problema resuelto de mecanica

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August 14, 2003

PHYSICS 44 MECHANICS

Homework Assignment VI

SOLUTIONS

Problem 1

(a) Consider the case involving motion on the (x; y)-plane perpendicular to the angularvelocity vector ! = ! bz with the potential energy

U (r) = 1

2 k

³x2 + y2

´:

Using the Euler-Lagrange equations

m Är = ¡ rU (r) ¡ m [ 2!£ _r + !£ (!£ r) ] ;

derive the equations of motion for x and y.

(b) By using the equations of motion derived in Part (a), show that the canonical angularmomentum ` = bz ¢ (r£p) is a constant of the motion.

Solution

(a) Using the vector identity j!£ rj2 = !2 r2 ¡ (! ¢ r)2, we ¯nd

@L

@ _r = m (_r + !£ r) ! d

dt

Ã@L

@ _r

! = m (Är + !£ _r)

@L

@ r = ¡ rU (r) + m

h !2 r ¡ ! (! ¢r)

i + m _r£!

or, using the identity ! 2 r¡! (! ¢ r) = ¡ !£ (!£ r), the Euler-Lagrange equation is

m Är = ¡ rU ¡ !£ (!£ r) ¡ 2m !£ _r: (1.1)

We now consider two-dimensional motion under the in°uence of the e®ective potential

U eff (x;y) = k

2

³x2 + y2

´ ¡ m !2

2

³r2¡ z 2

´ =

K

2

³x2 + y2

´;

where K = k ¡ m !2. The Lagrangian in this case is

L(x; _x; y; _y) = m

2

³_x2 + _y2

´ + m! (x _y ¡ y _x) ¡ K

2

³x2 + y2

´: (1.2)

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The Euler-Lagrange equations are

m Äx = 2 m! _y ¡ K x; (1.3)

m Äy = ¡ 2 m! _x ¡ K y: (1.4)

(b) The canonical momentum p for this problem is de¯ned as

p = @L

@ _r = m ( _r + !£ r) ;

so that the canonical angular momentum ` = bz ¢ r£p is expressed as

` = bz ¢ r£ (m _r + m!£r) = m (x _y ¡ y _x) + m!³

x2 + y2´

:

Using Eqs. (1.3) and (1.4), we ¯nd

d`

dt = m (x Äy

¡y Äx) + 2 m! (x _x + y _y) = 0:

That the canonical angular momentum ` is a constant of the motion can also be seen bytransforming the Lagrangian (1.2) in Cartesian coordinates (x; y) into the Lagrangian

L(r; _r; _') = m

2

³_r2 + r2 _'2

´ + m! r2 _' ¡ K

2 r2; (1.5)

in polar coordinates (r; '), where x = r cos ' and y = r sin '. Since the Lagrangian (1.5)is independent of the polar angle ' , the canonical angular momentum

p' = @L

@ _' = mr2 ( _' + !) (1.6)

is a constant of the motion. It is easy to verify that ` = p'.

Problem 2

If a particle is projected vertically upward to a height h above a point on the Earth'ssurface at a northern latitude ¸, show that it strikes the ground at a point

4!

3 cos ¸

s 8 h3

g

to the west. (Neglect air resistance, and consider only small vertical heights.)

Solution

The Coriolis acceleration equations are

Äx = 2 ! sin ¸ _yÄy = ¡ 2 ! (sin ¸ _x + cos ¸ _z )Äz = ¡ g + 2 ! cos ¸ _y

9>=>; : (2.1)

2



A ¯rst integration of Eq. (2.1) yields

_x = 2 ! sin ¸ y + C x

_y = ¡ 2 ! (sin ¸ x + cos ¸ z ) + C y_z = ¡ g t + 2 ! cos ¸ y + C z9>=>; ; (2.2)

where (C x; C y; C z) are constants de¯ned from initial conditions (x0; y0; z 0) and (_x0; _y0; _z 0).

Here, the initial conditions are (x0; y0; z 0) = (0; 0; 0) and (_x0; _y0; _z 0) = (0; 0; v0). Thus,the constants are (C x; C y; C z) = (0; 0; v0) and integration of Eqs. (2.2) yields

x(t) = 2 ! sin ¸Z t

0

y(¿ ) d¿; (2.3)

y(t) = ¡ 2 ! sin ¸Z t0

x(¿ ) d¿ ¡ 2 ! cos ¸Z t0

z (¿ ) d¿; (2.4)

z (t) = v0 t ¡ g t2

2 + 2 ! cos ¸

Z t0

y(¿ ) d¿: (2.5)

To ¯rst order in !, the equations (2.3)-(2.5) become

x(t) = 0; (2.6)

y(t) = ¡ 2 ! cos ¸Z t0

µv0 ¿ ¡ g

2 ¿ 2¶

d¿ = ¡ ! cos ¸µ

v0 t2 ¡ g

3 t3¶

; (2.7)

z (t) = v0 t ¡ g t2

2: (2.8)

The projectile reaches its maximum height h after a time T = v0=g and thus v0 = p 2 hg.

After a time 2T =q

8h=g, the projectile is back on the ground and has drifted by

y(2T ) = ¡ ! cos ¸µ

4 v0 T 2 ¡ 8g

3 T 3

¶ = ¡ 4

3 ! cos ¸

v30

g2 = ¡ 4

3 ! cos ¸

s 8 h3

g :

Since y(2T ) < 0, the drift is westward .

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problema resuelto de mecanica

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