8. Lagrangian equations of motion

LAST TIME

• Introduced generalized coordinates, velocities and forces.
• Defined holonomic, scleronomic and rheonomic constraints.
• Showed how to determine number of degrees of freedom.

TODAY
we will use these concepts and define the Lagrangian and derive the Lagrangian equations of motion.

Let us assume that we are dealing with a system which can be described by the Cartesian coordinates

$$x_1,x_2\cdots x_M$$

in an inertial reference frame. The kinetic energy is

$$T=\frac{1}{2}\sum_{i=1}^Mm_i\dot{x}_i^2$$

The equations of motion are given by Newton's second law

$$\dot{p}_i=m_i\ddot{x}_i=F_i$$

where $p_i$ is the $i$th component of the momentum, and $F_i$ is the component of the force acting in the direction of $x_i$. These forces may include holonomic forces of constraint. Let us assume that when these have been eliminated we have $N$ degrees of freedom described by generalized coordinates

$$x_1=x_1(q_1,q_2\cdots q_N,t)$$

$$x_2=x_2(q_1,q_2\cdots q_N,t)$$

$$\cdots$$

$$\cdots$$

$$x_M=x_M(q_1,q_2\cdots q_N,t)$$

We now imagine that these equations have been substituted into the expression for the kinetic energy

$$T(\cdots\dot{x}_i\cdots)={\cal T}(\cdots q_k\cdots,\cdots\dot{q}_k,t)$$

We have

$$\frac{\partial \cal T}{\partial q_k}=\sum_{i=1}^Mm_i\dot{x}_i... ...l q_k} =\sum_{i=1}^Mp_i\frac{\partial \dot{x}_i}{\partial q_k}$$

Next

$$\dot{x}_i(\cdots q_k\cdots,\cdots\dot{q}_k\cdots,t)\equiv\fra... ...ial x_i}{\partial q_k}\dot{q}_k+\frac{\partial x_i}{\partial t}$$

and we find

$$\frac{\partial \dot{x}_i}{\partial \dot{q}_k}=\frac{\partial x_i(q_1\cdots q_N,t)}{\partial q_k}$$

and

$$\frac{d}{dt}\frac{\partial x_i}{\partial q_k}=\sum_{j=1}^N\fr... ...\partial q_k\partial t}=\frac{\partial \dot{x}_i}{\partial q_k}$$

We use these result to write

$$\frac{\partial \cal T}{\partial q_k}=\sum_{i=1}^Mp_i\frac{d}{dt}\frac{\partial x_i}{\partial q_k}$$ (1)

and

$$\frac{\partial \cal T}{\partial \dot{q}_k}=\sum_{i=1}^Mm_i\d... ...l \dot{q}_k}=\sum_{i=1}^Mp_i\frac{\partial x_i}{\partial q_k}$$ (2)

Next take the time derivative of (2)

$$\frac{d}{dt}\left(\frac{\partial \cal T}{\partial \dot{q}_k}... ... q_k}+p_i\frac{d}{dt}\frac{\partial x_i} {\partial q_k}\right)$$ (3)

For the first term inside the sum on the right hand side of(3) we use the definition of generalized force from lecture 7

$${\cal F}_k=\sum_{i=1}^M F_i\frac{\partial x_i}{\partial q_k}$$

while for the second term we use (1) to find

$$\frac{d}{dt}\left(\frac{\partial \cal T}{\partial \dot{q}_k}\right)-\frac{\partial \cal T}{\partial q_k}= {\cal F}_k$$

We know assume that the generalized force can be split up into two contributions

$${\cal F}_k=-\frac{{\cal V}(\cdots q_k, \cdots)}{\partial q_k}+ Q_k$$

The first contribution is a conservative force derived from a velocity independent potential. The second term represents "left-overs" such as e.g. friction or drag forces.

We also define the Lagrangian

$$\cal L=T-V$$

Since we have assumed that the potential and the constraints are velocity independent

$$\frac{\partial \cal V}{\partial \dot{q}_k}=0$$

we find Lagrangian equation of motion



Note that there is one equation for each degree of freedom. In the special case that all forces are conservative and derived from a velocity independent potential



You might think that we have only taken something relatively simple (Newton's equations) and made it into something complicated (Lagrange's equations). Actually, this is not the case: we will find that in most cases the Lagrangian approach is the easiest to work with, when solving problems of even moderate difficulty. Furthermore, we will find that the more complicated the problem the greater the advantage of using the Lagrangian approach.

EXAMPLE
As our first example consider the slider block problem of the last lecture. The kinetic energy is in terms of the generalized coordinates $s$ and $X$ is

$${\cal T}=\frac{M\dot{X}^2}{2}+\frac{m}{2}[(\dot{X}+\dot{s}\cos\alpha)^2 +\dot{s}^2\sin^2\alpha]$$

The potential energy is

$${\cal V}=mg(h-s\sin\alpha)$$

We have

$$\frac{\partial {\cal L}}{\partial \dot{s}}=m\cos\alpha\dot{X}+m\dot{s}$$

$$\frac{\partial {\cal L}}{\partial \dot{X}}=(m+M)\dot{X}+m\cos\alpha\dot{s}$$

$$\frac{\partial {\cal L}}{\partial X}=0$$

$$\frac{\partial {\cal L}}{\partial s}=mg\sin\alpha$$

The equations of motion are thus

$$(m+M)\ddot{X}+m\cos\alpha\;\ddot{s}=0$$

$$m\cos\alpha\ddot{X}+m\ddot{s}-mg\sin\alpha=0$$

These equations of motion are the same as we found last time using Newtonian mechanics. The main difference is that there is no need to bother about the free body diagram or the normal forces $N$ and $n$.

EXAMPLE
The pendulum

Instead of using the Cartesian coordinates of the mass it is convenient describe the motion by the angle $\theta$. The length of the pendulum, $r$, is assumed to be constant (i.e. is a holonomic constraint).

The kinetic and potential energies are

$${\cal T}=\frac{mr^2\dot{\theta}^2}{2}; {\cal V}(\theta)=-mgr\cos\theta$$

With $\cal L=T-V$ the equation of motion becomes

$$\frac{\partial {\cal L}}{\partial \theta}-\frac{d}{dt}(\frac{... ...al \dot{\theta}})= -mgr\sin\theta-mr^2\frac{d\dot\theta}{dt}=0$$

giving the equation of motion:

$$\ddot{\theta}+\frac{g}{r}\sin\theta=0$$

We will discuss the properties of the solutions to the equation of motion in lecture 13 and lecture 14

SUMMARY
We have defined the Lagrangian

$$\cal L=T-V$$

and derived the Lagrangian equations of motion

$$\frac{d}{dt}\frac{\partial \cal L}{\partial \dot{q}_k}-\frac{\partial \cal L}{\partial q_k}= Q_k,\;\;k=1\cdots N$$

where $Q_k$ are generalized nonconservative forces. If all forces are conservative $Q_k=0$.

Example problems: See problems 1 and 2 of 1999 midterm, with solution, problem set 3, 2000, with solution, problem set 4 2001, with solution, question 2 of 2001 midterm, with solution.

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