July 21, 2003

PHYSICS 44 MECHANICS

Homework Assignment II

SOLUTION

Problem 1

A cart of mass M is placed on rails and attached to a wall with the help of a masslessspring with constant k (as shown in the Figure below); the spring is in its equilibrium statewhen the cart is at a distance x0 from the wall. A pendulum of mass m and length ` isattached to the cart (as shown).

(a) Write the Lagrangian L(x; _x; µ; _µ) for the cart-pendulum system, where x denotes theposition of the cart (as measured from a suitable origin) and µ denotes the angular positionof the pendulum.

(b) From your Lagrangian, write the Euler-Lagrange equations for the generalized coordi-nates x and µ.

Solution

(a) Using the generalized coordinates x (the displacement of the cart from the equilibriumpoint of the spring) and µ (the displacement of the pendulum from the vertical) shown inthe Figure, the coordinates of the cart are xM = x and yM = 0 while the coordinates ofthe pendulum are xm = x + ` sin µ and ym = ¡ ` cos µ and thus the squared velocities are

v2M = _x2M + _y2M = _x2

v2m = _x2m + _y2m = ( _x + ` _µ cosµ)2 + (` _µ sin µ)2

= _x2 + `2 _µ2 + 2 ` _x _µ cos µ:

The expression for the kinetic energy of the cart-pendulum system is therefore

K = (m + M )_x2

2+ m`2

_µ2

2+ m ` _x _µ cos µ:

The potential energy U of the cart-pendulum system is broken into two parts: thegravitational potential energy ¡mg` cosµ of the pendulum and the elastic potential energyk x2=2 stored in the spring. The Lagrangian L = K ¡ U of the cart-pendulum system istherefore

L(x; _x; µ; _µ) = (m + M)_x2

2+ m `2

_µ2

2+ m ` cosµ

³_x _µ + g

´¡ k

2x2:

(b) The Euler-Lagrange equation for x is

@L

@ _x= (m+M) _x + m` _µ cosµ !

d

dt

Ã@L

@ _x

!= (m+M) Äx + m`

³Äµ cos µ ¡ _µ2 sin µ

´@L

@x= ¡ k x

or

(m+M ) Äx + m`³Äµ cos µ ¡ _µ2 sin µ

´+ k x = 0

The Euler-Lagrange equation for µ is

@L

@ _µ= m`

³` _µ + _x cos µ

´!

d

dt

Ã@L

@ _µ

!= m`

³` ĵ + Äx cos µ ¡ _x _µ sin µ

´@L

@µ= ¡m ` _x _µ sin µ ¡ mg` sin µ

or

` ĵ + Äx cos µ + g sin µ = 0

Problem 2

Show that the two Lagrangians

L(q; _q; t) and L0(q; _q; t) = L(q; _q; t) +dF (q; t)

dt;

where F (q; t) is an arbitrary function of the generalized coordinates q(t), yield the sameEuler-Lagrange equations. Hence, two Lagrangians which di®er only by an exact timederivative are said to be equivalent.

Solution

We call L0 = L + dF=dt the new Lagrangian and L the old Lagrangian. The Euler-Lagrange equations for the new Lagrangian are

d

dt

Ã@L0

@ _qi

!=@L0

@qi;

wheredF (q; t)

dt=@F

@t+

Xj

_qj@F

@qj:

Let us begin with

@L0

@ _qi=

@

@ _qi

0@L +@F

@t+

Xj

_qj@F

@qj

1A =@L

@ _qi+@F

@qi;

so thatd

dt

Ã@L0

@ _qi

!=

d

dt

Ã@L

@ _qi

!+

@2F

@t@qi+

Xk

_qk@2F

@qk@qi:

Next, we ¯nd

@L0

@qi=

@

@qi

0@L + @F

@t+

Xj

_qj@F

@qj

1A =@L

@qi+

@2F

@qi@t+

Xj

_qj@2F

@qi@qj:

Using the symmetry properties

_qj@2F

@qi@qj= _qj

@2F

@qj@qiand

@2F

@t@qi=

@2F

@qi@t;

we easily verifyd

dt

Ã@L0

@ _qi

!¡ @L0

@qi=

d

dt

Ã@L

@ _qi

!¡ @L

@qi= 0;

and thus since L and L0 = L+ dF=dt lead to the same Euler-Lagrange equations, they aresaid to be equivalent.

Problem 3

An Atwood machine is composed of two masses m and M attached by means of amassless rope into which a massless spring (with constant k) is inserted (as shown in theFigure below). When the spring is in a relaxed state, the spring-rope length is `.

(a) Find suitable generalized coordinates to describe the motion of the two masses (allowingfor elongation or compression of the spring).

(b) Using these generalized coordinates, construct the Lagrangian and derive the appropri-ate Euler-Lagrange equations.

Solution

(a) The Figure below shows a suitable set of generalized coordinates

where x denotes the distance of mass M from the top of the pulley, ` ¡ x denotes thedistance of the equilibrium point of the spring from the top of the pulley, and y denotesthe distance of mass m from the equilibrium point of the spring (i.e., its elongation).

(b) Using the generalized coordinates (x; y), the coordinate of mass M is xM = x while thecoordinate of mass m is xm = `¡ x+ y and thus the squared velocities are

v2M = _x2 and v2m = ( _y ¡ _x)2 :

The expression for the kinetic energy of the system is therefore

K = (m + M )_x2

2+ m

_y2

2¡ m _x _y:

The potential energy U of the system is broken into two parts: the gravitational poten-tial energy ¡Mgx¡ mg(y¡x) and the elastic potential energy k y2=2 stored in the spring.The Lagrangian L =K ¡ U of the system is therefore

L(x; _x; y; _y) = (m + M )_x2

2+ m

_y2

2¡ m _x _y + (M ¡m)g x + mg y ¡ k

2y2:

(c) The Euler-Lagrange equation for x is

@L

@ _x= (m+M) _x ¡ m _y ! d

dt

Ã@L

@ _x

!= (m+M) Äx ¡ m Äy

@L

@x= (M ¡m) g

or

(m+M) Äx ¡ m Äy = (M ¡m) g

The Euler-Lagrange equation for y is

@L

@ _y= m ( _y ¡ _x) ! d

dt

Ã@L

@ _y

!= m (Äy ¡ Äx)

@L

@x= m g ¡ k y

or

m (Äy ¡ Äx) + k y = m g