Solve Lagrangian Homework: Bead on Rotating Bicycle Wheel

  • Thread starter halukalutalu
  • Start date
  • Tags
    Lagrangian
In summary, the bead of mass m on a rotating wheel is subject to only the force of constraint from the weight of the bead and the force of gravity. If the wheel is free to rotate, then the Lagrangian must include the weight of the bead and the rotation of the wheel.
  • #1
halukalutalu
3
0

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

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

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

[itex]\Rightarrow \ddot{r} = r\dot{\theta}^{2} - gcosθ[/itex]

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

[itex]\Rightarrow (-mgrsinθ) - mr^{2}\ddot{θ} = 0[/itex]

[itex]\Rightarrow \ddot{θ} = - \frac{g}{r}sinθ [/itex]

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?
 
Physics news on Phys.org
  • #2
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, [itex]T = 0.5m(\dot{r}^{^2} + r^{2}\dot{\theta}^{2})[/itex] and [itex]U = mgrcosθ[/itex].

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

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

[itex]\Rightarrow \ddot{r} = r\dot{\theta}^{2} - gcosθ[/itex]

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

[itex]\Rightarrow (-mgrsinθ) - mr^{2}\ddot{θ} = 0[/itex]

[itex]\Rightarrow \ddot{θ} = - \frac{g}{r}sinθ [/itex]

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
 
  • #3
To be honest, I don't see how I've got things wrong with the time derivative. Of course, [itex]\dot{θ}[/itex] differentiated wrt time becomes [itex]\ddot{θ}[/itex], right?

Also, isn't r constrained to be smaller than the radius of the wheel. Doesn't this have an effect on the solution?
 
  • #4
[itex] \frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = \frac{d}{dt}(mr^2\dot{θ})=2mr\dot r\dot {θ}+mr^2 \ddot {θ}[/itex]

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
 
  • #5
ehild said:
[itex] \frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = \frac{d}{dt}(mr^2\dot{θ})=2mr\dot r\dot {θ}+mr^2 \ddot {θ}[/itex]

You ignored the time dependence of r.

Thanks for the pointer. :)

ehild said:
[itex]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

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't this also why we can safely say that the wheel rotates freely?

So, if we assume that the radius of the wheel is arbitrarily large, would you say that under those assumptions, my solution is correct?
 
  • #6
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:

FAQ: Solve Lagrangian Homework: Bead on Rotating Bicycle Wheel

What is the Lagrangian equation and how is it used in the context of a bead on a rotating bicycle wheel?

The Lagrangian equation is a mathematical formula used in classical mechanics to describe the motion of a system. It takes into account both the kinetic and potential energies of the system and is used to derive the equations of motion. In the context of a bead on a rotating bicycle wheel, the Lagrangian equation can be used to describe the motion of the bead as it moves along the wheel.

What are the assumptions made in the Lagrangian equation for a bead on a rotating bicycle wheel?

The Lagrangian equation for a bead on a rotating bicycle wheel assumes that the wheel is rotating with a constant angular velocity, the bead is constrained to move along the wheel without slipping, and there is no friction between the bead and the wheel. It also assumes that the bead is a point mass, meaning its size and shape are not taken into account.

What are the steps to solving a Lagrangian homework problem for a bead on a rotating bicycle wheel?

The first step is to define the system and identify all relevant variables, such as the position, velocity, and acceleration of the bead. Next, write out the Lagrangian equation, which is the difference between the kinetic and potential energy of the system. Then, use the Euler-Lagrange equation to derive the equations of motion for the system. Finally, solve the equations to find the position, velocity, and acceleration of the bead as a function of time.

What are the common mistakes made when solving a Lagrangian homework problem for a bead on a rotating bicycle wheel?

One common mistake is not properly defining the system and including all relevant variables. Another mistake is not correctly calculating the kinetic and potential energies of the system. It is also important to be careful when using the Euler-Lagrange equation, as it can be easy to make mistakes in the differentiation process. Lastly, not checking the units and dimensions of the final solutions can also result in errors.

What are some practical applications of the Lagrangian equation for a bead on a rotating bicycle wheel?

The Lagrangian equation for a bead on a rotating bicycle wheel has applications in various fields, such as physics, engineering, and robotics. It can be used to study the motion of a bead on a rotating gear or pulley system, which is commonly used in machines and vehicles. It can also be applied to the design and analysis of rotating systems, such as gyroscopes and centrifuges. Additionally, the principles of the Lagrangian equation can be extended to more complex systems, making it a powerful tool in theoretical and applied physics.

Back
Top