📘 CodingMarble Learn

Moment of Inertia, Radius of Gyration and Rotational Motion

Every idea of straight-line motion has a turning twin: θ for x, ω for v, α for a, I for m, τ for F, L = Iω for p. Moment of inertia I = Σmr² tells how hard it is to change a body's spin; it grows fast when mass sits far from the axis. Radius of gyration k is the distance at which all mass could sit to give the same I: I = Mk². Standard values: ring MR², disc ½MR², solid sphere ⅖MR², rod about centre ML²/12. With constant α: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ, and τ = Iα.

🎬 Step-by-step story

  1. On the left a block slides; on the right a wheel spins. Each linear idea has a rotational twin: x → θ, v → ω, a → α, m → I, F → τ.
  2. A rod carries two masses. Slide them outward. The same torque now speeds the rod up much more slowly. Moment of inertia I = Σmr² went up.
  3. Now imagine all the mass placed on one green ring. At the right radius k the ring has the same I as the real body. That k is the radius of gyration: I = Mk².
  4. Same mass, same size, different shapes: ring, disc, solid sphere, rod. The more mass is far from the axis, the bigger I. Ring MR² is biggest; sphere ⅖MR² is smaller.
  5. Give the wheel a steady twist. Its speed grows by the same amount every second: ω = ω₀ + αt. The angle counter follows θ = ω₀t + ½αt².
  6. Your turn. Pick a mode, move the masses, choose a shape, or change ω₀ and α and run the wheel.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why is it called moment of inertia and not just mass?

Mass resists being pushed in a line. For turning, what matters is mass and how far it is from the axis. Same mass farther out resists turning much more, as the rod with sliding masses shows.

Why does r come squared in I = mr²?

A bit far from the axis moves faster (v = rω) and also needs more torque (τ = rF). Two factors of r give r². Move the masses out to 2× and I becomes 4×.

Is the radius of gyration a real part of the body?

No. It is a make-believe radius: if all mass sat at k, I would stay the same. The green ring shows this imaginary ring.

Why does a ring have more I than a disc of the same mass and radius?

In a ring, all mass is at R. In a disc, a lot of mass is near the centre, where it adds little to I. So ring MR² > disc ½MR².

Can I use rpm directly in ω = ω₀ + αt?

Only if α is also in rpm-type units. Safer: convert to rad/s. 1 rpm = 2π/60 rad/s.

Is τ = Iα always true?

It holds for rotation about a fixed axis (or about the CM) with I constant. When I changes, use τ = dL/dt instead.

Linear vs rotational motion

In translation, every point of a body moves the same way. In rotation about a fixed axis, every point moves in a circle round the axis, and all points turn through the same angle.

LinearRotational
displacement xangular displacement θ (rad)
velocity vangular velocity ω (rad/s)
acceleration aangular acceleration α (rad/s²)
mass mmoment of inertia I
force F = matorque τ = Iα
momentum p = mvangular momentum L = Iω
KE = ½mv²KE = ½Iω²
work = Fs, power = Fvwork = τθ, power = τω

Links: v = rω, a_t = rα.

Moment of inertia

Moment of inertia (I) is rotational inertia: how much a body resists a change in its spin about an axis.

I = Σmᵢrᵢ² (for a solid body, I = ∫r² dm), where r is the perpendicular distance of each bit from the axis. Unit kg m², dimensions [M L²].

I depends on (1) the mass, (2) how the mass is spread about the axis, and (3) which axis you choose. It is not fixed for a body the way mass is. Doubling r of a bit makes its share four times bigger.

Rotational Newton's law: τ = Iα. Same torque, bigger I → smaller α.

Radius of gyration

The radius of gyration k is the distance from the axis at which the whole mass M could be placed (as a thin ring) to give the same moment of inertia.

I = Mk², so k = √(I/M). Unit: metre.

It is a root-mean-square distance: k = √((m₁r₁² + m₂r₂² + …)/M). For n equal particles, k = √((r₁² + r₂² + … + rₙ²)/n).

Examples: ring k = R, disc k = R/√2, solid sphere k = √(2/5) R, rod about centre k = L/√12.

Moment of inertia of simple shapes (no derivation)

Learn these values (M = mass, R = radius, L = length). The syllabus does not ask for their derivation.

BodyAxisI
Thin ringthrough centre, ⟂ to planeMR²
Thin ringa diameter½MR²
Uniform discthrough centre, ⟂ to plane½MR²
Uniform disca diameter¼MR²
Hollow cylinderits own axisMR²
Solid cylinderits own axis½MR²
Solid spherea diameter⅖MR²
Thin spherical shella diameter⅔MR²
Thin rodthrough centre, ⟂ to rodML²/12
Thin rodthrough one end, ⟂ to rodML²/3

Pattern: the more mass lies far from the axis, the larger the fraction. A ring has all mass at R, so it gets the full MR².

Equations of rotational motion

For rotation about a fixed axis with constant angular acceleration α, the equations copy the linear ones exactly:

Here ω₀ is the starting angular velocity. Always use radians: 1 revolution = 2π rad; n rev/s gives ω = 2πn rad/s; rpm ÷ 60 gives rev/s.

Dynamics: τ = Iα, work W = τθ, power P = τω, rotational KE = ½Iω².

Try it at home

Tape two equal coin stacks on a ruler, first 2 cm from the centre, then at the two ends. Hold the middle and twist the ruler back and forth. Predict which is harder, then feel it: the far coins make I bigger. Also spin a pencil holding it at the centre, then at one end: the end is harder to swing (ML²/3 is four times ML²/12).

Key formulas and definitions

Worked examples

1. Two 2 kg masses sit 0.5 m on either side of a light rod's centre. Find I about the centre.

I = Σmr² = 2 × 0.5² + 2 × 0.5² = 0.5 + 0.5 = 1 kg m².

2. The same masses are moved to 1 m from the centre. By what factor does I change?

I = 2 × 1² + 2 × 1² = 4 kg m². Distance doubled, I became 4 times (r²).

3. A disc of mass 2 kg and radius 0.3 m spins about its axis. Find I and k.

I = ½MR² = ½ × 2 × 0.09 = 0.09 kg m². k = √(I/M) = √0.045 = 0.212 m (= R/√2).

4. A 1.2 m thin rod of mass 0.5 kg turns about an axis through its centre. Find I; then find I about one end.

Centre: ML²/12 = 0.5 × 1.44/12 = 0.06 kg m². End: ML²/3 = 0.24 kg m² (4 times more).

5. A fan starts from rest with α = 2 rad/s². Find ω after 5 s and the angle turned.

ω = 0 + 2 × 5 = 10 rad/s. θ = ½ × 2 × 25 = 25 rad ≈ 4 rev.

6. A wheel at 1200 rpm is braked uniformly to rest in 20 s. Find α and the number of turns.

ω₀ = 1200/60 × 2π = 40π rad/s. α = (0 − 40π)/20 = −2π rad/s² ≈ −6.28. θ = (ω₀ + ω)t/2 = 20π × 20 = 400π rad = 200 rev.

7. A torque of 6 N m acts on a solid sphere (M = 5 kg, R = 0.2 m). Find α.

I = ⅖ × 5 × 0.04 = 0.08 kg m². α = τ/I = 6/0.08 = 75 rad/s².

8. A flywheel (I = 2 kg m²) spins at 30 rad/s. Find its KE and the constant torque needed to stop it in 10 s.

KE = ½Iω² = ½ × 2 × 900 = 900 J. α = 30/10 = 3 rad/s², τ = Iα = 6 N m.

Common mistakes

Practice quiz

1. Moment of inertia plays the role of which linear quantity?
2. I of a uniform disc about its axis is:
3. Radius of gyration k is given by:
4. The rotational twin of v = u + at is:
5. Same mass and radius. Which has the largest I about its central axis?

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is moment of inertia in Class 11?

It is rotational inertia, I = Σmr², the resistance of a body to change in its rotation about an axis. Unit kg m².

What is radius of gyration?

The distance k from the axis where the whole mass could be placed to give the same I: I = Mk², k = √(I/M).

What are the equations of rotational motion?

For constant α: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ.

Where this is taught

PolandLiceum ogólnokształcące, klasa IRigid body mechanics
Ukraine10 класMechanics
CBSE (India)Class 11Motion of System of Particles and Rigid Body
England (GCSE, A level)Year 133.11 Engineering physics
USA (Common Core, NGSS, AP)Grade 11Torque and Rotational Dynamics
USA (Common Core, NGSS, AP)Grade 11Energy and Momentum of Rotating Systems
USA (Common Core, NGSS, AP)Grade 12Torque and Rotational Dynamics
USA (Common Core, NGSS, AP)Grade 12Energy and Momentum of Rotating Systems

Learn first

Learn next

Related lessons

All Physics lessons