Question

A bead of mass m slides along a frictionless wire under the influence of gravity. The shape of the wire is given by the equation y = axa, where x denotes the horizontal co-ordinate, y denotes the vertical co-ordinate, and a is a constant. (a) Use Lagrange's equation to determine the equation of motion. (b) Compute Hamilton's equations of motion and show that they are equivalent to your result for item (a).

Answer 1

KE of bead = 1/2 m ( ( dx/dt )² + ( dy/dt )² )

= 1/2 m ( dx/dt )² ( 1 + 4 a² x² )

PE of bead = m g y

= m g a x²

a )

L = 1/2 m ( dx/dt )² ( 1 + 4 a² x² ) - m g y

d($\partial$L/$\partial$x) / dt - $\partial$L/$\partial$x = 0

d($\partial$(1/2 m ( dx/dt )² ( 1 + 4 a² x² ) - m g y )/$\partial$x) / dt - $\partial$(1/2 m ( dx/dt )² (1+4a²x² )-m g y )/$\partial$x = 0

m ( 1 + 4 a² x² ) d²x/dt² + 4 m a² x (dx/dt)² + 2 m g a x = 0 ------- 1

b )

the hamiltonion equation

H = dx/dt P_x - L

P_x = $\partial$L/$\partial$x

= m (dx/dt) ( 1 + 4 a² x² )

H = ( P_x² / m ( 1 + 4 a² x² ) - 2 P_x² / m ( 1 + 4 a² x² ) + m g a x²

dP_x/dt = - $\partial$H/$\partial$x

= 4 a² x P_x² / m ( 1 + 4 a² x² )² - 2m g a x

(dx/dt) = $\partial$H/$\partial$P_x

= P_x / m ( ( 1 + 4 a² x² )

( ( 1 + 4 a² x² ) ( dx/dt ) = P_x  

m ( 8 a² x (dx/dt) ) dx/dt + m ( 1 + 4 a² x² ) d²x/dt²= dP_x/dt

m ( 1 + 4 a² x² ) d²x/dt² + 4 m a² x (dx/dt)² + 2 m g a x = 0 ------- 2

equation 1 and 2 are equal.