Solve Lagrangian Ques: Point Mass on Rotating Hoop

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In summary, the problem involves a point mass constrained to move on a massless hoop of radius a in a vertical plane rotating at a constant angular speed \omega. The Lagrange's equations of motion are obtained assuming that the only external forces are from gravity. The kinetic energy term includes separate components for linear velocities and the angular velocity. The moment of inertia can be determined using the mass and geometry of the system. The Lagrangian can be expressed using two generalized coordinates, representing the angular position and velocity of the particle. The solution for this problem will involve solving for the equations of motion and determining the trajectory of the point mass on the hoop.
  • #1
Hypnotoad
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A point mass is constrained to move on a massless hoop of radius a fixed in a vertical plane that rotates about its vertical symmetry axis with constant angular speed [tex] \omega [/tex].

a. Obtain the Lagrange's equations of motion assuming that the only external forces arise from gravity.

Should I have separate KE components for the linear velocities as well as the angular velocity? I have this so far (with separate x,y and z velocity components written in spherical coordinates) [tex]T=\frac{m}{2} v^2 + I\omega^2[/tex]. I'm pretty sure that is correct, but I don't know what to use for the moment of inertia? Can I just use the moment of inertia for a spherical shell, or would I use that of a ring or something else entirely?

EDIT:
Since [tex]\omega[/tex] is the rate of change of the angle [tex]\theta[/tex] in spherical coordinates, could I set that equal to [tex]\frac{d}{dt}\theta[/tex] in the kinetic energy term? Or can I not ignore the moment of inertia like that?
 
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  • #2
I just had a similar question in an assignment, and I'm curious if my approach was correct:

First of all, note that the hoop is massless. So where is this term for kinetic energy due to the hoop's angular velocity coming from? My reasoning was that the only particle in the system with any mass is the one with mass m moving around the wire. So the total kinetic energy of the system comes entirely from this particle's kinetic energy. Correct?

[tex] \sum{T} = \frac{1}{2}m|\vec{v}|^2 [/tex]

You are also told (as were we) that external forces are limited to graviational forces. More specifically, it was spelled out for us that the particle's potential energy is [itex] V = mgz [/itex], where z is the height above some reference point (wherever you want). Of course, it would seem that we are done:

[tex] \mathcal{L} = T - V [/tex]

However, when introducing the Lagrangian formulation of dynamics, my prof stressed that any system of particles can be expressed completely using a certain number of generalized coordinates. I don't quite understand the fine aspects of this point, but in this case I used two generalized coordinates:

[tex] \phi \ \ \text{and} \ \ \dot{\phi} [/tex] representing the angular position and angular velocity of the particle respectively.

Note that the velocity, although in some crazy direction, is the resultant of two components. The first, [itex] \vec{v_1} [/itex], is the tangential component of the velocity due to it's revolution around the wire.. The second, [itex] \vec{v_2} [/itex], is the component of the velocity in a plane perpendicular to the first (in which the hoop lies). If you have trouble seeing this, draw a "top" view, which just ends up being a circle of radius [itex] a\cos\phi [/itex] on the verge of being swept out by the particle on the wire at that instant.

[tex] |\vec{v_1}| = \dot\phi a [/tex]

[tex] |\vec{v_2}| = \omega a\cos\phi [/tex]

[tex] z = a \sin\phi [/tex]

Now the Lagrangian can be expressed entirely in terms of the two generalized coordinates. Is this method essentially correct?
 
  • #3
I was thinking that the moment of inertia would come from the rotating mass, not the hoop itself, but then I realized that I had two terms that were representing the same angular velocity of the mass. I also completely missed the fact that since [tex]\omega[/tex] is constant, the height of the mass in the hoop is a constant and the number ov velocity components reduces to two. I'm pretty sure I can get the rest of the problem now.
 
  • #4
Hypnotoad said:
A point mass is constrained to move on a massless hoop of radius a fixed in a vertical plane that rotates about its vertical symmetry axis with constant angular speed [tex] \omega [/tex].

a. Obtain the Lagrange's equations of motion assuming that the only external forces arise from gravity.

Should I have separate KE components for the linear velocities as well as the angular velocity? I have this so far (with separate x,y and z velocity components written in spherical coordinates) [tex]T=\frac{m}{2} v^2 + I\omega^2[/tex]. I'm pretty sure that is correct, but I don't know what to use for the moment of inertia? Can I just use the moment of inertia for a spherical shell, or would I use that of a ring or something else entirely?

EDIT:
Since [tex]\omega[/tex] is the rate of change of the angle [tex]\theta[/tex] in spherical coordinates, could I set that equal to [tex]\frac{d}{dt}\theta[/tex] in the kinetic energy term? Or can I not ignore the moment of inertia like that?

what will the solution look like for this problem? I would appreciate to see the entire solution to this problem
 
  • #5


The Lagrangian method is a powerful tool for solving problems in classical mechanics. In this case, we are dealing with a point mass that is constrained to move on a rotating hoop. The only external force acting on the mass is gravity. We can use the Lagrangian method to obtain the equations of motion for this system.

First, we need to define the generalized coordinates for our system. Since the hoop is rotating in a vertical plane, we can use spherical coordinates to describe the position of the point mass. Let's define our generalized coordinates as \theta, the angle between the point mass and the vertical axis, and \phi, the azimuthal angle around the hoop.

Next, we need to determine the kinetic energy of the system. As you correctly pointed out, we need to consider both the linear and angular velocities of the point mass. The kinetic energy can be written as:

T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\dot{\phi}^2) + \frac{1}{2}I\dot{\phi}^2

where r is the distance of the point mass from the center of the hoop and I is the moment of inertia of the hoop. As for the moment of inertia, we can use the moment of inertia for a thin ring since the hoop can be approximated as a thin ring.

Now, we can construct the Lagrangian for our system as L = T - V, where V is the potential energy due to gravity. Since the only external force is gravity, the potential energy can be written as V = mgr\cos\theta.

Finally, we can apply the Euler-Lagrange equations to obtain the equations of motion:

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right) - \frac{\partial L}{\partial \phi} = 0

Solving these equations will give us the equations of motion for the system. By solving these equations, we can determine the behavior of the point mass on the rotating hoop and how it is affected by gravity.

In summary, we can use the Lagrangian method
 

1. What is a Lagrangian equation?

A Lagrangian equation is a mathematical equation used to describe the motion of a system by taking into account both the kinetic and potential energy of the system. It is named after the mathematician Joseph-Louis Lagrange and is commonly used in physics and engineering.

2. How does a point mass on a rotating hoop relate to the Lagrangian equation?

A point mass on a rotating hoop is a common example used to illustrate the principles of the Lagrangian equation. In this scenario, the point mass represents the system and the rotating hoop represents the external forces acting on the system. By applying the Lagrangian equation, we can determine the equations of motion for the point mass.

3. What are the key components of the Lagrangian equation for a point mass on a rotating hoop?

The key components of the Lagrangian equation for a point mass on a rotating hoop include the kinetic energy of the mass, given by its mass and velocity, and the potential energy of the system, which is determined by the position of the mass and the forces acting on it (such as gravity or a tension force from the hoop).

4. How does the Lagrangian equation help solve for the motion of the point mass on a rotating hoop?

By using the Lagrangian equation, we can derive the equations of motion for the system, which describe how the point mass will move on the rotating hoop based on its initial conditions. These equations can then be solved to determine the position, velocity, and acceleration of the point mass at any given time.

5. Are there any limitations to using the Lagrangian equation for this scenario?

Yes, there are a few limitations to using the Lagrangian equation for a point mass on a rotating hoop. One limitation is that it assumes the hoop is perfectly circular and rigid, which may not be the case in real-world situations. Additionally, the equation does not take into account any dissipative forces (such as friction) that may affect the motion of the point mass.

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