Radius of Gyration Solved Examples

Example 1: Radius of gyration of a compound pendulum about the point of suspension is 100 mm. The distance between the point suspension and the centre of mass is 250 mm. Considering the acceleration due to gravity is 9.81 m/s². What will be the natural frequency (in radian/s) of the compound pendulum.

Solution:

Given, Radius of Gyration(k) = 100mm = 0.1 m

We know that Moment of Inertia(I) = mk²

Put k = 0.1 in the above equation, we get

I = m(0.1)² kg m²

Also, we know

[Tex]I\ddot{θ} + mg. asinθ = 0 [/Tex]

Here sinθ ≈ θ

[Tex]I\ddot{θ} + mg. aθ = 0 [/Tex]

[Tex]\ddot{θ} + \frac{mga.θ}{I} = 0 [/Tex]

we have ,

I = m(0.1)²

a = 250 mm = 2.5 m

g = 9.81 m/s²

[Tex]\ddot{θ} + \frac{m9.81\times 2.5 \times θ}{m(0.1)^2} = 0 [/Tex]

[Tex]\ddot{θ} + (245.25) θ= 0 [/Tex]

ω_n=√245.25=15.660 rad/s

Hence, the natural frequency of the compound pendulum is 15.660 rad/s.

Example 2: Find the radius of gyration of a disc of mass M and radius 2m rotating about an axis passing through the center of mass and perpendicular to the plane of the disc.

Solution:

We know that, the radius of gyration of a disc is given by K = R/√2.

Given R = 2 m, putting value in the above equation we get

K = 2/ √2

K = √2 m

Hence, the radius of gyration of the given disc is √2 m.

Radius of gyration, R, is a measure used in mechanics and engineering to describe the distribution of mass or inertia of an object relative to its axis of rotation. Radius of Gyration, or the radius of a body, is always centered on its rotational axis. It is a geometric characteristic of a rigid body and is described as the distance between the axis of the body to the point where the body’s moment of inertia is the same as the body’s total moment of inertia. The S.I. unit of the gyration radius is a meter denoted by ‘m’.

In this article, we will discuss the Radius of Gyration, its derivation, formulas, and radius of gyration of a thin rod, circle, and disc along with some applications and significance of radius of gyration.