Question

Find the Lagrangian and the Lagrange equation for the pendulum shown. The vertical circle is fixed. The string winds up or unwinds as the massswings back and forth. Assume that the unwound part of the string at any time is in a straight-line tangent to the circle. Letbe the length of the unwound string when the pendulum hangs straight down.

Short answer

The Lagrangian of the masses will be:

$\mathrm{L}=\frac{1}{2}\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2{\dot{\mathrm{\theta }}}^2-\mathrm{mg}(\mathrm{asin\theta }-(\mathrm{l}+\mathrm{a\theta }\left)\mathrm{cos\theta }\right)$

Lagrange’s equation is:

$(\mathrm{l}+\mathrm{a\theta })\ddot{\mathrm{\theta }}+\mathrm{a}{\dot{\mathrm{\theta }}}^2+\mathrm{gsin\theta }=0.$

Step-by-step solution

Meaning of the Lagrange’s equation and Lagrangian

An ordinary first-order differential equation that is linear in the independent variable and unknown function but not solved for the derivative is termed as Lagrange's equation.

A function that equals the difference between potential and kinetic energy and characterizes the state of a dynamic system in terms of position coordinates and their time derivatives is termed as Lagrangian.

 Step 2: Given parameter

Given the mass is and the length of the unwound string is .

Find the Kinetic energy

The length of pendulum changes during the motion and so does the point of rotation.

This implies that the length of the pendulum changes with angle$\mathrm{\theta }$ as:

$\mathrm{L}=\mathrm{l}+\mathrm{a\theta }$

Now the angle between the vertical line and the pendulum in motion is$\mathrm{\theta }$ as well.

Also, the horizontal axis is axis and the vertical z-axis is axis.

Then the position of the mass will be given by:

$\begin{array}{l}\mathrm{x}=\mathrm{acos\theta }+(\mathrm{l}+\mathrm{a\theta })\mathrm{sin\theta }\\ \mathrm{z}=\mathrm{asin\theta }-(\mathrm{l}+\mathrm{a\theta })\mathrm{cos\theta }\end{array}$

Now the velocity will be the time derivative of the position which will be given by:

$\begin{array}{c}\dot{\mathrm{x}}=-\mathrm{a}\dot{\mathrm{\theta }}\mathrm{sin\theta }+\dot{\mathrm{\theta }}(\mathrm{l}+\mathrm{a\theta })\mathrm{cos\theta }+\mathrm{a}\dot{\mathrm{\theta }}\mathrm{sin\theta }\\ =\dot{\mathrm{\theta }}(\mathrm{l}+\mathrm{a\theta })\mathrm{cos\theta }\\ \dot{\mathrm{z}}=\mathrm{a}\dot{\mathrm{\theta }}\mathrm{cos\theta }+\dot{\mathrm{\theta }}(\mathrm{l}+\mathrm{a\theta })\mathrm{sin\theta }-\mathrm{a}\dot{\mathrm{\theta }}\mathrm{cos\theta }\\ =\dot{\mathrm{\theta }}(\mathrm{l}+\mathrm{a\theta })\mathrm{sin\theta }\end{array}$

Then the kinetic energy is given by:

$\begin{array}{c}\mathrm{T}=\frac{1}{2}\mathrm{m}\left({\dot{\mathrm{x}}}^2+{\dot{\mathrm{z}}}^2\right)\\ =\frac{1}{2}\mathrm{m}\left({\dot{\mathrm{\theta }}}^2{(\mathrm{l}+\mathrm{a\theta })}^2{\cos }^2\mathrm{\theta }+{\dot{\mathrm{\theta }}}^2{(\mathrm{l}+\mathrm{a\theta })}^2{\sin }^2\mathrm{\theta }\right)\\ =\frac{1}{2}\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2{\dot{\mathrm{\theta }}}^2\end{array}$

So, the Kinetic energy is:

$T=\frac{1}{2}\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2{\dot{\mathrm{\theta }}}^2$

Find the Potential energy

The potential energy will be due to the gravity and thus the potential energy will be given by

$\mathrm{V}=\mathrm{mgz}$

Since the position of the kinetic energy is obtained the z-coordinate. Thus:

$\mathrm{V}=\mathrm{mg}\left(\mathrm{asin\theta }-(\mathrm{l}+\mathrm{a\theta })\mathrm{cos\theta }\right)$

So, the potential energy will be$\mathrm{V}=\mathrm{mg}\left(\mathrm{asin\theta }-(\mathrm{l}+\mathrm{a\theta })\mathrm{cos\theta }\right)$

Find the Lagrangian of the two masses

$\begin{array}{c}\mathrm{L}=\mathrm{T}-\mathrm{V}\\ \mathrm{L}=\frac{1}{2}\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2{\dot{\mathrm{\theta }}}^2-\mathrm{mg}(\mathrm{asin\theta }-(\mathrm{l}+\mathrm{a\theta }\left)\mathrm{cos\theta }\right)\end{array}$

So, the Lagrangian is$\mathrm{L}=\frac{1}{2}\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2{\dot{\mathrm{\theta }}}^2-\mathrm{mg}(\mathrm{asin\theta }-(\mathrm{l}+\mathrm{a\theta }\left)\mathrm{cos\theta }\right)$

Find the Lagrange’s equation

Since the Lagrangian only has one degrees of freedom.

Thus, the Euler equation of degree of freedom will be given by:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta }=0$

Then the calculation of the required derivation will be:

$\begin{array}{c}\frac{\partial L}{\partial \dot{\theta }}=\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2\dot{\mathrm{\theta }}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2\ddot{\mathrm{\theta }}+2\mathrm{m}(\mathrm{l}+\mathrm{a\theta }){\dot{\mathrm{\theta }}}^2\\ \frac{\partial L}{\partial \theta }=\mathrm{m}(\mathrm{l}+\mathrm{a\theta })\mathrm{a}{\dot{\mathrm{\theta }}}^2-\mathrm{mg}(\mathrm{l}+\mathrm{a\theta })\mathrm{sin\theta }\end{array}$

Then the Euler equation will be:

$\mathrm{m}(\mathrm{l}+\mathrm{a\theta }{)}^2\ddot{\mathrm{\theta }}+m(l+a\mathrm{\theta })\mathrm{a}{\dot{\mathrm{\theta }}}^2+m\mathrm{g}(\mathrm{l}+\mathrm{a\theta })\mathrm{sin\theta }=0$

Now dividing the above Euler equation by , then the result will be:

$(\mathrm{l}+\mathrm{a\theta })\ddot{\mathrm{\theta }}+\mathrm{a}{\dot{\mathrm{\theta }}}^2+\mathrm{gsin\theta }=0$

So, the Lagrange equation is $(\mathrm{l}+\mathrm{a\theta })\ddot{\mathrm{\theta }}+\mathrm{a}{\dot{\mathrm{\theta }}}^2+\mathrm{gsin\theta }=0$.