 Solution

Let $\mathbf{q}=(x,y)$ be the centre displacement and $\theta$ the small disk rotation. Put $d=l-l_0>0$. The moment of inertia of a uniform disc is $I=m{a}^2/2$, obtained from $I={\int }_0^a\rho r^2(2\pi rdr)$ and $m=\rho \pi a^2$. Thus the kinetic energy is $T=m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$.

For one spring, let ${\mathbf{n}}_j$ point radially towards its fixed anchor and let ${\mathbf{t}}_j$ be the perpendicular unit vector. The moving attachment is $\mathbf{q}+aR(\theta ){\mathbf{n}}_j$, while the anchor is $(a+l){\mathbf{n}}_j$. Expanding their separation to second order gives the spring extension beyond natural length:Insert this in the elastic potential energy $V=(k/2){\sum }_j(s_j-l_0{)}^2$. For the three symmetric directions, ${\sum }_j{\mathbf{n}}_j={\sum }_j{\mathbf{t}}_j=0$ and ${\sum }_j({\mathbf{n}}_j\cdot \mathbf{q}{)}^2={\sum }_j({\mathbf{t}}_j\cdot \mathbf{q}{)}^2=3|\mathbf{q}{|}^2/2$. Dropping the constant equilibrium energy givesTherefore the quadratic Lagrangian isThe prestretch terms cannot be omitted: they supply transverse spring stiffness and the entire rotational restoring torque, as derived generally in small oscillations of a disk attached to prestretched radial springs.