- #1
Rick16
- 75
- 24
- TL;DR Summary
- getting the sign right for the potential energy in the Lagrangian
I want to use the Lagrangian approach to find the equation of motion for a mass sliding down a frictionless inclined plane. I call the length of the incline a and the angle that the incline makes with the horizontal b. Then the mass has kinetic energy 1/2m(da/dt)2 and the potential energy should be -- as I see it -- mgh = mgasin(b). So the Lagrangian is 1/2m(da/dt)2 - mgasin(b), which leads to the equation of motion
md2s/dt2 = -mgsin(b). I am bothered by the minus sign in front of the right-hand term. The Newtonian method leads to md2s/dt2 = +mgsin(b).
I searched the internet and looked at 3 solutions to this problem, none of which was helpful. The first internet expert sets up the Lagrangian the way I do, he also gets the same equation of motion, and he does not see anything wrong with it. The second expert also sets up the Lagrangian the way I do, but he makes a mistake solving the Lagrange equation and ends up with md2s/dt2 = +mgsin(b). The third expert simply writes down the potential energy as -mgh, without any further comment. With a negative potential energy he naturally gets the solution md2s/dt2 = +mgsin(b), but I can't see how the potential energy could be negative.
Could someone help me see clearer? Thanks ahead.
md2s/dt2 = -mgsin(b). I am bothered by the minus sign in front of the right-hand term. The Newtonian method leads to md2s/dt2 = +mgsin(b).
I searched the internet and looked at 3 solutions to this problem, none of which was helpful. The first internet expert sets up the Lagrangian the way I do, he also gets the same equation of motion, and he does not see anything wrong with it. The second expert also sets up the Lagrangian the way I do, but he makes a mistake solving the Lagrange equation and ends up with md2s/dt2 = +mgsin(b). The third expert simply writes down the potential energy as -mgh, without any further comment. With a negative potential energy he naturally gets the solution md2s/dt2 = +mgsin(b), but I can't see how the potential energy could be negative.
Could someone help me see clearer? Thanks ahead.