# Lagrangian: Bead on a rotating hoop with mass

• raggle
In summary: Then I don't see the difference between the mass in the Lagrangian and the mass of the bead.In summary, the conversation discusses the system of a bead and a smooth circular wire hoop, where the hoop is free to rotate about a vertical axis. The equations for the kinetic energy of the bead and the hoop are given, and the Lagrangian is obtained by going through the usual steps. The discussion also covers the three equilibria of the system, one at θ=0, one at θ=π, and one at cosθ=𝑔2𝑅ω2. There is a slight discrepancy in the equations due to a typo and a mistake in the moment of inertia of the hoop, but
raggle

## Homework Statement

'Consider the system consisting of a bead of mass m sliding on a smooth circular wire hoop of mass 2m and radius R in a vertical plane, and the vertical plane containing the hoop is free to rotate about the vertical axis. Determine all relative equilibria of the bead.'

## Homework Equations

$$T_r= \frac{1}{2}I\omega^2$$
The kinetic energy of the bead is
$$T=\frac{1}{2}m(\dot\theta^2+\dot\phi^2\sin^2\theta) + \frac{1}{2}mR^2\sin^2\theta\dot\phi^2$$
The kinetic energy of the hoop is
$$T=\frac{1}{2}mR^2\dot\phi^2$$

## The Attempt at a Solution

First of all I said the bead is at position
$$r=R(\sin\theta\cos\phi, \sin\theta\sin\phi, 1-\cos\theta)$$
the equation for the z coordinate looks a bit odd but I think it's right (using this mean that the bead is at height 0 when it's at the bottom of the hoop and at 2R when it's at the top of the hoop). Going through the usual steps to get the Lagrangian gives (I can go into more detail on this if anyone wants)
$$L=\frac{1}{2}[\dot\theta^2+\dot\phi^2(1+2\sin^2\theta)]+mgR\cos\theta$$

Since I'm looking for solutions where ##\theta## is constant I rewrite this as
$$L=\frac{1}{mR^2}[\frac{1}{2}\dot\theta^2 -V]$$
where
$$V=-\frac{1}{mR^2}[\frac{1}{2}\dot\phi^2(1+2\sin^2\theta)+mgR\cos\theta]$$

This gives
$$\ddot\theta=-\frac{\partial V}{\partial\theta}$$
Setting this equal to zero gives
$$2\dot\phi^2\sin\theta\cos\theta-mgR\sin\theta=0$$

Giving 3 equilibria, one at ##\theta=0##, one at ##\theta=\pi## and one at ##\cos\theta=\frac{g}{2R\dot\phi^2}##.
I'm not sure about the third solution at all. The closest I could find was a similar problem in Tong's lecture notes, where the third solution is ##\cos\theta=\frac{g}{R\dot\phi^2}## (page 27). The factor of 2 is bothering me, is it from me including the moment of inertia of the hoop?

raggle said:
The kinetic energy of the bead is
$$T=\frac{1}{2}m(\dot\theta^2+\dot\phi^2\sin^2\theta) + \frac{1}{2}mR^2\sin^2\theta\dot\phi^2$$
The kinetic energy of the hoop is $$T=\frac{1}{2}mR^2\dot\phi^2$$
For the first equation: typo? or error in the dimensions ?
For te second: wasn't the ring mass equal to 2m ?

Since you made the invitation "I can go into more detail on this if anyone wants", why not show the steps ?

--

BvU said:
For the first equation: typo? or error in the dimensions ?
For te second: wasn't the ring mass equal to 2m ?

Since you made the invitation "I can go into more detail on this if anyone wants", why not show the steps ?

--

You're right about the first equation, it should be
$$T=\frac{1}{2}mR^2(\dot\theta^2+\dot\phi^2\sin^2\theta) + \frac{1}{2}mR^2\sin^2\theta\dot\phi^2$$

You're also right about the second equation, I thought the moment of inertia of a hoop was ##\frac{1}{2}mR^2## but it's actually ##mR^2##.

I figured out where I was going wrong, I accidentally count the rotational kinetic energy of the bead twice. That gives me the factor of 2 in the equation of motion. The actual Lagrangian is
$$L=\frac{1}{2}[\dot\theta^2+\dot\phi^2(1+\sin^2\theta)]+mgR\cos\theta$$

the moment of inertia of a hoop is ## \tfrac{1}{2}mR^2## for rotation about a middle line. But the mass is 2mbead.

## 1. What is a Lagrangian?

A Lagrangian is a mathematical function that describes the dynamics of a physical system in terms of its position and velocity.

## 2. What is a bead on a rotating hoop with mass?

A bead on a rotating hoop with mass is a physical system in which a bead is constrained to move along a hoop that is rotating around a fixed point.

## 3. How is the Lagrangian for a bead on a rotating hoop with mass derived?

The Lagrangian for a bead on a rotating hoop with mass is derived by considering the kinetic and potential energy of the system and using the Euler-Lagrange equations to find the equations of motion.

## 4. What are the key variables in the Lagrangian for a bead on a rotating hoop with mass?

The key variables in the Lagrangian for a bead on a rotating hoop with mass are the position and velocity of the bead along the hoop, the angular velocity of the hoop, and the mass of the bead.

## 5. How does the Lagrangian for a bead on a rotating hoop with mass explain the motion of the system?

The Lagrangian for a bead on a rotating hoop with mass contains all the information necessary to determine the equations of motion for the system. By solving these equations, we can understand how the bead moves along the hoop as the hoop rotates.

Replies
7
Views
500
Replies
6
Views
384
Replies
8
Views
719
Replies
7
Views
1K
Replies
3
Views
518
Replies
9
Views
1K
Replies
2
Views
1K
Replies
3
Views
675
Replies
1
Views
335
Replies
11
Views
440