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Next: Hamilton's Equations Up: Hamiltonian Dynamics Previous: Hamilton's Principle

Constrained Lagrangian Dynamics

Suppose that we have a dynamical system described by two generalized coordinates, $q_1$ and $q_2$. Suppose, further, that $q_1$ and $q_2$ are not independent variables. In other words, $q_1$ and $q_2$ are connected via some constraint equation of the form
f(q_1,q_2,t) = 0,
\end{displaymath} (714)

where $f$ is some function of three variables. This type of constraint is called a holonomic. [A general holonomic constraint is of the form $f(q_1,q_2,\cdots,q_{\cal F},t)=0$.] Let $L(q_1,q_2,\dot{q}_1,\dot{q}_2,t)$ be the Lagrangian. How do we write the Lagrangian equations of motion of the system?

Well, according to Hamilton's principle,

\delta\int_{t_1}^{t_2} L\,dt = \int_{t_1}^{t_2}\left\{\left[...
... L}{\partial \dot{q}_2}\right)\right]\delta q_2 \right\} dt=0.
\end{displaymath} (715)

However, at any given instant in time, $\delta q_1$ and $\delta q_2$ are not independent. Indeed, Equation (714) yields
\delta f = \frac{\partial f}{\partial q_1}\,\delta q_1 + \frac{\partial f}{\partial
q_2}\,\delta q_2 = 0
\end{displaymath} (716)

at a fixed time. Eliminating $\delta q_2$ from Equation (715), we obtain
\frac{\partial L}{\partial q_1...]\frac{1}{\partial f/\partial q_2}\right\} \delta q_1\,dt=0.
\end{displaymath} (717)

This equation must be satisfied for all possible perturbations $\delta q_1(t)$, which implies that the term enclosed in curly brackets is zero. Hence, we obtain
\frac{\partial L/\partial q_1-(d/dt)\,(\partial L/\partial \...
...t)\,(\partial L/\partial \dot{q}_2)}{\partial f/\partial q_2}.
\end{displaymath} (718)

One obvious way in which we can solve this equation is to separately set both sides equal to the same function of time, which we shall denote $-\lambda(t)$. It follows that the Lagrangian equations of motion of the system can be written
$\displaystyle \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_1}\right)-\frac{\partial L}{\partial q_1} - \lambda(t)\,\frac{\partial f}{\partial q_1}$ $\textstyle =$ $\displaystyle 0,$ (719)
$\displaystyle \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_2}\right)-\frac{\partial L}{\partial q_2} - \lambda(t)\,\frac{\partial f}{\partial q_2}$ $\textstyle =$ $\displaystyle 0.$ (720)

In principle, the above two equations can be solved, together with the constraint equation (714), to give $q_1(t)$, $q_2(t)$, and the so-called Lagrange multiplier $\lambda(t)$. Equation (719) can be rewritten
\frac{d}{dt}\!\left(\frac{\partial K}{\partial \dot{q}_1}\ri...
...U}{\partial q_1}+ \lambda(t)\,\frac{\partial f}{\partial q_1}.
\end{displaymath} (721)

Now, the generalized force conjugate to the generalized coordinate $q_1$ is [see Equation (599)]
Q_1 = -\frac{\partial U}{\partial q_1}.
\end{displaymath} (722)

By analogy, it seems clear from Equation (721) that the generalized constraint force [i.e., the generalized force responsible for maintaining the constraint (714)] conjugate to $q_1$ takes the form
\tilde{Q}_1 = \lambda(t)\,\frac{\partial f}{\partial q_1},
\end{displaymath} (723)

with a similar expression for the generalized constraint force conjugate to $q_2$.

Suppose, now, that we have a dynamical system described by ${\cal F}$ generalized coordinates $q_i$, for $i=1,{\cal F}$, which is subject to the holonomic constraint

f(q_1,q_2,\cdots,q_{\cal F},t) = 0.
\end{displaymath} (724)

A simple extension of above analysis yields following the Lagrangian equations of motion of the system,
\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_i}\ri...
...artial q_i} - \lambda(t)\,\frac{\partial f}{\partial q_i} = 0,
\end{displaymath} (725)

for $i=1,{\cal F}$. As before,
\tilde{Q}_i = \lambda(t)\,\frac{\partial f}{\partial q_i}
\end{displaymath} (726)

is the generalized constraint force conjugate to $q_i$. Finally, the generalization to multiple holonomic constraints is straightforward.

Figure 36: A cylinder rolling down an inclined plane.
\epsfysize =2.5in

Consider the following example. A cylinder of radius $a$ rolls without slipping down a plane inclined at an angle $\theta $ to the horizontal. Let $x$ represent the downward displacement of the center of mass of the cylinder parallel to the surface of the plane, and let $\phi$ represent the angle of rotation of the cylinder about its symmetry axis. The fact that the cylinder is rolling without slipping implies that $x$ and $\phi$ are interrelated via the well-known constraint

f(x,\phi) = x - a\,\phi = 0.
\end{displaymath} (727)

The Lagrangian of the cylinder takes the form
L = \frac{1}{2}\,m\,\dot{x}^{\,2} + \frac{1}{2}\,I\,\dot{\phi}^{\,2} + m\,g\,x\,\sin\theta,
\end{displaymath} (728)

where $m$ is the cylinder's mass, $I$ its moment of inertia, and $g$ the acceleration due to gravity.

Note that $\partial f/\partial x = 1$ and $\partial f/\partial\phi = -a$. Hence, Equation (725) yields the following Lagrangian equations of motion:

$\displaystyle m\,\ddot{x} - m\,g\,\sin\theta - \lambda$ $\textstyle =$ $\displaystyle 0,$ (729)
$\displaystyle I\,\ddot{\phi} + \lambda\,a$ $\textstyle =$ $\displaystyle 0.$ (730)

Equations (727), (729), and (730) can be solved to give
$\displaystyle \ddot{x}$ $\textstyle =$ $\displaystyle \frac{g\,\sin\theta}{1+ I/m\,a^2},$ (731)
$\displaystyle a\,\ddot{\phi}$ $\textstyle =$ $\displaystyle \frac{g\,\sin\theta}{1+ I/m\,a^2},$ (732)
$\displaystyle \lambda$ $\textstyle =$ $\displaystyle -\frac{m\,g\,\sin\theta}{1+ m\,a^2/I}.$ (733)

The generalized constraint force conjugate to $x$ is
\tilde{Q}_x = \lambda\,\frac{\partial f}{\partial x} = - \frac{m\,g\,\sin\theta}{1+ m\,a^2/I}.
\end{displaymath} (734)

This represents the frictional force acting parallel to the plane which impedes the downward acceleration of the cylinder, causing it to be less than the standard value $m\,g\,\sin\theta$. The generalized constraint force conjugate to $\phi$ is
\tilde{Q}_\phi = \lambda\,\frac{\partial f}{\partial\phi} =
\frac{m\,g\,a\,\sin\theta}{1+ m\,a^2/I}.
\end{displaymath} (735)

This represents the frictional torque acting on the cylinder which forces the cylinder to rotate in such a manner that the constraint (727) is always satisfied.

Consider a second example. A bead of mass $m$ slides without friction on a vertical circular hoop of radius $a$. Let $r$ be the radial coordinate of the bead, and let $\theta $ be its angular coordinate, with the lowest point on the hoop corresponding to $\theta=0$. Both coordinates are measured relative to the center of the hoop. Now, the bead is constrained to slide along the wire, which implies that

f(r,\theta) = r - a = 0.
\end{displaymath} (736)

Note that $\partial f/\partial r =1$ and $\partial f/\partial\theta = 0$. The Lagrangian of the system takes the form
L = \frac{1}{2}\,m\,(\dot{r}^{\,2} + r^2\,\dot{\theta}^{\,2})
+ m\,g\,r\,\cos\theta.
\end{displaymath} (737)

Hence, according to Equation (725), the Lagrangian equations of motion of the system are written
$\displaystyle m\,\ddot{r} -m\,r\,\dot{\theta}^{\,2} - m\,g\,\cos\theta - \lambda$ $\textstyle =$ $\displaystyle 0,$ (738)
$\displaystyle m\,r^2\,\ddot{\theta} + m\,g\,r\,\sin\theta$ $\textstyle =$ $\displaystyle 0.$ (739)

The second of these equations can be integrated (by multiplying by $\dot{\theta}$), subject to the constraint (736), to give
\dot{\theta}^{\,2} = \frac{2\,g}{a}\,\cos\theta + c,
\end{displaymath} (740)

where $c$ is a constant. Let $v_0 $ be the tangential velocity of the bead at the bottom of the hoop (i.e., at $\theta=0$). It follows that
\dot{\theta}^{\,2} = \frac{2\,g}{a}\,(\cos\theta-1) + \frac{v_0^{\,2}}{a^2}.
\end{displaymath} (741)

Equations (736), (738), and (741) can be combined to give
\lambda = - m \left(3\,g\,\cos\theta - 2\,g + \frac{v_0^{\,2}}{a}\right).
\end{displaymath} (742)

Finally, the constraint force conjugate to $r$ is given by
\tilde{Q}_r = \lambda\,\frac{\partial f}{\partial r} = -m \left(3\,g\,\cos\theta - 2\,g + \frac{v_0^{\,2}}{a}\right).
\end{displaymath} (743)

This represents the radial reaction exerted on the bead by the hoop. Of course, there is no constraint force conjugate to $\theta $ (since $\partial f/\partial\theta = 0$) because the bead slides without friction.

next up previous
Next: Hamilton's Equations Up: Hamiltonian Dynamics Previous: Hamilton's Principle
Richard Fitzpatrick 2011-03-31