Question

A uniform disc of mass $m$ and radius $R=\frac{80}{23{\pi }^2}$ is pivoted smoothly at point $P$. If another uniform ring of mass $m$ and radius $R$ is welded at the lowest point of the disc, the time period (in seconds) of the SHM of the system is

Answer 1

COM of the and ring are at $A$ and $B$ respectively.
Hence, centre of mass of system lies at $C$.
$[\because y_{cm}=\frac{m\left(R\right)+m\left(3R\right)}{m+m}=2R]$
The whole system behaves as a physical pendulum.
So, $T=2\pi \sqrt{\frac{I}{mgd}}$
where $d=2R$
$I=I_P=$ moment of inertia about hinged point
& $d=$ distance of COM from hinge point.
So, $I_P={\left(I_{disc}\right)}_P+{\left(I_{ring}\right)}_P$
$I_P=\left(\frac{3}{2}m{R}^2\right)+(m{R}^2+m\times (3R{)}^2)=\frac{23}{2}m{R}^2$
$\Rightarrow T=2\pi \sqrt{\frac{\frac{23}{2}m{R}^2}{\left(2m\right)g\left(2R\right)}}=2\pi \sqrt{\frac{23R}{8g}}$
$\Rightarrow T=2\pi \sqrt{\frac{23}{8\times 10}.\frac{80}{23{\pi }^2}}=2\text{sec}$