91Ó°ÊÓ

The distance between the center of the disk and the pivot point is $d=1.75\text{cm}  \text{or}  1.75×{10}^{-2}\text{m}$

The moment of inertia about an axis of rotation is equal to the sum of the moment of inertial about a parallel axis passing through the center of mass and the product of mass and a square of perpendicular distance between two axes. The time period of the physical pendulum can be defined in terms of its moment of inertia, mass, gravitational acceleration, and height.

Use the concept of parallel axis theorem and expression of the period for the physical pendulum.

Formulae:

$I=I_{com}+m{h}^2$ …(¾±)

Here, I is a moment of inertia about any axis, I_com is a moment of inertia about a parallel axis passing through the center of mass, is mass, and is the perpendicular distance between the two axes.

$T=2\mathrm{Ï€}\sqrt{\frac{I}{mgh}}$

…(¾±¾±)

Here, T is the time period g is the gravitational acceleration

According to the parallel axis theorem

$I=I_{com}+m{h}^2$

Here, the distance between the center of the disk and the pivot point is

The axis of rotation passing through the center of the solid disk and perpendicular to the plane is

$I_{com}=\frac{1}{2}m{R}^2$

Therefore, equation (i) becomes

$I=\frac{1}{2}m{R}^2+m{d}^2$

The period of oscillation of the physical pendulum is

$$\begin{gathered} T=2\mathrm{π}\sqrt{\frac{I}{mgh}} \\ =2\mathrm{π}\sqrt{\frac{\frac{1}{2}m{R}^2+m{d}^2}{mgd}} \\ =2\mathrm{π}\sqrt{\frac{R^2+2d^2}{2gd}} \\ =2×3.14\sqrt{\frac{{\left(2.35×{10}^{-2}\text{m}\right)}^2+2×{\left(1.75×{10}^{-2}\text{m}\right)}^2}{2×9.8{\text{m/s}}^{\text{2}}×1.75×{10}^{-2}\text{m}}} \\ =0.366\text{s} \end{gathered}$$

Therefore, the time period of the resulting simple harmonic motion is 0.366 s.