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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. Table of Content What is the Radius of Gyration?Radius of Gyration is the imaginary distance from the centroid of the given axis to a point where the whole mass of a body is supposed to be concentrated. Its moment of inertia about the given axis is equal to the actual distribution of mass. It is represented by ‘k’. Mathematically, it can be described as the root mean square distance from the given axis,
Radius of Gyration Definition
Radius of Gyration FormulaThe radius of gyration can be expressed numerically as the root mean square distance of particles from the axis of rotation. The Formula to calculate the radius of gyration can be derived as follows: The moment of inertia of a body with mass (m) can be given by: I = mk² ….(1) Here, k denotes the radius of gyration. So, the formula for radius of gyration can be given by: K = √I/m ….(2) Note: Equation (1) can be used to determine the moment of inertia of any rigid body by knowing its radius of gyration. Now consider a body that consists of n particles, each with mass m, and the perpendicular distance of particles from the axis of rotation be r1, r2, r3,…, rn. Substituting these values in equation 1 we get: I = m1r1² + m2r2² + m3r3² + ……. + mnrn² ……(3) But we have to consider the mass of each particle as ‘m’ so our equation becomes: I = m(r1²+ r2² + r3² + …….. + rn²) Multiply the right-hand side by n/n we get, I = mn(r1²+ r2² + r3² + …….. + rn²)/n Now, we can write mn as the M, which represents the body’s overall mass. So, the equation will become: I = M(r1²+ r2² + r3² + …….. + rn²)/n ……..(4) Substituting I = mk² in equation (4) we get, Mk² = M(r²1 + r²2 + r²3 + …….. + r²n)/n ⇒ k² = (r²1 + r²2 + r²3 + …….. + r²n)/n ⇒ k = √r²1 + r²2 + r²3 + …….. + r²n/n From the above equation, we can say that the root-mean-square distance of the various body particles is referred as the radius of gyration of that body. Unit of Radius of GyrationThe radius of gyration represents a distance, so the SI unit of the radius of gyration is meter denoted by m. Dimension of Radius of GyrationThe dimensional formula of the radius of gyration is M0 L1 T0. Factors Affecting Radius of GyrationThe radius of gyration depends on three key factors: Shape and size of the object: Different shapes have different mass distributions, causing changes in the mass elements’ average distance from the rotational axis. E.g., Even though their overall sizes may be comparable, a sphere’s radius of gyration is smaller than that of a long, thin rod with the same mass. Configuration and Position of the axis of rotation: The axis that is chosen significantly affects the calculated radius of gyration. E.g., A cylinder’s value will vary depending on whether it is rotated around its short or long axis. Mass distribution within the object: Uneven distribution of mass will influence the radius of gyration. The radius of gyration of an object will be greater for denser or heavier parts that are positioned farther from the axis of rotation than for objects with a more uniform mass distribution. Radius of Gyration of Various ObjectsThe radius of gyration varies with distinct objects. The explanation of few of them is below:
Radius of Gyration of Compound PendulumThe radius of gyration of a compound pendulum depends upon its mass distribution. It is described as the distance from the axis of rotation to the point where the mass of the whole body is concentrated to produce the same moment of inertia as the actual mass distribution. The formula to calculate the radius of gyration of a compound pendulum is given by:
Radius of Gyration of Solid SphereMoment of Inertia of a solid sphere about an axis passing through its center and perpendicular to its plane with mass M and radius R is given by:
![]() Radius of Gyration of a Sphere But we know I = MK2, so equation 1 becomes: MK2 = 2 MR2 / 5 On canceling M from both sides and solving, we get K2 = 2R2/5 By taking the square root, we get: K = √2/5 R
Radius of Gyration of Circular RingThe Moment of Inertia of a Circular ring about an axis passing through the tangent of the ring in its plane with radius R can be calculated by the parallel axis theorem. According to this, we have: I = Icom + Md2 , Here d = R, ⇒ I = MR²/2 + MR2
![]() Radius of Gyration of a Circular Ring We know I = MK2, so equation 1 becomes: MK2 = 3/2 MR2 On canceling M from both sides and solving, we get K2 = 3/2 . R2 By taking the square root on both sides, we get: K = √3/2 . R
Radius of Gyration of DiscThe Moment of Inertia of a disc about an axis passing through its center and perpendicular to its plane with radius R and mass M is:
![]() Radius of Gyration of a Disc We know I = MK2, so equation 1 becomes: MK2 = MR2 / 2 On canceling M from both sides and solving, we get K2 = R2 / 2 By taking the square root, we get: K = R / √2
Radius of Gyration of Thin RodThe moment of inertia of a uniform rod about an axis passing through its center and perpendicular to its plane with mass M and length L is given by:
![]() Radius of Gyration of a Thin Rod If K is the radius of the thin rod about an axis, then the equation will be I = Mk² By equating the value of the moment of Inertia, we get Mk² = ML²/12 On canceling M from both sides and solving, we get k² = L²/12 By taking the square root, we get: K = L/√12
Radius of Gyration Solved ExamplesExample 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:
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:
Significance of Radius of GyrationThe significance of the radius of gyration is listed below:
Applications of Radius of GyrationThe radius of gyration is a crucial parameter in various fields and applications due to its ability to characterize the distribution of mass in a given object or system.
Also, Check Radius of Gyration FAQsWhat is meant by Radius of Gyration?
On what Factors Does Radius of Gyration Depend?
What is the SI Unit of Radius of Gyration?
What are the Uses of Radius of Gyration?
What is the Difference Between Radius of Gyration and Moment of Inertia?
What is the Symbol for Radius of Gyration?
What is Radius of Gyration Range?
How does Radius of Gyration Increase?
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