Parameterizing Motion along a Curve 

by Shawn Hedman, Florida Southern College, shedman@flsouthern.edu

Introduction

We animate a point mass moving along a parabola, a helix, and a piecewise defined combination of these two curves.  

The animations model the motion of a bead sliding along a wire under a couple of assumptions.

Initialization

Part 1  A point mass on a parabola.

Suppose a point mass of m kg is released from the point (0,9) with an initial velocity of 0 m/s.

As the bead falls along the curve, it loses potential energy and gains kinetic energy.

The motion of the point mass from time t=a to t=b minimizes .  

The difference is called the Lagrangian and is denoted by L:

To minimize this integral, the point mass will move fastest at points where the kinetic energy is greatest and slowest at points where the potential energy is greatest.

Because we are assuming the total energy remains constant:   for some C. 

Initially, we have: , and so .

In general, when the point mass is located at we have:

Solving for  yields:

Our goal is to parameterize the curve as with the required speed.

The velocity of this parameterization is .  

The speed v is the magnitude of the velocity   

Squaring both sides:  

Note:  If we set (4) equal to (6) we get a nonlinear first order ordinary differential equation.   Because this differential equation can be expressed in terms of (or in terms of ).  The initial value problem with initial condition (or ) has constant solution (or ).  Because v has been canceled out of the equation, we cannot give the object a nonzero initial velocity.  This is not helpful.  

To find the required parameterization , we use the fact that the functions  and  minimize  where L is the Lagrangian:

To minimize L, we first express L in terms of a single variable. 

Substituting these quantitites into (7) gives L in terms of :

To minimize L, the above expression must satisfy the Euler-Lagrange equation:

   
d/dt (∂L/∂X)=∂L/∂x.  
 

In this equation, X represents dx/dt, but is treated as an independent variable.  To compute the left and right sides of this equation, we express the Lagrangian as:

Because m occurs on both sides of the Euler-Lagrange equation, it cancels out.  For this reason, we have omitted it from the definition of L1.  We have also replaced g with its numerical value.   

EL1R is the right side of the Euler-Lagrange equation.  We call the left side EL1L.  

EL1L is the derivative of the previous output with respect to t.  

The Euler-Lagrange equation (EL) is a second order nonlinear ordinary differential equation for x1.  We solve this with the initial conditions given by x=0 and v=0:

The output is given as a procedure.  Given input t, this procedure outputs the numerical solution to the equation.  For instance:

The output is a list with three items and the function x1(t) is second in this list.

To extract x1(t) from this output, we use the `op' function:  

To see the point mass go back and forth, view on "continuous cycle."

For future reference, we note that the object first passes the point (3.5,0.35) at approximately t =1.46777 and again passes this point (moving in the opposite direction) at approximately t=4.22958.  

Part 2 (in 3D)  A point mass on a helix.

Suppose a point mass is released from rest at the point (cos(35), sin(35), 35) and travels along the helix to the xy-plane.  To parameterize the motion of the point mass along the helix, we again look at the Lagrangian and solve the corresponding Euler-Lagrange equation.    

Let r2(t)=[x2(t),y2(t),z2(t)]  be the required parameterization.

      Let Z denote dz/dt.  The Euler-Lagrange equation d/dt (∂L/∂Z)=∂L/∂z is simply:

We see that the height of an object falling along a helix, like an object in free fall, is a quadratic

function of time:

The initial height is

The initial velocity is  

And

Part 3  Motion along a piecewise defined curve.

We attach a helix to the parabola z=f1.

The motion of the point mass along the parabola is given by [x1sol(t),0,y1sol(t)].  When x1sol(t)=3.5, the point mass begins motion along the helix.  As previously mentioned, this occurs at approximately t = 1.46777 (from line (22)).  

The helix is parameterized by [sin(z3-.25)+3.5, -cos(z3-.25)+1, z3].  The initial point of motion along the helix is (3.5,0,.25).  

The initial rates of change are (approximately) dx1/dt=dz3/dt = 9.26008073820564.   

As in Part 2, z3(t) is quadratic:

So the object will first pass the point (3.5,0.25) at approximately t=1.46777, and again (in the opposite direction) at t=5.247394792.  When the object returns to the parabola,  it is located at the point (x1(t),y1(t)) for t = 4.22958 (from (23)).

    To see the point mass go back and forth, view on "continuous cycle."

Legal Notice: Maplesoft, a division of Waterloo Maple Inc. 2009. Maplesoft and Maple are trademarks of Waterloo Maple Inc. This application may contain errors and Maplesoft is not liable for any damages resulting from the use of this material. This application is intended for non-commercial, non-profit use only. Contact Maplesoft for permission if you wish to use this application in for-profit activities.