Motion of a particle in a frictionless cone (C.M.)

[Physics_Enjay]

Homework Statement

A particle slide on the frictionless surface of the interior of a 45 degree cone $x^2 + y^2 = z^2$

a) Find the 2D Lagrangian in terms of the vertical coordinate $z$ and an angular coordinate $\theta$.
b) Find the Hamiltonian $H$.
c) Show that $p_\theta$ and $H$ are constant.
d) If the particle starts at height $z_0$ with no vertical component of velocity, show that the particle stays between $z = z_0$ and $$ z = {\frac { p^2_\theta + \sqrt {p^2_\theta ( p^2_\theta + 8 m^2 g z^3_0)}}{4 m^2 g z^2_0}} $$

Homework Equations

Note in what follows $q(t)'=\dot{q}$.

Lagrangian:

$L = T - U$

E-L Equation:

$\frac{d}{dt}\frac{\partial L}{\partial q'(t)} = \frac{\partial L}{\partial q(t)}$

Conjugate momenta and their time derivatives:

$p(t) = \left[\frac{\partial L}{\partial q'(t)}\right]$
$p'(t) = \left[\frac{\partial L}{\partial q(t)}\right]$

Hamiltonian:

$H = \sum_{i} p_i q_i' - L$

If $\frac{\partial H}{\partial t} = 0$ then $H = T + U = E$ and energy is conserved.

Hamilton's Equations:

$p'(t) = - \left[\frac{\partial H}{\partial q(t)}\right]$
$q'(t) = \left[\frac{\partial H}{\partial p(t)}\right]$

The Attempt at a Solution

Coordinate vector and its total time derivative are:

$\vec{r}=[z(t) \cos (\phi (t)),z(t) \sin (\phi (t)),z(t)]$
$\vec{r'}=[z'(t) \cos (\phi (t))-z(t) \phi '(t) \sin (\phi (t)),z'(t) \sin (\phi (t))+z(t) \phi '(t) \cos (\phi (t)),z'(t)]$

Therefore:

$L=\frac{1}{2} m \left(2 z'(t)^2+z(t)^2 \phi '(t)^2-2 g z(t)\right)$

Conjugate momenta:

$p_z (t) = 2 m z'(t)$
$p_z '(t) = m z(t) \phi '(t)^2-g m$
$p_\phi (t) = m z(t)^2 \phi '(t)$
$p_\phi '(t) = 0$

Therefore:

$H = \frac{4 g m^2 z(t)+\frac{2 p_{\phi }^2}{z(t)^2}+p_z^2}{4 m}$

And since H does not have an explicit time dependency, H is conserved, and therefore H=T+U = E.

Hamilton's equations =
$z'(t) = \frac{p_z}{2 m}$
$\phi '(t) = \frac{p_{\phi }}{m z(t)^2}$
$p_z '(t) = g m-\frac{p_{\phi }^2}{m z(t)^3}$
$p_\phi '(t) = 0$

Equations of motion:

$2 m z''(t)=m z(t) \phi '(t)^2-g m$

Substituting in for $\phi '(t)$, we get:

$2 m z''(t)=m z(t) \left(\frac{p_{\phi }}{m z(t)^2}\right){}^2-g m$

Simplifying:

$m \left(g+2 z''(t)\right)-\frac{p_{\phi }^2}{m z(t)^3} = 0$

This is where I got stuck. Evaluating this is a nightmare. Mathematica doesn't help. Plus I don't know where I can figure in the initial conditions that are given. Any help would be greatly appreciated.

 

[mfb]

You are not supposed to find an explicit solution for this equation - if it would be reasonable, the problem statement would ask for it, but it does not.
If the particle stays between two values of z, what do you know about the motion (both for phi and z) at those extremal z-values?

 

[Physics_Enjay]

Since energy is conserved in the system the particle doesn't come to rest. If the particle moves between those values for z, then they are extremas of z, so I know that $z'(t)$ is zero. I'm not sure what this says for $\phi$ but I know that since angular momentum is conserved then so is angular velocity.

 

[Physics_Enjay]

Sorry, what I meant to say was that because angular velocity is constant, then $\phi '(t)$ is constant

 

[mfb]

z'(t)=0, right.
ϕ'(t) is "constant" - well, its derivative is zero: ϕ''(t)=0.

You can find p_z and plug everything into the Hamiltonian, that should work (you know it is conserved).

 

[Physics_Enjay]

$\phi''(t) = \frac{p_{\phi }'(t)}{m z(t)^2}-\frac{2 p_{\phi }(t) z'(t)}{m z(t)^3} = 0$ implies that z'(t) =0 always, not just at extrema, right?

This then implies that $p_z (t) = 0$ which gets rid of that term in the Hamiltonian.

Also have that $p_z '(t) = g m-\frac{p_{\phi }^2}{m z(t)^3} = 0$ or $z(t) = (-\frac{p_{\phi }^2}{gm^2})^{1/3}$

I'm not sure what I'm supposed to do with the Hamiltonian like you suggested, or how any of these substitutions lead to the result:

$z = {\frac { p^2_\phi + \sqrt {p^2_\phi ( p^2_\phi + 8 m^2 g z^3_0)}}{4 m^2 g z^2_0}}$

Thank you for your help so far though.

 

[Physics_Enjay]

Sorry I forgot to include the minus for $p_z '(t)$ so $p_z '(t) = - (g m-\frac{p_{\phi }^2}{m z(t)^3})$ and hence the previous result is $z(t) = (\frac{p_{\phi }^2}{gm^2})^{1/3}$.

I'm still not sure how to get further with this. I already used the fact that the Hamiltonian is conserved to arrive at these results.

 

[mfb]

implies that z'(t) =0 always, not just at extrema, right?

Why? This condition is satisfied at the extrema only.

You know all initial conditions and one term in the Hamiltonian vanishes for the two extremal points.
g, m and p_ϕ do not change, so only z can be different for those points. Solve for z, it will give you two solutions, one is the initial z-value.

 

[Physics_Enjay]

Why? This condition is satisfied at the extrema only.

This is my reasoning:

We said that angular momentum is constant, which implies that angular velocity is constant, which implies that angular acceleration $\phi''(t)$ is zero, i.e. $\phi''(t) = \frac{p_{\phi }'(t)}{m z(t)^2}-\frac{2 p_{\phi }(t) z'(t)}{m z(t)^3} = 0$. The first term is zero since it contains $p_{\phi }'(t)=0$. The second term is zero only if $2 p_{\phi }(t) z'(t)$. So either angular momentum is zero, or z'(t)=0.

I'm confused because I must be getting something basic wrong. I don't doubt your approach.

 

[TSny]

We said that angular momentum is constant, which implies that angular velocity is constant, ...

How do you show that constant angular momentum implies constant angular velocity? (A spinning ice skater who pulls in her arms will increase her angular velocity while her angular momentum remains constant.)

 

[Physics_Enjay]

(A spinning ice skater who pulls in her arms will increase her angular velocity while her angular momentum remains constant.)

Isn't it because Ice skater changes his or her moment of inertia? In this problem we have a point particle.

 

[mfb]

Isn't it because Ice skater changes his or her moment of inertia? In this problem we have a point particle.

It still has a moment of inertia around the central axis, and that changes.

 

[Madelynne]

Quick question, you say to solve for z, which equation are we solving for z? is it Φ''(t)?

 

[TSny]

Quick question, you say to solve for z, which equation are we solving for z? is it Φ''(t)?

Quick answer, see mfb's post #8

 

[Madelynne]

Yeah, that's the post I was referring to, I'm not sure what exactly I'm solving for z?

 

[Madelynne]

So we know that at the extrema z'(t)=0 and Φ''(t)=0 but I have no idea how to use those...

 

[TSny]

Can you express the Hamiltonian at the initial time in terms of z_o? Note that $\phi''$ is irrelevant in the Hamiltonian.

z_o is one of the extreme values of z.

Let z₁ correspond to another extreme value at a later time. Can you express the Hamiltonian at this time in terms of z₁?

What can you say about the values of the Hamiltonian at these two times?

 

[Physics_Enjay]

Thanks for your help guys. H(z0)=H(z1), where z1 and z0 are extrema of z (ie z' and hence p_z are zero), and solve for z1 in terms of z0. I found the question confusing because I misread it as though z in the question was a general z.