- #1

Rick16

- 81

- 25

- 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)

md

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 md

Could someone help me see clearer? Thanks ahead.

^{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 motionmd

^{2}s/dt^{2}= -mgsin(b). I am bothered by the minus sign in front of the right-hand term. The Newtonian method leads to md^{2}s/dt^{2}= +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 md

^{2}s/dt^{2}= +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 md^{2}s/dt^{2}= +mgsin(b), but I can't see how the potential energy could be negative.Could someone help me see clearer? Thanks ahead.