Lagrangian of two masses connected by a spring in semicircle

[Klas]

Homework Statement

Two masses are connected by a weightless spring in a friction-less semicircular well (Picture included). Derive the equations of motion with help of lagrange

Homework Equations

L = T - U = kinetic energy - potential energy

The Attempt at a Solution

$L = \frac{1}{2}m_1(\dot{x_1}^2 + \dot{y_1}^2) + \frac{1}{2}m_2(\dot{x_2}^2 + \dot{y_2}^2) - (m_1gx_1 + m_2gx_2 + \frac{1}{2}k(\sqrt{x^2-y^2} -l_0)^2)$

Where:
$x_1 = R\cos\theta_1, \;\;x_2 = R\cos(\theta_1+\theta_2) \\y_1 = R\sin\theta_1, \;\;y_2 = R\sin(\theta_1+\theta_2)$
$x=x_2-x_1$
$y=y_2-y_1$

I think I'm going to be able to get the equation of motion from here if only L is right...
And if it's correct it feels like the EOM will be a pain in the ass from here but it feels like I'm missing something rather essential here?

I'll be thankful for any help

 

[DocZaius]

Your gravitational potential energy is of the form "mgy" yet your y seems to point down in your definition.

 

[Klas]

I guess you mean it should rather be $mgRcos(\theta)$. I'm have put potential zero at the bottom so at $mgRcos0$ it should be the max potential energy.

Did an edit on the original post

 

[DocZaius]

No I meant that your gravitational potential terms should be of the form -mgy instead of mgy since you have y increasing downwards in your definition.

 

[Klas]

I see what you mean. If I rewrite it as $L = \frac{1}{2}m_1(\dot{x_1}^2 + \dot{y_1}^2) + \frac{1}{2}m_2(\dot{x_2}^2 + \dot{y_2}^2) - (-m_1gy_1 - m_2gy_2 + \frac{1}{2}k(\sqrt{x^2-y^2} -l_0)^2)$
should it be correct then?

And if, when I rewrite it with $\theta_1, \theta_2$ this part is going to be hard to partial derivative with respect to $\theta1, \theta2$ $(\sqrt{x^2-y^2} -l_0)^2) = (\sqrt{(Rcos(\theta_1+ \theta_2)-Rcos(\theta_1))^2+(Rsin(\theta_1+\theta_2)+Rsin(\theta_2))^2)}-l_0)^2$

 

[DocZaius]

You still have x's in your potential energy. You should put the y's back in. Just make sure to keep the negatives. The rest of the Lagrangian looks good to me.

 

[Klas]

Fixed it. Any tips on how to deal with $(\sqrt{x^2-y^2} -l_0)^2) = (\sqrt{(Rcos(\theta_1+ \theta_2)-Rcos(\theta_1))^2+(Rsin(\theta_1+\theta_2)+Rsin(\theta_2))^2)}-l_0)^2$ when I'm going to partial derivative it later?

Thank you for all your help!

 

[TSny]

Express $T$ and $U$ in terms of $\theta_1$ and $\theta_2$ and their time derivatives. For the distance between the masses, consider the right triangle shown below where one leg of the triangle bisects $\theta_2$.

 

[Klas]

$L = \frac{1}{2}m_1R^2\dot{\theta_1}^2 + \frac{1}{2}m_2R^2(\dot{\theta_1} + \dot{\theta_2})^2 + m_1gRcos\theta_1 + m_2gRcos(\theta_1 + \theta_2) -(\sqrt{(Rcos(\theta_1+ \theta_2)-Rcos(\theta_1))^2+(Rsin(\theta_1+\theta_2)+Rsin(\theta_2))^2)}-l_0)^2$
Is what I get. Since:
$\dot{x_1}^2 + \dot{y_1}^2 = (-R\dot{\theta_1}sin\theta_1)^2 + (R\dot{\theta_1}cos\theta_1)^2 = R^2\dot{\theta_1}^2(cos^2\theta_1 + sin^2\theta_2) = R^2\dot{\theta_1}^2*1$.

TSny. I'm still unsure what to do with that.. For the partial derivative when coming to the sqrt.. Is there anything I'm missing to make my life easier with that part or is it just time to get my hands dirty?

 

[TSny]

You should be able to avoid the square root. Don't use Cartesian coordinates at all.