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Rotational Motion

Section 01
Interactive Simulation
Rotational Motion — SciSim
Ready
α
rad/s²
ω
rad/s
θ
rad
τ
N·m
L
kg·m²/s
KE_rot
J
Controls
Parameters
Moment of Inertia I1.00kg·m²
Torque τ5.00N·m
Damping b0.10Nms/rad
Radius R0.50m
Display
Grid
Vectors
Trail
Labels
Section 02
The Idea, Step by Step

Push on a door right next to its hinge and almost nothing happens — but push with the same hand out near the handle and it swings open easily. For spinning things, what matters isn't just how hard you push, but how far from the pivot you push. That twisting effect is called torque. The same idea explains why a long wrench loosens a stubborn bolt that your fingers never could.

Naming the pieces

Three quantities describe a spin. The torque $\tau$ is the twisting effort. The moment of inertia $I$ is the object's "rotational laziness" — how stubbornly it resists changing its spin. And the angular acceleration $\alpha$ is how quickly the spin rate speeds up or slows down. They lock together in the rotational version of Newton's second law:

Rotational Newton's 2nd Law
$$\tau = I\,\alpha$$

It reads just like $F=ma$, with torque in place of force, $I$ in place of mass, and $\alpha$ in place of acceleration. Worked number: a wheel with $I = 2\ \text{kg·m}^2$ driven by a torque of $\tau = 10\ \text{N·m}$ gains spin at $\alpha = \tau/I = 5\ \text{rad/s}^2$ — every second its spin rate climbs by 5 rad/s.

Going deeper

Why does shape matter so much? Because $I=\sum m_i r_i^2$: mass counts more the farther it sits from the axis. Pack the same mass into a solid disk ($I=\tfrac12 MR^2$) and it spins up easily; spread it onto a thin ring ($I=MR^2$) and the very same torque produces only half the angular acceleration. The deepest statement uses angular momentum $L=I\omega$, where $\omega$ is the spin rate: torque is its rate of change, $\tau = dL/dt$. When the net torque is zero, $L$ can't change — so if a skater pulls mass inward and shrinks $I$, the spin rate $\omega$ must shoot up to keep $L$ fixed. In the sim above, the I slider sets that rotational inertia, τ the driving torque, b a damping torque that fights the spin (so the net becomes $\alpha=(\tau-b\,\omega)/I$), and R the radius that sets rolling speed $v=\omega R$.

Try this in the sim above

Set damping b = 0 with a small torque and watch $\omega$ climb in a perfectly straight line — that's constant $\alpha$, pure $\tau=I\alpha$. Next, keep the torque fixed but slide I up: the same push now spins the disk up far more slowly, so you can feel inertia resisting. Finally, open the L(t) graph with a steady torque: the angular-momentum line rises at a constant slope, a direct picture of $\tau = dL/dt$.

Section 03
Equations & Derivation
Rotational 2nd Law
$$\tau_{\text{net}} = I\alpha$$
Angular Kinematics
$$\omega=\omega_0+\alpha t,\quad\theta=\omega_0 t+\tfrac{1}{2}\alpha t^2,\quad\omega^2=\omega_0^2+2\alpha\theta$$
Rotational KE & Angular Momentum
$$KE_{\text{rot}}=\tfrac{1}{2}I\omega^2,\quad L=I\omega,\quad\tau=\frac{dL}{dt}$$
Moment of Inertia (common shapes)
$$I_{\text{disk}}=\tfrac{1}{2}MR^2,\quad I_{\text{ring}}=MR^2,\quad I_{\text{rod,end}}=\tfrac{1}{3}ML^2$$
Rolling Without Slipping
$$v_{\text{cm}}=\omega R,\quad KE_{\text{total}}=\tfrac{1}{2}Mv^2+\tfrac{1}{2}I\omega^2$$

Symbol Definitions

SymbolQuantitySI Unit
$I$Moment of inertiakg m²
$\tau$Net torqueN m
$\alpha$Angular accelerationrad s⁻²
$\omega$Angular velocityrad s⁻¹
$L=I\omega$Angular momentumkg m² s⁻¹
$KE_{\text{rot}}$Rotational kinetic energyJ
1
Rotational analogy of Newton's 2nd Law. $\tau=I\alpha$ mirrors $F=ma$: torque replaces force, moment of inertia replaces mass, angular acceleration replaces linear acceleration. Every linear law has a rotational counterpart.
2
Moment of inertia. $I=\sum m_i r_i^2$ (discrete) or $\int r^2\,dm$ (continuous). Distributing mass farther from the axis increases $I$ — a hollow cylinder ($I=MR^2$) is harder to spin than a solid disk ($I=\tfrac{1}{2}MR^2$) of same mass and radius.
3
Angular momentum conservation. $dL/dt=\tau_{\text{net}}$. If $\tau_{\text{net}}=0$, then $L=I\omega=\text{const}$. Skater pulling arms inward reduces $I$ → $\omega$ increases to conserve $L$.
4
Rolling without slipping. Contact point is instantaneously at rest; $v_{\text{cm}}=\omega R$. Total KE = translational + rotational. A hollow cylinder accelerates more slowly down a ramp than a solid cylinder of same mass — more KE goes into rotation.
Ref: Halliday, Resnick & Walker, 10th Ed., Ch. 10–11; Serway & Jewett, 8th Ed., Ch. 10–11; Kleppner & Kolenkow, 2nd Ed., Ch. 7.
Section 04
Frequently Asked Questions
All the mass of a hollow cylinder is at maximum distance $R$, giving $I=MR^2$. For a solid disk, mass is spread from centre to edge, averaging to $I=\frac{1}{2}MR^2$. Higher $I$ requires more torque for the same angular acceleration, by $\tau=I\alpha$.
Key takeaway: Moment of inertia depends on where mass is distributed, not just total mass.
Angular momentum $L=I\omega$ is conserved when no external torque acts. Pulling arms in reduces $I$ (mass closer to axis). Since $L$ is constant, $\omega$ must increase. This works for any spinning object — a collapsing star becomes a rapidly spinning neutron star by the same principle.
Key takeaway: $L=I\omega=$ const when $\tau_{\text{net}}=0$: reducing $I$ increases $\omega$.
Flywheels store energy in power plants and hybrid vehicles. Gyroscopes stabilize aircraft, ships, and smartphones. Bicycle wheels resist tipping due to angular momentum. Spinning tops and yo-yos. Car engines and all rotating machinery. Divers and gymnasts control spin by tucking.
Key takeaway: From engines to gyroscopes, rotational dynamics governs all spinning systems.
$I=I_{\text{cm}}+Md^2$, where $I_{\text{cm}}$ is the moment of inertia about the centre of mass and $d$ is the perpendicular distance to the new axis. This lets you find $I$ about any parallel axis without re-integrating.
Key takeaway: Parallel axis theorem: $I=I_{\text{cm}}+Md^2$.
Resources: Khan Academy; HyperPhysics (hyperphysics.phy-astr.gsu.edu); MIT OCW 8.01/8.02.
Section 05
Common Misconceptions
❌ Torque equals force times distance regardless of angle.
✅ Torque $\tau=rF\sin\theta$ — only the component of force perpendicular to the radius vector contributes. A force along the line joining pivot and point of application produces zero torque, regardless of magnitude.
📖 HRW 10th Ed., §10-2.
❌ Angular momentum is conserved whenever no external forces act.
✅ Angular momentum is conserved when net external torque is zero — not merely when no forces act. A force can act without torque if it passes through the axis of rotation.
📖 HRW 10th Ed., §11-4.
❌ Rolling without slipping means there is no friction.
✅ Friction is what causes rolling without slipping — it applies the torque that makes the object rotate. The contact point has zero velocity (no sliding), but static friction force is not zero. Without friction a wheel would slide without rotating.
📖 HRW 10th Ed., §11-2.
Misconception research: Arons — A Guide to Introductory Physics Teaching; Halloun & Hestenes (1985), AJP 53.