Lagrangians Pt. 2: Complex Systems

In an extension of the previous post, Hamilton’s Principle and Lagrangians, the discussed principles can greatly simplify solving for the motion of otherwise complicated, mechanical systems.  If one can determine the time dependent formulas describing the object of interest’s various degrees of freedom, the system’s kinetic energy, potential energy, Lagrangian equation, and Euler-Lagrange equations can be formulated.  By applying the basic concepts established in the previous post, we will determine the motion undergone by two otherwise complicated systems.

Example 1:

Consider a pendulum of length ℓ.  On one end, the pendulum is attached to a bob of mass m.  On the other end, the pendulum is attached to a peg on a disk of radius R that rotates with an angular velocity ω.   Determine the equation of motion of the bob.  The system is illustrated in Figure 1.

Figure 1

Solution:

To begin, we formulate time dependent equations representing the x and y coordinates of our bob.  For our system,

and

where φ is a function of time representing the angle from the vertical of the pendulum.  For each of these equations, the first term represents the respective component of the vector from the center of the disk to the peg.  Likewise, the second term represents the respective components of the vector from the peg to the bob.

Next, we can determine the kinetic energy, potential energy, and, subsequently, the Lagrangian of the system.  As discussed in the previous post, the kinetic energy T and potential energy U of the bob are

and

where g is gravity.  Furthermore, since

we determine our Lagrangian to be

Finally, by applying the Euler-Lagrange equation, we determine our system’s equation of motion:

The motion of the system given initial conditions is simulated below.  For this simulation, R = 0.75 m, ℓ=0.75 m, ω=1.0 rad/sec, g=9.8 m/s², m=1.0 kg, φ(0)=π/4, and φ’(0)=0.

Example 2:

Consider the same system as in example 1, but with an additional pendulum and bob attached to the original bob, see Figure 2.  Determine the equations of motion of the system.

Figure 2

Solution:

As our system now has an additional degree of freedom, we must determine an additional equation of motion.  Applying the same procedure as before, we begin by determining the time dependent equations representing the x and y coordinates of our two masses.  For this system,

and

Next, we determine the kinetic energy, potential energy, Lagrangian, and Euler-Lagrange equations of the system.  As before, the kinetic energy T and potential energy U of the system are

and

After plugging the energies into our Lagrangian equation and setting up Euler-Lagrange equations for both φ and θ, we obtain our equations of motion:

and

The motion of the system given initial conditions is simulated below.  For this simulation: R = 0.75 m, ℓ₁=0.5 m, ℓ₂=0.5 m, ω=1.0 rad/sec, g=9.8 m/s², m₁=1.0 kg, m₂=1.0 kg, θ(0)=π/2, θ’(0)=0, φ(0)=π/4, and φ’(0)=0.