1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data
    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 $$ z = {\frac { p^2_\theta + \sqrt {p^2_\theta ( p^2_\theta + 8 m^2 g z^3_0)}}{4 m^2 g z^2_0}} $$

    2. Relevant 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]##

    3. 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.
     
  2. jcsd
  3. Mar 6, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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?
     
  4. Mar 6, 2015 #3
    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.
     
  5. Mar 6, 2015 #4
    Sorry, what I meant to say was that because angular velocity is constant, then ##\phi '(t)## is constant
     
  6. Mar 6, 2015 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

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

    You can find pz and plug everything into the Hamiltonian, that should work (you know it is conserved).
     
  7. Mar 6, 2015 #6
    ##\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.
     
  8. Mar 6, 2015 #7
    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.
     
  9. Mar 7, 2015 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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 pϕ 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.
     
  10. Mar 7, 2015 #9
    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.
     
  11. Mar 7, 2015 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.)
     
  12. Mar 7, 2015 #11
    Isn't it because Ice skater changes his or her moment of inertia? In this problem we have a point particle.
     
  13. Mar 7, 2015 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It still has a moment of inertia around the central axis, and that changes.
     
  14. Mar 8, 2015 #13
    Quick question, you say to solve for z, which equation are we solving for z? is it Φ''(t)?
     
  15. Mar 8, 2015 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Quick answer, see mfb's post #8 :smile:
     
  16. Mar 8, 2015 #15
    Yeah, that's the post I was referring to, i'm not sure what exactly i'm solving for z?
     
  17. Mar 8, 2015 #16
    So we know that at the extrema z'(t)=0 and Φ''(t)=0 but I have no idea how to use those.....
     
  18. Mar 8, 2015 #17

    TSny

    User Avatar
    Homework Helper
    Gold Member

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

    zo 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?
     
  19. Mar 10, 2015 #18
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Motion of a particle in a frictionless cone (C.M.)
  1. Motion of a particle (Replies: 1)

  2. Motion of a particle (Replies: 4)

Loading...