How Do You Calculate Potential Energy Along a Curved Wire?

[maria clara]

Homework Statement

a bead is constrained to move on a wire: y=acosh(x/a).
write the kinetic energy and the potential energy using the coordinates:
1) x
2) l - the length of a curve

2. The attempt at a solution

1) is simple:
the kinetic energy in cartesian coordinates is 0.5($$\dot{x}$$^2+$$\dot{y}$$^2)
and using y=acosh(x/a) I get the following expression
0.5m$$\dot{x}$$^2(cosh(x/a)^2
the potential energy is mgh = mgy=mgcosh(x/a).
2) l=integral(dl)=integral[0-->x]{sqrt[(1+(dy/dx)^2)]dx}=asinh(x/a).
so the kinetic energy is 0.5m$$\dot{l}$$^2=0.5m$$\dot{x}$$^2(cosh(x/a)^2
but what is the expression for the potential energy here? how do I express h?

 

[George Jones]

cosh^2(x/a) - sinh^2(x/a) = 1

 

[maria clara]

got it, thanks.