- #1
jacobrhcp
- 169
- 0
[SOLVED] Lagrange/Hamilton system
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 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.
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.
Last edited: