Exploring Mechanics / 7. Rotation and Rolling
Every straight-line idea from Kinematics and Newton's Laws has a rotational twin — this section is that translation, one concept at a time.
Make a prediction first: a spinning fan blade has every part turning at the exact same angular speed. Does a point near the tip move through space faster, slower, or at the same speed as a point near the hub?
v = rω · ω = Δθ/Δt · α = Δω/Δt
Every point on a spinning rigid body shares the exact same ω — that's what "rigid" and "spinning together" mean. But tangential speed v = rω depends on r too, so a point farther from the axis sweeps through more actual distance per second, moving faster in the ordinary sense, even while turning at the identical angular rate.
Every linear kinematics equation from Section 1 has a rotational twin, symbol for symbol: x → θ, v → ω, a → α. So v = u + at becomes ω = ω₀ + αt, and s = ut + ½at² becomes θ = θ₀ + ω₀t + ½αt². Nothing new needs deriving — the mathematics is identical, only the physical meaning of each symbol has changed.
Drag any marker to a new radius, or spin the disk faster — every marker always shares the same angular speed, but their tangential speed readouts (and the length of their velocity arrows) scale directly with how far out they sit.
Set an initial spin and an angular acceleration and watch the θ–t, ω–t, and α–t graphs trace out — exactly the same shapes as the linear motion graphs from Section 1, just relabeled.
Make a prediction first: you push a heavy door with the exact same force, first right next to the hinge, then out at the handle, then finally at the handle but pushing at a shallow angle instead of straight-on. Do all three pushes turn the door equally well?
τ = Frsinθ
Force alone doesn't determine how well something turns — where and at what angle you push matters just as much. Pushing right at the hinge (small r) barely turns the door no matter how hard you push; pushing at a shallow angle (small sinθ) wastes most of the force on a component that just presses along the door instead of turning it.
Move the push point and rotate the push angle. The torque readout (and the door's swing speed) responds to both independently — try finding a spot and angle that produces exactly zero torque despite a nonzero force.
τ = Frsinθ is the magnitude of the vector cross product τ⃗ = r⃗ × F⃗ — the sinθ factor is exactly what makes a force pointed directly along the lever arm (θ = 0° or 180°) produce zero torque no matter how large F is, since that force has no component actually perpendicular to the arm to do any turning.
Two equal, opposite forces at different points on a beam can balance in torque (no net spin) while the beam still accelerates sideways — or balance in force while still spinning faster and faster. They're two separate conditions, and satisfying one says nothing about the other.
Two equal, opposite, parallel forces acting at different points — set both sliders to opposite equal values — form what's called a "couple": net force is always exactly zero (nothing accelerates the beam's center of mass anywhere), yet the net torque is not zero, so the beam spins in place, faster and faster, around its own center forever. It's the cleanest possible illustration that force balance and torque balance are independent bookkeeping.
Apply the exact same torque to a hoop, a disk, and a sphere of the same mass and radius, all at once, and watch the sphere spin up fastest — same τ, but very different I = τ/α.
Make a prediction first: a hoop and a solid disk, made of the same material with the same total mass and the same radius, race down the same ramp. Do they tie?
I = Σmiri² (sums up every bit of mass, weighted by how far it sits from the axis)
Unlike ordinary mass, I depends on how that mass is arranged, not just how much there is. A hoop puts every bit of its mass out at the rim (all of it far from the axis); a solid disk spreads mass all the way from the center out to the rim. Same total mass, same radius, very different I — which is why they won't tie down that ramp (the full race is in the next tabs).
Same mass, same radius, four different shapes. Watch how differently "hard to spin up" each one is.
Every one of these formulas has the form I = kmr² for some shape-dependent fraction k ≤ 1. The hoop has k = 1 — the theoretical maximum — because literally all of its mass sits at the full radius r. Every other shape has some mass closer to the axis than r, which can only pull the average r² (and so k) down, never up.
A person on a spinning stool pulls their arms in. Their mass hasn't changed, but pulling it closer to the spin axis shrinks I — and (as the next tab on angular momentum shows) their spin rate has to jump to compensate.
Make a prediction first: a solid sphere and a hoop, same mass and radius, are released from rest at the top of the same ramp and roll (without slipping) to the bottom. Do they arrive at the same speed?
KErot = ½Iω² · Rolling object's total KE = ½mv² + ½Iω²
A flywheel is a purely rotational battery: no chemistry, just spin. Some buses and trains genuinely use flywheels to store braking energy and give it back on acceleration.
Split a spinning object into tiny mass pieces mi, each moving at its own tangential speed vi = riω. Total kinetic energy is Σ½mivi² = Σ½miri²ω² = ½ω²Σmiri² = ½Iω², using the moment-of-inertia definition directly. It's not a new law — it's ordinary KE = ½mv², added up piece by piece around the object.
Release different shapes from the same height and let energy conservation (now split between translation and rotation) predict the finishing order before you ever build the ramp.
Energy conservation: mgh = ½mv² + ½Iω². For rolling without slipping, ω = v/r, and for a solid sphere I = (2/5)mr². Substituting: mgh = ½mv² + ½(2/5)mr²(v/r)² = ½mv²(1 + 2/5) = (7/10)mv². Solving for v gives √(10gh/7) — mass and radius both cancel completely, so the result only depends on shape (through that 2/5), never on size or weight.
Make a prediction first: a bowling ball is thrown down the lane sliding, with almost no spin at all. Does it slide all the way to the pins, or does something change partway down?
Rolling condition (no slipping): v = rω
Now watch the Rotational Energy tab's prediction actually play out — same height, same ramp, released together.
A falling yo-yo is "rolling without slipping" along its own string — the string's contact point on the axle can't slip, which slows the fall dramatically compared to a plain dropped weight.
Treating the yo-yo as a uniform disk, I = ½mR², so the fall acceleration is g/(1 + R²/2r²). Because the axle radius r is so much smaller than the body radius R, that ratio R²/r² is huge, making the whole denominator huge and the fall dramatically slower than free fall — exactly why yo-yos hang and spin for a satisfying few seconds instead of just dropping.
Answering the prediction: friction between the ball and the lane keeps decelerating its slide and speeding up its spin until v = rω is finally satisfied — at which point the ball switches to true rolling and friction's job is done.
Friction acts exactly at the contact point, so it exerts zero torque about that specific point (zero lever arm) — meaning angular momentum measured about the contact point stays constant throughout the skid, even though the ball's linear momentum and spin are each changing. Setting initial and final angular momentum about that point equal, then imposing the final rolling condition vf = rωf, gives the (5/7)v₀ + (2/7)rω₀ formula directly for a solid sphere.
Pull a spool's string at a shallow angle and it rolls toward you; pull it steeper and it reverses and rolls away — the exact same critical-angle idea (θc = arccos(r/R)) that made the double cone and other counterintuitive rolling problems tick.
Make a prediction first: a spinning bicycle wheel is held by its axle by someone sitting on a stool that spins freely. They flip the wheel upside down. Does anything happen to them, even though they never touched the stool?
L = Iω (conserved whenever no external torque acts)
Section 7's Moment of Inertia tab showed I dropping when arms pull in. Here's the payoff: L = Iω has to stay fixed with no external torque, so if I drops, ω must climb to compensate — exactly, not approximately.
Just like linear momentum p = mv stays fixed with no external force, angular momentum L = Iω stays fixed with no external torque. Pulling arms in doesn't apply any external torque about the spin axis (the pulling force acts along a line through that axis), so L before and after must match exactly: Ioutωout = Iinωin.
Answering the prediction: yes — flipping the wheel reverses its angular momentum vector. Since nothing external is touching the stool+person+wheel system, they must start spinning to keep the system's total angular momentum unchanged.
A fast-spinning top under gravity doesn't fall over — its spin axis instead sweeps slowly around in a circle. Torque changes the direction of angular momentum here, not just its size, and the two vectors being perpendicular is exactly what turns "falling" into "precessing."
Precession rate Ω = τ/L — for a fixed tipping torque, a larger spin angular momentum L gives a smaller Ω. That's why a fast top precesses slowly and looks almost stationary, while a nearly-stopped top wobbles wildly right before it topples — as L shrinks toward zero, precession rate blows up, and the simple "torque only changes direction" picture breaks down entirely.
Spin a rigid body about its smallest or largest principal axis and it wobbles a little, stably, forever. Spin it about the intermediate axis instead, and even a tiny wobble grows until the whole body flips — the exact same "no external torque" angular-momentum bookkeeping as the rest of this tab, just with three axes instead of one.