Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
Question
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Chapter 7, Problem 7.1P
To determine

The lagrangian for a projectile.

Expert Solution & Answer
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Answer to Problem 7.1P

The three lagrangian for a projectile are, mx¨=0, my¨=0, z¨=g

Explanation of Solution

Classical Mechanics, Chapter 7, Problem 7.1P

The lagrangian of the projectile system is,

  L=TU        (I)

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

For this projectile the kinetic energy is,

  T=12mr˙2        (II)

Here, m is the mass of the projectile and r˙ is the velocity of the projectile.

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

  r˙=x˙2+y˙2+z˙2        (III)

Substitute the expression (III) in (II)

  T=12m(x˙2+y˙2+z˙2)        (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=12m(x˙2+y˙2+z˙2)mgz        (VI)

The lagrangian equation for the projectile along x axis is,

  Lx=ddt(Lx˙)        (VII)

Calculate Lx and Lx˙

  Lx=x(12m(x˙2+y˙2+z˙2)mgz)=0Lx˙=x˙(12m(x˙2+y˙2+z˙2)mgz)=12m2x˙=mx˙        (VIII)

Calculate ddt(Lx˙)

  ddt(Lx˙)=ddtmx˙=mx¨

Conclusion:

Substitute 0 for Lx and mx¨ for ddt(Lx˙) in expression (VII)

  mx¨=0        (IX)

Therefore, the first lagrangian equation for the projectile is mx¨=0

The lagrangian equation for the projectile along y axis is,

  Ly=ddt(Ly˙)        (X)

Calculate Ly and Ly˙

  Ly=y(12m(x˙2+y˙2+z˙2)mgz)=0Ly˙=y˙(12m(x˙2+y˙2+z˙2)mgz)=12m2y˙=my˙

Calculate ddt(Ly˙)

  ddt(Ly˙)=ddtmy˙=my¨

Substitute 0 for Ly and my¨ for ddt(Ly˙) in expression (X)

  my¨=0        (XI)

Therefore, the second lagrangian equation for the projectile is my¨=0

The lagrangian equation for the projectile along z axis is,

  Lz=ddt(Lz˙)        (XII)

Calculate Lz and Lz˙

  Lz=z(12m(x˙2+y˙2+z˙2)mgz)=mgLz˙=z˙(12m(x˙2+y˙2+z˙2)mgz)=12m2z˙=mz¨

Calculate ddt(Lz˙)

  ddt(Lz˙)=ddtmz˙=mz¨

Substitute mg for Lz and mz¨ for ddt(Lz˙) in expression (XII)

  mg=mz¨z¨=g        (XIII)

Therefore, the third lagrangian equation for the projectile is z¨=g

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