- #1
Rob Hal
- 13
- 0
Hi,
I posted this yesterday, but I figured I'd better post it again since the other thread can't be replied to at this time since its in the archives.
I'm looking for some advice on whether or not I'm doing a problem correctly.
The problem is:
A particle of mass m rests on a smooth plane. (the particle starts at r) The plane is raised to an inclination [tex]\theta[/tex], at a constant rate [tex]\alpha[/tex], with [tex]\theta = 0[/tex] at t=0, causing the particle to move down the plane.
So, I'm taking the x to be the distance the particle travels down the slope.
I come up with the following as the Lagrangian:
[tex]L = \frac{1}{2} m\dot{x}^2 - mg(r-x)sin\theta[/tex]
I'm not sure if this is correct.
I would then get the equations of motion to be [tex]mgsin\theta - m\ddot{x}=0[/tex] and [tex]-mgsin(r-x)cos\theta=0[/tex].
I posted this yesterday, but I figured I'd better post it again since the other thread can't be replied to at this time since its in the archives.
I'm looking for some advice on whether or not I'm doing a problem correctly.
The problem is:
A particle of mass m rests on a smooth plane. (the particle starts at r) The plane is raised to an inclination [tex]\theta[/tex], at a constant rate [tex]\alpha[/tex], with [tex]\theta = 0[/tex] at t=0, causing the particle to move down the plane.
So, I'm taking the x to be the distance the particle travels down the slope.
I come up with the following as the Lagrangian:
[tex]L = \frac{1}{2} m\dot{x}^2 - mg(r-x)sin\theta[/tex]
I'm not sure if this is correct.
I would then get the equations of motion to be [tex]mgsin\theta - m\ddot{x}=0[/tex] and [tex]-mgsin(r-x)cos\theta=0[/tex].