Mathematical Pendulum
Plain Pendulum
We will illustrate the application of Lagrangian mechanics by analyzing the movement of a mathematical pendulum - a problem we have already discussed in the Physics 4 Teens - Mechanics - Pendulum, Spring - Pendulum using the Newtonian approach.
We recommend reviewing the lecture mentioned above and refresh the Newtonian method of deriving the main equation of motion of the pendulum:
α"(t) = −(g/l)·sin(α(t))
which was obtained from properly determining the force F that moves a pendulum as a vector sum of the gravity force directed vertically down and the tension of an unstretchable thread that keeps an object at the free end of a thread on a constant distance from the fixed end of a thread.
The two forces involved in formation of a resulting force F, gravity P=m·g and tension of a thread T, had to be combined using the rules for addition of vectors, which required some thinking.
Let's apply the Lagrangian mechanics to this problem using an angle of a thread with a vertical α as the one and only parameter that determines a position of an object.
This way of identification of a position is more convenient than Cartesian coordinates originated at the fixed end of a thread because both of them can be easily derived from α
x = l·sin(α)
y = −l·cos(α)
The Lagrangian is the difference between kinetic and potential energies.
L(α(t),α'(t)) =
= Ekin(α'(t)) − Epot(α(t))
Kinetic energy depends on mass M and linear speed of an object along its circular trajectory v=l·α'(t)
Ekin = ½M·v² =
= ½M·l²·[α'(t)]²
Potential energy depends on a mass of an object M, its height over the ground h(t) and an acceleration of free fall g.
Epot(t) = M·g·h(t)
If the origin of our coordinates, the fixed end of a thread, is at height H over the ground,
h = H − l·cos(α)
and, therefore,
Epot(t) = M·g·[H−l·cos(α(t))]
Now we can construct the Euler-Lagrange equation
(∂/∂α)L(α(t),α'(t)) =
= (d/dt)(∂/∂α')L(α(t),α'(t))
Calculate left and right sides separately.
(∂/∂α)L(α(t),α'(t)) =
= (∂/∂α)[Ekin(α'(t)) − Epot(α(t))] =
= (∂/∂α)[−Epot(α(t))] =
= (∂/∂α)[−M·g·[H−l·cos(α(t))]] =
= −M·g·l·sin(α(t))
(d/dt)(∂/∂α')L(α(t),α'(t)) =
=(d/dt)(∂/∂α')[Ekin(α'(t))−Epot(α(t))]=
= (d/dt)(∂/∂α')Ekin(α'(t)) =
= (d/dt)(∂/∂α')½M·l²·[α'(t)]² =
= (d/dt)M·l²·α'(t) =
= M·l²·α"(t)
The Euler-Lagrange equation is
−M·g·l·sin(α(t)) = M·l²·α"(t)
or
α"(t) = −(g/l)·sin(α(t))
which is exactly as applying Newtonian mechanics.
If you don't think the Lagrangian mechanics is simpler than Newtonian for those who are familiar with Calculus of partial derivatives, consider the next example.
Spring Pendulum
A weightless spring replaces an unstretchable thread of the previous problem.
The spring and an object on its end are in a weightless frictionless tube that maintains a straight form, so an object has two degrees of freedom - radial inside a tube stretching and squeezing a spring and pseudo-circular as it moves together with a tube in a pendulum like motion.
The problem of specifying the motion of an object is much more complex here because the spring tension is changing not only with an angle α(t) but also because of the movement of an object within a tube.However, using the Langrangian mechanics, this problem can be analyzed with much less efforts and the corresponding differential equation can be constructed relatively easy.
As before, let's calculate the kinetic and potential energies of an object.
The object's kinetic energy can be calculated as a sum of its radial movement's kinetic energy inside the tube and kinetic energy of its pseudo-circular movement perpendicularly to the tube.
The reason is simple. The object's linear velocity vector can be represented as a sum of two perpendicular to each other vectors, one is inside the tube and another perpendicular to it.
v = v|| + v⊥
Since kinetic energy depends on a square of the linear speed, according to Pythagorean Theorem
v² = v||² + v⊥²
from which follows that
Ekin = ½M·v² =
= ½M·v||² + ½M·v⊥²
Distance of an object from the fixed point of oscillation l is variable and depends on time: l=l(t).
Therefore, v||(t)=l'(t).
Perpendicular to l component of an object speed is v⊥(t)=l(t)·α'(t).
Therefore, kinetic energy of an object is
Ekin = ½M·[l'(t)²+l(t)²·α'(t)²]
The object's potential energy can be calculated as a sum of its potential energy due to gravity and potential energy of a stretched or a squeezed spring.
If the fixed point of oscillation is at the height H above the ground, an object is at height h(t)=H−l(t)·cos(α(t)) above the ground, and potential energy of an object related to its position in the gravitational field is
Egrav = M·g·[H−l(t)·cos(α(t))]
Potential energy of a spring depends on the degree of its stretching or squeezing.
Assume, the length of a spring in a neutral state is l0. Then the length of stretching or squeezing at time t is l(t)−l0.
Therefore, potential energy of an object related to a spring is
Espr = ½k·[l(t)−l0]²
where k is a coefficient of elasticity of a spring.
Total potential energy of an object is
Epot = Egrav + Espr =
= M·g·[H−l(t)·cos(α(t))] +
+ ½k·[l(t)−l0]²
Construct Lagrangian
L(l,l',α,α') = Ekin − Epot =
= ½M·[l'(t)²+l(t)²·α'(t)²] −
− M·g·[H−l(t)·cos(α(t))] −
− ½k·[l(t)−l0]²
All which remains is to write the Euler-Lagrange equation for this Lagrangian.
The problem is, we are familiar only with the Euler-Lagrange equation for a system with one degree of freedom, like x(t). Now we have two degrees of freedom - l(t) and α(t).
Fortunately, the Euler-Lagrange equation can be specified for each degree of freedom independently, which will be proven in the next lecture.
Therefore, we can write two independent Euler-Lagrange equations (skipping (t) for brevity):
(∂/∂l)L(l,l',α,α') =
= (d/dt)(∂/∂l')L(l,l',α,α')
and
(∂/∂α)L(l,l',α,α') =
= (d/dt)(∂/∂α')L(l,l',α,α')
The first equation is
M·l·α'²+M·g·cos(α)−k·(l−l0) =
= M·l"
The second equation is
−M·g·l·sin(α) = M·l²·α"
or
−g·sin(α) = l·α"
or
−(g/l)·sin(α) = α"
which looks exactly the same as in the case above with unstretchable thread instead of a spring.
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