PlanetPhysics/Example of Generalized Coordinates for Constrained Motion on a 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 $$q_{1}$$ the angle $$\theta$$ which the particle has described about the center of its path in the time $$t$$.

For the kinetic energy $$ T=\frac{m}{2}a^{2}\dot{\theta}^{2}, $$

and we have

$$ \frac{\partial T}{\partial \dot{\theta}}=ma^{2}\dot{\theta}. $$

\quad Our differential equation is

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

which reduces to

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

or

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

Separating the variables,

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

Integrating,

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

$$ \frac{1}{\dot{\theta}}=\frac{ma+k V a t}{m V}, $$

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

$$ \theta=\frac{m}{ka}\log\left[m+kVt\right]+C, $$

$$ \theta=\frac{m}{ka}\log\left[1+\frac{kVt}{m}\right]; $$

and the problem of the motion is completely solved.