Chapter 4.P4, Problem 1P

Problems $1$ through $3$ are concerned with one degree-of-freedom systems.

Falling Bodies. Let $y\left(t\right)$ be the height above Earth’s surface of a body of mass $m$ subject only to Earth’s gravitational acceleration $g$. Find the action integral and Euler-Lagrange equation for $y\left(t\right)$

To determine

The action integral and Euler-Lagrange equation for the height above the Earth’s surface of a body of mass $m$ subject only to gravitational acceleration $g$.

Answer to Problem 1P

Solution:

The action integral for $y\left(t\right)$ is $J\left(y\right)={\displaystyle \underset{0}{\overset{T}{\int }}\left(\frac{1}{2}my{\text{'}}^2-mgy\right)}dt$.

The Euler-Lagrange equation for $y\left(t\right)$ is $y\text{'}\text{'}=-g$.

Explanation of Solution

Given information:

Let $y\left(t\right)$ be the height above the Earth’s surface of a body of a mass $m$ subject only to gravitational acceleration $g$.

Explanation:

The action of the integral of a system is defined by ${\displaystyle \underset{0}{\overset{T}{\int }}\left(K-V\right)}dt$ where $K$ is the kinetic energy and $V$ is the potential energy.

The kinetic energy of a falling body above the earth’s surface subject only to Earth’s gravitational acceleration is,

$K=\frac{1}{2}m{v}^2$ where $v$ is the velocity of the body.

By using velocity as the derivative of height,

$K=\frac{1}{2}m{\left(\frac{dy}{dt}\right)}^2$

$K=\frac{1}{2}my{\text{'}}^2$

The potential energy of a falling body above the earth’s surface subject to the gravitational acceleration is,

$V=mgy$

Thus the action integral is $J\left(y\right)={\displaystyle \underset{0}{\overset{T}{\int }}\left(\frac{1}{2}my{\text{'}}^2-mgy\right)}dt$.

The Euler-Lagrange equation for the functional $J\left(y\right)={\displaystyle \underset{a}{\overset{b}{\int }}F\left(t,y,y\text{'}\right)}dt$ is defined as,

$\frac{\partial }{\partial y}F\left(t,y,y\text{'}\right)-\frac{{\partial }^2}{\partial t\partial y\text{'}}F\left(t,y,y\text{'}\right)=0$

From the action integral $F\left(t,y,y\text{'}\right)=\frac{1}{2}my{\text{'}}^2-mgy$.

Hence $\frac{\partial }{\partial y}F\left(t,y,y\text{'}\right)=\frac{\partial }{\partial y}\left(\frac{1}{2}my{\text{'}}^2-mgy\right)=-mg$

$\frac{\partial }{\partial y\text{'}}F\left(t,y,y\text{'}\right)=\frac{\partial }{\partial y\text{'}}\left(\frac{1}{2}my{\text{'}}^2-mgy\right)=my\text{'}$ and $\frac{{\partial }^2}{\partial t\partial y\text{'}}F\left(t,y,y\text{'}\right)=\frac{\partial }{\partial t}\left(my\text{'}\right)=my\text{'}\text{'}$

Consider $\frac{\partial }{\partial y}F\left(t,y,y\text{'}\right)-\frac{{\partial }^2}{\partial t\partial y\text{'}}F\left(t,y,y\text{'}\right)=0$

$\Rightarrow -mg-my\text{'}\text{'}=0$

$\Rightarrow -g-y\text{'}\text{'}=0$

$\Rightarrow y\text{'}\text{'}=-g$

Therefore, the Euler-Lagrange equation for $y\left(t\right)$ is $y\text{'}\text{'}=-g$.