Question

Question: In Figure, a physical pendulum consists of a uniform solid disk (of radius R = 2.35 cm ) supported in a vertical plane by a pivot located a distance d = 1.75 cm from the center of the disk. The disk is displaced by a small angle and released. What is the period of the resulting simple harmonic motion?

Short answer

Answer

The period of the resulting simple harmonic motion is 0.366s

Step-by-step solution

Identification of given data 

1. The radius of the uniform solid disk is $R=2.35\text{cm} \text{or} 2.35\times {10}^{-2}\text{m}$

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

Understanding the concept of physical pendulum

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:

$I=I_{com}+m{h}^2$ …(i)

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.

$T=2\pi \sqrt{\frac{I}{mgh}}$

…(ii)

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

Determining the period of the resulting simple harmonic motion

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

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

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