Circular motion — centripetal acceleration derivation: Worked examples

This page is the "throw anything at me" companion to the parent topic. The physics is settled: an object turning at speed $v$ on a circle of radius $r$ has an inward acceleration $a_{c} = \frac{v ^{2}}{r} = ω^{2} r = v ω .$ Here $v$ is the speed (metres per second, always positive), $r$ is the radius (metres, always positive), and $ω$ (the Greek letter "omega") is the angular speed — how many radians of angle the object sweeps per second. If any of those three words feels shaky, read Angular velocity and period first, then come back.

One more input word we'll need: frequency $f$ is how many full turns the object completes each second, measured in hertz ( $Hz$ , i.e. turns per second). Since one turn is $2 π$ radians, angular speed and frequency are tied by $ω = 2 π f$ . Frequency and period are reciprocals: $f = 1/ T$ .

Our job now: enumerate every kind of problem this one formula can be dressed up as, then work one example for each so you never meet a scenario you haven't already seen.

The scenario matrix

Every centripetal-acceleration problem is really "you are handed two of the quantities and asked for a third". The rows below list the class of input you get; the last three rows are the traps that exams add on top.

Case	Case class	What you're given	Trick you must know
A	Speed + radius	$v, r$	plug into $v^{2} / r$ directly
B	Angular speed + radius	$ω, r$	use $ω^{2} r$ , do not find $v$ first
C	Period or frequency	$T$ or $f$ , and $r$	convert to $ω = 2 π / T = 2 π f$
D	Solve backwards	$a_{c}$ known, find $v$ or $r$	rearrange the formula
E	Degenerate: zero input	$v = 0$ or $r \to \infty$	acceleration $\to 0$ (straight line)
F	Limiting behaviour	scale $v$ or $r$	$a_{c} \propto v^{2}$ , $a_{c} \propto 1/ r$
G	Real-world word problem	mixed units (km/h, rpm)	convert units before plugging
H	Exam twist: force + Newton II	mass, string breaks	link to $F_{c} = m a_{c}$ and Newton's Second Law

The nine examples below hit cells A through H in order (Case C gets two: one from a period, one from a frequency). Each says which cell it covers.

The one picture behind all of them

Figure — Circular motion — centripetal acceleration derivation

Look at the figure. The object sits on the circle. Two arrows leave it: the coral arrow is the velocity $v$ , always tangent (grazing the circle, pointing the way it's headed). The lavender arrow is the acceleration $a_{c}$ , always pointing straight to the centre, perpendicular to $v$ . Every example is just asking "how long is the lavender arrow?"

Case A — speed and radius given

Case B — angular speed and radius given

Worked example Whirling stone

A stone whirls at $ω = 10 rad/s$ on a string of length $r = 0.5 m$ . Find $a_{c}$ .

Forecast: Will using $ω^{2} r$ give a bigger or smaller number than the car above?

Choose $a_{c} = ω^{2} r$ . Why this step? We were given the angular speed $ω$ directly. Converting to $v$ first would just be extra work (and a chance to slip).
Square $ω$ : $ω^{2} = 1 0^{2} = 100 rad^{2} / s^{2}$ . Why this step? Same "squared" structure — spinning twice as fast quadruples the pull.
Multiply by radius: $a_{c} = 100 \times 0.5 = 50 m/s^{2}$ . Why this step? Here a bigger radius makes $a_{c}$ larger (because $v = ω r$ also grows) — the opposite of Case A, so don't blindly assume "big $r$ = small $a_{c}$ ".

Verify: Cross-check via $v$ : $v = ω r = 10 \times 0.5 = 5 m/s$ , then $v^{2} / r = 25/0.5 = 50 m/s^{2}$ ✓. Both roads meet. Units: radians are dimensionless, so $s^{- 2} \cdot m = m/s^{2}$ ✓.

Case C — period or frequency given

Worked example C1 — Satellite from its period

A satellite orbits at radius $r = 7.0 \times 1 0^{6} m$ with period $T = 5800 s$ . Find $a_{c}$ .

Forecast: Will $a_{c}$ be around $9$ , or tiny like $0.001$ ?

Convert period to angular speed: $ω = \frac{2 π}{T}$ . Why this step? One full loop is $2 π$ radians; doing it once every $T$ seconds means $2 π / T$ radians each second. See Angular velocity and period.
Compute $ω = \frac{2 π}{5800} \approx 1.0834 \times 1 0^{- 3} rad/s$ . Why this step? A slow orbit ⇒ tiny $ω$ , as expected for something that takes ~1.6 hours per loop.
Apply $a_{c} = ω^{2} r = (1.0834 \times 1 0^{- 3})^{2} \times 7.0 \times 1 0^{6} \approx 8.22 m/s^{2}$ . Why this step? We can chain the period straight in: $a_{c} = \frac{4 π ^{2} r}{T ^{2}}$ , which is just $ω^{2} r$ with $ω = 2 π / T$ substituted.

Verify: $\frac{4 π ^{2} r}{T ^{2}} = \frac{4 π ^{2} ( 7.0 \times 1 0 ^{6} )}{580 0 ^{2}} \approx 8.22 m/s^{2}$ ✓. Sanity: it's close to $g = 9.8$ — correct, low-orbit gravity is nearly full strength (see Gravitation — orbital motion).

Worked example C2 — Fan blade tip from its frequency

A ceiling fan spins at frequency $f = 4 Hz$ (four turns per second). A point on the blade is $r = 0.60 m$ from the axis. Find $a_{c}$ .

Forecast: Bigger or smaller than the satellite's $8 m/s^{2}$ ?

Convert frequency to angular speed: $ω = 2 π f$ . Why this step? Each turn is $2 π$ radians, and there are $f$ turns per second, so the radians swept per second is $2 π f$ . (Equivalently $ω = 2 π / T$ with $T = 1/ f$ .)
Compute $ω = 2 π \times 4 = 8 π \approx 25.13 rad/s$ . Why this step? A fast, small spin ⇒ a large $ω$ , unlike the slow satellite.
Apply $a_{c} = ω^{2} r = (8 π)^{2} \times 0.60 = 4 π^{2} f^{2} r \approx 379 m/s^{2}$ . Why this step? Substituting $ω = 2 π f$ into $ω^{2} r$ gives the frequency form $a_{c} = 4 π^{2} f^{2} r$ — plug $f$ straight in.

Verify: $4 π^{2} f^{2} r = 4 π^{2} (4)^{2} (0.60) \approx 379 m/s^{2}$ ✓. Far larger than $8$ — a fast spin near a modest radius wins big, as forecast.

Case D — solve backwards for a missing quantity

Case E — degenerate inputs (zero and infinity)

Worked example The two "boring" limits

(a) A merry-go-round is stopped: $v = 0$ , $r = 3 m$ . Find $a_{c}$ . (b) A car drives on a dead-straight road — treat it as a circle of radius $r \to \infty$ at $v = 30 m/s$ . Find $a_{c}$ .

Forecast: Both should give the same physically-obvious number. Which?

Case (a): $a_{c} = v^{2} / r = 0^{2} /3 = 0 m/s^{2}$ . Why this step? Not moving ⇒ velocity vector isn't turning ⇒ nothing to accelerate. The formula agrees.
Case (b): $a_{c} = v^{2} / r = 3 0^{2} / r$ ; let $r$ grow without bound. Why this step? A straight line is a circle of infinite radius. As $r \to \infty$ , any fixed numerator over a huge denominator $\to 0$ .
So $r \to \infty lim \frac{900}{r} = 0 m/s^{2}$ . Why this step? This is exactly Newton's First Law (Inertia): no turning force ⇒ straight line ⇒ zero centripetal acceleration.

Verify: Both give $0$ . Consistent: centripetal acceleration exists only when the path actually curves. A stopped object and an infinitely-gentle curve both fail to curve, so both give $0$ ✓.

Case F — limiting / scaling behaviour

Worked example Double the speed, halve the radius

A cyclist rounds a corner at $a_{c} = 4 m/s^{2}$ . In a second run she doubles her speed and takes a corner of half the radius. What is the new $a_{c}$ ?

Forecast: ×2? ×4? ×8?

Write the ratio, not the numbers: $\frac{a _{c, new}}{a _{c, old}} = \frac{v _{new}^{2} / r _{new}}{v _{old}^{2} / r _{old}}$ . Why this step? We're never told the actual $v$ or $r$ , only how they change — so work with ratios and everything unknown cancels.
Speed doubles ⇒ $v_{new} = 2 v_{old}$ ⇒ $v^{2}$ factor is $2^{2} = 4$ . Radius halves ⇒ $r_{new} = \frac{1}{2} r_{old}$ ⇒ dividing by half means ×2. Why this step? $a_{c} \propto v^{2}$ and $a_{c} \propto 1/ r$ act independently; multiply their factors.
Total factor $= 4 \times 2 = 8$ , so $a_{c, new} = 8 \times 4 = 32 m/s^{2}$ . Why this step? Both changes make the turn harder, so they pile up — this is why "just a bit faster on a tighter bend" spins cars out.

Verify: Take a concrete case: $v_{old} = 4 m/s, r_{old} = 4 m \Rightarrow a = 16/4 = 4$ ✓. New $v = 8, r = 2 \Rightarrow 64/2 = 32 m/s^{2}$ ✓. Ratio $32/4 = 8$ ✓.

Case G — real-world word problem (messy units)

Worked example Turntable in rpm, kitchen-table radius

A vinyl record spins at $33 \frac{1}{3} rpm$ (revolutions per minute). A speck of dust sits $r = 0.15 m$ from the centre. Find its $a_{c}$ .

Forecast: Bigger or smaller than $2 m/s^{2}$ ?

Convert rpm to rad/s: $ω = 33.333 \frac{rev}{min} \times \frac{2 π rad}{1 rev} \times \frac{1 min}{60 s}$ . Why this step? The formula only speaks radians-per-second. Convert first, plug second — never feed rpm straight in.
Compute: $ω = \frac{33.333 \times 2 π}{60} \approx 3.491 rad/s$ . Why this step? One revolution is $2 π$ radians; sixty seconds per minute turns "per minute" into "per second".
Apply $a_{c} = ω^{2} r = (3.491)^{2} \times 0.15 \approx 1.828 m/s^{2}$ . Why this step? We were given a rate of spinning (rpm ⇒ $ω$ ) and a radius, so Case-B form $ω^{2} r$ is natural.

Verify: Alternative route: $v = ω r = 3.491 \times 0.15 = 0.5237 m/s$ , then $v^{2} / r = 0.523 7^{2} /0.15 \approx 1.828 m/s^{2}$ ✓. Under $2 m/s^{2}$ as forecast — gentle, dust stays put.

Case H — exam twist: force, Newton II, and a breaking string

Read the figure before the algebra. The butter-yellow line is the string; the coral dot is the ball. Only one arrow points inward — the lavender tension arrow $T$ — and that single inward force is the centripetal force $F_{c}$ (there is no separate "centripetal force" to add). The mint arrow is the ball's velocity $v$ , tangent to the dashed circle. The whole example is: the tension can only get so big before the string snaps, and that caps how fast we can whirl.

Worked example How fast until the string snaps?

A ball of mass $m = 0.20 kg$ is whirled on a horizontal string of length $r = 0.80 m$ . The string breaks when the tension exceeds $F_{m a x} = 40 N$ . What is the maximum speed $v_{m a x}$ ?

Forecast: Around $8$ , $12$ , or $40 m/s$ ?

Identify what provides the centripetal force. Here the tension is the only inward force, so $F_{c} = T$ . Why this step? Centripetal force is never a new force — it's the net inward force supplied by real ones (see the parent note's mistake list). On this horizontal whirl, that's the string tension (the lavender arrow in the figure).
Apply Newton's Second Law in the radial direction: $T = m a_{c} = \frac{m v ^{2}}{r}$ . Why this step? Newton II says net force $=$ mass × acceleration; the inward net force equals the tension, the acceleration is $v^{2} / r$ .
The string breaks when $T = F_{m a x}$ , so set $\frac{m v _{m a x}^{2}}{r} = F_{m a x}$ and solve: $v_{m a x} = \frac{F _{m a x} r}{m}$ . Why this step? Rearrange for the unknown speed; the largest tension the string can hold sets the largest speed it can whirl at.
Numbers: $v_{m a x} = \frac{40 \times 0.80}{0.20} = 160 \approx 12.65 m/s$ . Why this step? Plug in; the square root undoes the $v^{2}$ .

Verify: Back-substitute: $a_{c} = v_{m a x}^{2} / r = 160/0.80 = 200 m/s^{2}$ , then $F = m a_{c} = 0.20 \times 200 = 40 N = F_{m a x}$ ✓. Units of $v$ : $\frac{N \cdot m}{kg} = \frac{kg m/s ^{2} \cdot m}{kg} = m^{2} / s^{2} = m/s$ ✓.

Wrap-up recall

Recall Which formula for which given input?

Given $v$ and $r$ ::: use $a_{c} = v^{2} / r$ . Given $ω$ and $r$ ::: use $a_{c} = ω^{2} r$ (don't convert to $v$ first). Given period $T$ ::: use $a_{c} = 4 π^{2} r / T^{2}$ (since $ω = 2 π / T$ ). Given frequency $f$ ::: use $a_{c} = 4 π^{2} f^{2} r$ (since $ω = 2 π f$ ). Given a breaking-string tension ::: set that tension equal to $m v^{2} / r$ via Newton II. Object at rest or on a straight road ::: $a_{c} = 0$ (no curving path).