Effective potential of a particle sliding on an inclined rod

In summary, the conversation discusses a system with a pointlike mass sliding along a rigid rod that is rotating at a constant angular velocity. The system has one degree of freedom and the Lagrangian and Lagrange equations are used to determine the motion of the mass. The conversation also mentions finding the effective potential, equilibrium positions, and stability of the system, which can be done by differentiating the Lagrange equation.
  • #1
Dixanadu
254
2

Homework Statement



Hey guys,

Here is the question:

A pointlike mass [itex]m[/itex] can slide along a rigid rod of length [itex]l[/itex] and negligible mass. One extremity of the rod is fixed at the origin [itex]O[/itex] of an inertial system [itex](x,y,z)[/itex], and the rod forms a constant angle [itex]α[/itex] with the [itex]z[/itex]-axis. The rod rotates about the [itex]z[/itex]-axis with constant angular velocity [itex]ω[/itex]. Gravity acts in the negative [itex]z[/itex]-direction.

(i) How many degrees of freedom does the system have?

(ii) Write down the Lagrangian and the Lagrange equations.

(ii) Recast the system as a 1-dimensional motion in an effective potential. Find an expression for the effective potential and determine the equilibrium positions as well as their stability

Homework Equations


[itex]T=1/2mv^{2}[/itex]
Lagrangian: [itex]L = T - V[/itex]
Lagrange Equation: [itex]\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}[/itex]

The Attempt at a Solution


Here are my solutions:

(i) one degree of freedom - the distance from the origin to the point mass. call this distance [itex]r[/itex].

(ii) I won't put the calculations here, I'll just give my results:

Lagrangian [itex]L = T - V = \frac{1}{2}m[\dot{r}^{2}+r^{2}ω^{2}sin^{2}α] - mgrcosα[/itex]


Lagrange equation: [itex]\ddot{r}-rω^{2}sin^{2}α + gcosα = 0 [/itex]

(iii)

Okay, so here is the issue. How do you get the effective potential from this? and do I just differentiate that to get the equilibrium positions and then differentiate again to determine stability?
 
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  • #2
Okay, so here is the issue. How do you get the effective potential from this?
If you had a particle of mass m confined to 1-dimension of space (a straight line) and you let r be the distance from the origin, how would you express the kinetic energy T? How would you construct a potential energy function V(r) for this particle so that it would have the same Lagrangian L as your three-dimensional problem of the particle on the rotating wire?

and do I just differentiate that to get the equilibrium positions and then differentiate again to determine stability?
Sounds good.
 

Related to Effective potential of a particle sliding on an inclined rod

1. What is the effective potential of a particle sliding on an inclined rod?

The effective potential of a particle sliding on an inclined rod is the energy required to move the particle from one point to another along the rod. It takes into account the gravitational potential energy and the kinetic energy of the particle.

2. How is the effective potential of a particle sliding on an inclined rod calculated?

The effective potential can be calculated using the equation: V_eff = mgh + 1/2mv^2, where m is the mass of the particle, g is the acceleration due to gravity, h is the height of the particle on the rod, and v is the velocity of the particle.

3. What factors affect the effective potential of a particle sliding on an inclined rod?

The effective potential is affected by the mass of the particle, the angle of the rod, and the initial height and velocity of the particle. Additionally, any external forces acting on the particle, such as friction, will also affect the effective potential.

4. How does the effective potential change as the particle slides down the rod?

As the particle slides down the rod, its effective potential decreases due to the decrease in height and increase in velocity. This is because the particle's kinetic energy increases while its gravitational potential energy decreases.

5. Can the effective potential of a particle sliding on an inclined rod be negative?

Yes, the effective potential can be negative if the initial height of the particle is greater than its final height. This means that the particle has enough kinetic energy to overcome the force of gravity and move to a higher position on the rod.

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