Motion on a cone surface

An alternate solution to that provided by @Steven Chase . One may refer to his solution for the derivation of energy expressions. Consider the expressions for kinetic and potential energy derived by him. Using that and plugging in $\alpha = \sqrt{3}$ one gets:

$T = 2m\dot{r}^2 + \frac{mr^2\dot{\theta}^2}{2}$ $V = \sqrt{3}mgr$

$r(0) = 10 \ ; \ \theta(0) = \dot{r}(0) = 0 \ ; \ \dot{\theta}(0) = 1$

Using the Lagrange's equation with respect to $\theta$ , one can easily prove that:

$r^2 \dot{\theta} = 100$

The reader is requested to prove the above by him/her self.

Plugging the above expression in $T$ gives:

$T = 2m\dot{r}^2 + \frac{5000m}{r^2}$

Applying conservation of energy gives:

$T + V = T_{initial} + V_{initial}$

Simplifying by plugging in initial conditions and solving for $\dot{r}^2$ gives:

$2\dot{r}^2 = \frac{(50 + 10g\sqrt{3}) r^2- \sqrt{3}g r^3 -5000}{r^2}$

Now, since:

$z = \sqrt{3}r$

THe minimum value of $z$ is achieved when $\dot{z} = \sqrt{3}\dot{r} = 0$ , which means $\dot{r}=0$ . This means that one must find the roots of the numerator of the expression:

$2\dot{r}^2 = \frac{(50 + 10g\sqrt{3}) r^2- \sqrt{3}g r^3 -5000}{r^2}$ $\implies (50 + 10g\sqrt{3}) r^2- \sqrt{3}g r^3 -5000=0$

Solving using a computer/ Wolfram Alpha gives one of the roots as:

$r_o \approx 7.092$ $\boxed{z_{min} = \sqrt{3}r_o \approx 12.2836}$

Express the coordinates in cylindrical form $(\alpha = \sqrt{3})$ :

$x = r \cos \theta \\ y = r \sin \theta \\ z = \alpha r$

Lagrangian:

$L = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2 ) - m g z \\ = \frac{1}{2} m \Big(r^2 \dot{\theta}^2 + \dot{r}^2 (1 + \alpha^2) \Big) - mg \alpha r$

Equations of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{r}}} = \frac{\partial{L}}{\partial{r}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}$

Evaluating these results in the following expressions for the double-dot terms:

$\ddot{r} = \frac{r \dot{\theta}^2 - g \alpha}{1 + \alpha^2} \\ \ddot{\theta} = -\frac{2 \dot{r} \dot{\theta}}{r}$

Initial conditions:

$r = 10 \\ \theta = 0 \\ \dot{r} = 0 \\ \dot{\theta} = 1$

Running a numerical simulation results in $z_{min} \approx 12.283$ . Some plots of the trajectory are included below (up until 100 seconds), along with simulation code.

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import math

dt = 10.0**(-5.0)

g = 9.81
alpha = math.sqrt(3.0)

##################################################

zmin = 9999999.0

t = 0.0
count = 0

r = 10.0
theta = 0.0

rd = 0.0
thetad = 1.0

rdd = (r*(thetad**2.0) - g*alpha)/(1.0 + alpha**2.0)
thetadd = -2.0*rd*thetad/r

while t <= 20.0:

    r = r + rd*dt
    theta = theta + thetad*dt

    rd = rd + rdd*dt
    thetad = thetad + thetadd*dt

    rdd = (r*(thetad**2.0) - g*alpha)/(1.0 + alpha**2.0)
    thetadd = -2.0*rd*thetad/r

    x = r*math.cos(theta)
    y = r*math.sin(theta)
    z = alpha*r

    if z < zmin:
        zmin = z

    t = t + dt
    count = count + 1

    #if count % 100000 == 0:
        #print t,x,y,z

##################################################

print ""
print ""
print dt
print t
print zmin

#1e-05
#20.0000099994
#12.2832541399
#>>>