How Does a Mouse's Motion on a Rotating Circle Relate to Lagrangian Mechanics?

[jacobrhcp]

[SOLVED] Lagrange/Hamilton system

Homework Statement

There is a circle without mass, radius r. On the edge of the circle there is a mouse, being forced to move around the circle.

The angle the mouse makes with respect to the centre of the circle is called [tex]\theta(t)[/tex]

At the same time, the circle is hold at it's place at a point Q on the edge and is rotated around this point Q with constant angular velocity [tex]\omega[/tex]. Forget about friction.

write down the equation of motion of the mouse.

The Attempt at a Solution

the number of degrees of freedom for this system is 2, [tex]\omega[/tex] and [tex]\theta[/tex]

The Lagrangian L=K-V
[tex]V=0,
K=\frac{m v^2}{2} [/tex]

where [tex] v=v_c+v_\theta[/tex]
and [tex]v_c[/tex] is the velocity of the centre of the circle, and [tex]v_\theta[/tex] is the velocity of the mouse around the circle.

[tex]v_c=\omega l [/tex]
[tex]v_\theta=\theta' l[/tex]

so [tex] \delta J = \delta {\int_{t_1}}^{t_2} \frac{m (\omega l + \theta' l)^2}{2} dt = \delta {\int_{t_1}}^{t_2} \frac{m (\omega^2 l^2 + 2 \omega \theta' l^2 + \theta'^2 l^2)}{2} dt = 0[/tex]

Now I have two problems.

1) is this a good way to write down the Lagrangian?
2) how do I get the [tex] "\delta" [/tex] inside the integral in a good way?

I suppose after that it's integration by parts and setting the integrand equal to zero.

EDIT: I solved the problem myselve today =)thanks for anyone who was willing to look at it.

 

[Jhero]

Hello,

As a fellow scientist, I would like to offer some comments on your solution.

Firstly, your approach to writing down the Lagrangian is correct, and it is a good way to approach this problem. However, there is one small mistake in your expression for the kinetic energy. The velocity of the mouse around the circle should be v_\theta = r\theta' instead of v_\theta = \theta' l. This is because the distance travelled by the mouse around the circle is r\theta, not l\theta. So the correct expression for the kinetic energy would be:

K=\frac{m(\omega l + r\theta')^2}{2}

Secondly, to get the "\delta" inside the integral, we can use the fact that the Lagrangian is a function of the generalized coordinates and their derivatives. So we can write:

\delta J = \delta \int_{t_1}^{t_2} L(\omega, \theta, \omega', \theta') dt

where L is the Lagrangian. Now, we can use the chain rule to write:

\delta J = \int_{t_1}^{t_2} \frac{\partial L}{\partial \omega} \delta \omega + \frac{\partial L}{\partial \theta} \delta \theta + \frac{\partial L}{\partial \omega'} \delta \omega' + \frac{\partial L}{\partial \theta'} \delta \theta' dt

Now, we can use the Euler-Lagrange equations to write:

\frac{\partial L}{\partial \omega} - \frac{d}{dt} \frac{\partial L}{\partial \omega'} = 0
\frac{\partial L}{\partial \theta} - \frac{d}{dt} \frac{\partial L}{\partial \theta'} = 0

Substituting these into our expression for \delta J, we get:

\delta J = \int_{t_1}^{t_2} \left(\frac{\partial L}{\partial \omega'} \delta \omega' + \frac{\partial L}{\partial \theta'} \delta \theta'\right) dt

Now we can substitute our expression for the Lagrangian and the values of \delta \omega' and \delta \theta' that we get from the Euler-Lag