=nt-a\sin\theta, \qquad Y=a\cos\theta,$

$\displaystyle \dot X=n-a\cos\theta\,\dot\theta, \qquad \dot Y=-a\sin\theta\,\dot\theta,$

and

$\displaystyle T = \frac{m}{2}(\dot X^2+\dot Y^2) = \frac{m}{2} \left[ n^2+a^2\dot\theta^2-2an\cos\theta\,\dot\theta \right].$

Then

$\displaystyle \frac{\partial T}{\partial\dot\theta} = m(a^2\dot\theta-an\cos\theta), \qquad \frac{\partial T}{\partial\theta} = man\sin\theta\,\dot\theta,$

and

$\displaystyle m\left[ \frac{d}{dt}(a^2\dot\theta-an\cos\theta) -an\sin\theta\,\dot\theta \right]\delta\theta = 0.$

Therefore

$\displaystyle \ddot\theta=0, \qquad \dot\theta=\frac{n}{a},$

as before.

For (b),

$\displaystyle T=\frac{m}{2}\left(\dot r^2+\omega^2r^2\right),$

$\displaystyle \frac{\partial T}{\partial\dot r}=m\dot r, \qquad \frac{\partial T}{\partial r}=m\omega^2r.$

Thus

$\displaystyle m(\ddot r-\omega^2r)\delta r=0,$

and

$\displaystyle r=a\cosh\omega t,$

as before.

Examples

1. A Particle on a Horizontal Whirling Table

A particle rests on a smooth horizontal whirling table and is attached by a string of length $a$ to a point fixed in the table at a distance $b$ from the center. The particle, the point, and the center are initially in the same straight line. The table is then made to rotate with uniform angular velocity $\omega$ . Find the motion of the particle.

Suggestion. Take as the single coordinate $\theta$ the angle made by the string with the radius of the point. Let $X,Y$ be the rectangular coordinates of the particle, referred to the line initially joining it with the center and to a perpendicular thereto through the center as axes.

Then

$\displaystyle X=b\cos\omega t+a\cos(\theta+\omega t),$

$\displaystyle Y=b\sin\omega t+a\sin(\theta+\omega t),$

and

$\displaystyle T = \frac{m}{2} \left[ b^2\omega^2+a^2(\omega+\dot\theta)^2 +2ab\omega(\omega+\dot\theta)\cos\theta \right].$

The equation of motion is

$\displaystyle \ddot\theta+\frac{b\omega^2}{a}\sin\theta=0,$

and the relative motion on the table is simple pendulum motion, the length of the equivalent pendulum being

$\displaystyle l=\frac{ag}{b\omega^2}.$

2. A Particle Attracted toward a Point on a Rotating Table

A particle is attracted toward a fixed point in a horizontal whirling table with a force proportional to the distance. It is initially at rest at the center. The table is then made to rotate with uniform angular velocity $\omega$ . Find the path traced on the table by the particle.

Suggestion. Take as coordinates $x,y$ , rectangular coordinates referred to the moving radius of the fixed point as axis of abscissas and to the center of the table as origin. Let $X,Y$ be the rectangular coordinates referred to fixed axes coinciding with the initial positions of the moving axes.

$\displaystyle X=x\cos\omega t-y\sin\omega t, \qquad Y=x\sin\omega t+y\cos\omega t.$

$\displaystyle T = \frac{m}{2} \left[ \dot x^2+\omega^2x^2+\dot y^2+\omega^2y^2 -2\omega y\dot x+2\omega x\dot y \right].$

Whence come

$\displaystyle m(\ddot x-2\omega\dot y-\omega^2x)=-\mu(x-a),$

$\displaystyle m(\ddot y+2\omega\dot x-\omega^2y)=-\mu y.$

$\displaystyle \omega^2=\frac{\mu}{m},$

the solution is easy and interesting:

$\displaystyle \ddot x-2\omega\dot y=a\omega^2, \tag{1}$

$\displaystyle \ddot y+2\omega\dot x=0. \tag{2}$

Integrating (2),

$\displaystyle \dot y+2\omega x=0.$

Substituting in (1),

$\displaystyle \ddot x+4\omega^2x=a\omega^2. \tag{3}$

Multiplying (3) by $2\dot x$ , and integrating,

$\displaystyle \dot x^2+4\omega^2x^2=2a\omega^2x.$

Hence

$\displaystyle \dot x = 2\omega\sqrt{\frac{ax}{2}-x^2}.$

Whence

$\displaystyle 2\omega t=\operatorname{vers}^{-1}\frac{4x}{a},$

$\displaystyle x=\frac{a}{4}(1-\cos 2\omega t),$

$\displaystyle y=-\frac{a}{4}(2\omega t-\sin2\omega t).$

Replacing $2\omega t$ by $\theta$ ,

$\displaystyle x=\frac{a}{4}(1-\cos\theta), \qquad y=-\frac{a}{4}(\theta-\sin\theta),$

and the curve traced on the table is the cycloid generated by a circle of radius $a/4$

rolling backward along the moving axis of $Y$

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.

"example 2 of dynamics of a particle: constrained motion" is owned by bloftin.
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