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

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Answer

The period of the resulting simple harmonic motion is 0.366s

Step by step solution

01

Identification of given data 

  1. The radius of the uniform solid disk is R=2.35cmor2.35×10-2m

The distance between the center of the disk and the pivot point is d=1.75cmor1.75×10-2m

02

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=Icom+mh2 …(i)

Here, I is a moment of inertia about any axis, Icom 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=2πImgh

…(ii)

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

03

Determining the period of the resulting simple harmonic motion

According to the parallel axis theorem

I=Icom+mh2

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

Icom=12mR2

Therefore, equation (i) becomes

I=12mR2+md2

The period of oscillation of the physical pendulum is

T=2πImgh=2π12mR2+md2mgd=2πR2+2d22gd=2×3.142.35×10-2m2+2×1.75×10-2m22×9.8m/s2×1.75×10-2m=0.366s

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

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