Module 1: Mathematical Foundations · Practice · assumes Foundations 3 and 8
Six harder examples on one operation: differentiating when the basis vectors themselves are turning.
What this set is for
Foundation 8 derived the two results everything here rests on: dr̂/dt = θ̇θ̂ and dθ̂/dt = −θ̇r̂, and from them the velocity and acceleration in polar form. This post uses them six times, on cases of increasing difficulty, until the terms stop looking arbitrary.
The two expressions being exercised throughout:
v = ṙ r̂ + rθ̇ θ̂
a = (r̈ − rθ̇²) r̂ + (rθ̈ + 2ṙθ̇) θ̂
Four terms. Most problems switch off two or three of them, and the examples below are ordered so that each one lights up something the previous one did not. Same three-part structure as the earlier sets.
On this page
1. The same circle, computed twice
Problem. A particle moves on a circle of radius R at constant angular speed ω, so that its Cartesian position is r(t) = R cos ωt î + R sin ωt ĵ. Find the acceleration in Cartesian coordinates, then again in polar coordinates, and show the two agree.
The physics
One particle, one acceleration. The two calculations must agree, and the interest lies in how differently they get there. In Cartesian coordinates the circular geometry is buried inside sines and cosines that have to be differentiated twice. In polar coordinates the geometry is a single statement, r = R, and the trigonometry never appears.
The mathematics
Cartesian. The basis is constant, so only the components differentiate:
v = −Rω sin ωt î + Rω cos ωt ĵ
a = −Rω² cos ωt î − Rω² sin ωt ĵ
= −ω² r
Polar. The trajectory is r = R and θ = ωt, so ṙ = 0, r̈ = 0, θ̇ = ω, θ̈ = 0. Substituting into the two standard expressions, three of the four terms vanish:
v = 0 · r̂ + Rω θ̂ = Rω θ̂
a = (0 − Rω²) r̂ + (0 + 0) θ̂ = −Rω² r̂
Agreement. Since r = R r̂, the Cartesian answer −ω²r is −ω²(R r̂) = −Rω² r̂. The same vector.
The contemplation
Notice where the work went in each version. The Cartesian route did the differentiating and the polar route did the vanishing — three of its four terms were zero before any calculation began, because the constraints ṙ = 0 and θ̈ = 0 had already been stated in the description of the motion. That is what a well-chosen coordinate system does: it converts a physical constraint into a term that is zero from the outset, rather than into trigonometry that has to be carried and then cancelled.
2. A circle that speeds up
Problem. A particle moves on a circle of radius R, starting from rest, with angular position θ(t) = ½αt² for constant α. Find the velocity and acceleration, and describe how the direction of the acceleration changes with time.
The physics
The particle is going round and getting faster. Two things are therefore changing about its velocity — its direction, as before, and now its magnitude too. Expect two acceleration terms rather than one: something inward for the turning, and something along the motion for the speeding up. Since the speed starts at zero, the inward term should start at zero and grow, while the tangential term has no obvious reason to change.
The mathematics
From θ = ½αt² we get θ̇ = αt and θ̈ = α. The radius is fixed, so ṙ = r̈ = 0.
v = Rαt θ̂ so the speed is Rαt, growing linearly
a = −Rα²t² r̂ + Rα θ̂
Both predictions hold. The tangential part is constant at Rα; the radial part grows as t², and it is worth checking that it matches the familiar form: with speed v = Rαt, the quantity v²/R = Rα²t², exactly the radial term.
At t = 0 the acceleration is purely tangential — the particle is not yet turning because it is not yet moving. As time passes the radial part overwhelms the tangential one, and the acceleration vector swings round from tangential toward inward, approaching the radial direction without reaching it.
The contemplation
Students often learn “centripetal acceleration” as though it were the acceleration in circular motion. Here it is one of two terms, and at t = 0 it is the one that is absent. The formula did not have to be extended for this case — the tangential term was always there, sitting at zero whenever θ̈ = 0. A general expression that reduces to the familiar one is worth more than the familiar one, because it also tells you when the familiar one applies.
3. A spiral, with all four terms alive
Problem. A particle moves so that r = bt and θ = ωt, with b and ω constant — an outward spiral traced at steady angular rate. Find its velocity and acceleration, and account physically for each surviving term.
The physics
The particle is doing two things at once: moving outward at a steady rate, and sweeping round at a steady rate. Neither rate is changing, so a first guess might be that the acceleration is small or zero. That guess is wrong, and seeing why is the point of the example — the direction of “outward” is itself rotating, and the distance over which the rotation acts is itself growing.
The mathematics
Here ṙ = b, r̈ = 0, θ̇ = ω, θ̈ = 0.
v = b r̂ + btω θ̂
speed = b√(1 + ω²t²), which grows without bound
Now the acceleration. The two zero quantities remove r̈ and rθ̈, leaving:
a = (0 − btω²) r̂ + (0 + 2bω) θ̂
= −btω² r̂ + 2bω θ̂
The radial term is the familiar centripetal one, growing because the particle is getting further out. The transverse term is 2bω, and it is not zero despite θ̈ = 0 — the angular rate never changes, yet there is a steady acceleration around the circle. It comes entirely from the cross term 2ṙθ̇, which requires both a radial drift and a rotation, and vanishes if either stops.
The contemplation
The 2ṙθ̇ term is the one students most often drop, because nothing in the problem statement announces it — no rate is changing, and yet an acceleration is required. The reason is that the transverse velocity rθ̇ is growing even at constant θ̇, simply because r is. This term is the Coriolis acceleration, met here in its simplest form and in a purely inertial frame. It is not an artefact of rotating reference frames, whatever its reputation. It is what the product rule produces when two things are changing at once.
4. A bead on a spinning rod
Problem. A bead slides without friction on a straight rigid rod that rotates in a horizontal plane at constant angular speed ω about one end. At t = 0 the bead is at r = r0 and at rest relative to the rod. Find the equation governing r(t), solve it, and find the force the rod exerts on the bead.
The physics
The rod is smooth, so it cannot push the bead along its own length — the only force it can exert is perpendicular to itself, in the θ̂ direction. That single sentence is the whole setup: the radial equation has zero on the left-hand side. Whatever happens to r, no radial force causes it. Physically we should expect the bead to work its way outward, since nothing is holding it in.
The mathematics
Radial equation. With no radial force and θ̇ = ω constant:
m(r̈ − rω²) = 0
r̈ = ω²r
Compare this with the SHM condition from Foundation 8, which was ẍ = −ω²x. The sign is opposite, and it changes everything: instead of a restoring pull toward the centre, this drives the bead away in proportion to how far out it already is. The solution is exponential rather than oscillatory. With r(0) = r0 and ṙ(0) = 0:
r(t) = r0 cosh(ωt)
You can verify it directly: differentiating twice returns ω²r0 cosh(ωt), and at t = 0 the position is r0 with zero velocity, since sinh 0 = 0.
The force from the rod. This is the transverse equation, with θ̈ = 0:
N = m(rθ̈ + 2ṙθ̇) = 2mωṙ
= 2m r0ω² sinh(ωt)
The rod must push harder and harder sideways as the bead moves out — and that force exists only because the bead is moving radially. A bead held at fixed r would need no transverse force at all.
The contemplation
The equation r̈ = ω²r differs from the SHM equation by one sign, and the physics differs completely — bounded oscillation in one case, exponential escape in the other. It is worth pausing on how much a sign can carry. Note also that no centrifugal force was invoked anywhere. In this inertial description there is no outward force on the bead at all; there is only the fact that going straight, in a frame where “outward” keeps turning, requires r to increase. The bead is not being flung out. It is failing to be pulled in.
5. Reading the geometry before the algebra
Problem. A particle moves on a spherical surface of radius R. In spherical coordinates its motion satisfies r = R, θ = θ0 (both constant), and φ(t) = ωt. Identify the path and find the acceleration — without expanding the full spherical derivative expressions.
The physics
Two of the three coordinates are frozen, which means the motion has only one degree of freedom, and one degree of freedom in three dimensions is a curve. Before differentiating anything, work out which curve. Fixing r = R puts the particle on a sphere; fixing θ = θ0 puts it on a cone with its apex at the centre. A sphere and a cone sharing a centre intersect in a circle — a line of latitude. So this is circular motion, and everything already known about circular motion applies.
The mathematics
The circle does not have radius R. Its radius is the perpendicular distance from the particle to the z-axis, which the geometry gives immediately:
ρ = R sin θ0
Since φ = ωt, the particle sweeps round this circle at constant angular speed ω. That is uniform circular motion of radius ρ, so example 1 already gives the answer:
speed = ρω = Rω sin θ0
|a| = ρω² = Rω² sin θ0, directed horizontally inward toward the z-axis
Two checks. At θ0 = 90° the path is the equator, ρ = R, and the acceleration is Rω² pointing at the centre of the sphere. At θ0 → 0 the circle shrinks toward the pole and the acceleration tends to zero. Both are what they should be.
Note particularly that the acceleration does not point at the centre of the sphere except in the equatorial case. It points at the centre of the circle, which lies on the axis, above or below the sphere’s centre.
The contemplation
Expanding the spherical acceleration formula would have taken a page and produced the same answer. What replaced it was two sentences of geometry: sphere meets cone in a circle, and the circle has radius R sin θ0. The constraints in the problem were not obstacles to be carried through the algebra — they were information about the shape of the motion, available before any differentiating began. This is the habit worth building from this whole set: read what the equations are describing before deciding what to do to them.
6. Why you cannot just differentiate the components
Problem. A vector A = Aî is constant — fixed in magnitude and direction for all time. An observer moving with θ = ωt writes it in the local polar basis, obtains components that vary with time, and differentiates them. Show what goes wrong, and what the correct derivative is.
The physics
The vector does not change. Its derivative must therefore be exactly zero, and any calculation producing anything else has an error in it. This makes the example a test: we know the answer in advance, so we can see precisely which step of the naive method fails and by how much.
The mathematics
The polar components of a constant Aî at angle θ are Ar = A cos θ and Aθ = −A sin θ. With θ = ωt, differentiating the components alone gives:
Ȧr = −Aω sin θ Ȧθ = −Aω cos θ
Naive answer: −Aω sin θ r̂ − Aω cos θ θ̂, of magnitude Aω — not zero
The error is that r̂ and θ̂ were treated as constants. Differentiating properly means applying the product rule to each term, including the basis vectors:
dA/dt = Ȧr r̂ + Ar (dr̂/dt) + Ȧθ θ̂ + Aθ (dθ̂/dt)
= (Ȧr − Aθθ̇) r̂ + (Ȧθ + Arθ̇) θ̂
Now substitute:
Radial: −Aω sin θ − (−A sin θ)(ω) = −Aω sin θ + Aω sin θ = 0
Transverse: −Aω cos θ + (A cos θ)(ω) = 0
Both components cancel exactly, and dA/dt = 0 as it had to be. The terms the naive method omitted were precisely equal and opposite to the terms it kept.
The contemplation
The naive answer had magnitude Aω — a definite, plausible-looking rate of change for something that was not changing at all. Every term in it was correctly computed; what was wrong was the assumption that a vector is its components. In Cartesian coordinates that assumption is harmless, because the basis never moves, and years of practice can install it invisibly. This is the sharpest reason the distinction matters: not that it is philosophically tidy, but that the moment your basis starts turning, treating components as the whole story produces confident wrong answers.
What the four terms are for
Across these six examples each term earned its place. r̈ is ordinary radial acceleration. −rθ̇² is centripetal, and appeared everywhere something went round. rθ̈ is angular speeding-up, which only example 2 switched on. And 2ṙθ̇ is the one nobody expects — it needs a radial drift and a rotation together, and it is the term the spiral and the bead both turned on.
None of these was postulated. All four dropped out of differentiating r = r r̂ twice, with the product rule applied honestly to a basis that moves. That is the whole content of this post: the machinery is one derivative, taken carefully.
Assumes: Foundations 3 and 8 · Type: Worked examples · Reading: 30 min · Difficulty: Advanced