A uniform thin disk of mass 1 kg and radius 0.1 m is kept on a surface as shown in the figure. A spring of stiffness k₁ = 400 N/m is connected to the disk center A and another spring of stiffness k₂ = 100 N/m is connected at point B just above point A on the circumference of the disk. Initially, both the springs are unstretched. Assume pure rolling of the disk. For small disturbance from the equilibrium, the natural frequency of vibration of the system is _______ rad/s (round off to one decimal place)

The correct answer is 23.1 (or 23.094).

Based on the provided system diagram, we have a uniform thin disk of mass m = 1 kg and radius R = 0.1 m executing pure rolling on a flat horizontal surface. Point A represents the center of the disk, and point B is at the top circumference directly above A. A spring of stiffness k₁ = 400 N/m is attached to the center A, and another spring of stiffness k₂ = 100 N/m is attached to point B.

Let $\theta$ be a small angular displacement of the disk about its contact point with the surface (which acts as the instantaneous center of rotation due to pure rolling).
The horizontal displacement of the disk center A is:
$x_1=R\theta$
The horizontal displacement of the top point B is:
$x_2=2R\theta$

The kinetic energy (KE) of the disk in pure rolling is:
$\text{KE}=\frac{1}{2}{I}_c{\dot{\theta }}^2$
where $I_c$ is the mass moment of inertia of the disk about the instantaneous center of rotation at the contact point:
$I_c=I_g+m{R}^2=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2$
Substituting $I_c$ into the kinetic energy equation yields:
$\text{KE}=\frac{3}{4}m{R}^2{\dot{\theta }}^2$

The potential energy (PE) stored in the springs is:
$\text{PE}=\frac{1}{2}{k}_1{x}_1^2+\frac{1}{2}{k}_2{x}_2^2$
Substituting $x_1$ and $x_2$:
$\text{PE}=\frac{1}{2}{k}_1{\left(R\theta \right)}^2+\frac{1}{2}{k}_2{\left(2R\theta \right)}^2=\frac{1}{2}\left(k_1+4k_2\right)R^2{\theta }^2$

Using the conservation of total mechanical energy:
$\frac{d}{dt}\left(\text{KE}+\text{PE}\right)=0$
$\frac{d}{dt}\left(\frac{3}{4}m{R}^2{\dot{\theta }}^2+\frac{1}{2}\left(k_1+4k_2\right)R^2{\theta }^2\right)=0$
Differentiating with respect to time:
$\frac{3}{2}m{R}^2\dot{\theta }\ddot{\theta }+\left(k_1+4k_2\right)R^2\theta \dot{\theta }=0$
Simplifying by dividing through by $R^2\dot{\theta }$:
$\frac{3}{2}m\ddot{\theta }+\left(k_1+4k_2\right)\theta =0$
$\ddot{\theta }+\frac{2\left(k_1+4k_2\right)}{3m}\theta =0$

Comparing this with the standard equation of angular simple harmonic motion $\ddot{\theta }+{\omega }_n^2\theta =0$, the natural frequency of the system is:
${\omega }_n=\sqrt{\frac{2\left(k_1+4k_2\right)}{3m}}$

Substituting the given numerical values:
${\omega }_n=\sqrt{\frac{2\left(400+4\times 100\right)}{3\times 1}}$
${\omega }_n=\sqrt{\frac{2\times 800}{3}}=\sqrt{\frac{1600}{3}}=\sqrt{533.33}\approx 23.094\text{rad/s}$

Rounding to one decimal place, the natural frequency of the system is 23.1 rad/s.