Lagrange equations of motion for hoop rolling down moving ramp.

In summary, Homework Statement states that a hoop of mass M and radius R rolls without slipping down an inclined plane of mass M, which makes an angle \alpha with the horizontal. The Lagrange equations and the integrals of the motion are found if the plane can slide without friction along a horizontal surface. If the hoop is on a non-moving ramp as a reference, the equations are: T = 1/2 M \dot{x}^{2} + 1/2 m \dot{s}^{2} + 1/2 m r^{2} \dot{\theta}^{2}. U = m g s sin\alpha.
  • #1
Strukus
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Homework Statement


A hoop of mass m and radius R rolls without slipping down an inclined plane of mass M, which makes an angle [tex]\alpha[/tex] with the horizontal. Find the Lagrange equations and the integrals of the motion if the plane can slide without friction along a horizontal surface.

Homework Equations


L = T - U

T = kinetic energy
U = potential energy

Inertia tensor of hoop = mr[tex]^{2}[/tex]

The Attempt at a Solution


Here's what I believe to be the situation.
homework3-1.jpg


I know that I have to break up the kinetic energy into its separate components, but I'm not entirely sure how. I have 2 examples to work with, one being a block sliding down a moving ramp and the other being a hoop rolling down a stationary ramp.

Here's the block.

homework3-1b.jpg


Here's the hoop.
homework3-1a.jpg


With the block, we break it down into velocity along the x-axis and the y-axis for both the block and the ramp (with the ramp's y-velocity being 0). But with the hoop, we break up the velocity into an x'-axis (axis at the same angle as the ramp) and a [tex]\theta[/tex]. Both of the examples make sense to me, though combining them in this way leaves me a little confused.

If I were to just solve with what I knew now, I would combine the rotational velocity, the velocity along the x-axis, and the velocity along the y-axis into the kinetic energy.
 
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  • #2
When you do Lagrangian problems, a good starting point is to decide on your generalized coordinates and then write any constraints relating them. So what are your generalized coordinates and what are your constraints?
 
  • #3
I'm going to use the hoop on the non-moving ramp as my reference for this.

It has the equation of constraint being what describes the rolling without slipping, so
f = dx' - r d[tex]\theta[/tex] = 0
f = [tex]\dot{x}[/tex]' - r [tex]\dot{\theta}[/tex] = 0

And y-ramp = 0.

And the generalized coordinates are x' and [tex]\theta[/tex].

That makes sense to me, but since the ramp is moving I'll have to describe the x-axis movement in terms of the x'-axis movement or describe the x'-axis movement in terms of the x-axis and y-axis, which is where I'm getting stuck.

And I'm not sure why but whenever I preview the post it's not what I typed (in terms of the greek letters).
 
  • #4
It is better to use x for the horizontal position of the ramp (more conventional) and s for the coordinate down the plane, relative to the plane. Keep angle θ as you have it. Can you write expressions for the kinetic energy T and the potential energy U in terms of these coordinates?
 
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  • #5
Wow, that really helps. I forget what example it was when I was taking notes in class but I forgot with this approach you can kind of just use whichever axes you wish and it will still work.

That being said, this is what I have come up with.

T = 1/2 M [tex]\dot{x}[/tex][tex]^{2}[/tex] + 1/2 m [tex]\dot{s}[/tex][tex]^{2}[/tex] + 1/2 m r[tex]^{2}[/tex] [tex]\dot{\theta}[/tex][tex]^{2}[/tex]
U = m g s sin[tex]\alpha[/tex]
 
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  • #6
After much working it through, looking at sample problems, and asking for help, I finished it! Thank you for your help kuruman!
 
  • #7
I hope you wrote the translational kinetic energy of the center of mass of the rolling hoop correctly. It is not simply

[tex]\frac{1}{2}m\dot{s}^2[/tex]

It is

[tex]\frac{1}{2}m(\dot{s}cos\alpha+\dot{x})^2+\frac{1}{2}m(\dot{s}sin\alpha)^2[/tex]

:wink:
 
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1. What are Lagrange equations of motion for hoop rolling down a moving ramp?

The Lagrange equations of motion for a hoop rolling down a moving ramp are mathematical equations used to describe the motion of a hoop as it rolls down a ramp that is itself in motion. These equations take into account the rotational and translational motion of the hoop, as well as the motion of the ramp.

2. What are the variables involved in the Lagrange equations of motion for hoop rolling down a moving ramp?

The variables involved in the Lagrange equations of motion for hoop rolling down a moving ramp include the mass and radius of the hoop, the mass and angle of the ramp, the acceleration due to gravity, and the velocity of the ramp. These variables are used to calculate the forces and energy involved in the motion of the hoop.

3. How are the Lagrange equations of motion for hoop rolling down a moving ramp derived?

The Lagrange equations of motion for hoop rolling down a moving ramp are derived using the principles of classical mechanics, specifically the Lagrangian approach. This approach involves defining the kinetic and potential energy of the system and using the Euler-Lagrange equations to determine the equations of motion.

4. What are the applications of the Lagrange equations of motion for hoop rolling down a moving ramp?

The Lagrange equations of motion for hoop rolling down a moving ramp are commonly used in fields such as physics and engineering to study the motion of objects on inclined planes or ramps. They can also be applied to other systems involving rotational and translational motion, such as rolling spheres or cylinders.

5. Are there any limitations to the Lagrange equations of motion for hoop rolling down a moving ramp?

Like any mathematical model, the Lagrange equations of motion for hoop rolling down a moving ramp have limitations. They assume that the hoop is a perfect rigid body and that there is no slipping or energy loss during the motion. In reality, these assumptions may not hold true, leading to discrepancies between the predicted and actual motion of the hoop.

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