A particle on the edge of an inclined spinning disc

In summary, the conversation discusses finding the equations of motion for a particle fixed to the edge of a disc and free to rotate in a uniform gravitational field. The Lagrangian is defined in spherical coordinates, and the attempts at finding the equations of motion involve using cylindrical coordinates and imposing constraints.
  • #1
jack476
328
125

Homework Statement


A particle is fixed to the edge of a disc of negligible mass making an angle θ0 with the ground in a uniform gravitational field, and is free to rotate about the center of the disc. I need to find the equations of motion.

Homework Equations


The definition of the Lagrangian in spherical coordinates,

##\frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + r^2sin^2{\theta}\dot{\phi}^2) - U(r, \phi, \theta)##

where ##\theta## is given to be the angle of altitude and ##\phi## is the azimuthal angle.

The Attempt at a Solution


[/B]
It was easy to find the Lagrangian to be

##L = \frac{R^2}{2}m(\dot{\theta}^2 + sin^2{\theta}\dot{\phi}^2) + mgrcos{\theta}##

It is from here that I do not know how to proceed. Wouldn't using that Lagrangian to find the equations of motion for ##\phi## and ##\theta## just get me the equations of motion for a particle constrained to move?

The first thing I've thought to try is to treat the particle as though it is moving on the surface of a sphere, fixed to move along the great circle produced by the plane passing through the center of the sphere and making an angle θ0 with its equator, then I could use the equation of the great circle as a constraint relating ##\phi## and ##\theta##, but I'm struggling to find the parametric equations for a great circle.

The other thing I tried was to use cylindrical coordinates and impose the constraint that the height ##z## is some function of ##\phi##. My guess is that this function would be something of the form ##z = Rcos{\theta_0}cos{\phi}##, then use ## f = z - Rcos{\theta_0}cos{\phi} = 0## as my equation of constraint. This gets me the equations

##\ddot{\phi} + \frac{\lambda}{R}cos{\theta_0}sin{\phi} = 0##
##\ddot{z} - g + \frac{\lambda}{m} = 0##

It is from there that I do not know how to proceed.
 
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  • #2
How you write the Lagrangian for horizontal disk on external field by angle ##\theta_0##?
This Lagrangian produce the same equation of motion? Must be?
 
  • #3
Are you taking the z axis as vertical? Doesn't that make phi and its derivatives a bit tricky?
Have you tried using cylindrical coordinates but taking the z axis perpendicular to the disk?
 

Related to A particle on the edge of an inclined spinning disc

1. What is a particle on the edge of an inclined spinning disc?

A particle on the edge of an inclined spinning disc refers to a small object or particle that is placed on the outer edge of a disc that is spinning at an angle or inclination.

2. What causes a particle on the edge of an inclined spinning disc to move?

The motion of the inclined spinning disc causes a centrifugal force on the particle, making it move outward from the center of the disc.

3. How does the angle of inclination affect the movement of the particle?

The steeper the angle of inclination, the greater the centrifugal force and the faster the particle will move away from the center of the disc.

4. Can a particle on the edge of an inclined spinning disc change direction?

Yes, if the angle of inclination is changed or if there is an external force acting on the particle, it can change direction and move towards or away from the center of the disc.

5. What are some real-life examples of a particle on the edge of an inclined spinning disc?

A common example is a coin or a toy car placed on the edge of a spinning record or turntable. Another example is a particle on the edge of a spinning centrifuge in a laboratory.

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