6.3 Moment of Inertia and Rotation about a Fixed Axis
In rotation, the moment of inertia $I=\sum m_ir_i^2$ plays the part of mass: $K=\tfrac12I\omega^2$, $\tau=I\alpha$ and $L=I\omega$ โ and with no external torque, $I\omega$ stays constant.
A potter kicks a heavy stone wheel to set it turning, and once it is going it keeps turning steadily for minutes while she shapes the clay. A bicycle wheel spins up after a flick of the hand, but a solid wheel of the same mass and size takes noticeably less effort to get going than a heavy-rimmed one. A figure skater turning slowly with arms outstretched pulls them in and suddenly whirls several times a second โ with nothing pushing her round at all.
All three stories turn on one quantity. For a body rotating about a fixed axis, what resists changes in rotation is not just its mass but how that mass is spread out around the axis: the moment of inertia. With it, the rotational versions of the laws of motion read exactly like the ones you already know: kinetic energy $\tfrac12I\omega^2$ instead of $\tfrac12mv^2$, $\tau=I\alpha$ instead of $F=ma$, and angular momentum $L=I\omega$ instead of $p=mv$. And just as momentum is conserved when no external force acts, $I\omega$ is conserved when no external torque acts โ which is the skater's secret, and the reason a collapsing star ends up spinning hundreds of times a second.
Kinematics of rotation about a fixed axis
For rotation about a fixed axis the whole body is described by one angle $\theta$, its angular velocity $\omega=\dfrac{d\theta}{dt}$ and its angular acceleration $\alpha=\dfrac{d\omega}{dt}$ (Definition 6.4). These are related exactly as $x$, $v$ and $a$ are for motion in a straight line, so when $\alpha$ is constant the derivation of Section 2.3 goes through word for word with the letters changed.
For rotation about a fixed axis with constant angular acceleration $\alpha$, starting with angular velocity $\omega_0$ at $\theta=0$, $t=0$:
$$\omega=\omega_0+\alpha t,$$ $$\theta=\omega_0t+\tfrac12\alpha t^2,$$ $$\omega^2=\omega_0^2+2\alpha\theta,$$ $$\theta=\tfrac12(\omega_0+\omega)\,t .$$Angles must be in radians ($1\ \text{rev}=2\pi\ \text{rad}$; $\ 1\ \text{rpm}=\tfrac{2\pi}{60}\ \text{rad/s}$).
Moment of inertia and rotational kinetic energy
A spinning wheel has kinetic energy even though its centre of mass is at rest. Add up the kinetic energies of its particles: particle $i$, at distance $r_i$ from the axis, moves at $v_i=\omega r_i$, so
$$K=\sum\tfrac12m_iv_i^2=\tfrac12\Big(\sum m_ir_i^2\Big)\omega^2 .$$Every particle shares the same $\omega$, so the bracket depends only on the body and the axis. It plays the role that mass plays in $\tfrac12mv^2$.
The moment of inertia of a body about a given axis is
$$I=\sum m_ir_i^2 ,$$where $r_i$ is the perpendicular distance of particle $i$ from the axis. SI unit: $\text{kg m}^2$. It depends on the mass, on how the mass is distributed, and on the choice of axis โ the same body has different moments of inertia about different axes.
The radius of gyration $k$ of a body of mass $M$ about an axis is defined by
$$I=Mk^2,\qquad k=\sqrt{\frac{I}{M}} .$$It is the distance from the axis at which the whole mass, concentrated in a thin ring, would have the same moment of inertia.
A rigid body rotating with angular velocity $\omega$ about a fixed axis has kinetic energy
$$K=\tfrac12I\omega^2 ,$$where $I$ is its moment of inertia about that axis.
Because $r_i$ is squared, mass far from the axis counts for much more than mass near it. That is why flywheels are made with heavy rims, why a tightrope walker carries a long pole, and why a hammer is far easier to twirl about an axis through its heavy head than about the end of its handle. For continuous bodies the sum becomes an integral; for the shapes you will meet, the results are standard and are simply quoted.
For uniform bodies of mass $M$ (quoted results โ they come from integrating $r^2\,dm$):
| Body | Axis | $I$ | $k^2$ |
|---|---|---|---|
| thin ring or thin-walled hollow cylinder, radius $R$ | central axis, โ to the plane of the ring | $MR^2$ | $R^2$ |
| thin ring, radius $R$ | a diameter | $\tfrac12MR^2$ | $\tfrac12R^2$ |
| disc or solid cylinder, radius $R$ | central axis | $\tfrac12MR^2$ | $\tfrac12R^2$ |
| thin disc, radius $R$ | a diameter | $\tfrac14MR^2$ | $\tfrac14R^2$ |
| thin rod, length $L$ | โ to the rod, through its centre | $\tfrac1{12}ML^2$ | $\tfrac1{12}L^2$ |
| thin rod, length $L$ | โ to the rod, through one end | $\tfrac13ML^2$ | $\tfrac13L^2$ |
| solid sphere, radius $R$ | a diameter | $\tfrac25MR^2$ | $\tfrac25R^2$ |
| thin spherical shell, radius $R$ | a diameter | $\tfrac23MR^2$ | $\tfrac23R^2$ |
For a ring all the mass is at distance $R$, so $I=MR^2$ is the largest possible for a given $M$ and $R$; a disc has mass spread from $0$ to $R$ and only half as much.
Dynamics of rotation about a fixed axis
Why does the ring in Interactive 1 lag behind? Apply Newton's second law to each particle of a body rotating about a fixed axis. Particle $i$ has tangential acceleration $a_{t,i}=\alpha r_i$, so the tangential component of the total force on it is $F_{t,i}=m_i\alpha r_i$, and its torque about the axis is $F_{t,i}r_i=m_ir_i^2\alpha$. (Radial force components point at the axis and have no torque about it.) Add over all particles: the internal torques cancel in pairs (Result 6.5), leaving only the external torque.
For a rigid body rotating about a fixed axis, with $\tau$ the total external torque about that axis and $I$ the moment of inertia about the same axis,
$$\tau=I\alpha .$$The moment of inertia measures rotational inertia: the same torque gives a smaller angular acceleration to a body with larger $I$.
Work and power. A force $F$ applied tangentially at radius $r$ turns the body through a small angle $\Delta\theta$, moving its point of application a distance $r\,\Delta\theta$. The work done is $F\,r\,\Delta\theta=\tau\,\Delta\theta$, and dividing by the time taken gives the power.
A torque $\tau$ about a fixed axis does work $W=\tau\theta$ when the body turns through angle $\theta$ (for constant $\tau$; otherwise the area under the $\tau$โ$\theta$ graph) and delivers power
$$P=\tau\omega .$$The workโenergy theorem becomes $W_\text{net}=\tfrac12I\omega^2-\tfrac12I\omega_0^2$.
| Translation (straight line) | Rotation (fixed axis) |
|---|---|
| displacement $x$, velocity $v$, acceleration $a$ | angle $\theta$, angular velocity $\omega$, angular acceleration $\alpha$ |
| mass $m$ | moment of inertia $I=\sum m_ir_i^2$ |
| force $F=ma$ | torque $\tau=I\alpha$ |
| kinetic energy $\tfrac12mv^2$ | kinetic energy $\tfrac12I\omega^2$ |
| work $Fx$, power $Fv$ | work $\tau\theta$, power $\tau\omega$ |
| momentum $p=mv$ | angular momentum $L=I\omega$ |
- Draw a free-body diagram for each translating body (a hanging mass: $mg$ down, $T$ up) and write $F=ma$.
- For each rotating body write $\tau=I\alpha$ about its axle: a string pulling tangentially at radius $R$ gives torque $TR$.
- Link them: a string that does not slip on the drum gives $a=\alpha R$.
- Solve the equations together; check with energy (loss of $mgh$ = gain of $\tfrac12mv^2+\tfrac12I\omega^2$ if the axle is frictionless).
Angular momentum about a fixed axis and its conservation
Each particle of a body rotating about a fixed axis moves in a circle of radius $r_i$ with speed $\omega r_i$, so its angular momentum about the axis is $m_i(\omega r_i)r_i=m_ir_i^2\omega$ (Definition 6.6). Adding them up gives $I\omega$. Since $I$ is constant for a rigid body, $\dfrac{d(I\omega)}{dt}=I\alpha=\tau$ โ Result 6.12 is just $\dfrac{dL}{dt}=\tau$ for this case. But a body that can change its shape (a skater, a diver, a collapsing star) can change $I$ while it spins, and then the angular-momentum form is the one that holds.
The angular momentum of a body rotating with angular velocity $\omega$ about a fixed axis, with moment of inertia $I$ about that axis, is (the component along the axis)
$$L=I\omega ,\qquad \frac{dL}{dt}=\tau_\text{ext}.$$If the total external torque about the axis is zero, $L$ is conserved, even if the body changes shape:
$$I_1\omega_1=I_2\omega_2 .$$Gravity and the normal force from the ice act through the skater's axis of rotation (or parallel to it) and exert no torque about it, so, apart from small friction, her angular momentum about the vertical axis is constant. Pulling her arms in reduces $I$, so $\omega$ must rise. Her kinetic energy, $K=\tfrac12I\omega^2=\dfrac{L^2}{2I}$, increases as $I$ falls: the extra energy is the work done by her arm muscles pulling her hands inward against their tendency to fly outward. A diver tucks to spin faster and opens out to slow down before entering the water; a cat twists in mid-air; a pulsar โ the collapsed core of a star โ spins hundreds of times a second because its radius shrank enormously while its angular momentum stayed the same.
1. "Moment of inertia is a property of the body." It is a property of the body and an axis. Always say which axis; Example 2 gave $0.25$ and $0.50\ \text{kg m}^2$ for the same dumbbell.
2. Distance from a point instead of from the axis. In $I=\sum m_ir_i^2$, $r_i$ is the perpendicular distance from the axis line.
3. Using rpm or degrees in formulas. $\tau=I\alpha$, $K=\tfrac12I\omega^2$, $v=\omega r$ and $W=\tau\theta$ all need radians.
4. Tension equals weight. For a bucket on a drum (Example 4), $T\lt mg$ while the bucket accelerates; setting $T=mg$ would give the drum the wrong torque.
5. Conserving kinetic energy instead of angular momentum. When a skater pulls her arms in, or a lump of clay lands on a turntable, $I\omega$ is conserved but $\tfrac12I\omega^2$ is not.
6. Conservation of $L$ with a torque acting. $I_1\omega_1=I_2\omega_2$ needs zero external torque about the axis. Friction at a bearing, or a motor, breaks it.
Both topics have been removed from the NCERT chapter and are not in the CBSE syllabus. The parallel- and perpendicular-axes theorems are in JEE Main, JEE Advanced and NEET 2026; rolling motion is in JEE Advanced 2026 only.
Parallel-axes theorem. If $I_\text{cm}$ is the moment of inertia of a body of mass $M$ about an axis through its centre of mass, its moment of inertia about any parallel axis a distance $d$ away is
$$I=I_\text{cm}+Md^2 .$$So the moment of inertia is least about an axis through the centre of mass. Example: a rod about a perpendicular axis through one end: $\tfrac1{12}ML^2+M\left(\tfrac L2\right)^2=\tfrac13ML^2$, as quoted in Result 6.11. A ring about a tangent in its own plane: $\tfrac12MR^2+MR^2=\tfrac32MR^2$.
Perpendicular-axes theorem (plane laminas only). For a flat body lying in the $xy$-plane, $I_z=I_x+I_y$, where the three axes are mutually perpendicular and meet at a point of the lamina. Example: for a disc, $I_z=\tfrac12MR^2$ and by symmetry $I_x=I_y$, so the moment of inertia about a diameter is $\tfrac14MR^2$.
Rolling down a slope (JEE Advanced only). A body rolling without slipping (Section 6.2 box: $v=R\omega$) has kinetic energy of translation and rotation about its centre of mass:
$$K=\tfrac12Mv^2+\tfrac12I_\text{cm}\omega^2=\tfrac12Mv^2\left(1+\frac{k^2}{R^2}\right).$$Static friction at the contact point does no work (that point is momentarily at rest), so rolling from rest down a height $h$: $Mgh=\tfrac12Mv^2\left(1+\tfrac{k^2}{R^2}\right)$, giving $v=\sqrt{\dfrac{2gh}{1+k^2/R^2}}$ and acceleration $a=\dfrac{g\sin\theta}{1+k^2/R^2}$. With $\dfrac{k^2}{R^2}=\tfrac25$ (solid sphere), $\tfrac12$ (disc or solid cylinder), $1$ (ring), the sphere wins the race, then the disc, then the ring โ whatever their masses and radii. For $h=1.4\ \text{m}$: sphere $4.4\ \text{m/s}$, disc $4.3\ \text{m/s}$, ring $3.7\ \text{m/s}$ (a frictionless sliding block: $5.2\ \text{m/s}$). The fraction of the kinetic energy that is rotational is $\dfrac{k^2/R^2}{1+k^2/R^2}$: $\tfrac27$, $\tfrac13$ and $\tfrac12$.
Exercises
Take $g=9.8\ \text{m/s}^2$. Axles are frictionless and strings light and non-slipping unless stated otherwise. Use the moments of inertia of Result 6.11.
- 1 A wheel starts from rest with $\alpha=3.0\ \text{rad/s}^2$. Find its angular velocity after $4.0\ \text{s}$ and the number of revolutions made.
- 2 A fan turning at $300\ \text{rpm}$ is switched off and stops uniformly in $15\ \text{s}$. Find its angular acceleration and the number of revolutions it makes while stopping.
- 3 A flywheel turning at $20\ \text{rad/s}$ slows down at $0.50\ \text{rad/s}^2$. How long does it take to stop, and through what angle does it turn?
- 4 A disc speeds up uniformly from $10\ \text{rad/s}$ to $30\ \text{rad/s}$ while turning through $100\ \text{rad}$. Find $\alpha$ and the time taken.
- 5 A drill bit reaches $2400\ \text{rpm}$ from rest in $0.50\ \text{s}$. Find its average angular acceleration and the tangential acceleration of a point $3.0\ \text{mm}$ from its axis.
- 6 Show from $\omega=\omega_0+\alpha t$ and $\theta=\omega_0t+\tfrac12\alpha t^2$ that $\theta=\tfrac12(\omega_0+\omega)t$.
- 7 A wheel of radius $0.40\ \text{m}$ has $\alpha=2.0\ \text{rad/s}^2$. At the instant when $\omega=3.0\ \text{rad/s}$, find the tangential, centripetal and total acceleration of a point on the rim.
- 1 Four $0.20\ \text{kg}$ particles sit at the corners of a light square frame of side $0.50\ \text{m}$. Find the moment of inertia about an axis perpendicular to the square through its centre.
- 2 For the frame of part 1, find the moment of inertia about an axis along one side.
- 3 Find the moment of inertia about the central axis of (a) a $2.0\ \text{kg}$ thin ring of radius $0.30\ \text{m}$, (b) a $2.0\ \text{kg}$ uniform disc of radius $0.30\ \text{m}$.
- 4 Find the radius of gyration of a solid sphere of radius $10\ \text{cm}$ about a diameter.
- 5 A uniform rod is $1.2\ \text{m}$ long and has mass $0.50\ \text{kg}$. Find its moment of inertia about a perpendicular axis (a) through its centre, (b) through one end.
- 6 A $0.50\ \text{kg}$ solid sphere of radius $0.10\ \text{m}$ spins at $20\ \text{rad/s}$ about a diameter. Find its kinetic energy.
- 7 A ring and a disc have the same mass and radius and spin at the same rate about their central axes. Which has more kinetic energy, and by what factor?
- 8 The Earth is denser at the centre; its moment of inertia about its axis is about $0.33M_ER_E^2$. Estimate the kinetic energy of the Earth's daily rotation ($M_E=5.97\times10^{24}\ \text{kg}$, $R_E=6.37\times10^6\ \text{m}$, $\omega=7.27\times10^{-5}\ \text{rad/s}$). Why is the factor less than $0.4$?
- 1 A torque of $6.0\ \text{N m}$ acts on a wheel with $I=1.5\ \text{kg m}^2$, starting from rest. Find $\alpha$ and $\omega$ after $3.0\ \text{s}$.
- 2 What constant torque stops a flywheel ($I=8.0\ \text{kg m}^2$) spinning at $50\ \text{rad/s}$ in $20\ \text{s}$?
- 3 A rope wound round a pulley โ a uniform solid cylinder of mass $10\ \text{kg}$ and radius $0.20\ \text{m}$ โ is pulled with a steady $40\ \text{N}$, starting from rest. Find $\alpha$, the angular velocity after $2.0\ \text{s}$ and the length of rope unwound. Check that the work done by the pull equals the kinetic energy gained.
- 4 A $2.0\ \text{kg}$ block hangs from a string wound round a uniform disc pulley of mass $4.0\ \text{kg}$ and radius $0.10\ \text{m}$. Find the block's acceleration and the tension.
- 5 A motor delivers $2.0\ \text{kW}$ at $1500\ \text{rpm}$. Find the torque it supplies.
- 6 A constant torque of $15\ \text{N m}$ turns a wheel through $8$ revolutions. How much work does it do?
- 7 A grinding wheel ($I=0.80\ \text{kg m}^2$) spinning at $100\ \text{rad/s}$ is switched off and a friction torque of $3.0\ \text{N m}$ brings it to rest. Find its initial kinetic energy, the time to stop, and the number of revolutions.
- 8 Starting from Newton's second law for each particle, show that $\tau=I\alpha$ for a rigid body rotating about a fixed axis.
- 1 Find the angular momentum of a $2.0\ \text{kg}$ uniform disc of radius $0.25\ \text{m}$ spinning at $120\ \text{rpm}$ about its central axis.
- 2 A turntable ($I=0.020\ \text{kg m}^2$) turns freely at $3.0\ \text{rad/s}$. A $0.20\ \text{kg}$ lump of putty is dropped on it and sticks $0.15\ \text{m}$ from the axis. Find the new angular velocity and the kinetic energy lost.
- 3 A skater's moment of inertia falls from $4.0$ to $1.6\ \text{kg m}^2$ as she pulls her arms in. She was turning at $2.0\ \text{rev/s}$. Find her new rate and the ratio of her final to initial kinetic energy.
- 4 (Modelling) A star of radius $7.0\times10^8\ \text{m}$ rotates once every $25$ days. Its core collapses to a neutron star of radius $12\ \text{km}$. Treating both as uniform spheres of the same mass with no loss of angular momentum, estimate the new period of rotation.
- 5 A diver leaves the board rotating at $1.0\ \text{rev/s}$ in a straight position, then tucks, reducing her moment of inertia by a factor of $3.5$. How many turns does she make in $1.0\ \text{s}$ of tucked flight?
- 6 Disc A ($I=0.10\ \text{kg m}^2$, spinning at $20\ \text{rad/s}$) is pressed against disc B ($I=0.30\ \text{kg m}^2$, at rest) on the same axle until they turn together, as in a car's clutch. Find the common angular velocity and the fraction of kinetic energy lost.
E1. Board ยท 3 marks Show that the kinetic energy of a rigid body rotating with angular velocity $\omega$ about a fixed axis is $\tfrac12I\omega^2$, defining $I$. Why is the moment of inertia called the rotational inertia of the body?
E2. NEET A uniform disc and a thin ring have the same mass and radius and can turn about their central axes. The same torque is applied to each, starting from rest. The ratio of their angular accelerations, $\alpha_\text{disc}:\alpha_\text{ring}$, is (a) $1:2$ (b) $2:1$ (c) $1:1$ (d) $4:1$
E3. JEE Main (Numerical value) A horizontal disc with moment of inertia $0.40\ \text{kg m}^2$ about its vertical axis rotates freely at $10\ \text{rad/s}$. A ring of moment of inertia $0.10\ \text{kg m}^2$ about the same axis is gently placed on it, coaxially, from rest, and the two soon turn together. Find their common angular velocity in rad/s.
E4. JEE Advanced (One or more correct) A solid sphere, a uniform disc and a thin ring, all of the same mass $M$ and radius $R$, are released together from rest at the top of the same rough incline and roll down without slipping. Which statements are correct? (A) The sphere reaches the bottom first. (B) All three have the same total kinetic energy at the bottom. (C) The ring has the largest translational kinetic energy at the bottom. (D) For the ring, half its kinetic energy at the bottom is rotational.
E5. JEE Main NEET The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis perpendicular to it, through a point at distance $L/4$ from one end, is (a) $\dfrac{ML^2}{12}$ (b) $\dfrac{7ML^2}{48}$ (c) $\dfrac{ML^2}{3}$ (d) $\dfrac{5ML^2}{48}$
E6. Board ยท Case study A $4.0\ \text{kg}$ bucket of water hangs from a rope wound round a windlass drum, a uniform solid cylinder of mass $12\ \text{kg}$ and radius $0.10\ \text{m}$ on a frictionless axle. The handle is released and the bucket falls from rest. (i) Find the acceleration of the bucket. [2] (ii) Find the tension in the rope. [1] (iii) Find the bucket's speed after it has fallen $3.0\ \text{m}$. [1]
Research question. How much energy can a spinning flywheel store, what limits it, and how does it compare with a battery?
Why it matters. Flywheels store energy as $\tfrac12I\omega^2$. They are used to smooth the output of wind farms and to steady electricity grids, as uninterruptible power supplies for hospitals and data centres, to recover braking energy in buses, trams and racing cars, and in some experimental spacecraft. Unlike chemical batteries they can be charged and discharged millions of times, and very quickly.
What to do. Consider a grid-storage flywheel whose rotor is a carbon-fibre rim of mass $100\ \text{kg}$ and radius $0.30\ \text{m}$; treat it as a thin ring.
- Find its moment of inertia, and its kinetic energy at $30\,000\ \text{rpm}$, in joules and in kilowatt-hours ($1\ \text{kWh}=3.6\times10^6\ \text{J}$).
- Find the speed of the rim. Compare it with the speed of sound in air. Why are such rotors sealed in a vacuum chamber?
- The flywheel supplies a steady $100\ \text{kW}$. How long can it do so before its angular velocity has halved? (What fraction of the energy has been used by then?)
- Which raises the stored energy more: doubling the mass, doubling the radius, or doubling the speed? Why can't engineers simply keep increasing the speed? (Think about the tension needed to hold the rim in a circle.)
- Compare the energy stored per kilogram of rotor with a typical value for a lithium-ion battery, about $0.7\ \text{MJ/kg}$. Where do flywheels win, and where do batteries win?
Tools: Calculator or spreadsheet; manufacturers' and energy-agency data sheets for real flywheel storage systems.
Hand in: Your calculations, a short table comparing flywheel and battery (energy per kg, power, lifetime, cost if you can find it), and a recommendation for one application.
Going further. A spinning flywheel has a large angular momentum. Explain, using $\boldsymbol\tau=\dfrac{d\mathbf{L}}{dt}$, why a flywheel mounted in a vehicle resists being turned, and why some designs use two flywheels spinning in opposite directions.
๐งช Self-check quiz
Q1. The moment of inertia of a rigid body does not depend on
Q2. A torque of $2.0\ \text{N m}$ acts on a wheel of moment of inertia $0.50\ \text{kg m}^2$ about its fixed axle. Its angular acceleration is
Q3. A ring and a disc of the same mass and radius spin at the same angular velocity about their central axes. The ratio $K_\text{ring}:K_\text{disc}$ is
Q4. A skater spinning on smooth ice pulls her arms in. Which statement is correct?
Q5. A flywheel turning at $10\ \text{rad/s}$ slows uniformly and stops after turning through $25\ \text{rad}$. Its angular acceleration is