Circular Motion: Centripetal & Tangential Acceleration

An Olympic hammer thrower spins a $7.26\ \mathrm{kg}$ ball on a wire in a circle about $2$ metres across, and at the moment of release the ball is moving at nearly $28\ \mathrm{m/s}$ . Do the arithmetic this article teaches and you find the ball was being accelerated at roughly $400\ \mathrm{m/s^2}$ — about forty times the acceleration of free fall — while moving at constant speed. That sentence sounds like a contradiction, and resolving it is the whole point of circular-motion kinematics: velocity is a vector, so changing direction counts as accelerating, even when the speedometer never flickers. Every board meets this chapter head-on. CBSE and ISC want the angular vocabulary — radians, angular velocity, period, frequency — and the centerpiece result $a_c = v^2/r$ . AP Physics 1, IB and A-Level phrase it through satellites and cornering cars. JEE-Main and JEE-Advanced push further: motion on a circle where the speed also changes, so a tangential piece must be combined with the centripetal one as perpendicular components. NEET tests formula fluency — rpm conversions and ratio questions appear nearly every year. This module builds the kinematics only: what circular motion is, how fast the velocity turns, and what acceleration that turning demands. The question of which force pays that bill — tension, friction, gravity — belongs to the Centripetal Force module that follows. Learn the geometry here, and the dynamics there becomes bookkeeping.

Key takeaways

Uniform circular motion means constant speed on a circle — yet the velocity vector turns continuously, so the motion is accelerated at every instant.
Angular language is the circle's natural clock: $\omega = \theta/t$ in $\mathrm{rad/s}$ , with $v = \omega r$ , $T = 2\pi/\omega$ , $f = 1/T$ .
The centerpiece: centripetal acceleration $a_c = \dfrac{v^2}{r} = \omega^2 r$ , pointing at the center — always, at every instant.
Convert rotation rates first: $1\ \mathrm{rpm} = \pi/30\ \mathrm{rad/s}$ .
If the speed changes too (JEE territory), add a tangential piece $a_t = \Delta v/\Delta t$ along the path; the total acceleration is $a = \sqrt{a_c^2 + a_t^2}$ because the two components are perpendicular.
Key idea: turning is accelerating — a body on a circle accelerates toward the center at $v^2/r$ even when its speed never changes.

The core insight

A particle moving on a circle of radius $r$ at speed $v$ has an acceleration of magnitude $v^2/r = \omega^2 r$ directed toward the center at every instant, because the velocity vector's direction is being turned continuously; if the speed changes as well, a perpendicular tangential component $a_t = \Delta v/\Delta t$ adds to it.

Where this lives today

Centrifuges are the purest industrial use of $a_c = \omega^2 r$ : a benchtop machine at $6000$ rpm on a $10$ cm arm produces $(628)^2 \times 0.10 \approx 39{,}000\ \mathrm{m/s^2}$ — about $4000g$ — and ultracentrifuges pass $100{,}000g$ , enough to separate protein molecules by mass. The same scaling runs the astronaut-training centrifuge (a long arm at modest rpm delivers a controlled $8g$ ) and a washing machine's spin cycle (about $100g$ at the drum wall). At the other end of the scale, the Large Hadron Collider holds protons on a circle of radius $4.2\ \mathrm{km}$ at effectively the speed of light: $a_c \approx (3\times10^8)^2/4200 \approx 2\times10^{13}\ \mathrm{m/s^2}$ . Relativity modifies the dynamics, but the geometry — velocity forever tangent, acceleration forever inward — is exactly this article's first figure, and the MEMS gyroscope in your phone runs the same mathematics in silicon.

The physical picture

Whirl a stone on a string and freeze one instant. The velocity vector lies along the tangent — the direction the stone would fly if everything else vanished. A moment later the velocity, still the same length, points along a new tangent. Same speed, different direction: the velocity has changed, and change of velocity per unit time is acceleration. Let the string snap and the stone confirms it — departing along the tangent, not outward along the radius. Nothing ever pushed it outward; something was always pulling it in, and that pull's acceleration is what we are about to measure.

Figure

Uniform circular motion — velocity rides the tangent, acceleration points at the center

Why should the answer be $v^2/r$ , with the speed appearing twice? Here is the intuition before the derivation makes it exact. In a short time the velocity vector swings through the same small angle that the stone sweeps around the circle. A faster stone sweeps that angle more quickly — one factor of $v$ . And a faster stone has a longer velocity vector, so turning it through the same angle changes it by more — a second factor of $v$ . Meanwhile a larger circle makes the turning more leisurely, putting $r$ downstairs. Fast motion on a tight circle is punished on both counts — the hammer thrower's price.

The math: the angular toolkit

Angles in physics are measured in radians: the angle whose arc length equals the radius. A full turn is $2\pi$ radians, so $360° = 2\pi\ \mathrm{rad}$ and $1\ \mathrm{rad} \approx 57.3°$ . Radians earn their keep through one clean statement — arc length $s = r\theta$ — and everything below silently assumes them. The angular velocity $\omega$ is the angle swept per unit time. For uniform motion it is the circle's built-in clock, and it converts to the ordinary speed of any point through the radius. One full revolution ( $2\pi$ radians) takes one period $T$ , and the frequency $f$ counts revolutions per second:

\omega = \frac{\theta}{t} = \frac{2\pi}{T} = 2\pi f, \qquad v = \omega r \qquad (1)

where: $\omega$ — angular velocity in $\mathrm{rad/s}$ ; $T$ — period in seconds; $f$ — frequency in $\mathrm{Hz}$ (revolutions per second); $v$ — the linear (tangential) speed of a point at radius $r$ . Read $v = \omega r$ as an exchange rate between the angular and linear languages: on one spinning disc, $\omega$ is the same everywhere, so the speed grows in proportion to the radius — the rim of a merry-go-round moves fast while a child near the hub barely drifts. Exams love to quote rotation rates in revolutions per minute. The conversion is worth making reflexive, because a wrong factor here poisons everything downstream:

1\ \mathrm{rpm} = \frac{2\pi\ \mathrm{rad}}{60\ \mathrm{s}} = \frac{\pi}{30}\ \mathrm{rad/s} \qquad (2)

where: the $2\pi$ converts revolutions to radians and the $60$ converts minutes to seconds. So $300\ \mathrm{rpm} = 10\pi \approx 31.4\ \mathrm{rad/s}$ . NEET and JEE-Main hide this conversion inside half their circular-motion numericals.

you are given	reach for	to get
period $T$ or frequency $f$	$\omega = 2\pi/T = 2\pi f$	angular velocity
rotation rate in rpm	$\omega = \mathrm{rpm} \times \pi/30$	angular velocity in rad/s
$\omega$ and radius $r$	$v = \omega r$ , $a_c = \omega^2 r$	speed and centripetal acceleration
speed $v$ and radius $r$	$a_c = v^2/r$	centripetal acceleration
speed changing at rate $a_t$	$a = \sqrt{a_c^2 + a_t^2}$	total acceleration (JEE)

Derivation ACentripetal acceleration from similar triangles

We derive the centerpiece $a_c = v^2/r$ using nothing but geometry — the route every board accepts and the one that shows why the speed enters twice. The strategy: draw the position vectors at two nearby instants and the velocity vectors at the same instants, notice the two triangles are similar, and convert the similarity into a ratio between velocity change and distance travelled. One subtlety to watch: the velocity triangle is built by sliding both vectors to a common tail — legal, because a vector is its magnitude and direction, not its location.

Step 1. Freeze the particle at two nearby moments, a short time $\Delta t$ apart. The two position vectors from the center, each of length $r$ , enclose a small angle $\Delta\theta$ , and the particle has travelled an arc of length $v\,\Delta t$ between them. For a small enough angle, the arc and the chord are indistinguishable, so the triangle of the two radii and the chord has two sides $r$ and a base of length approximately $v\,\Delta t$ .

Step 2. Now draw the two velocity vectors tail-to-tail. Each has length $v$ , and because each is perpendicular to its own radius, the angle between them is the same $\Delta\theta$ . Their difference $\Delta\vec{v}$ closes this second triangle: two sides $v$ , base $|\Delta\vec{v}|$ , apex angle $\Delta\theta$ . Why: rotating a radius by some angle rotates its perpendicular by exactly the same angle — the velocity vector is chained to the radius at a fixed 90°, so it turns in lockstep.

Step 3. The two triangles are isosceles with the same apex angle, hence similar. Matching bases to sides:

\frac{|\Delta\vec{v}|}{v} = \frac{v\,\Delta t}{r}

Step 4. Divide by $\Delta t$ to extract the acceleration magnitude:

a_c = \frac{|\Delta\vec{v}|}{\Delta t} = \frac{v^2}{r}

Step 5. Direction: as $\Delta\theta$ shrinks, the base $\Delta\vec{v}$ of the isosceles velocity triangle becomes perpendicular to the velocity itself — and perpendicular-to-tangent means along the radius, pointing inward. Substituting $v = \omega r$ gives the twin form $a_c = \omega^2 r$ . Why: both forms matter in exams — $v^2/r$ when you know the speed, $\omega^2 r$ when you know the rotation rate; they are the same statement translated through equation (1). *Verified: limits behave — $r \to \infty$ at fixed $v$ gives $a_c \to 0$ (a straight line is a circle of infinite radius); doubling $v$ quadruples $a_c$ . The result matches the calculus route (differentiate $\vec r = r(\cos\omega t, \sin\omega t)$ twice, giving $\vec a = -\omega^2 \vec r$ ) and NCERT Ch. 4 / Halliday–Resnick Ch. 6.*

Figure

The similar-triangles argument — the geometry behind v squared over r

A note on the calculus route, purely as a preview: differentiating $x = r\cos\omega t$ , $y = r\sin\omega t$ twice returns exactly $a = \omega^2 r$ inward. A satisfying check if you know derivatives — but nothing in this chapter needs it; the similar-triangles argument is the one board marking schemes reward.

When the speed changes too: tangential acceleration

Uniform circular motion is the pure case, but JEE-Main and JEE-Advanced routinely put a particle on a circle and change its speed — a car accelerating out of a bend, a fan spinning up after the switch flips. Two different things are then happening to the velocity vector: its direction is turning (that is the centripetal part, still $v^2/r$ , still inward) and its length is growing or shrinking. The rate of length-change is the tangential acceleration, and it points along the tangent:

a_t = \frac{\Delta v}{\Delta t} \qquad (3)

where: $a_t$ — tangential acceleration, positive when speeding up (along the motion) and negative when braking (against it). This component is exactly the ordinary one-dimensional acceleration you know from straight-line kinematics, wrapped onto the curve. In uniform circular motion $a_t = 0$ — that is what "uniform" means.

Go deeper · Derivation B · 1 minTotal acceleration on a circle — combining the two components

Derivation BTotal acceleration on a circle — combining the two components

We show that when a particle on a circle changes speed, its total acceleration has magnitude $a = \sqrt{a_c^2 + a_t^2}$ , and we locate its direction. The strategy is decomposition: split the change in the velocity vector into a piece that turns it and a piece that stretches it, identify each with one component, and add the perpendicular pieces by Pythagoras. The thing to watch: perpendicular components never mix — never add $a_c + a_t$ as plain numbers.

Step 1. Over a short time $\Delta t$ , the velocity vector changes from $\vec v_1$ to $\vec v_2$ . Resolve the change $\Delta \vec v$ into two perpendicular directions at the particle's location: along the tangent (the direction of $\vec v_1$ ) and along the inward radius.

Step 2. The radial piece comes from the turning of the direction alone. By the similar-triangles argument of the previous derivation, its rate is $a_c = v^2/r$ , pointing at the center — with $v$ the current speed. Why: the turning rate cares only about how fast the direction is swinging right now, so the formula survives unchanged even while $v$ drifts; you simply feed it the instantaneous speed.

Step 3. The tangential piece comes from the stretching of the length alone: $a_t = \Delta v/\Delta t$ , along the tangent (or against it while braking).

Step 4. The two pieces sit at $90°$ to each other — one along the tangent, one along the radius. Perpendicular components combine by Pythagoras:

a = \sqrt{a_c^2 + a_t^2}

and the total acceleration leans away from the pure-inward direction by an angle $\phi$ with $\tan\phi = a_t/a_c$ measured from the radius. Why: a classic JEE distractor offers $a_c + a_t$ — correct only if the components were parallel, and they never are. *Verified: limits — $a_t = 0$ recovers uniform circular motion ( $a = a_c$ , purely inward); $r \to \infty$ kills $a_c$ and recovers straight-line kinematics ( $a = a_t$ ). Numbers from Worked Example 3 check: $a_c = 18$ , $a_t = 3$ , $a = \sqrt{333} \approx 18.2\ \mathrm{m/s^2}$ , not $21$ .*

Limiting cases

case	result	meaning
$r \to \infty$ at fixed $v$	$a_c \to 0$	a straight line is a circle of infinite radius — straight-line kinematics recovered
$a_t = 0$	$a = a_c$ , purely inward	uniform circular motion: speed constant, direction forever turning
$a_c = a_t$	$a = \sqrt{2}\,a_c$ , at $45°$ to the radius	the crossover moment for a particle speeding up from rest (a JEE favourite)
same $\omega$ , different $r$ (rigid disc)	$v \propto r$ , $a_c \propto r$	rim points move faster and accelerate harder than hub points
doubling $\omega$ at fixed $r$	$a_c$ quadruples	$\omega^2$ scaling: spin a centrifuge twice as fast, get four times the effect

Worked examples

1 · Board level — the ceiling fan (CBSE/ISC pattern, 3 marks)

A ceiling fan's blade tips move on a circle of radius $0.30\ \mathrm{m}$ at $300\ \mathrm{rpm}$ . Find the angular velocity, the tip speed, and the centripetal acceleration of a tip.

Step 1. Convert the rotation rate first, using equation (2): $\omega = 300 \times \dfrac{\pi}{30} = 10\pi \approx 31.4\ \mathrm{rad/s}$ .

Step 2. Tip speed from equation (1): $v = \omega r = 31.4 \times 0.30 = \boxed{9.42\ \mathrm{m/s}}$ .

Step 3. Centripetal acceleration, angular form: $a_c = \omega^2 r = (10\pi)^2 \times 0.30 = 987 \times 0.30 = \boxed{296\ \mathrm{m/s^2}}$ .

Step 4. Sanity check by the linear form: $a_c = v^2/r = (9.42)^2/0.30 = 88.8/0.30 = 296\ \mathrm{m/s^2}$ ✓ — thirty times $g$ from a household appliance.

2 · Board level — the record player (ISC/board pattern)

An insect sits $10\ \mathrm{cm}$ from the center of a gramophone record rotating at $33\tfrac{1}{3}$ rpm. Find the period, the insect's speed, and its centripetal acceleration. If the insect walks out to $20\ \mathrm{cm}$ , what happens to each quantity?

Step 1. Angular velocity: $\omega = \dfrac{100}{3} \times \dfrac{\pi}{30} = \dfrac{10\pi}{9} \approx 3.49\ \mathrm{rad/s}$ . Period: $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi \times 9}{10\pi} = 1.8\ \mathrm{s}$ .

Step 2. Speed at $r = 0.10\ \mathrm{m}$ : $v = \omega r = 3.49 \times 0.10 = \boxed{0.349\ \mathrm{m/s}}$ .

Step 3. Centripetal acceleration: $a_c = \omega^2 r = (3.49)^2 \times 0.10 = 12.2 \times 0.10 = \boxed{1.22\ \mathrm{m/s^2}}$ .

Step 4. At $r = 0.20\ \mathrm{m}$ the record turns at the same $\omega$ and $T$ — the disc is rigid — but $v \propto r$ and $a_c \propto r$ , so each doubles: $v = 0.698\ \mathrm{m/s}$ , $a_c = 2.44\ \mathrm{m/s^2}$ .

3 · Competitive level — JEE-Main pattern (~2.5 minutes)

A particle starts from rest on a circle of radius $2.0\ \mathrm{m}$ and speeds up with constant tangential acceleration $a_t = 3.0\ \mathrm{m/s^2}$ . Find the magnitude and direction of its total acceleration at $t = 2.0\ \mathrm{s}$ .

Step 1. Speed at $t = 2.0$ s, by straight-line kinematics along the arc: $v = a_t t = 3.0 \times 2.0 = 6.0\ \mathrm{m/s}$ .

Step 2. Centripetal component at that instant: $a_c = \dfrac{v^2}{r} = \dfrac{36}{2.0} = 18\ \mathrm{m/s^2}$ , pointing at the center.

Step 3. Combine perpendicular components: $a = \sqrt{a_c^2 + a_t^2} = \sqrt{18^2 + 3^2} = \sqrt{324 + 9} = \sqrt{333} = \boxed{18.2\ \mathrm{m/s^2}}$ .

Step 4. Direction: $\tan\phi = \dfrac{a_t}{a_c} = \dfrac{3}{18} = \dfrac{1}{6}$ , so the acceleration leans only $\phi \approx 9.5°$ off the inward radius — the turning job already dwarfs the speeding-up job, since $a_c$ grows like $t^2$ while $a_t$ stays fixed. Timing note: budget ~2.5 min. The only real work is Steps 1–3; the trap is the distractor $18 + 3 = 21\ \mathrm{m/s^2}$ , which is always on offer. If you catch yourself adding the components as plain numbers, stop — they are perpendicular.

4 · Competitive level — NEET/JEE-Main pattern (~2 minutes)

A particle moves on a circle of radius $4.0\ \mathrm{m}$ with its speed growing as $v = 2t$ (in $\mathrm{m/s}$ , $t$ in seconds). At what time are the centripetal and tangential accelerations equal in magnitude, and what is the total acceleration then?

Step 1. The tangential acceleration is the rate of speed growth: $a_t = 2.0\ \mathrm{m/s^2}$ , constant.

Step 2. The centripetal component grows with the speed: $a_c = \dfrac{v^2}{r} = \dfrac{(2t)^2}{4.0} = \dfrac{4t^2}{4.0} = t^2$ .

Step 3. Set them equal: $t^2 = 2.0 \Rightarrow t = \sqrt{2} \approx \boxed{1.41\ \mathrm{s}}$ .

Step 4. At that moment both components equal $2.0\ \mathrm{m/s^2}$ , so $a = \sqrt{2^2 + 2^2} = 2\sqrt{2} \approx \boxed{2.83\ \mathrm{m/s^2}}$ , at $45°$ to the radius — the crossover geometry from the limiting-cases table. Timing note: ~2 min. Recognize the template — " $v = kt$ , when does $a_c$ catch $a_t$ ?" — and the answer is always $t = \sqrt{r/k}$ : check, $\sqrt{4/2} = \sqrt2$ ✓.

Exam traps

Feeding rpm into formulas raw. $\omega$ in $a_c = \omega^2 r$ must be in $\mathrm{rad/s}$ : convert via equation (2) first. Forgetting the $2\pi$ (or the $60$ ) scales the answer by $39$ or $3600$ — and examiners plant those among the options.

Calculator in degree mode. Anything with $\sin$ or $\cos$ of $\omega t$ takes radians; a calculator left in degree mode is a silent killer.

Adding $a_c$ and $a_t$ as plain numbers. They are perpendicular; the total is $\sqrt{a_c^2 + a_t^2}$ , never $a_c + a_t$ . The plain sum is a standard distractor (Worked Example 3's $21\ \mathrm{m/s^2}$ ).

Radius vs diameter. "A circle of diameter $80$ cm" means $r = 0.40\ \mathrm{m}$ — halve before you square.

"Uniform" does not mean constant velocity. In uniform circular motion the constant quantities are speed, $\omega$ , $T$ , and $|a_c|$ ; the vectors $\vec v$ and $\vec a_c$ change direction at every instant.

Mixing the disc's quantities with the point's. On one rigid body, $\omega$ , $T$ , $f$ are shared by every point; $v$ and $a_c$ belong to each point and grow with radius (Example 2's doubling).

Which $g$ ? Circular-motion kinematics itself has no $g$ , but the moment a question compares $a_c$ to gravity ("how many g's?"), use the paper's convention: CBSE/ISC $9.8\ \mathrm{m/s^2}$ , JEE/NEET as stated (usually $10$ ), A-Level $9.81$ .

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1642–1727 · Cambridge & London

Newton named the subject. In the Principia (1687) he coined the term vis centripeta — "center-seeking force" — deliberately inverting Christiaan Huygens's vis centrifuga, and proved that a body swept toward a fixed center obeys Kepler's equal-areas law. He had already used the $v^2/r$ rule in the 1660s for his "Moon test" — comparing the Moon's centripetal acceleration to an apple's fall and finding the inverse-square ratio $1/60^2$ — the calculation that convinced him terrestrial and celestial gravity are one law. The kinematics you have just learned was, in his hands, the key that unlocked the solar system.

History

1659

— Christiaan Huygens derives the quantitative rule of circular motion — in modern notation

a = v^2/r

— and later publishes theorems on his "centrifugal force" in Horologium Oscillatorium (1673).

1666

— Newton runs the Moon test: the Moon's inward acceleration is smaller than an apple's by 60², exactly the inverse-square prediction at 60 Earth radii.

1687

— Newton's Principia coins "centripetal" and builds planetary motion from the same turning-velocity geometry.

1851

— Foucault's pendulum makes the Earth's own rotation visible in a Paris museum hall.

1959

— Luna 1 becomes the first spacecraft to reach escape speed; every orbit flown since is centripetal bookkeeping.

Going further

Texts. NCERT Physics Class 11, Ch. 4 (Motion in a Plane, uniform circular motion section) · HC Verma, Concepts of Physics Vol. 1, Ch. 7 · Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 4 and Ch. 6.

Provenance. All results are standard kinematics. A verifier should recompute: the rpm conversions ( $300\ \mathrm{rpm} = 10\pi = 31.4\ \mathrm{rad/s}$ ; $600\ \mathrm{rpm} = 20\pi\ \mathrm{rad/s}$ ); Example 1 ( $v = 9.42\ \mathrm{m/s}$ , $a_c = 296\ \mathrm{m/s^2}$ by both forms); Example 2 ( $T = 1.8$ s, $v = 0.349\ \mathrm{m/s}$ , $a_c = 1.22\ \mathrm{m/s^2}$ , both doubling at $0.20$ m); Example 3 ( $v = 6.0$ , $a_c = 18$ , $a = \sqrt{333} = 18.2\ \mathrm{m/s^2}$ , $\tan\phi = 1/6$ ); Example 4 ( $t = \sqrt 2 = 1.41$ s, $a = 2\sqrt 2 = 2.83\ \mathrm{m/s^2}$ ); the hammer-throw estimate ( $28^2/1.9 \approx 410\ \mathrm{m/s^2}$ ); the centrifuge figure ( $(628)^2 \times 0.10 \approx 3.9\times10^4\ \mathrm{m/s^2}$ ).

Prerequisites. Speed & Velocity · Components & Resolution of Vectors · Motion in a Plane. Leads to. Centripetal Force: Dynamics of Circular Motion · Turning on a Flat Road · Banked Curves — and, for the deep dive, the university-tier treatment in Circular Motion & Centripetal Acceleration (Advanced Mechanics).

If the speed never changes, what exactly is accelerating?

The direction. Velocity is a package of magnitude and direction, and acceleration is the rate of change of the whole package. On a circle at constant speed the package changes purely by rotating: over each short interval the velocity vector swings through a small angle, and that swing, divided by the time, is a genuine acceleration of magnitude $v^2/r$ pointing at the center. Physical test: accelerating things require net force, and the string of a whirled stone really does pull.

Why do we bother with radians instead of degrees?

Because radians make arc length trivial: $s = r\theta$ works only with radians, and everything downstream — $v = \omega r$ , $a_c = \omega^2 r$ — inherits its cleanliness from that one formula. In degrees every equation would sprout factors of $\pi/180$ . The price is one habit: full turn $= 2\pi$ rad, rpm $\to$ rad/s via $\pi/30$ — and exams quote rpm precisely to test that you pay it.

Is centripetal acceleration constant in uniform circular motion?

Its magnitude is constant ( $v^2/r$ never changes), but its direction rotates continuously — it always aims at the center, whose direction from the particle keeps changing. As a vector, $\vec a_c$ is not constant. Favourite one-mark trap: "uniform circular motion is motion with constant acceleration — true or false?" False.

Where does the outward force I feel on a merry-go-round fit into this?

Your own inertia, not an outward force. Your body tries to continue along the tangent (Newton's first law) while the ride keeps pulling you onto the circle; "flung outward" is the ride accelerating inward underneath you. An inertial-frame free-body diagram contains only real inward-pointing agents. The outward "centrifugal force" earns legitimate meaning later, in the Non-Inertial Frames module — no inertial-frame analysis here ever needs it.

How can uniform circular motion be accelerated if the kinetic energy never changes?

Because the acceleration is always perpendicular to the velocity. Kinetic energy responds to work, and work needs a force component along the motion. The centripetal pull has none — it redirects the velocity without stretching it, like steering a car without touching the pedals. The moment a tangential component appears, work is done and the kinetic energy changes — and this is why a satellite in circular orbit keeps its speed forever: gravity does zero work on it.

For the JEE-style problems, when do I use v²/r and when ω²r?

They are the same quantity in two currencies, linked by $v = \omega r$ . Use $v^2/r$ when the problem hands you a speed; use $\omega^2 r$ when it hands you a rotation rate, period, or frequency. Converting to the wrong currency first is legal but wastes time and invites slips — a two-second glance at what is given should pick the form.