Pulleys and spring (?)

1. Jan 28, 2014

ShayanJ

Consider the system shown in the picture.
http://updata.ir/images/m9wcnahubdfanhq9edok.jpg [Broken]
What are its degrees of freedom?
And is the Lagrangian below,the correct one for this system?
$\mathfrak{L}=\frac{1}{2} M_2 \dot{x}^2+\frac{1}{2}m_1\dot{x}_1^2+\frac{1}{2}m_2\dot{x}_2^2+\frac{1}{2}I_1\omega_1^2+\frac{1}{2}I_2\omega_2^2-\frac{1}{2}kx^2+M_2gx+m_1g(x+x_1)+m_2g(x+x_2)$
Where x is the distance from the center of the lower pulley to the static platform that the spring is connected to directly and $x_1$ and $x_2$ are the distances from masses to the center of the lower pulley and the Is are the moments of inertia of the pulleys and I have taken the lower platform to be the zero point for gravitational potential energy.
What are the constraints?
I can think of the constancy of the length of ropes and the relation between $\omega$s and the velocity of masses.But I have problem finding the relation between the change of length in the spring and the distance from the lower platform to the center of the lower pulley and I just assumed that they're equal.
Any idea is welcome.
Thanks

Last edited by a moderator: May 6, 2017
2. Jan 28, 2014

Staff: Mentor

Hi Shyan,

I see only two degrees of freedom in the figure (adequate for establishing all the required features of the kinematics): The angular rotation of the upper pulley (as a function of time) and the angular rotation of the lower pulley (as a function of time).

The angular rotation of the upper pulley determines the amount of stretch of the spring, and it also determines how far down the center of the lower pulley moves.

The angular rotation of the lower pulley determines how far mass 1 moves downward and how far mass 2 moves upward relative to the center of the lower pulley.

I hope this helps.

Chet

3. Jan 28, 2014

ShayanJ

Hi chet
Yeah,it was helpful...and after reading it,I was like: " Oohhh...Of course! "...I should have noticed it.But I think I was a little confused.
Thanks