July 21, 2003

PHYSICS 44 MECHANICS

Homework Assignment III

SOLUTION

Problem 1

A block of mass m rests on the inclined plane (with angle µ) of a triangular block ofmass M as shown in the Figure below. Here, we consider the case where both blocks slidewithout friction (i.e., m slides on the inclined plane without friction andM slides withoutfriction on the horizontal plane).

(a) Using the generalized coordinates (x; y) shown in the Figure above, construct the La-grangian L(x; _x; y; _y).

(b) Derive the Euler-Lagrange equations for x and y.

(c) Calculate the canonical momenta

px(x; _x; y; _y) =@L

@ _xand py(x; _x; y; _y) =

@L

@ _y;

and invert these expressions to ¯nd the functions _x(x; px; y; py) and _y(x; px; y; py).

(d) Calculate the Hamiltonian H(x; px; y; py) for this system by using the Legendre trans-formation

H(x; px; y; py) = px _x + py _y ¡ L(x; _x; y; _y);

where the functions _x(x; px; y; py) and _y(x; px; y; py) are used.

(e) Find which of the two momenta found in Part (c) is a constant of the motion anddiscuss why it is so. If the two blocks start from rest, what is the value of this constant ofmotion?

Solution

(a) The total kinetic energy of the system is

K = (m+M)_x2

2+ m

_y2

2+ m cos µ _x _y;

while the gravitational potential energy is

U = ¡ mg sin µ y;

where the zero potential energy point is located at the top of the inclined plane. TheLagrangian L =K ¡ U is therefore

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

2+ m

_y2

2+ m cos µ _x _y + mg sin µ y:

(b) The Euler-Lagrange equation for x is

@L

@ _x= (m+M) _x + m cos µ _y ! d

dt

Ã@L

@ _x

!= (m+M) Äx + m cosµ Äy

@L

@x= 0

or

(m+M ) Äx + m cos µ Äy = 0

The Euler-Lagrange equation for y is

@L

@ _y= m ( _y + cosµ _x) ! d

dt

Ã@L

@ _y

!= m (Äy + cos µ Äx)

@L

@x= mg sin µ

or

Äy + cos µ Äx = g sin µ

(c) The canonical momenta are

px =@L

@ _x= (m+M ) _x + m cos µ _y

py =@L

@ _y= m ( _y + cos µ _x) ;

which can be expressed in matrix form as0B@ px

py

1CA =

0B@ (m+M ) m cos µ

m cos µ m

1CA ¢0B@ _x

_y

1CA ! p = M ¢ _r;

where M denotes the mass matrix. The determinant of the mass matrix is

detM = m (m+M) ¡ m2 cos2 µ = m³M + m sin2 µ

´= ¢;

and we note that ¢ never vanishes and thus the Lagrangian of the system is regular.

The inversion p! _r is expressed in matrix form as0B@ _x

_y

1CA =1

¢

0B@ m ¡m cos µ

¡ m cos µ (m+M )

1CA ¢0B@ px

py

1CA ! _r = M¡1 ¢ p

or

_x =px ¡ cos µ pyM +m sin2 µ

and _y =(m+M )py ¡m cos µ pxm (M +m sin2 µ)

:

(d) The Hamiltonian function is

H(r;p) =1

2pT ¢ M¡1 ¢ p + U(r)

=1

2¢

hm p2x + (m+M ) p2y ¡ 2m cosµ px py

i¡ mg sin µ y:

(e) Since the Lagrangian is independent of the coordinate x, the canonical momentum

px = (m+M ) _x + m cos µ _y

is a constant of the motion (i.e., _px = 0). If the two blocks are initially at rest, we¯nd px = 0 and thus as the rectangular block starts to slide down the inclined plane thetriangular block starts to move in the opposite direction.

The equations of motion in this case become

Äy =(m+M) g sin µ

M +m sin2 µand Äx = ¡ m cos µ Äy

m+M= ¡ m g sin µ cos µ

M +m sin2 µ: