Chapter 7, Problem 7.1P

To determine

The lagrangian for a projectile.

Answer to Problem 7.1P

The three lagrangian for a projectile are, $m\ddot{x}=0$, $m\ddot{y}=0$, $\ddot{z}=-g$

Explanation of Solution

The lagrangian of the projectile system is,

  $L=T-U$        (I)

Here, $T$ is the kinetic energy, $U$ is the potential energy of the system.

For this projectile the kinetic energy is,

  $T=\frac{1}{2}m{\dot{r}}^2$        (II)

Here, $m$ is the mass of the projectile and $\dot{r}$ is the velocity of the projectile.

The velocity of the projectile in terms of Cartesian coordinates is,

  $\dot{r}=\sqrt{{\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2}$        (III)

Substitute the expression (III) in (II)

  $T=\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)$        (IV)

The potential energy of the projectile is,

  $U=mgz$        (V)

Here, $g$ is the acceleration due to gravity and the gravity has no effect along $x$ and $y$ directions.

Substitute the expression (IV) and (V) in expression (I)

  $L=\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)-mgz$        (VI)

The lagrangian equation for the projectile along $x$ axis is,

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

Calculate $\frac{\partial L}{\partial x}$ and $\frac{\partial L}{\partial \dot{x}}$

  $\begin{array}{c}\frac{\partial L}{\partial x}=\frac{\partial }{\partial x}\left(\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)-mgz\right)\\ =0\\ \frac{\partial L}{\partial \dot{x}}=\frac{\partial }{\partial \dot{x}}\left(\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)-mgz\right)\\ =\frac{1}{2}m2\dot{x}\\ =m\dot{x}\end{array}$        (VIII)

Calculate $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$

  $\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=\frac{d}{dt}m\dot{x}\\ =m\ddot{x}\end{array}$

Conclusion:

Substitute $0$ for $\frac{\partial L}{\partial x}$ and $m\ddot{x}$ for $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$ in expression (VII)

  $m\ddot{x}=0$        (IX)

Therefore, the first lagrangian equation for the projectile is $m\ddot{x}=0$

The lagrangian equation for the projectile along $y$ axis is,

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

Calculate $\frac{\partial L}{\partial y}$ and $\frac{\partial L}{\partial \dot{y}}$

  $\begin{array}{c}\frac{\partial L}{\partial y}=\frac{\partial }{\partial y}\left(\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)-mgz\right)\\ =0\\ \frac{\partial L}{\partial \dot{y}}=\frac{\partial }{\partial \dot{y}}\left(\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)-mgz\right)\\ =\frac{1}{2}m2\dot{y}\\ =m\dot{y}\end{array}$

Calculate $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)$

  $\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)=\frac{d}{dt}m\dot{y}\\ =m\ddot{y}\end{array}$

Substitute $0$ for $\frac{\partial L}{\partial y}$ and $m\ddot{y}$ for $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)$ in expression (X)

  $m\ddot{y}=0$        (XI)

Therefore, the second lagrangian equation for the projectile is $m\ddot{y}=0$

The lagrangian equation for the projectile along $z$ axis is,

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

Calculate $\frac{\partial L}{\partial z}$ and $\frac{\partial L}{\partial \dot{z}}$

  $\begin{array}{c}\frac{\partial L}{\partial z}=\frac{\partial }{\partial z}\left(\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)-mgz\right)\\ =-mg\\ \frac{\partial L}{\partial \dot{z}}=\frac{\partial }{\partial \dot{z}}\left(\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right)-mgz\right)\\ =\frac{1}{2}m2\dot{z}\\ =m\ddot{z}\end{array}$

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

  $\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{z}}\right)=\frac{d}{dt}m\dot{z}\\ =m\ddot{z}\end{array}$

Substitute $-mg$ for $\frac{\partial L}{\partial z}$ and $m\ddot{z}$ for $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{z}}\right)$ in expression (XII)

  $\begin{array}{c}-mg=m\ddot{z}\\ \ddot{z}=-g\end{array}$        (XIII)

Therefore, the third lagrangian equation for the projectile is $\ddot{z}=-g$