Find the Lagrangian of a System of Particles

[CM Longhorns]

Homework Statement

Ok, so in this system, there are two point particles of mass M connected by massless levers of length L. The pair of masses pivots about the upper point and rotates about the axis at an angular frequency ω. The lower mass is constrained to slide on the vertical axis. The system is illustrated below:

Find the Lagrangian in terms of Θ.

Homework Equations

I know that the Lagrangian is represented by the difference in kinetic and potential energies,
L = T - U
T = ½Mv^2,
and U should be derived from the gravitational force only,
U = Mgh,
where h is the vertical distance from the origin.

The Attempt at a Solution

I have a fundamental lack of understanding regarding finding the kinetic energy of this system. It has been suggested that I view this in cylindrical coordinates, where
x=LsinΘ
y=LcosΘ
z= z.
I intuit that the origin of my cylindrical coordinate system should reside at the pivot point above the two masses. Could someone please illuminate the subtleties in finding T? I understand conceptually the Lagrangian and the parts of the problem to follow.

 

[BvU]

Hi CM, and welcome to PF :)

Does U increase if z (h) increases ?

For T you have to express v for each of the masses in terms of ##\theta## and ##\omega##. x and y you already have, so ##\dot x## and ##\dot y## aren't too difficult. You will have to do something about z, though. I can see that if ##\theta## is given, z is determined; i.e. there is a way to express z in terms of ##\theta##

 

[CM Longhorns]

I worked out this problem in the mean time. You are correct!
I see that the potential increases with z, which can be determined to be 2Lcosθ.
The gravitational potential energy is thus U = 3Lcosθ*mg when we account for g's effect on each mass.
The kinetic energy I have determined as well.

Then, L = T - U, and voila, we have a Lagrangian! I can now find the equation of motion in the generalized coordinate θ.
Thanks for your help; hope to see you here again :)

 

[BvU]

However, your expression for U decreases when ##\theta## increases from zero. I would expect it to increase; wouldn't you ?

 

[paranormal]

The Lagrangian of a system of particles can be written as:
L = T - U
where T is the total kinetic energy of the system and U is the total potential energy of the system.

In this case, we have two point masses connected by a massless lever. The lever is rotating about the upper point at an angular frequency ω and the lower mass is constrained to slide on the vertical axis. To find the kinetic energy of the system, we need to consider the motion of each individual particle and the motion of the lever.

Let's start by considering the kinetic energy of each particle. The kinetic energy of a single particle can be written as:
T = ½mv^2
where m is the mass of the particle and v is its velocity.

For the first particle, located at position (LsinΘ, LcosΘ, z), the velocity can be written as:
v = ω(LcosΘ, -LsinΘ, 0)
since it is rotating about the upper point at an angular frequency ω and its position is changing with time.

For the second particle, located at position (0, 0, z), the velocity is simply:
v = (0, 0, 0)
since it is constrained to slide on the vertical axis and its position is not changing with time.

Now, let's consider the motion of the lever. The lever is rotating about the upper point at an angular frequency ω, so its kinetic energy can be written as:
T = ½Iω^2
where I is the moment of inertia of the lever.

To find the moment of inertia, we can use the parallel axis theorem. Since the lever is rotating about the upper point, its moment of inertia can be written as:
I = Icm + md^2
where Icm is the moment of inertia about the center of mass of the lever and d is the distance between the upper point and the center of mass of the lever.

Since the lever is massless, its center of mass is located at the midpoint between the two particles, at (LsinΘ/2, LcosΘ/2, z). Therefore, we can write the moment of inertia as:
I = ½mL^2sin^2Θ + ½mL^2cos^2Θ = ½mL^2
where we have used the fact that sin^