Is My Lagrangian Setup for a Particle on an Inclined Plane Correct?

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SUMMARY

The discussion centers on the Lagrangian formulation for a particle on an inclined plane. The proposed Lagrangian is L = (1/2) m˙x² - mg(r-x)sin(θ). However, it is concluded that the kinetic energy (KE) term is incorrect as it should include two terms, one of which must incorporate the constant rate of inclination, α. The potential energy (PE) term is deemed acceptable, leading to the equations of motion mgsin(θ) - m&ddot;x = 0 and -mgsin(r-x)cos(θ) = 0.

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Rob Hal
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Hi,

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].
 
Last edited:
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Sorry for the late reply. In case you're still interested, here's my response to this question.

Rob Hal said:
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]

The KE term isn't right. It should have 2 terms, and one should contain an [itex]\alpha[/itex]. The PE term is OK.
 

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