Question

A uniform disc of mass $m$ and radius $R$ is pivoted smoothly at its centre of mass. A light spring of stiffness $k$ is attached with the disc tangentially as shown in the figure. If the disc is rotated through a small angle and released, the angular frequency of oscillation of the disc is (in $\text{rad/s}$)
[(Take $m=5\text{kg}$ and $k=10\text{N/m}$]

Answer 1

Let us twist (rotate) the disc through a small clockwise angle $\theta$. Then, the spring will be deformed (compressed) by a distance $x=R\theta$. Hence, the spring force $F_s=kx=k\left(R\theta \right)$ will produce a restoring torque.


Restoring torque, $\tau =-F_{s}R$ where $F_s=kR\theta$
This gives, $\tau =-k{R}^2\theta .....\left(1\right)$
It means that after removing the external, (applied) torque, the restoring torque rotates the disc with an angular acceleration $\alpha$ which will bring the spring disc system back to its original state.
Applying Newton's second law for rotational motion, we have
$\tau =I\alpha$
Using $\left(1\right)$ in the above formula, we get
$-k{R}^2\theta =I\alpha$
This gives $\alpha =-\frac{k{R}^2\theta }{I}$
Moment of inertia of a disc about an axis passing through its COM is given by
$I=\frac{m{R}^2}{2}$
Thus, $\alpha =-\frac{2k}{m}\theta$
Comparing the above equation with $\alpha =-{\omega }^2\theta$, we get
$\omega =\sqrt{\frac{2k}{m}}$
From the data given in the question,
$\omega =\sqrt{\frac{2\times 10}{5}}\Rightarrow \omega =2\text{rad/sec}$