Bead on a rotating stick and the Lagrangian

In summary, the conversation discusses a stick pivoted at the origin and swinging at a constant angular speed. A bead on the stick moves without friction and the conserved quantity E is found to be ##E = \frac{m(r'^2 - (r\theta')^2)}{2}##, not the energy of the bead. This is due to the fact that the angular velocity, ##\dot{\theta}##, is constant and only the radial position, r, affects the Lagrangian, resulting in the conserved quantity Q.
  • #1
LCSphysicist
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Homework Statement
A stick is pivoted at the origin and is arranged to swing around in a
horizontal plane at constant angular speed ω. A bead of mass m slides
frictionlessly along the stick. Let r be the radial position of the bead.
Find the conserved quantity E given in Eq. (6.52). Explain why this
quantity is not the energy of the bead.
Relevant Equations
All below...
A stick is pivoted at the origin and is arranged to swing around in a
horizontal plane at constant angular speed ω. A bead of mass m slides
frictionlessly along the stick. Let r be the radial position of the bead.
Find the conserved quantity E given in Eq. (6.52). Explain why this
quantity is not the energy of the bead.

1602869687378.png


I came across with such question and i am a little confuse with the answer.
$$L = \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$
$$E = \frac{m(r'^{2} + (r\theta ')^{2})}{1}- \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$
$$E = \frac{m(r'^{2} + (r\theta ')^{2})}{2}$$

But the answer should be... $$E = \frac{m(r'^{2} - (r\theta ')^{2})}{2}$$
 

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  • #2
Since ##\dot{\theta} = \omega## is constant, ##L = L(r)## only. So ##\frac{\partial L}{\partial \dot{r}} \dot{r} = m\dot{r}^2## but ##\frac{\partial L}{\partial \dot{\theta}} = 0##. Then, since ##L = T - V = T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2}mr^2 \omega^2##, you get the desired result, i.e. the conserved quantity is$$Q = \frac{1}{2}m\dot{r}^2 - \frac{1}{2}mr^2 \omega^2$$but this is not the energy.
 
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  • #3
etotheipi said:
Since ##\dot{\theta} = \omega## is constant, ##L = L(r)## only. So ##\frac{\partial L}{\partial \dot{r}} \dot{r} = m\dot{r}^2## but ##\frac{\partial L}{\partial \dot{\theta}} = 0##. Then, since ##L = T - V = T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2}mr^2 \omega^2##, you get the desired result, i.e. the conserved quantity is$$Q = \frac{1}{2}m\dot{r}^2 - \frac{1}{2}mr^2 \omega^2$$but this is not the energy.
Oh, My partial derivative with respect to the theta dot was wrong. Thx
 
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FAQ: Bead on a rotating stick and the Lagrangian

1. What is a bead on a rotating stick?

A bead on a rotating stick is a physical system in which a bead is attached to a stick that is rotating about a fixed axis. This system is often used in physics to demonstrate concepts related to rotational motion and conservation of energy.

2. What is the Lagrangian in this system?

The Lagrangian is a mathematical function that describes the dynamics of a physical system. In the case of a bead on a rotating stick, the Lagrangian takes into account the rotational kinetic energy of the stick and the potential energy of the bead due to its position on the stick.

3. How is the Lagrangian used in this system?

The Lagrangian is used in this system to derive the equations of motion for the bead on the rotating stick. By finding the minimum value of the Lagrangian, known as the action, the equations of motion can be determined and used to analyze the behavior of the system.

4. What is the significance of the Lagrangian in this system?

The Lagrangian is significant in this system because it allows for a more elegant and efficient way to describe the dynamics of the system compared to using traditional Newtonian mechanics. It also allows for the incorporation of constraints and generalized coordinates, making it useful for more complex systems.

5. How does the Lagrangian approach differ from the Newtonian approach in this system?

The Lagrangian approach differs from the Newtonian approach in that it takes into account all possible paths that the system can take, rather than just the one that satisfies Newton's laws. This allows for a more comprehensive understanding of the system's behavior and can also be applied to non-conservative systems.

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