(a) A Particle on a Smooth Horizontal Circle

Let a particle of mass $m$ , constrained to move on a smooth horizontal circle of radius $a$ , be given an initial velocity $V$ , and let it be resisted by the air with a force proportional to the square of its velocity.

Here we have one degree of freedom. Let us take as our coordinate the angle $\theta$ which the particle has described about the center of its path in the time $t$ .

$\displaystyle T=\frac{m}{2}a^2\dot\theta^{\,2}, \qquad \frac{\partial T}{\partial\dot\theta}=ma^2\dot\theta.$

Our differential equation is

$\displaystyle ma^2\ddot\theta\,\delta\theta = -ka^2\dot\theta^{\,2}\,a\,\delta\theta,$

which reduces to

$\displaystyle \ddot\theta+\frac{k}{m}a\dot\theta^{\,2}=0,$

$\displaystyle \frac{d\dot\theta}{dt} + \frac{k}{m}a\dot\theta^{\,2} = 0.$

Separating the variables,

$\displaystyle \frac{d\dot\theta}{\dot\theta^{\,2}} +\frac{k}{m}a\,dt=0.$

Integrating,

$\displaystyle -\frac{1}{\dot\theta} +\frac{k}{m}at = C = -\frac{a}{V}.$

Hence

$\displaystyle \frac{1}{\dot\theta} = \frac{ma+kVat}{mV},$

$\displaystyle \frac{d\theta}{dt} = \frac{mV}{ma+kVat},$

and

$\displaystyle \theta = \frac{m}{ka}\log[m+kVt]+C = \frac{m}{ka} \log\left(1+\frac{kVt}{m}\right). \tag{1}$

The problem of the motion is completely solved.

(b) The Pressure of the Constraining Curve

If, however, we are interested in $R$ , the pressure of the constraining curve, we must proceed somewhat differently. We have only to replace the constraint by a force $R$ directed toward the center of the path. There are now two degrees of freedom, and we shall take $\theta$ and the radius vector $r$ as our coordinates and form two differential equations of motion.

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

$\displaystyle \frac{\partial T}{\partial\dot\theta}=mr^2\dot\theta, \qquad \fra... ...rtial\dot r}=m\dot r, \qquad \frac{\partial T}{\partial r}=mr\dot\theta^{\,2}.$

Thus

$\displaystyle m\frac{d}{dt}\left(r^2\dot\theta\right)\delta\theta = -kr^2\dot\theta^{\,2}\,r\,\delta\theta, \tag{1}$

$\displaystyle m\left(\ddot r-r\dot\theta^{\,2}\right)\delta r = -R\delta r. \tag{2}$

To these we may add

$\displaystyle r=a.$

Whence

$\displaystyle \ddot\theta+\frac{ka}{m}\dot\theta^{\,2}=0, \tag{3}$

as before, and

$\displaystyle R=ma\dot\theta^{\,2}. \tag{4}$

(c) The Constraining Circle Rough

Let us now suppose that the constraining circle is rough. Here, since the friction is $\mu R$ (the coefficient of friction multiplied by the normal pressure), $R$ will be needed, and we must replace the constraint by $R$ as before.

We have now

$\displaystyle m\frac{d}{dt}\left(r^2\dot\theta\right)\delta\theta = -kr^2\dot\theta^{\,2}\,r\,\delta\theta -\mu Rr\,\delta\theta,$

$\displaystyle m\left(\ddot r-r\dot\theta^{\,2}\right)\delta r = -R\delta r,$

and

$\displaystyle r=a.$

Whence

$\displaystyle R=ma\dot\theta^{\,2},$

as before, and

$\displaystyle \ddot\theta+\frac{ka}{m}\dot\theta^{\,2}+\frac{\mu}{ma}R=0,$

$\displaystyle \ddot\theta+ \left( \frac{ka}{m}+\mu \right)\dot\theta^{\,2}=0.$

Replacing $ka/m$ in (1) by $ka/m+\mu$ , we have

$\displaystyle \theta = \frac{1}{\dfrac{ka}{m}+\mu} \log\left[ 1+ \left( \frac{ka}{m}+\mu \right)\frac{Vt}{a} \right]. \tag{1}$

Examples

Obtain the familiar equation

$\displaystyle \frac{d^2\theta}{dt^2}+\frac{g}{a}\sin\theta=0$
for the simple pendulum.
Find the tension of the string in the simple pendulum.
Answer.

$\displaystyle R=m\left[ g\cos\theta+a\left(\frac{d\theta}{dt}\right)^2 \right].$
Obtain the equations of the spherical pendulum in terms of the spherical coordinates $\theta$ and $\phi$ .
Answer.

$\displaystyle \ddot\theta-\sin\theta\cos\theta\,\dot\phi^{\,2} +\frac{g}{a}\sin\theta=0, \qquad \sin^2\theta\,\dot\phi=C.$

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 of dynamics of a particle: constrained motion" is owned by bloftin.
(view preamble)