Write down the equations of motion in polar co-ordinates for a particle of unit mass moving in a plane under a force with potential energy function \(V=-k \ln r+c r+g r \cos \theta\), where \(k, c\) and \(g\) are positive constants. Find the positions of equilibrium (a) if \(c>g\), and (b) if \(c

The kinetic energy (T) of the moving particle of unit mass in polar coordinates can be calculated as \(T = \frac{1}{2}( \dot{r}^2 + r^2 \dot{\theta}^2 )\) where \(\dot{r}\) and \(\dot{\theta}\) are the time derivatives of the radial (r) and angular (\(\theta\)) coordinates, respectively. The potential energy function (V) is given as \(V = -k \ln r + cr + gr \cos \theta\) Step 2: Find the Lagrangian

The Lagrangian (L) of the system is the difference between the kinetic and potential energies: \(L = T - V = \frac{1}{2}( \dot{r}^2 + r^2 \dot{\theta}^2 ) - (-k \ln r + cr + gr \cos \theta)\) Step 3: Obtain the equations of motion using Lagrange's equations

For a system with two generalized coordinates, the Lagrange's equations can be written as: \(\frac{d}{dt} (\frac{\partial L}{\partial \dot{q}_i}) - \frac{\partial L}{\partial q_i} = 0\), where \(q_i = r,\theta\) Applying Lagrange's equations for \(r\) and \(\theta\) gives us the following differential equations for the motion: Equation for r: \(\ddot{r} - r \dot{\theta}^2 + k/r - c - g \cos \theta = 0\) Equation for \(\theta\): \(r \ddot{\theta} + 2 \dot{r} \dot{\theta} + g \sin \theta = 0\) Step 4: Find the equilibrium points

The equilibrium points are the positions where the particle's velocity and acceleration are both zero. Therefore, the equilibrium conditions are: \(\dot{r} = 0, \dot{\theta} = 0, \ddot{r} = 0,\) and \(\ddot{\theta} = 0\) From the equations of motion, we can find the equilibrium conditions as follows: (a) If \(c > g\): \(-c + k/r = g \cos \theta\) and \(g \sin \theta = 0\) (b) If \(c < g\): \(-c + k/r = -g \cos \theta\) and \(g \sin \theta = 0\) Step 5: Determine the stability of the equilibrium

To determine the stability, we'll study the behavior of the system near the equilibrium points. For a stable equilibrium, the acceleration terms near the equilibrium points should be positive for radial and angular directions. (a) If \(c > g\): From the equation of motion solution, we can check that, near the equilibrium points, the second derivative of r is positive, and the second derivative of \(\theta\) is positive. Hence, the particle will tend to return repeatedly to its equilibrium position, and the equilibrium is stable. (b) If \(c < g\): From the equation of motion solution, we can check that, near the equilibrium points, the second derivative of r is negative, and the second derivative of \(\theta\) is positive. Therefore, the equilibrium position is not stable, as the particle will not tend to return repeatedly to its equilibrium position.