= Small oscillations of a disk attached to prestretched radial springs
{title2=$\omega_T^2=\frac{Nk}{2m}(1+\tau/l),\quad\omega_R^2=\frac{2Nk\tau(a+l)}{mal},\quad\tau=l-l_0>0$}
A uniform <rigid body> disk of <mass> $m$ and radius $a$ moves in a frictionless plane. Let $N\ge3$ identical <springs> attach equally spaced rim points to fixed anchors in their radial directions. The equilibrium spring length is $l$, natural length $l_0<l$, and <spring constant> $k$. Write the small centre <displacement> as $\mathbf q$ and the rotation as $\theta$. For a spring with radial and tangential unit <vectors> $\mathbf n,\mathbf t$, its length is
$$
s=l-\mathbf n\cdot\mathbf q+\frac a2\theta^2+
\frac{(\mathbf t\cdot\mathbf q+a\theta)^2}{2l}+O((|\mathbf q|+|\theta|)^3).
$$
This follows by a second-order <Taylor series> for the distance between $\mathbf q+aR(\theta)\mathbf n$ and $(a+l)\mathbf n$. In the <elastic potential energy> $k(s-l_0)^2/2$, the equilibrium tension $k\tau$ multiplies the second-order change of length. Thus it contributes to the linearized stiffness even though the displacement is small.
The equally spaced directions satisfy $\sum\mathbf n=\sum\mathbf t=0$ and $\sum\mathbf n\mathbf n^T=\sum\mathbf t\mathbf t^T=(N/2)I$. Summing the spring energies gives
$$
V_2=\frac{Nk}{4}(1+\tau/l)|\mathbf q|^2+\frac{Nk\tau a(a+l)}{2l}\theta^2.
$$
The <kinetic energy> is $m|\dot{\mathbf q}|^2/2+ma^2\dot\theta^2/4$, using the <moment of inertia of a uniform disc>. The <Euler-Lagrange equations> give the displayed squared <angular frequencies>. The translational <normal modes> are degenerate; rotation is independent. At zero prestretch the rotational frequency vanishes at quadratic order, and finite-angle spring stretching first contributes at higher order.
\Image[/radially-sprung-disk-normal-modes.png]
{title=Translations and rotation of a disk attached to symmetric prestretched radial springs}
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