Answer to Question #120465 in Classical Mechanics for Jia

Question

Equation of motion of free falling object using lagrangian mechanics

Answer 1

Let us consider the object is under free falling motion from height $H$ .Mass is $m$ .

At certain time the position of the object will be at $\begin{pmatrix} 0\\ y\\ 0 \end{pmatrix}$ and velocity will be $\begin{pmatrix} 0 \\ \dot y\\0 \end{pmatrix}$ .

Now, Kinetic energy of the object will be

$$K=\frac{1}{2}m\dot y^2$$

And, Potential energy will be

$$U=mgy$$

Thus, Lagrangian is given by

$$\mathscr{L}=K-U\\\implies \mathscr{L}=\frac{1}{2}m\dot y^2-mgy$$

Therefore, from Euler Lagrange Equation of motion we get,

$$\frac{\partial \mathscr{L}}{\partial y}-\frac{d}{dt}\bigg(\frac{\partial \mathscr{L}}{\partial \dot y}\bigg)=0$$

Hence,

$$-mg-\frac{d}{dt}(m\dot y)=0\\ \implies m\ddot y=-mg\\ \implies \boxed{ \ddot y=-g}$$

Which is what exactly Newton's law predicted.