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

$\displaystyle T=\frac{m}{2}a^{2}\dot{\theta}^{2},
$

and we have

$\displaystyle \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,
$

or

$\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} a t= C =-\frac{a}{V}.
$

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

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

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

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

and the problem of the motion is completely solved.



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