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Formula of Radius of Gyration

Rotation and circular motion formulasLengthClass 11Updated 1 Nov 2026

K=IMK = \sqrt{\frac{I}{M}}
SI unit
m
Dimensions
[M⁰ L¹ T⁰]
where
  • K radius of gyration about the axis m
  • I moment of inertia about the same axis kg m²
  • M total mass of the body kg

The formula of radius of gyration is K=IMK = \sqrt{\frac{I}{M}}, so that I=MK2I = MK^2. It is the distance from the axis at which you could put the whole mass of a body, as a single point or a thin ring, and get the same moment of inertia. It is a length, measured in metres.

K turns a messy moment of inertia into one easy number. A ring of radius R has all its mass at R, so its K is R. A disc of the same radius has mass spread from the centre outward, so its K is smaller, R2\frac{R}{\sqrt{2}}. The topic sits in the Class 11 rotation chapter, and students also meet it as radius of gyration formula Class 12 revision for JEE and NEET. It depends entirely on the moment of inertia about the chosen axis.

Quick facts

  • Formula

    K=IMK = \sqrt{\frac{I}{M}}

  • Rearranged

    I=MK2I = MK^2

  • For particles

    K=m1r12+m2r22+…m1+m2+…K = \sqrt{\frac{m_1r_1^2 + m_2r_2^2 + \dots}{m_1 + m_2 + \dots}}

  • SI unit

    metre (m)

  • Dimensional formula

    [M⁰ L¹ T⁰]

  • Depends on

    Axis and mass spread, not total mass

Formula of radius of gyration for common bodies

K for uniform bodies of mass M
Body and axisIK
Ring, radius R, central axisMR2MR^2RR
Disc or solid cylinder, central axis12MR2\frac{1}{2}MR^2R2≈0.707R\frac{R}{\sqrt{2}} \approx 0.707R
Rod, length L, axis through centre112ML2\frac{1}{12}ML^2L23≈0.289L\frac{L}{2\sqrt{3}} \approx 0.289L
Rod, length L, axis through one end13ML2\frac{1}{3}ML^2L3≈0.577L\frac{L}{\sqrt{3}} \approx 0.577L
Solid sphere, diameter25MR2\frac{2}{5}MR^225R≈0.632R\sqrt{\frac{2}{5}}R \approx 0.632R
Hollow sphere, diameter23MR2\frac{2}{3}MR^223R≈0.816R\sqrt{\frac{2}{3}}R \approx 0.816R

Notice that K never exceeds the largest distance of any mass from the axis, and it equals that distance only when all mass is there, as in the ring.

Worked examples

Example 1: from a known I

A flywheel has I = 0.5 kg m² and mass 2 kg. Find K.

K=0.52=0.25=0.5 mK = \sqrt{\frac{0.5}{2}} = \sqrt{0.25} = 0.5\ \text{m}

Example 2: a disc

Find K for a uniform disc of radius 20 cm about its central axis.

K=R2=0.201.414≈0.141 mK = \frac{R}{\sqrt{2}} = \frac{0.20}{1.414} \approx 0.141\ \text{m}. The mass does not matter.

Example 3: rod about its centre, then its end

A uniform rod is 1.2 m long. Find K about its centre and about one end.

Centre: K=1.223=1.23.464≈0.346 mK = \frac{1.2}{2\sqrt{3}} = \frac{1.2}{3.464} \approx 0.346\ \text{m}.

End: K=1.23≈0.693 mK = \frac{1.2}{\sqrt{3}} \approx 0.693\ \text{m}. The shifted axis rule confirms it: Kend2=Kcm2+(L/2)2=0.12+0.36=0.48K_{end}^2 = K_{cm}^2 + (L/2)^2 = 0.12 + 0.36 = 0.48, and 0.48≈0.693\sqrt{0.48} \approx 0.693.

Where K is useful

Once you know K, rotational formulas look like linear ones. Kinetic energy of rolling, for example, is 12Mv2(1+K2R2)\frac{1}{2}Mv^2\left(1 + \frac{K^2}{R^2}\right). That gives a famous result: for a body rolling without slipping down a slope of angle θ, the acceleration is a=gsin⁡θ1+K2/R2a = \frac{g\sin\theta}{1 + K^2/R^2}. A solid sphere (K²/R² = 0.4) beats a disc (0.5), which beats a ring (1).

Engineers use K in column buckling, where a slender column with a small K buckles at a lower load. The area based version links to the polar moment of inertia. Experimentally, K can be found from a bifilar suspension timing.

Units, dimensions and traps

kg m2/kg=m\sqrt{\text{kg m}^2 / \text{kg}} = \text{m}, which in base quantities is pure length, [M⁰ L¹ T⁰]. K is a length, not an area, and carries no mass.

  • Forgetting the square root and quoting I/M, which has unit m².
  • Quoting K without the axis. A rod has a different K about its centre and its end.
  • Assuming K is the distance to the center of mass. It is not; for a rod about its centre, the center of mass is on the axis but K = 0.289L.

For links to torque, angular momentum and the rest, use the circular motion and rotation hub.

Radius of gyration of a group of particles

For separate masses, find I first and then divide by the total mass: K=∑miri2∑miK = \sqrt{\frac{\sum m_i r_i^2}{\sum m_i}}. This is a root mean square distance, weighted by mass.

Example 4: three beads on a light wire

Beads of 1 kg, 2 kg and 1 kg sit 0.1 m, 0.2 m and 0.3 m from an axis. Find K.

I=1(0.01)+2(0.04)+1(0.09)=0.18 kg m2I = 1(0.01) + 2(0.04) + 1(0.09) = 0.18\ \text{kg m}^2, total mass 4 kg.

K=0.184=0.045≈0.212 mK = \sqrt{\frac{0.18}{4}} = \sqrt{0.045} \approx 0.212\ \text{m}

K is a little more than the average distance of 0.2 m, because squaring favours the outer bead.

Frequently asked questions

What is the formula for radius of gyration?

K = √(I / M), where I is the moment of inertia about the axis and M is the total mass. Rearranged, I = MK².

What is the SI unit of radius of gyration?

Being a length, K is measured in metres. Grams and kilograms play no part in its unit.

Does radius of gyration depend on mass?

Not on the total mass, since mass cancels. It depends on how the mass is spread around the axis and on which axis you choose.

What is the radius of gyration of a disc?

About its central axis K = R / √2, about 0.707R. About a diameter it is R / 2.

Is radius of gyration a Class 11 or a Class 12 topic?

Indian boards teach it with rotational motion in Class 11, and students revise it again in Class 12 while preparing for entrance exams.