Classical Mechanics · Chapter 7

Rotational Motion

Every rotational quantity has a linear analogue — once you see the correspondence, rotating systems become as natural as sliding ones.

PrerequisitesNewton's Laws \cdot Energy & Work \cdot Vectors and cross products
Learning Goals
  • Translate between linear and angular kinematic quantities.
  • Compute moment of inertia and explain why mass distribution matters.
  • Usetorqueasavectorproductandapplyτnet=IαUse torque as a vector product and apply \tau_net = I\alpha.
  • Apply angular momentum conservation when external torque is negligible.
  • Analyze rolling motion by combining translational and rotational kinetic energy.

7.1 Angular Kinematics

When a rigid body rotates, every point traces a circular arc. Rather than tracking individual points, we describe the entire body with three scalar quantities: angular position θ (radians), angular velocity ω = dθ/dt (rad/s), and angular acceleration α = dω/dt (rad/s²).

Definition 7.1 — Angular Kinematic Equations
Forconstantangularaccelerationα,therotationalequationsmirrorthelinearonesexactFor constant angular acceleration \alpha, the rotational equations mirror the linear ones exactly:ω=ω0+\alphatθ=θ0+ω0t+12\alphat2ω2=ω02+2α(θ−θ0\omega = \omega_{0} + \alphat \qquad \theta = \theta_{0} + \omega_{0}t + \frac{1}{2}\alphat^{2} \qquad \omega^{2} = \omega_{0}^{2} + 2\alpha(\theta - \theta_{0}Thecorrespondence:θ↔x,ω↔v,α↔a.EverylinearkinematicresulthasadirectrotatiThe correspondence: \theta ↔ x, \omega ↔ v, \alpha ↔ a. Every linear kinematic result has a direct rotational translation.

A point at radius r from the rotation axis has linear speed v = rω, tangential acceleration aₜ = rα, and centripetal acceleration aᶜ = rω² directed toward the axis. The two accelerations are perpendicular.

7.2 Moment of Inertia

In linear motion, mass resists changes in velocity (F = ma). In rotation, moment of inertia I resists changes in angular velocity (τ = Iα). But unlike mass, I depends on how mass is distributed relative to the rotation axis — not just how much there is.

I=∑imiri2=∫r2 dmI = \sum_i m_ir_i^2 = \int r^2\,dm(7.1)

This integral gives different results for different shapes. A ring concentrates all mass at radius R, giving I = MR². A solid disk distributes mass inward, so I = ½MR² — it's easier to spin. This is why figure skaters pull their arms in to spin faster: they reduce I, and angular momentum L = Iω is conserved.

Definition 7.2 — Common Moments of Inertia
  • Solid disk or cylinder: I=12MR2I = \frac{1}{2}MR^{2}
  • Ring or thin hoop: I=MR2I = MR^{2}
  • Solid sphere: I=2/5MR2I = ^{2}/_{5}MR^{2}
  • Rod about center: I=1/12ML2I = ^{1}/_{12}ML^{2}
  • Rod about end: I=1/3ML2I = ^{1}/_{3}ML^{2}
The parallel axis theorem shiftsanyaxis:I=Ishifts any axis: I = Icm + Md2,wheredisthedistancefromthecenterofmasstothenewaxisMd^{2}, where d is the distance from the center of mass to the new axis
Figure 7.1. Rotating body simulation. Switch between disk, ring, rod, and sphere to compare how shape affectsspin.Applytorquetoseeangularacceleration.Thegoldarrowshowsthedirectionofωts spin. Apply torque to see angular acceleration. The gold arrow shows the direction of \omega

7.3 Torque and Newton's Second Law for Rotation

Torque τ is the rotational analogue of force — it is the tendency of a force to cause angular acceleration. Torque depends not just on the magnitude of the force but on where and in what direction it is applied:

τ=r×F∣τ∣=rFsin⁡ϕ\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} \qquad |\boldsymbol{\tau}| = rF\sin\phi(7.2)

where r is the position vector from the axis to the point of application and φ is the angle between r and F. The maximum torque occurs when F is perpendicular to r (φ = 90°). Newton's second law for rotation is then:

τnet=Iα\tau_\mathrm{net} = I\alpha(7.3)
Example 7.1 — Spinning Down a Flywheel

Aflywheel(soliddisk,M=20kg,R=0.5m)spinsat120rpm.Abrakeapplies15N\cdotmofA flywheel (solid disk, M = 20 kg, R = 0.5 m) spins at 120 rpm. A brake applies 15 N\cdotm of torque. How long until it stops?

Moment of inertia:I=12MR2=12(20)(0.5)2=I = \frac{1}{2}MR^{2} = \frac{1}{2}(20)(0.5)^{2} = 2.5kg\cdotm22.5 kg\cdotm^{2}
Angular deceleration:α=−τ/I=−15/2.5=\alpha = -\tau/I = -15/2.5 = −6rad/s2-6 rad/s^{2}
Initial ω:ω0=120rpm×2π/60=\omega_{0} = 120 rpm \times 2\pi/60 = 12.57 rad/s
Time to stop:t=−ω0/α=12.57/6=t = -\omega_{0}/\alpha = 12.57/6 = 2.09 s

7.4 Angular Momentum and Conservation

Just as linear momentum p = mv is conserved when no external force acts, angular momentumL = Iω is conserved when no external torque acts. This is one of the most powerful conservation laws in physics — it governs everything from spinning tops to collapsing stars.

Theorem 7.1 — Conservation of Angular Momentum
If the net external torque on a system is zero, angular momentum is constant:L=Iω=constant→I1ω1=I2ω2L = I\omega = constant \qquad \to \qquad I_{1}\omega_{1} = I_{2}\omega_{2}Whenafigureskaterpullstheirarmsin,Idecreases,soωincreases—thespinspeedsupWhen a figure skater pulls their arms in, I decreases, so \omega increases — the spin speeds up. When a gas cloud collapses into a star, the same principle causes rapid rotation.

7.5 Rotational Kinetic Energy and Rolling

Krot=12Iω2K_\mathrm{rot} = \frac{1}{2}I\omega^2(7.4)

For a rolling body (no slipping), the total kinetic energy combines translational and rotational:

Ktotal=12mv2+12Iω2=12mv2(1+ImR2)K_\mathrm{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)(7.5)

A solid disk (I = ½mR²) rolls slower down a ramp than a sliding block because some energy goes into rotation. A hollow ring (I = mR²) rolls even slower. The shape of the object determines its rolling speed — not its mass.

Example 7.2 — Rolling Down a Ramp

Asolidsphere(I=2/5mR2)rollsfromrestdownarampofheighth=2m.FindthespeedA solid sphere (I = ^{2}/_{5}mR^{2}) rolls from rest down a ramp of height h = 2 m. Find the speed at the bottom.

Energy conservation:mgh=12mv2+12Iω2=12mv2+12(2/5mR2)(v/R)2=12mv2(1+2/5)=7/10mv2mgh = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} = \frac{1}{2}mv^{2} + \frac{1}{2}(^{2}/_{5}mR^{2})(v/R)^{2} = \frac{1}{2}mv^{2}(1 + 2/5) = ^{7}/_{10}mv^{2}
Solve for v:v=(10gh/7)=(10×9.81×2/7)=v = \sqrt(10gh/7) = \sqrt(10 \times 9.81 \times 2/7) = 5.29 m/s
Compare:Aslidingblock(norotation):v=(2gh)=6.26m/s—18A sliding block (no rotation): v = \sqrt(2gh) = 6.26 m/s — 18% faster.
Definition 7.3 — Common Traps
  • Radians are dimensionless but essential: angular equations assume angles are in radians.
  • Moment of inertia is axis-dependent: changing the rotation axis changes I.
  • Torque depends on lever arm: only the perpendicular component produces rotation.
  • Angular momentum conservation needs zero external torque: internalrearrangementcanchangeIandω,notLinternal rearrangement can change I and \omega, not L
  • Rolling without slipping adds a constraint: usev=Rωonlywhenthereisnoslipuse v = R\omega only when there is no slip
Exercises — 7.1–7.5 Rotational Motion
1.
AsoliddiskhasmassM=5kgandradiusR=1m.FinditsmomentofinertiaabouttheceA solid disk has mass M = 5 kg and radius R = 1 m. Find its moment of inertia about the central axis.
kg·m²
Straightforward
2.
A wheel spins at 30 rpm. What is its angular velocity in rad/s?
rad/s
Straightforward
3.
Askaterspinsat2.5rad/swithI=2.0kg\cdotm2.Shepullsherarmsin,reducingIto1.0kA skater spins at 2.5 rad/s with I = 2.0 kg\cdotm^{2}. She pulls her arms in, reducing I to 1.0 kg\cdotm2.Findthenewangularvelocityg\cdotm^{2}. Find the new angular velocity.
rad/s
Intermediate
4.
A thin ring (hoop) rolls without slipping from rest down a ramp of height 3 m. Find the speed at the bottom.
m/s
Intermediate
5.Agrindingwheel(soliddisk,I=0.25kg\cdotm2,R=15cm)spinsat3rev/s.AbrakepadisA grinding wheel (solid disk, I = 0.25 kg\cdotm^{2}, R = 15 cm) spins at 3 rev/s. A brake pad is pressed against the rim and brings it to rest in 3 seconds. Find the braking force applied at the rim.
Challenging
Key Takeaways
  • Angularkinematicsmirrorslinear:θ↔x,ω↔v,α↔a—allfourkinematicequationsapplydireAngular kinematics mirrors linear: \theta↔x, \omega↔v, \alpha↔a — all four kinematic equations apply directly.
  • MomentofinertiaI=\intr2dmdependsonmassdistribution,notjusttotalmassMoment of inertia I = \intr^{2}dm depends on mass distribution, not just total mass.
  • Newton′ssecondlawforrotation:τnet=Iα—torquecausesangularaccelerationNewton's second law for rotation: \tau_net = I\alpha — torque causes angular acceleration.
  • AngularmomentumL=Iωisconservedwhenexternaltorqueiszero(figureskater,collapsiAngular momentum L = I\omega is conserved when external torque is zero (figure skater, collapsing star).
  • Rolling objects split KE between translational and rotational — shape determines rolling speed.