Torque, Moment of Inertia and Angular Acceleration

τ = Iα is Newton’s second law with every quantity swapped for its rotational twin: torque for force, moment of inertia for mass, angular acceleration for acceleration. Pick a shape, give a mass, a radius and a torque, and this builds the moment of inertia, the angular acceleration, the speed after a given time, and the radius of gyration.

Last updated 11 Nov 2026

τ = Iα for a rigid body about a fixed axis

Shape to I, torque to angular acceleration

Shape, and the axis it turns aboutEach option carries the numerical coefficient in I = k mr², which is the list the CBSE Class XI syllabus asks for as “values of moments of inertia for simple geometrical objects (no derivation)”. Two warnings. For either rod option the length field below means the WHOLE length of the rod, not half of it. And if you already know I from some other source, choose the ring option — which is I = mr² — and type √(I/m) into the radius field: that is the radius of gyration, and it reproduces any moment of inertia you like.

Mass of the bodyIn kilograms. Unlike the straight-line case, the mass alone tells you very little here: the same 2 kg has four times the moment of inertia as a ring as it does as a disc of half the radius, and that is before you change the axis.

Radius, or the full length for a rodIn metres. This is the dimension that dominates everything, because it enters SQUARED: double it and the moment of inertia quadruples while the mass is unchanged. It is also the field most often filled in wrongly — for the two rod options it is the full length L and not the distance from the axis to the end, and a diameter typed where a radius belongs gives an answer four times too large.

Torque applied about the axisIn newton metres. If the question gives you a force and a distance instead, the torque is τ = rF sin θ, where θ is the angle between the force and the line from the axis to where it acts — so a force pointing straight at the axis has no turning effect at all, and the full rF only applies when the force is perpendicular. Zero is allowed and gives zero angular acceleration, which is to say a body spinning steadily or not spinning.

Time the torque acts for, from restIn seconds, used for the rows that follow the motion rather than just the acceleration. They assume the body starts from REST and that the torque is constant throughout, so they are ω = αt and θ = ½αt² — the rotational twins of v = at and s = ½at². Zero is allowed and means nothing has happened yet.

16.6667rad/s²Example

A 2 kg solid disc of radius 0.3 m, with 1.5 N m applied for 4 s from rest

Newton’s second law with every term replaced by its rotational twin

τ = Iα · I = k mr² · τ = rF sin θ · ω = αt · θ = ½αt² · L = Iω · E = ½Iω²

τ = Iα: the rotational form of F = ma, and the correspondence is complete: torque plays the part of force, moment of inertia the part of mass, angular acceleration the part of acceleration. Every straight-line result you know has a twin — momentum mv becomes angular momentum Iω, kinetic energy ½mv² becomes ½Iω², work Fs becomes τθ. The CBSE Class XI list asks for exactly this, as “comparison of linear and rotational motions”
I = k mr²: the moment of inertia, and the thing with no straight-line analogue: mass alone does not determine it, because WHERE the mass sits matters. The coefficient k is the shape factor, from ½ for a disc down to 1/12 for a rod about its centre. The r is squared, so it dominates
τ = rF sin θ: the torque a force produces, where θ is the angle between the force and the line from the axis to its point of application. A force aimed at the axis produces no torque whatever its size, which is why a door will not open if you push the hinge. Cambridge 9702 asks you to “define and apply the moment of a force” and separately “the torque of a couple” — a couple being two equal opposite forces, which turn a body without pushing it anywhere
k = √(I/m): the radius of gyration, named in the CBSE list in its own right: the single distance at which the whole mass could be placed to give the same moment of inertia. For a disc it is 0.707r, for a thin shell 0.816r, for a solid sphere 0.632r. It is also the practical way to use this page with a moment of inertia you already have from elsewhere
radians, always: every angular quantity here is in radians and radians per second, which is what makes ω = αt and v = ωr work without a conversion factor. Degrees and revolutions per minute are printed for comparison; never substitute either into a formula. One revolution is 2π radians, and one rad/s is 9.5493 rpm

Worked example

A 2 kg solid disc of radius 0.3 m, with 1.5 N m applied for 4 s from rest
THE MOMENT OF INERTIA FIRST, because nothing else can be done without it. For a solid disc about its own axis, I = ½mr² = 0.5 × 2 × 0.3² = 0.090000 kg m². Note how small that number is: the radius is squared, and 0.3² is 0.09.
THE ANGULAR ACCELERATION. α = τ/I = 1.5 / 0.09 = 16.6667 rad/s². In radians per second squared, which is the only unit the rest of the arithmetic will accept.
FOLLOW THE MOTION, from rest, using the rotational twins of the straight-line equations. ω = αt = 16.6667 × 4 = 66.6667 rad/s, which is 636.62 rpm. θ = ½αt² = 0.5 × 16.6667 × 16 = 133.3333 rad, which is 21.22 revolutions.
THE TWO ENERGY FIGURES, AND WHY BOTH ARE PRINTED. Rotational kinetic energy: ½Iω² = 0.5 × 0.09 × 66.6667² = 200.0000 J. Work done by the torque: τθ = 1.5 × 133.3333 = 200.0000 J. They agree because the work-energy theorem holds in rotation too, and seeing them agree is worth more than either number on its own.
AND THE SANITY CHECK ON THE SHAPE. Change the disc to a thin ring of the same mass and radius and I doubles to 0.18, so α halves to 8.33 rad/s² — same mass, same radius, same torque, half the response, because the ring keeps all its mass out at the rim. That sensitivity to where the mass sits, and not merely how much there is, is the one genuinely new idea in rotational dynamics.

The shape factor k in I = k mr², and the radius of gyration it implies

The last column is the same mass, the same radius and the same torque in every row, so the only thing changing is where the mass sits — and the angular acceleration varies by a factor of twelve across the list. That is the difference between rotational and straight-line dynamics in one column: in F = ma the mass is the whole story, and in τ = Iα it is barely half of it. Two rows deserve a second look. The ring and the point mass are identical because in both cases every bit of the mass is at the same distance r, which is also why k can never exceed 1 for a body contained within radius r. And the two rod rows differ by a factor of four for the same rod, because the axis has moved from the centre to the end: that is the parallel-axis theorem, I = I_centre + md², with d = L/2. The radius of gyration column is the same information read as a distance — the single place you could put the whole mass and get the same answer — and it is √k, which is why it is always closer to 1 than k is.
Body and axis	Moment of inertia	k	Radius of gyration, as a fraction of r	Angular acceleration for 1.5 N m on 2 kg at r = 0.3 m
Thin ring or hoop, about its axis	mr²	1.0000	1.000	8.3333 rad/s²
Point mass at distance r	mr²	1.0000	1.000	8.3333 rad/s²
Thin spherical shell, about a diameter	⅔mr²	0.6667	0.816	12.5000 rad/s²
Solid disc or cylinder, about its axis	½mr²	0.5000	0.707	16.6667 rad/s²
Solid sphere, about a diameter	⅖mr²	0.4000	0.632	20.8333 rad/s²
Uniform rod about one end, r = L	mL²/3	0.3333	0.577	25.0000 rad/s²
Uniform rod about its centre, r = L	mL²/12	0.0833	0.289	100.0000 rad/s²

Where the mass sits, not just how much of it there is

Rotational dynamics is the straight-line kind with every symbol replaced, and one genuinely new idea. The replacements are mechanical: force becomes torque, mass becomes moment of inertia, acceleration becomes angular acceleration, F = ma becomes τ = Iα. Momentum mv becomes angular momentum Iω; kinetic energy ½mv² becomes ½Iω²; work Fs becomes τθ; the constant-acceleration equations keep their shape with every letter swapped. The CBSE Class XI syllabus asks for precisely this as “comparison of linear and rotational motions”, and learning the table of correspondences is a better use of an evening than learning the rotational formulae separately.

The new idea is that mass alone no longer determines the response. A 2 kg disc and a 2 kg ring of the same radius have moments of inertia differing by a factor of two, so the same torque spins one up twice as fast as the other. Across the shapes in the table on this page the factor is twelve. What matters is the distance of each piece of mass from the axis, and it matters SQUARED, which is why the radius is the dominant input on this page and why typing a diameter where a radius belongs makes the answer four times wrong. It is also why a figure skater pulling their arms in spins faster: the angular momentum Iω is conserved, and reducing I has to increase ω.

Torque is not force, and a force aimed at the axis has none. τ = rF sin θ, with θ measured between the force and the line from the axis to its point of application. Push on a door at the handle and it swings; push with the same force at the hinge and nothing happens, because r is zero; push along the line towards the hinge and nothing happens either, because sin θ is zero. The perpendicular distance from the axis to the line of the force — the moment arm — is the only geometry that counts. A couple is the limiting case worth knowing: two equal and opposite forces at different points give a resultant force of zero and a resultant torque that is not zero, so the body spins up while its centre of mass stays put.

Radians, and why nothing else will do. Every angular quantity on this page is in radians or radians per second. That is not a preference: the relations v = ωr, a = αr and s = θr are true in radians and false in degrees, because the radian is defined as the angle whose arc equals its radius. The page prints revolutions and revolutions per minute beside the radian figures because that is how machinery is quoted, but those are output only. One revolution is 2π radians; one radian per second is 9.5493 rpm.

What this page is not. It is school rigid-body dynamics about a single fixed axis: a flywheel on a shaft, a pulley being spun up, a rod pivoted at one end. It is not an engineering torque calculation. It will not tell you what to tighten a bolt to, what a gear train does to a motor’s output, how much power a shaft is transmitting at a given speed, or what a spring’s torsional rate is — all of which are different questions with their own standards and safety factors, and all of which live elsewhere on this site under mechanical engineering rather than here. It also does not handle rolling, where a body translates and rotates together and the two motions are tied by v = ωr, nor an axis anywhere other than the one each shape in the list names; moving the axis needs the parallel-axis theorem, I = I_centre + md², and the two rod entries in the table are a worked example of it.

Frequently asked questions

What is the moment of inertia, in plain terms?

The rotational equivalent of mass: how hard it is to change a body’s rate of spin. Unlike mass, it is not a property of the body alone — it depends on the axis as well, because what counts is how far each piece of mass is from that axis, squared. The same rod has four times the moment of inertia about its end as about its centre. Its unit is kg m², which already tells you that a distance is involved.

Why does the same torque spin a disc up faster than a ring?

Because the ring keeps all of its mass at the rim, as far from the axis as it can be, and distance from the axis counts squared. A disc of the same mass and radius has much of its material close in, so its moment of inertia is ½mr² against the ring’s mr² — half as much, so twice the angular acceleration for the same torque. Nothing about the amount of material has changed; only where it is.

What is the radius of gyration for?

It is the single distance at which you could concentrate the whole mass and get the same moment of inertia: k = √(I/m). It turns an awkward integral into one number, so it is how engineers quote the rotational inertia of a real component whose shape has no formula. It is also the useful trick for this page: if you already know I, choose the ring option and type √(I/m) as the radius, and the calculation runs on your own value.

How do I get a torque from a force?

τ = rF sin θ, where r is the distance from the axis to where the force acts and θ is the angle between the force and that line. When the force is perpendicular to the arm, which is the usual case, it is simply rF. Equivalently, multiply the force by the perpendicular distance from the axis to the line of the force. A force whose line passes through the axis has no torque about it, however large it is.

Must I use radians?

Yes, in every formula. ω = αt, θ = ½αt², v = ωr and a = αr are all statements about radians, and substituting degrees or revolutions per minute into any of them gives a wrong answer with no warning. Convert at the end if the question wants revolutions: divide by 2π. The page prints both so you can quote either.

Is this the same as a bolt torque or a gear ratio calculation?

No. This page answers a dynamics question — a torque is applied to a rigid body, what angular acceleration results — using the moments of inertia a school syllabus lists. Tightening torques, gear ratios, shaft power at a given speed and torsional spring rates are engineering calculations with their own standards, tolerances and safety factors, and they belong to the mechanical engineering section of this site rather than to the physics syllabus.

Related calculators

References

PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26 (read 11 November 2026). Unit III, Laws of Motion, reads in part: “Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse”, “Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication” and “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Quoted to establish what is in scope, not to claim any endorsement by the board.
Cambridge International AS & A Level Physics 9702, syllabus for examination in 2025, 2026 and 2027, Cambridge Assessment International Education (read 11 November 2026). Two exclusions on this syllabus are worth more than most inclusions, because they explain why a question from one board looks unanswerable to a student taught on another. Topic 3.2 asks only for “a qualitative understanding of frictional forces and viscous/drag forces including air resistance” and states that “no treatment of the coefficients of friction and viscosity is required”. Topic 3.3 states that “knowledge of the concept of coefficient of restitution is not required”, while still requiring that for an elastic collision “the relative speed of approach is equal to the relative speed of separation”. Its data sheet gives the acceleration of free fall as 9.81 m s⁻².
The International System of Units (SI), 9th edition, 2019, Bureau International des Poids et Mesures (read 11 November 2026). Table 4 gives the newton as kg m s⁻², which is the only definition of force this site uses: a newton is not a unit of mass and a kilogram is not a unit of force. Appendix 1 lists the 3rd CGPM of 1901, whose declaration both separated mass from weight and fixed the conventional value of g_n — the two things a force question most often conflates. Cited, not reproduced.
Fundamental Physical Constants — Complete Listing, NIST Reference on Constants, Units and Uncertainty, 2022 CODATA adjustment (read 11 November 2026). The line this batch depends on is “standard acceleration of gravity 9.806 65 (exact) m s⁻²”. The word exact is the point: g_n is a conventional value, not a measured one, so it carries no uncertainty and it is not the g where you are standing. The same listing gives the Newtonian constant of gravitation as 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² with an uncertainty of 0.000 15 × 10⁻¹¹, which is 22 parts per million — the worst-known of the constants a school paper quotes, and a useful reminder that “constant” and “exact” are different claims.