Mass sliding down incline (incline is able to move)

[Sefrez]

Homework Statement

There is a frictionless incline with a "block" mass on it. The incline and the mass are able to move. The mass of the block is m and the mass of the incline is M.

I am supposed to find a suitable set of generalized coordinates and write the Lagrangian of the system, then from that, determine equations of motion.

2. The attempt at a solution

I chose coordinates in a cartesian frame to describe the position of the block mass. (x, y)
From this, I constrained the incline so that it is "in contact" with the block mass at all times (assuming the block to be a point mass):
(x2,y2) = (x1 + y1/tan(θ), 0)

where this is measured from the end of the inclination at some x and y = 0. θ is the angle from the horizontal.

I then constructed the Lagrange function:
L = (1/2)m((d/dt x)^2 + (d/dt y)^2) + (1/2)M(d/dt(x1 + y1/tan(θ)))^2 - mgy

where the last term is the potential of the block measured from the zero potential at y = 0. The inclination should have no potential term as it never moves along the direction of the conservative field, gravity (its height is constant.)

I computed the equations of motion, and they are incorrect. What am I doing wrong? Given (x, y) of the block mass, the position of the incline is fully defined and so it seems I have fully described the system. Any help is appreciated, this problem is driving me nuts!

 

[TSny]

In order to check your expressions, I think it would help if you described the geometry in more detail. What is the orientation of the inclined plane? Does it slope down to the right or down to the left? What point of the incline do the coordinates (x2, y2) refer to?

 

[haruspex]

I'm not promising to look into it, but it might help if you post your subsequent working. Maybe the problem lies therein.

 

[Sefrez]

In order to check your expressions, I think it would help if you described the geometry in more detail. What is the orientation of the inclined plane? Does it slope down to the right or down to the left? What point of the incline do the coordinates (x2, y2) refer to?

Yes, I see I was a little vague. I have attached a figure. The red represents the incline; the blue dot represents the block mass (taking it to be a point); and the black dot is the lower end of the ramp.

P is the coordinate points for the block, and Q is the coordinate points for the incline. I must either be missing some constraints, or improperly created these. Certainly it seems given P(a,b), we know Q(c,d). Is this not enough?

I'm not promising to look into it, but it might help if you post your subsequent working. Maybe the problem lies therein.

I will do this if the figure I attached is not enough to see my problem.

Thanks.

 

[TSny]

I chose coordinates in a cartesian frame to describe the position of the block mass. (x, y)
From this, I constrained the incline so that it is "in contact" with the block mass at all times (assuming the block to be a point mass):
(x2,y2) = (x1 + y1/tan(θ), 0)

where this is measured from the end of the inclination at some x and y = 0. θ is the angle from the horizontal.

I then constructed the Lagrange function:
L = (1/2)m((d/dt x)^2 + (d/dt y)^2) + (1/2)M(d/dt(x1 + y1/tan(θ)))^2 - mgy

where the last term is the potential of the block measured from the zero potential at y = 0. The inclination should have no potential term as it never moves along the direction of the conservative field, gravity (its height is constant.)

OK. Your Lagrangian looks good to me. It seems to yield correct equations of motion. So, as haruspex said, maybe the problem lies in working out the equations of motion from L. Can you show some of the work or at least state what you got for the equations of motion for x and y?

 

[Sefrez]

Your informing that my equations were correct made me further investigate what I was doing wrong. I come to find out I did nothing wrong, at least in arriving to my solutions.

I was comparing my solutions to those given here (at the bottom of the page):
http://farside.ph.utexas.edu/teaching/336k/Newton/node80.html

What I was doing wrong was assuming that d2/dt2(x') given here was the acceleration of the particle from an inertial frame of reference. Its measured from the top of the incline!

I was driving myself crazy for nothing. :)

Thanks people.