Circular motion — centripetal acceleration derivation: Question bank

Before we start, the only symbols used on this page:

$v$ = speed (how fast, a plain number, metres per second).
$v$ = velocity (speed with a direction — an arrow).
$r$ = radius of the circle (metres).
$m$ = mass of the moving object (kilograms).
$ω$ = angular speed (radians turned per second — see Angular velocity and period).
$a_{c} = \frac{v ^{2}}{r} = ω^{2} r = v ω$ = magnitude of the inward (centre-seeking) acceleration.
$F_{c} = m a_{c}$ = the net inward force needed to keep the turn going.

Two small pictures set up the geometry that several traps below refer to. Keep them in view.

The two triangles behind $a_{c} = v^{2} / r$ . On the left, the two radius arrows to points A and B (length $r$ ) enclose the swept angle $Δ θ$ ; the chord AB is their short side. On the right, the two velocity arrows (length $v$ ), each turned $9 0^{\circ}$ from its radius, enclose the same $Δ θ$ — so the two triangles are similar, and $∣Δ v ∣/ v = ∣ AB ∣/ r$ .

Figure — Circular motion — centripetal acceleration derivation

Chord vs. arc. The straight chord AB is always a little shorter than the curved arc it cuts off. Only as $Δ θ$ shrinks toward zero do the two lengths merge — which is exactly why the derivation must take a limit rather than claim they are equal.

True or false — justify

Each answer must give the reason, not a bare verdict.

At constant speed on a circle, the acceleration is zero.

False. Speed is constant but the direction of

v

turns every instant, so

Δ v \neq = 0

; acceleration points to the centre with magnitude

v^{2} / r

Centripetal acceleration does no work on the moving object.

True. It is always perpendicular to

v

, and work

= F \cdot d

vanishes for perpendicular vectors, so it changes direction only, never speed.

If the speed is constant, the magnitude of

a

is constant too.

True for a circle of fixed radius:

a_{c} = v^{2} / r

has all three quantities fixed, so

∣ a ∣

is constant even though the direction of

a

keeps rotating inward.

Centripetal force is an extra force you must add on top of tension or gravity.

False. It is the name for the net inward force, and that net force is supplied by tension, gravity, friction or normal force — drawing both double-counts (see Newton's Second Law).

Doubling the radius (same speed) halves the centripetal acceleration.

True.

a_{c} = v^{2} / r

, and

r

sits in the denominator to the first power, so

r \to 2 r

gives

a_{c} \to a_{c} /2

Doubling the speed (same radius) doubles the centripetal acceleration.

False.

a_{c} \propto v^{2}

, so doubling

v

quadruples

a_{c}

— the reason fast corners are so unforgiving.

The velocity and acceleration vectors point in the same direction during uniform circular motion.

False.

v

is tangent (along the motion);

a

is radial-inward. They are exactly

9 0^{\circ}

apart at every instant.

A satellite in a circular orbit is "falling" toward Earth.

True. Gravity supplies its centripetal acceleration, so it accelerates toward Earth's centre continuously — it simply moves sideways fast enough to keep missing (see Gravitation — orbital motion).

If you cut the string on a whirling stone, the stone flies radially outward.

False. With no inward force, inertia takes over and it flies off along the tangent (straight line), not outward along the radius.

Centrifugal force is a real force acting on the object in the ground frame.

False. In an inertial (ground) frame there is no outward force; centrifugal force is a bookkeeping term that only appears in a rotating frame (see Centrifugal force (non-inertial frames)).

The formulas

v^{2} / r

and

ω^{2} r

can give different numbers for the same motion.

False. They are algebraically identical via

v = ω r

; if you plug consistent values they must match — a mismatch means an arithmetic slip.

Spot the error

Find the flaw in each statement, then say what's correct.

a_{c} = v^{2} r

, so bigger radius means bigger acceleration."

The radius belongs in the denominator:

a_{c} = v^{2} / r

. Bigger radius (same speed) means a gentler turn, hence smaller acceleration.

"Since

a ⊥ v

, the acceleration can't change the velocity at all."

It doesn't change the magnitude (speed), but it does change the direction of

v

— that directional turning is the whole point of the acceleration.

a_{c} = ω^{2} / r

— angular speed squared, divided by radius."

Wrong placement of

r

: it is

ω^{2} r

(radius multiplies). Dimension check:

(rad/s)^{2} \cdot m = m/s^{2}

✓, whereas

ω^{2} / r

gives

1/ (s^{2} m)

, not an acceleration.

"In the similar-triangles derivation, the velocity vectors enclose a different angle than the radius vectors."

They enclose the same angle

Δ θ

— look at the two triangles in the figure above. Each

v

is perpendicular to its radius, and rotating both radii by

9 0^{\circ}

preserves the angle between them.

"The chord AB equals the arc, so

∣ AB ∣ = r Δ θ

exactly."

It is only approximately equal, and only for small

Δ θ

(see the chord-vs-arc figure above); the derivation is rescued by taking the limit

Δ t \to 0

where chord and arc coincide.

"A body on a circle at constant speed obeys Newton's first law — no net force."

A curved path is not straight, so there is a net force (the centripetal one). First law only guarantees straight-line motion when the net force is zero.

Why questions

Explain the mechanism, not just the label.

Why does centripetal acceleration point toward the centre and not along the motion?

Because the change in velocity,

Δ v = v_{B} - v_{A}

, in the limit of a tiny step points inward toward the centre — the only direction that bends the path without altering the speed.

Why is there acceleration even though the speedometer reads a constant number?

The speedometer reads

∣ v ∣

only; acceleration responds to any change in

v

, and the direction is changing continuously.

Why does the required centripetal force grow with the square of the speed?

Two effects each multiply the demand by a factor of

v

: (1) a faster object turns through the same small angle

Δ θ

in less time,

Δ t = Δ θ / ω

with

ω = v / r

, so

Δ t \propto 1/ v

; (2) from the velocity triangle the size of the velocity change is

∣Δ v ∣ = v Δ θ

, which itself grows with

v

. Dividing a

∣Δ v ∣

that scales as

v

by a

Δ t

that scales as

1/ v

gives

a_{c} \propto v^{2}

— the third figure below makes both scalings visible.

Why can gravity alone keep a planet in a circular orbit with no rope or track?

Gravity provides an inward pull toward the central body, and when the magnitude of the gravitational force equals

m v^{2} / r

the planet turns at just the right rate to trace a circle (see Gravitation — orbital motion).

Why do we bank (tilt) a road on a fast curve instead of relying on friction?

Tilting lets the normal force gain an inward horizontal component that supplies the centripetal force, so the turn works even with little or no friction (see Banking of roads).

Why is differentiating the position vector twice a valid second way to get

a_{c}

Acceleration is by definition the second time-derivative of position; applying it to

r = r (cos ω t, sin ω t)

yields

- ω^{2} r

, whose magnitude is

ω^{2} r = v^{2} / r

— matching the geometric result.

Why does the calculus method produce a minus sign in

a = - ω^{2} r

The minus signals that

a

points opposite to the outward radius vector

r

, i.e. straight toward the centre — the same "centre-seeking" conclusion as the geometry (see Vectors — derivative of a unit vector).

Edge cases

Boundary and degenerate scenarios the formulas still must handle.

What is

a_{c}

if the speed is exactly zero?

Zero:

a_{c} = v^{2} / r = 0

. A momentarily stationary object has no centripetal acceleration — there is no direction of motion to bend.

What happens to

a_{c}

as the radius grows without bound (straightening the path)?

a_{c} = v^{2} / r \to 0

r \to \infty

. An "infinite-radius circle" is a straight line, which needs no inward acceleration — consistent with Newton's First Law (Inertia).

Is centripetal acceleration defined when the radius is zero?

No:

a_{c} = v^{2} / r

blows up (division by zero). A point has no circle to move on, so the situation is physically meaningless.

If the speed itself is changing (speeding up while turning), is

a_{c} = v^{2} / r

still the whole acceleration?

No. There is now an additional tangential acceleration along

v

;

v^{2} / r

still gives only the inward (centripetal) part, and the total is the vector sum of the two.

At the exact top of a vertical loop, which direction does the centripetal acceleration point?

Straight down, toward the loop's centre — so gravity (also downward) can help supply it, which is why a minimum speed exists to keep contact.

Does a stone on a string need a minimum speed at the bottom of a vertical circle for the same reason as the top?

No — at the bottom, tension points up (inward) while gravity points down (outward), so tension must overcome gravity and supply

m v^{2} / r

; the tension is largest here, not the speed constraint.

Recall One-line self-test

If a friend says "the door pushes me outward in a fast turn," what's the one-sentence correction? Your body's inertia wants to continue straight; the door actually pushes you inward to force you around the curve — the "outward" sensation is the absence of an outward force, not its presence.