Home / Topics / Rotational Motion
Rotational Motion
Torque, moment of inertia, angular momentum, gyroscopes and rolling. 51 topics, from beginner to advanced. Open the interactive version with labs and quizzes
Rotational motion extends linear ideas to spinning and rolling bodies. Angular displacement, angular velocity and angular acceleration replace their linear counterparts, torque replaces force, and moment of inertia replaces mass. The area also covers angular momentum and why spinning objects such as gyroscopes and figure skaters behave the way they do.
How the topics connect
Rotational motion takes the ideas of linear motion and applies them to spinning bodies. Angular displacement, velocity and acceleration match their linear counterparts, while torque and moment of inertia replace force and mass. This area also explains rolling, gyroscopes and why skaters spin faster when they pull their arms in.
Real-world uses
Calculate the torque a motor must supply to accelerate a wheel, and how gear ratios trade speed for torque.
Explain why pulling the arms in increases spin rate, since angular momentum is conserved and moment of inertia decreases.
Understand how spinning rotors stabilise drones, satellites and some inertial navigation systems.
Compare how fast a solid ball, a hollow ball and a cylinder roll down a slope, using their moments of inertia.
Engineers use moment of inertia to choose flywheels, gears and motor shafts that give the required acceleration and energy storage.
Divers, skaters and gymnasts control their spin by changing their moment of inertia, which conserves angular momentum.
Rigid body
An idealised body whose parts keep fixed distances.
A rigid body is an idealised object whose parts never change their mutual distances. It lets us treat rotation and translation separately. Real objects are close to rigid for small forces.
A steel wheel is almost rigid; a jelly isn't.
Rigid bodies cannot vibrate or deform.
Axis of rotation
The line about which a body turns.
Rotation is described about an axis; every point moves in a circle around it with the same angular velocity. The axis can be fixed (door hinge) or moving (rolling wheel).
A spinning top about its tip axis.
Quantities like torque and I depend on the chosen axis.
Angular displacement
Angle swept, measured in radians.
θ = s/rθ = s/r in radians measures the turn. A full circle is 2π. The right-hand rule gives it a direction along the axis for small rotations.
An arc of 6 m on a radius of 3 m gives 2 rad.
Finite rotations in 3D don't add like vectors.
Angular velocity
Rate of change of angle.
ω = dθ/dtω = dθ/dt in rad/s; it is the same for all points in a rigid body. Direction along the axis by the right-hand rule.
1200 rpm = 125.7 rad/s.
Convert rpm using 2π/60.
Angular acceleration
Rate of change of angular velocity.
α = dω/dtα = dω/dt changes the spin. With constant α the angular equations mirror the linear ones.
Speeding up from 0 to 30 rad/s in 6 s: α = 5 rad/s².
rad/s² not m/s².
Radians and degrees
2π radians is one full turn.
180° = π radA radian is the angle for which arc length equals the radius; 2π rad = 360°. Formulas such as v = ωr and s = rθ need radians.
90° = π/2 rad ≈ 1.57.
Calculators must be in the right mode.
Relation between linear and angular quantities
Tangential speed is ω times radius.
v = ωrFor a point at radius r: v = ωr, a_t = αr, a_c = ω²r. Points farther from the axis move faster.
A fan blade tip at 0.5 m with ω = 40 rad/s has v = 20 m/s.
These hold at each instant for a rigid body about a fixed axis.
Rotational kinematics
Constant-α equations mirror linear ones.
θ = ω₀t + ½αt²With constant α: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ. They are used the same way as linear equations.
A wheel stops from 10 rad/s with α = −2: turns 25 rad.
Check signs of α and ω.
Torque
Turning effect of a force.
τ = r×F = rF sinθTorque τ = r × F = rF sinθ is the turning effect of a force about a point. Larger lever arm gives larger torque; a force through the axis gives none.
A 20 N force perpendicular at 0.3 m on a spanner: 6 N·m.
Only the perpendicular component of force counts.
Lever arm and moment
Torque depends on perpendicular distance from the axis.
The lever arm (moment arm) is the perpendicular distance from the axis to the line of action of the force. Torque = force × lever arm.
Door handle far from the hinge: easier to open.
Pushing along the line to the hinge gives zero torque.
Equilibrium of a rigid body
Net force and net torque both zero.
ΣF=0, Στ=0Static equilibrium needs both ΣF = 0 and Στ = 0. Taking moments about a clever point removes unknown forces.
A 6 m ladder leaning on a wall: use torque about the base.
Don't forget the weight acts at the centre of mass.
Centre of mass of shapes
Find by symmetry or integration.
Symmetry helps: a uniform rod's centre of mass is its midpoint, a triangle's is at one third of its height, a semicircle's disk at 4R/3π.
An L-shape: split into rectangles and use weighted averages.
Composite shapes with holes: subtract the hole.
Centre of gravity
The point where weight acts; equals centre of mass in uniform fields.
The point where the whole weight can be considered to act. In a uniform field it coincides with the centre of mass.
Balancing a ruler on a finger finds its centre.
On a huge body such as a mountain they differ slightly.
Stability and toppling
A body topples when the weight line falls outside its base.
A body is stable if the vertical line from its centre of gravity falls within its base. Lower centre and wider base are more stable.
A bus tilts safely until the weight line passes the wheel edge.
Tall, narrow objects topple easily.
Moment of inertia
Rotational inertia: resistance to angular acceleration.
I = Σm r²I = Σmr² measures the resistance to angular acceleration. Mass far from the axis counts a lot (r²). It is the rotational analogue of mass.
Two 1 kg masses at 0.5 m on a rod: I = 0.5 kg·m².
I depends on the axis chosen.
Moment of inertia of a rod
About the centre or the end of a uniform rod.
I = mL²/12; mL²/3About its centre I = mL²/12; about an end I = mL²/3 (parallel axis theorem with d = L/2).
1.2 kg, 1 m: 0.1 and 0.4 kg·m².
Use the right axis.
Moment of inertia of a ring
All mass at radius R.
I = mR²A thin ring about its centre axis: I = mR², because all mass is at distance R.
A 2 kg bicycle rim of 0.35 m: 0.245 kg·m².
About a diameter it is ½mR².
Moment of inertia of a disc
Uniform disc about its axis.
I = ½mR²A uniform disc about its centre axis: I = ½mR². It is smaller than the ring since the mass is closer to the axis.
m = 4 kg, R = 0.5: 0.5 kg·m².
Same mass and radius as a ring has twice the I.
Moment of inertia of a solid sphere
Uniform sphere about a diameter.
I = (2/5)mR²About a diameter I = (2/5)mR²; derived by integrating discs.
m = 5 kg, R = 0.1 m: 0.02 kg·m².
Hollow sphere has bigger I.
Moment of inertia of a hollow sphere
Thin spherical shell about a diameter.
I = (2/3)mR²A thin spherical shell about a diameter: I = (2/3)mR².
m = 3 kg, R = 0.2: 0.08.
Don't confuse shell with solid.
Parallel axis theorem
Shift axis by d from the centre of mass.
I = I_cm + md²I about any axis = I_cm + md², where d is the distance between the axes (parallel). The centre-of-mass axis gives the minimum I.
Disc about a point on its rim: ½mR² + mR² = 1.5 mR².
One of the axes must pass through the centre of mass.
Perpendicular axis theorem
For a flat lamina, I_z equals I_x plus I_y.
I_z = I_x + I_yFor a flat lamina in the xy-plane, I_z = I_x + I_y. Useful for discs and rings: I_x = I_z/2.
Ring I_z = mR², so I about a diameter = ½mR².
Only for flat (planar) bodies.
Radius of gyration
Distance at which all mass would give the same I.
I = mk²k = √(I/m) is the distance at which a point mass would have the same I. It's an alternative way to give rotational inertia.
I = 0.5 kg·m², m = 2 kg: k = 0.5 m.
k isn't a physical dimension of the body.
Newton's second law for rotation
Net torque equals I times α.
τ = IαThe net torque equals Iα. The larger the I, the smaller the angular acceleration for a given torque.
10 N·m on a 2 kg·m² wheel: 5 rad/s².
Torque and I must be about the same axis.
Rotational kinetic energy
Energy stored in spinning.
K = ½Iω²K = ½Iω². Sum of ½mv² of all particles.
A 3 kg·m² flywheel at 50 rad/s stores 3750 J.
Flywheels store energy.
Rolling without slipping
Contact point is momentarily at rest, so v = ωR.
v_cm = ωRThe point of contact is instantaneously at rest, so v_cm = ωR. Static friction makes it possible and does no work.
A 0.3 m radius wheel at 6 m/s has ω = 20 rad/s.
Top of the wheel moves at 2v.
Rolling down an incline
Rolling bodies accelerate slower than sliding ones due to rotational energy.
a = g sinθ/(1+I/mR²)a = g sinθ/(1 + I/mR²). Smaller I/mR² means faster descent.
Solid sphere (2/5): a = 0.71 g sinθ. Ring: 0.5 g sinθ.
Mass and radius cancel out.
Race of rolling bodies
Solid sphere beats disc, which beats ring.
On an incline the solid sphere beats the disc, then the hollow sphere, then the ring. A sliding block beats all of them.
Experiment: cans of soup rolling.
Winners don't depend on mass or size.
Work and power in rotation
Work equals torque times angle; power equals torque times ω.
P = τωW = τθ, P = τω. Motors give constant power: higher speed means lower torque.
A 2 kW motor at 100 rad/s gives 20 N·m.
Gear ratios trade speed for torque.
Angular momentum of a particle
Position cross momentum.
L = r×pL = r × p = mvr sinφ about an origin. A straight-line moving particle has angular momentum about points off its line.
1 kg at 3 m/s passing 2 m from the origin: L = 6.
Depends on the origin.
Angular momentum of rigid body
About a fixed axis, L equals Iω.
L = IωAbout a fixed axis L = Iω along the axis. For general rotation L = I·ω with a tensor I and need not be parallel to ω.
A 0.1 kg·m² wheel at 40 rad/s: 4 kg·m²/s.
L and ω can point in different directions for asymmetric bodies.
Torque and angular momentum
Net torque equals the rate of change of L.
τ = dL/dtτ_net = dL/dt. It is the rotational form of Newton's second law and also covers changing I.
A constant 2 N·m torque for 5 s adds 10 kg·m²/s.
Use the same origin for τ and L.
Conservation of angular momentum
No external torque keeps L constant.
I₁ω₁ = I₂ω₂With no net external torque, L stays constant: I₁ω₁ = I₂ω₂. It explains spinning skaters, divers, pulsars and the Moon's recession.
A star collapsing spins faster.
Check torque about the chosen point is zero.
Spinning skater and diver
Tucking in reduces I and raises ω.
Pulling limbs in reduces I, increasing ω. Rotational kinetic energy rises since the person does work.
Diver tucks: 3 rotations in under 2 s.
No external torque during flight.
Collisions in rotation
A bullet embedding in a rotating body conserves angular momentum.
A bullet hitting a rod or door conserves angular momentum about the pivot (not linear momentum, because the pivot exerts an impulse).
Bullet mv·d = Iω after embedding.
Energy is lost in the collision.
Precession of a gyroscope
A torque on a spinning wheel turns its axis sideways.
Ω = τ/LA torque from gravity on a spinning gyroscope changes the direction of L, so the axis sweeps round at Ω = τ/L = mgr/(Iω). The faster the spin, the slower the precession.
A spinning top at an angle doesn't fall at once.
Precession is perpendicular to the applied torque.
Nutation
Wobbling superimposed on precession.
Small nodding oscillations superimposed on precession, damped by friction. It occurs when the spin is initially not exactly at steady precession.
Earth's axis shows nutation of 9 arcseconds with an 18.6-year period.
Seen in tops that wobble.
Rotation of the Earth
Earth's spin causes day and night and the equatorial bulge.
Earth spins once per sidereal day, ω ≈ 7.29×10⁻⁵ rad/s. It causes day and night, flattening, trade winds and Coriolis deflection.
Equator speed is about 465 m/s.
Earth's spin is slowly decreasing due to tides.
Tops and bicycles
Gyroscopic effects stabilise spinning objects.
Spinning wheels have large angular momentum resisting changes of axis; this helps stability and steering. Bikes remain upright mainly by steering geometry, plus gyroscopic effects.
Faster bike rolls more stably.
Gyros alone don't explain bike stability.
Rotating frames
Centrifugal and Coriolis forces appear in rotating coordinates.
In a rotating frame add centrifugal force mω²r and Coriolis force −2mω×v. They explain weather patterns and artificial gravity in space stations.
A space station of radius 100 m at 3 rpm gives about 1 g.
Don't use fictitious forces in an inertial frame.
Euler's equations
Rigid-body rotation about principal axes.
I₁ω̇₁ = (I₂−I₃)ω₂ω₃The rotation of a rigid body about its principal axes: I₁ω̇₁ = (I₂ − I₃)ω₂ω₃ + τ₁, and cyclic permutations. They explain why a tossed book spins stably about two axes but tumbles about the middle one.
The tennis racket theorem.
Valid in the body frame.
Inertia tensor
A matrix generalising I to arbitrary axes.
L = I·ωA 3×3 symmetric matrix I_ij linking L and ω: L = I·ω. Its eigenvectors are the principal axes, with eigenvalues the principal moments.
A symmetric top has two equal principal moments.
L is not parallel to ω unless along a principal axis.
Principal axes
Axes along which the inertia tensor is diagonal.
Axes about which the inertia tensor is diagonal, so L ∥ ω. Symmetry axes are principal axes. A body rotates stably about the axes of largest and smallest I.
Cuboid: the three axes through the centre normal to its faces.
Rotation about the intermediate axis is unstable.
Torque on a dipole
Electric or magnetic dipoles feel torque in fields.
τ = p×EAn electric dipole p in a field E feels τ = p × E; a magnetic dipole feels τ = m × B. It aligns with the field, like a compass needle.
Compass needles align with Earth's field.
Torque is zero when aligned or anti-aligned, but only aligned is stable.
Pulley with mass
Pulley inertia reduces the acceleration of attached masses.
a = (m₁−m₂)g/(m₁+m₂+I/R²)A real pulley has moment of inertia I, so tensions on both sides differ: (T₁ − T₂)R = Iα with α = a/R. The acceleration becomes a = (m₁ − m₂)g/(m₁ + m₂ + I/R²).
Heavy pulleys slow the system.
Don't assume equal tension.
Atwood with real pulley
Include rotational inertia of the pulley in the equations.
Write ΣF = ma for each mass and τ = Iα for the pulley, then use a = αR (no slipping). Solve for a and the two tensions.
m₁ = 3, m₂ = 1, I/R² = 0.5 kg: a = 2g/4.5 ≈ 4.4 m/s².
The tensions differ by Ia/R².
Yo-yo and unwinding cylinder
Combine translation and rotation with tension torque.
For a cylinder unwinding from a string, a = g/(1 + I/mR²) down; for a solid disc a = 2g/3. Tension T = mg/3.
The yo-yo falls slower than free fall.
Rotation and translation are linked by a = αR.
Angular impulse
Torque integrated over time changes angular momentum.
∫τ dt = ΔLThe time integral of torque ∫τ dt changes angular momentum, the rotational version of impulse.
A 5 N·m torque for 2 s adds 10 kg·m²/s.
Use same axis for torque and L.
Static friction in rolling
Friction enables rolling but does no work in pure rolling.
Static friction at the contact point prevents slipping and may point either way depending on acceleration. In pure rolling at constant speed it is zero and does no work.
On an incline it points up the slope.
Rolling resistance is a different thing.
Slipping to rolling transition
A sliding ball begins to roll when v = ωR through kinetic friction.
A ball launched with backspin or sliding on a rough surface experiences kinetic friction until v = ωR, after which it rolls. For a solid sphere the final speed is 5/7 of the initial for pure forward sliding.
A bowling ball released skidding.
Energy is lost during skidding.
Bohr angular momentum
Atomic angular momentum is quantised.
L = nħBohr's model states that electron angular momentum is quantised in units of ħ: L = nħ. In quantum mechanics orbital angular momentum is √(l(l+1))ħ.
n = 2 gives L = 2ħ.
Modern theory modifies the Bohr numbers.
Frequently asked questions
What is torque?
Torque is the turning effect of a force about an axis. It equals the force multiplied by the perpendicular distance from the axis, so a longer spanner gives more torque for the same push.
Why are angles measured in radians in rotational physics?
A radian is the angle subtended by an arc equal in length to the radius. Using radians makes formulas such as v = rω and s = rθ simple, with no extra constants.
Why does a rolling object on a slope slow down compared with sliding?
Part of its energy goes into rotation, not just forward motion. A solid sphere rolls slower than a frictionless sliding block of the same height because its kinetic energy is shared between translation and rotation.
Why does the moment of inertia depend on the axis?
Mass farther from the axis contributes more, so the same object has different moments of inertia about different axes. Always state which axis you are using.
Why does a rolling ball beat a sliding one?
A rolling ball puts some of its energy into spinning, so for the same slope it accelerates more slowly. A sliding block has no spin, so it accelerates faster.
Other areas
Measurement & Math Tools · Kinematics · Newton's Laws & Forces · Work, Energy & Power · Momentum & Collisions · Gravitation & Orbits · Oscillations & SHM · Waves & Sound · Fluids & Elasticity · Heat & Thermodynamics · Kinetic Theory & Statistical Physics · Electrostatics · Current Electricity & Circuits · Magnetism & Induction · AC Circuits & EM Waves · Optics · Quantum Mechanics · Atomic & Nuclear Physics · Relativity · Particle Physics & Cosmology