Lagrangian of two masses connected by a pulley on inclined p

In summary, the problem involves two masses, m1 and m2, connected by a light string of length D. Mass m1 starts at rest on an inclined plane while mass m2 hangs from a frictionless pulley with a moment of inertia I and radius R. The Lagrangian of the system is found by taking into account the coordinates of mass m1 along the plane, mass m2's downward motion, and the rotation of the pulley. There are two constraints, one being the constant length of the string and the other the no-slip condition.
  • #1
Elvis 123456789
158
6

Homework Statement



Two masses, m1 and m2, are attached by a light string of length D. Mass m1 starts at rest on an inclined plane and mass m2 hangs as shown. The pulley is frictionless but has a moment of inertia I and radius R. Find the Lagrangian of the system and determine the acceleration of the masses using the Lagrangian. Though there are three coordinates of interest (along the plane for mass m1, down for mass m2, and an angle for the rotation of the pulley), there are two constraints.

Homework Equations


∂L/∂q - d/dt(∂L/∂(q-dot)2) = 0

The Attempt at a Solution


If I define the x-direction to be in the direction of the inclined plane then

L = 0.5*m1*(x-dot)2 + 0.5*m2*(y-dot)2 + 0.5*I*(phi-dot)2 -(m1*g*x*sin(θ) + m*g*y)

where phi is the angle that the pulley is rotating through

The length of the string is constant so the length of string on the plane plus the bit on the pulley plus the rest that is hanging holding up m2 is equal to D

so X + R*φ + y = D

i don't know what the other constraint is though, or if the one I have is even correct. Anybody care to give me a hand? The setup is shown in the attachment.
 

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  • #2
Elvis 123456789 said:
so X + R*φ + y = D
This is incorrect. The length of the thread over the pulley does not depend on the rotation angle.

A hint for the second constraint: no-slip condition.
 
  • #3
Orodruin said:
This is incorrect. The length of the thread over the pulley does not depend on the rotation angle.

A hint for the second constraint: no-slip condition.
Would this be correct for the length of the rope? the work is in the attachment. I drew the picture a bit exaggerated so you can see how I perceive the situation. I wrote "l" by mistake for the length of the string, I meant to write "D"
 

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Last edited:

Related to Lagrangian of two masses connected by a pulley on inclined p

1. What is the Lagrangian of two masses connected by a pulley on inclined plane?

The Lagrangian in this system is the difference between the kinetic energy and potential energy. It can be expressed as L = T - U, where T is the total kinetic energy of the two masses and U is the potential energy due to gravity and the tension in the string.

2. How is the Lagrangian derived for this system?

The Lagrangian is derived using the principle of least action, which states that the path taken by a system between two points in time is the one that minimizes the action integral. In this system, the action integral is the difference between the kinetic and potential energy, and the path that minimizes it is the one that satisfies the equations of motion.

3. What are the equations of motion for this system?

The equations of motion for this system can be derived from the Lagrangian using the Euler-Lagrange equations. They are given by:
m1a1 = T1 - U1
m2a2 = T2 - U2
Where m1 and m2 are the masses, a1 and a2 are the accelerations, T1 and T2 are the tensions in the string, and U1 and U2 are the potential energies of the masses.

4. How does the angle of inclination affect the Lagrangian in this system?

The angle of inclination affects the potential energy term in the Lagrangian. As the angle increases, the potential energy also increases, leading to a higher Lagrangian. This results in a more complex system with more difficult equations of motion.

5. Can the Lagrangian be used to find the equilibrium position for this system?

Yes, the Lagrangian can be used to find the equilibrium position of this system. The equilibrium position is where the Lagrangian is minimized, meaning the potential energy is equal to the kinetic energy. This can be solved by setting the derivatives of the Lagrangian with respect to the positions of the masses to zero and solving for the equilibrium positions.

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