Chapter 9, Problem 9.1E

For an object having mass $m$ falling in the z direction, the kinetic energy is $\frac{1}{2}mz$ and the potential energy is $mgz$, where $g$ is the gravitational acceleration constant (approximately $9.8{\text{m/s}}^{\text{2}}$ ) and $z$ is the position. For this one-dimensional motion, determine the Lagrangian function L and write the Lagrangian equation of motion.

Interpretation Introduction

Interpretation:

The Lagrangian function L for the given one dimensional motion is to be determined and Lagrangian equation of motion is to be stated.

Concept introduction:

The Lagrangian function is the formulation of the classical mechanics. According to mechanics, the Lagrangian function is the difference of kinetic energy and potential energy which is expressed as functions of position and velocity.

Answer to Problem 9.1E

The Lagrangian function L for the given one dimensional motion in the z direction is,

$L=\frac{1}{2}m\dot{z}-mgz$

The Lagrange equation of motion for this system is,

$mg=0$

Explanation of Solution

It is given that an object of mass $m$ is falling in the z direction, the kinetic energy and potential energy of the object is $\frac{1}{2}mz$ and $mgz$ respectively.

The Lagrangian function L for the given one dimensional motion in the z direction is,

$L\left(\dot{z},z\right)=K\left(\dot{z}\right)-V\left(z\right)$ …(1)

Where,

• $K$ is the kinetic energy.

• $V$ is the potential energy.

• $\dot{z}$ is the derivative of z with respect to time.

Substitute the values of kinetic energy and potential energy in the equation (1) as shown below.

$L=\frac{1}{2}m\dot{z}-mgz$

Newton’s second law of motion can be represented in the form of Lagrange’s equation of motion as shown below.

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{z}}\right)=\frac{\partial L}{\partial z}$

The Lagrange equation of motion for this system is,

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{z}}\right)-\frac{\partial L}{\partial z}=0$

The partial derivative of L with respect to first derivative of z and partial derivative of L with respect to z is shown below.

$\begin{array}{l}\frac{\partial L}{\partial \dot{z}}=\frac{1}{2}m\\ \frac{\partial L}{\partial z}=-mg\end{array}$

The Lagrange equation of motion can be rewritten using the partial derivative of L as,

$\frac{d}{dt}m-\left(-mg\right)=0$

This equation can be rearranged as shown below.

$mg=0$

Conclusion

The Lagrangian function L for the given one dimensional motion in the z direction is,

$L=\frac{1}{2}m\dot{z}-mgz$

The Lagrange equation of motion for this system is,

$mg=0$