1. Not finding help here? Sign up for a free 30min 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!

Hamilton's Principle Equations! Work Shown

  1. Oct 28, 2009 #1
    Hamilton's Principle Equations! Work Shown Please Help!

    A particle of mass m moves under the influence of gravity alng the helix z=k(theta), and r=R, where R and k are constants and z is vertical.

    a.) Using cartesian co-ordinates, write down the expressions for the kinetic energy of the system.

    b.) Change to cylindrical co-ordinate system using x= rcos(theta) y= rsin(theta)

    express your eq for the kinetic energy as a function of the new co-ordinates. Give also expressions for the potential energy and the Langrangian of the system (in cylindrical co-ordinates).

    c.) How many degrees of freedom do you have for this system? Name them (or it)

    d.) Calculate the equation(s) of motion using the Langrange equation for this system.



    my work:

    a.) T= 1/2 m(x*time-deriv.)^2 + 1/2(y*time-deriv)^2 + 1/2(k(theta)*time-deriv.)^2

    b.) if x= rcos(theta) and y=rsin(theta)

    equation for kinetic energy in terms of cylin. co-ord is:
    x= rcos(theta)
    y= rsin(theta)
    therefore, x-time deriv.= r*angle-timederiv. cos(theta)
    y-time deriv.= r*angle-timederiv.sin(theta)

    therefore, kinetic energy in cylin. co-ord.

    T= (1/2)*(m)[(r*time-deriv)+(r^2*angletimederiv^2)+(ztimederiv^2)]

    and therefore potential energy equals:
    U=mgz

    therefore the langrangian of the system in cylin. co-ord equals:

    since z=k(theta) and r=R

    therefore langrangian equals (L)= (1/2)*(m)[(R^2/k^2)(ztimederiv^2)+(ztimederiv^2)] -mgz


    c.) need some help here, i believe the degree of freedom is six:

    since in three dimensions, the six DOFs of a rigid body are sometimes described using these nautical names:
    Moving up and down (heaving);
    Moving left and right (swaying);
    Moving forward and backward (surging);
    Tilting forward and backward (pitching);
    Turning left and right (yawing);
    Tilting side to side (rolling).

    not sure if this is correct need help
    d.) using langrangian equation:

    z-2ndtime deriv= g/[(r^2/k^2)+1]

    need help proving that


    PLEASE HELP A LOT OF WORK SHOWN NEED HELP HERE!
     
  2. jcsd
  3. Oct 28, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    Re: Hamilton's Principle Equations! Work Shown Please Help!

    hi john - i (& others) will able to help more if things are written a little easier to understand, so i've given some tips below

    what is x*time-deriv ???
    do you mean dx/dt?, if so use x', it will make things clearer, or even better use tex, click below to see code (or the sum sign in the advanced tab)
    [tex] \frac{dx}{dt} = x' [/tex]

    also theta isn't a cartesian coordinate, if theta is to hard to write & you don;t want to use tex, give it a single letter varieable ie. q = theta

    those derivatves (if i'm reading them right) don't look correct, now you have assumed r is constant , but what is
    [tex] \frac{d}{dt} cos(\theta(t)) = ? [/tex]

    once again i can;t totally work out what you mean

    that sounds like all the DOF for a rigid mass. Here you have a particle follwing a constrained path. A better question is how many parameters does it take to characterise the particles motion, I think that is the number of DOF

     
    Last edited: Oct 28, 2009
  4. Oct 28, 2009 #3
    Re: Hamilton's Principle Equations! Work Shown Please Help!

    my work:

    a.) T= 1/2 m(x')^2 + 1/2(y')^2 + 1/2(k(theta)*z')^2 , I believe i made a mistake here need help with the cartesian eqn for the kinetic energy here

    b.) if x= rcos(theta) and y=rsin(theta)

    equation for kinetic energy in terms of cylin. co-ord is:
    x= rcos(theta)
    y= rsin(theta)
    therefore, x'= r*theta'. cos(theta)
    y'= r*theta'.sin(theta)

    therefore, kinetic energy in cylin. co-ord.

    T= (1/2)*(m)[(r')+(r^2*theta'^2)+(z'^2)]

    i believe i made a few mistakes here, need some help

    and therefore potential energy equals:
    U=mgz

    therefore the langrangian of the system in cylin. co-ord equals:

    since z=k(theta) and r=R which is a const.

    therefore langrangian equals (L)= (1/2)*(m)[(R^2/k^2)(z'^2)+(z'^2)] -mgz


    c.) need some help here, i believe the degree of freedom is 3:

    since in three dimensions and it is a rigid mass, and you said u have to use the parameters to explain the particle of motion, since it's three dimensional and it contains components for radius which is a constant R, and z which theta times a const. k, and z is vertical

    how do i describe the DOF's and how many are there?
    d.) using langrangian equation:

    z''= g/[(r^2/k^2)+1]

    I need help getting to this process


    PLEASE HELP I HOPE MY STEPS AND CLARIFICATION HELPS HERE.. NEED A LOT OF HELP
     
  5. Oct 28, 2009 #4

    lanedance

    User Avatar
    Homework Helper

    Re: Hamilton's Principle Equations! Work Shown Please Help!

    one other thing may help, if a function is a variable of something write it so, so you know what you're differentiating against eg. x = x(t)

    and the dashes here correspond only to time derivatives
    ok so if
    v(t) = (x(t),y(t),z(t)))

    the time derivative is
    v'(t) = (x'(t),y'(t),z'(t)))

    the kinetic energy is something like
    T= (1/2)m v'(t).v'(t) = (1/2)m (x'(t)^2 + y'(t)^2 + z'(t)^2)

    the only other thing you could do here is relate the x, y & z's together form the know equations, but that is much easier in cylindrical coordinates, so I would wait until next step

    i would include z(t) = k.theta(t) here and differentiate that as well
    ok so you need to note r is constant here before differntiating (ie r'(t) = 0)

    the derivtives are still not right, what is

    (d/dt) cos(theta(t)) = ?

    update below with you r new derivtives & show how you simplify
    this is not a rigid mass, it is a particle in sapce so has at maximum only has 3 DOF

    so at the end of the day - how many independent (and changing varibales are there)
     
  6. Oct 29, 2009 #5
    Re: Hamilton's Principle Equations! Work Shown Please Help!

    for a.) i got v(t)= 1/2*m*[x'(t)^2+ y'(t)^2 + z'(t)^2]

    for cartesian co-ord. expression for KE
    is this correct?

    b.) i got 1/2*m*[(theta'(t)^2*r^2)+ z'(t)^2]

    U=mgz

    since L= T-U
    since z=k(theta) z'(t)= k(theta'(t)

    therefore sub. L= 1/2*m[ ((z'(t)^2)/(k^2))*(r^2)+z'(t)^2]- mgz

    is this correct?

    c.) for degrees of freedom, since it's a particle of space it has 3 dimensions so degrees of
    freedom, how do i specify, do i say it's x,y, and z?

    d.) i got the equation of motion as: z''(t)= g/ ((r^2/k^2) + 1) is this correct


    please help work shown!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hamilton's Principle Equations! Work Shown
Loading...