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A simple question about coordinates

  • #1

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([tex]\dot{x}[/tex]^2+[tex]\dot{y}[/tex]^2)
and using y=acosh(x/a) I get the following expression
0.5m[tex]\dot{x}[/tex]^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[tex]\dot{l}[/tex]^2=0.5m[tex]\dot{x}[/tex]^2(cosh(x/a)^2
but what is the expression for the potential energy here? how do I express h?
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
cosh^2(x/a) - sinh^2(x/a) = 1
 
  • #3
got it, thanks.
 

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