Solve Lagrangian Homework: Bead on Rotating Bicycle Wheel

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Homework Help Overview

The discussion revolves around a bead of mass m moving on a spoke of a rotating bicycle wheel, focusing on deriving the Lagrangian and the equations of motion using generalized coordinates. Participants are exploring the implications of the forces acting on the system and the constraints involved.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of cylindrical coordinates and the formulation of kinetic and potential energy. There are attempts to derive the equations of motion using Lagrange's equations. Questions arise regarding the correctness of the time derivatives and the effects of constraints on the bead's motion.

Discussion Status

There is an ongoing examination of the equations of motion with some participants questioning the assumptions made about the forces and constraints. Clarifications are being sought regarding the nature of the wheel's rotation and the inclusion of external forces in the Lagrangian.

Contextual Notes

Participants note that the problem statement may lack clarity, particularly regarding the orientation of the wheel's rotation and the nature of the forces acting on the bead. There is also mention of the bead's radial constraint relative to the wheel's radius.

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Homework Statement



Consider a bead of mass m moving on a spoke of a rotating bicycle wheel. If there are no forces other than the constraint forces, then find the Lagrangian and the equation of motion in generalised coordinates. What is the possible solution of this motion?

Homework Equations



Lagrange's equation of motion

The Attempt at a Solution



Use cylindrical coordinates for the problem.

Let the origin be at the centre of the wheel and let the wheel lie on the rθ-plane such that θ=0 at the top of the wheel and θ>0 in the clockwise direction.

So, T = 0.5m(\dot{r}^{^2} + r^{2}\dot{\theta}^{2}) and U = mgrcosθ.

So, \frac{∂L}{∂r} - \frac{d}{dt}(\frac{∂L}{∂\dot{r}}) = 0

\Rightarrow (mr\dot{\theta}^{2} - mgcosθ) - m\ddot{r} = 0

\Rightarrow \ddot{r} = r\dot{\theta}^{2} - gcosθ

and \frac{∂L}{∂θ} - \frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = 0

\Rightarrow (-mgrsinθ) - mr^{2}\ddot{θ} = 0

\Rightarrow \ddot{θ} = - \frac{g}{r}sinθ

Do you think I've got the equations of motion right?

I have no idea how to work out the possible solution of the motion. Thoughts?
 
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halukalutalu said:

Homework Statement



Consider a bead of mass m moving on a spoke of a rotating bicycle wheel. If there are no forces other than the constraint forces, then find the Lagrangian and the equation of motion in generalised coordinates. What is the possible solution of this motion?


Homework Equations



Lagrange's equation of motion

The Attempt at a Solution



Use cylindrical coordinates for the problem.

Let the origin be at the centre of the wheel and let the wheel lie on the rθ-plane such that θ=0 at the top of the wheel and θ>0 in the clockwise direction.

So, T = 0.5m(\dot{r}^{^2} + r^{2}\dot{\theta}^{2}) and U = mgrcosθ.

So, \frac{∂L}{∂r} - \frac{d}{dt}(\frac{∂L}{∂\dot{r}}) = 0

\Rightarrow (mr\dot{\theta}^{2} - mgcosθ) - m\ddot{r} = 0

\Rightarrow \ddot{r} = r\dot{\theta}^{2} - gcosθ

and \frac{∂L}{∂θ} - \frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = 0

\Rightarrow (-mgrsinθ) - mr^{2}\ddot{θ} = 0

\Rightarrow \ddot{θ} = - \frac{g}{r}sinθ

Do you think I've got the equations of motion right?

I have no idea how to work out the possible solution of the motion. Thoughts?

The last one is not correct. Take care whit the time derivative.

ehild
 
To be honest, I don't see how I've got things wrong with the time derivative. Of course, \dot{θ} differentiated wrt time becomes \ddot{θ}, right?

Also, isn't r constrained to be smaller than the radius of the wheel. Doesn't this have an effect on the solution?
 
\frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = \frac{d}{dt}(mr^2\dot{θ})=2mr\dot r\dot {θ}+mr^2 \ddot {θ}

You ignored the time dependence of r.

The text of the problem is not clear.

Does that wheel rotate in the horizontal or in the vertical plane? mg is an outside force, not force of constraint.

Is there something making the wheel rotate with a given angular speed, or does the wheel rote freely about a fixed axis?

I think you can assume the radius of the wheel arbitrary long.

ehild
 
ehild said:
\frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = \frac{d}{dt}(mr^2\dot{θ})=2mr\dot r\dot {θ}+mr^2 \ddot {θ}

You ignored the time dependence of r.

Thanks for the pointer. :)

ehild said:
The text of the problem is not clear.<br /> <br /> Does that wheel rotate in the horizontal or in the vertical plane? mg is an outside force, not force of constraint. <br /> <br /> Is there something making the wheel rotate with a given angular speed, or does the wheel rote freely about a fixed axis? <br /> <br /> I think you can assume the radius of the wheel arbitrary long. <br /> <br /> ehild
<br /> <br /> I think the wheel rotates in the horizontal plane because there are no forces apart from the given force of constraint and because the force of gravity is not a force of constraint. What do you think? Isn&#039;t this also why we can safely say that the wheel rotates freely?<br /> <br /> So, if we assume that the radius of the wheel is arbitrarily large, would you say that under those assumptions, my solution is correct?
 
What is your solution?

If the wheel is free to rotate, it is part of the system. You have to include it into the Lagrangian.

Are you sure you copied the problem correctly?

How do you include the constraint?

ehild
 
Last edited:

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