Solving the Prism Mechanics Problem with Lagrangian & Constraints

• JohanL
In summary: Actually aren't the two first questions the same?In summary, the conversation discusses a problem involving two prisms on top of each other, with one gliding down to the ground causing the lower prism to move to the left. The problem is to find the distance the lower prism moves. There is a question about how to set up the Lagrangian and constraints for the problem, and a discussion about whether a Lagrangian approach is necessary. The correct solution is found using the Lagrangian equations, and another problem involving a particle attached to a spring on a horizontal rod and a pendulum is also discussed.
JohanL
If you have two prisms on top of each other, one smaller than the other, and the one on top glids down to the ground the lower prisms moves to the left.
The problem is to find how far the lower prism moves.

How do you set up the Lagrangian and the constraints for this problem?
Any ideas?

There is no friction between the prisms or between the lower prism and the ground.

You posted 2 questions in here. You do know that we have a Homework help section here in PF, don't you?

Zz.

Johan,

What exactly do you expect from us ? Haven't you read the PF-guidelines ?

What have you done up till now in order to solve your problem? What are your suggestions ? let us know and we will help you out as much as is needed. Just don't post such generalized questions. What you are really asking for is a plain complete solution...We cannot give you that according to the guidelines of this Forum

regards
marlon

I have posted in the homework help section before. I post in classical physics when its not homework. If you want i can post in here whenever i have questions about a problem.

I just wanted some hints on how to attack the problem. Not a complete solution.

I attached an image of the problem

If you choose an orthogonal coordinate system with the x-axis parallell to the ground
i can set up the Lagrangian. If Tpi is the kinetic energy for the i:th prism you get

L = Tp1 + Tp2 - mgh

where h is the height to the center of mass of prism 2, the prism on top of the other.
Then i guess you need a constraint because the upper prism glids down the lower prism. Here I am not sure about how to set up the equation for the constraint. The prisms are in contact all the time but i can't see an obvious relation between the center of masses or something like that. If you choose the coordinate system with the x-axis parallell to the "inclined plane" (the lower prism) the upper prism glids down on maybe its easier to set up the constraint equation but on the other hand this inclined plane moves because the lower prism moves to the left.

Attachments

• prisms.GIF
989 bytes · Views: 523
No idea if this is right but:
Call M the larger bottom mass, m the small top mass, L the distance the mass has to slide down, and theta the rightmost angle of the bottom prism.
So...
m exhibits a force downards on M of which a component of this force is perpendicular to the surface.
$$F_p = mgsin(\theta)$$
Of this force acting perpendicular to the surface there is a component of it that acts in the negative x direction.
$$F_x = mgsin(\theta)cos(\theta)$$
This causes an acceleration on M
$$F_x = Ma_x = mgsin(\theta)cos(\theta)$$
$$a_x = \frac{mgsin(\theta)cos(\theta)}{M}$$
There is also a component of the gravitational force acting parallel to the surface which directs m down M.
$$a_y = \frac{F_{||}}{m} = gcos(\theta)$$
The time it takes for m to get to the bottom can be found by the equation:
$$L=1/2gcos(\theta)t^2$$
The distance moved left by the larger prism can now be found.
$$d = 1/2a_x t^2 = 1/2 \frac{mgsin(\theta)cos(\theta)}{M} \frac{2sec(\theta)L}{g}$$
$$= \frac{MLsin(\theta)}{m}$$

But it doesn't work...i think.
I have attached a more correct image of the problem.

I know the answer. If b1 is the horizontal length of the upper prism and b2 is the horizontal length of the lower prism the distance to the left the lower prism moves when the upper prism hits the ground is: (m1 mass of the upper prism, m2 mass of the lower prism)

m1/(m1+m2) * (b2 - b1)

here is the more correct image of the prisms

Attachments

• prisms.GIF
1 KB · Views: 511
JohanL said:
If you have two prisms on top of each other, one smaller than the other, and the one on top glids down to the ground the lower prisms moves to the left.
The problem is to find how far the lower prism moves.

How do you set up the Lagrangian and the constraints for this problem?
Any ideas?

Are you required to use a Langrangian approach to solving this problem? If so, we need to focus on this aspect of the problem.

yes Doc Al, that is of course the easiest way to solve this problem! Thank you. But i still don't get the right answer tho.

with i for initial and f for final and 1 for upper prism and 2 for lower i get:

[m1x(1i) + m2x(2i)]/(m1 + m2) = [m1x(1f) + m2x(2f)]/(m1 + m2)

d = distance moved left by lower prism = x(2f) - x(2i)

x(1f) - x(1i) = b2 -b1

d = m1/m2 * (b2 - b1)

m1/(m1+m2) * (b2 - b1)

JohanL said:
with i for initial and f for final and 1 for upper prism and 2 for lower i get:

[m1x(1i) + m2x(2i)]/(m1 + m2) = [m1x(1f) + m2x(2f)]/(m1 + m2)

d = distance moved left by lower prism = x(2f) - x(2i)
No problem so far.

x(1f) - x(1i) = b2 -b1
This is an error. b2 -b1 is how much the top prism moves with respect to the bottom prism. Don't forget that the bottom prism moves also.

yes of course.

x(1f) - x(1i) = b2 - b1 - a

which gives the right answer! thx

But i still need a lot of training on using lagrange equations. Here is another problem where this approach is needed.

a "particle" with mass m is attached to a spring on a horizontal rod and another rod with mass m and length 2a, which can move as a pendulum.
See the image below.

Set up Lagrange equations for the systems movement. No friction
For small oscillations find the normal modes and x and phi as functions of time t, if the start is from rest.

solution:

If x is the particle's distance from the equillibrium position and phi is the angle for the pendulum you get.

$$T = \frac{1}{2}I\dot{\phi}^2 + \frac{m\dot{x}^2}{2}$$

$$V = -mgacos\phi + \frac{kx^2}{2}$$

$$L = T- V$$

But this isn't enough. If you solve lagrange equations for this you get a solution for a system where the spring is isolated from the pendulum. You also need this...i think

$$akx=I\ddot{\phi}$$

But how should you use this. You can't use it to eliminate one of the coordinates...that would be wrong i think.

The answer says that lagrange equations gives

$$4a\ddot{\phi} + 3\ddot{x}cos\phi + 3gsin\phi = 0$$

$$a\ddot{\phi}cos\phi + 2\ddot{x}- a\dot{\phi}^2sin\phi + 3gx/a = 0$$

Attachments

• springpendulum.GIF
1.2 KB · Views: 544
I'm thinking you are missing something in $T = \frac{1}{2}I\dot{\phi}^2 + \frac{m\dot{x}^2}{2}$. Since x is time varying, the angular velocity of the rod relative to the mass is not a complete description of its motion. It is going to have a translation component as well as a rotation component. I don't think you have that included.

yes i missed that.

$$T = \frac{1}{2}I\dot{\phi}^2 + \frac{m\dot{x}^2}{2} + \frac{m}{2}(\dot{x}+a\dot{\phi}cos\phi)^2$$

is the correct kinetic energy then...i think.

But i still don't get the right equations. With k=3mg/a and I=1/3*ma^2 the equations of motion are

$$4a\ddot{\phi} + 3\ddot{x}cos\phi + 3gsin\phi = 0$$

$$a\ddot{\phi}cos\phi + 2\ddot{x}- a\dot{\phi}^2sin\phi + 3gx/a = 0$$

Am I missing something else?

1. What is the "Prism Mechanics Problem"?

The Prism Mechanics Problem is a mathematical and physical problem that involves calculating the motion of a prism when subjected to external forces. The prism is assumed to have three degrees of freedom: rotation about its center of mass and translation in two directions.

2. What is the Lagrangian method used for in solving this problem?

The Lagrangian method is a mathematical approach to solving problems in classical mechanics. It involves writing the kinetic and potential energy of a system in terms of generalized coordinates and using the principle of least action to derive the equations of motion. In the context of the Prism Mechanics Problem, the Lagrangian method is used to find the equations of motion for the prism.

3. What are the constraints in the Prism Mechanics Problem?

The constraints in this problem refer to the limitations on the motion of the prism. In this case, the prism is constrained to move in two dimensions and rotate about its center of mass. These constraints are used in the Lagrangian method to derive the equations of motion for the prism.

4. How is the Lagrangian method applied to the Prism Mechanics Problem?

To apply the Lagrangian method to the Prism Mechanics Problem, the kinetic and potential energy of the prism are expressed in terms of the generalized coordinates. Then, the Lagrangian function is defined as the difference between the kinetic and potential energy. The equations of motion can then be derived using the principle of least action and the constraints of the problem.

5. What are some real-world applications of solving the Prism Mechanics Problem?

The Prism Mechanics Problem has applications in various fields such as robotics, aerospace engineering, and physics. It can be used to model the motion of objects subjected to external forces, such as satellites orbiting a planet or robots navigating through obstacles. It also has applications in studying the stability and control of mechanical systems.

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