Circular motion — centripetal acceleration derivation

WHAT / WHY / HOW

WHY does it exist? Newton's 1st law says a body keeps moving in a straight line unless a force acts. A circle is not a straight line, so the velocity vector must constantly be bent inward. Bending = acceleration. No inward acceleration ⇒ the object flies off tangentially.

WHAT direction? Always perpendicular to the velocity, pointing to the centre. Because it is perpendicular, it changes the direction of $\vec v$ but never its magnitude (it does no work, since $\vec a \perp \vec v$).

HOW big? That is the derivation below.

Derivation from first principles

We do it two ways so you can rebuild it from scratch.

Method 1 — Geometry of the velocity triangle (similar triangles)

Consider an object on a circle of radius $r$ moving at constant speed $v$.

At time $t_1$ it is at point A; a short time $\Delta t$ later it is at point B. It has swept a small angle $\Delta\theta$ at the centre.

Step 1 — Position triangle. The two radius vectors $\vec r_A$ and $\vec r_B$ have equal length $r$ and enclose angle $\Delta\theta$. The chord AB has length, for small angle, $|\text{AB}| \approx r\,\Delta\theta.$ Why this step? Arc length $= r\Delta\theta$, and for a small angle the chord ≈ the arc.

Step 2 — Velocity triangle. The velocity is tangent to the circle, so $\vec v_A \perp \vec r_A$ and $\vec v_B \perp \vec r_B$. Rotating both vectors by $90°$ does not change the angle between them — so $\vec v_A$ and $\vec v_B$ are also separated by the same angle $\Delta\theta$. Both have length $v$.

Why this step? This is the key insight: the velocity vectors form a triangle similar to the position triangle (two equal sides, same enclosed angle).

Step 3 — Similar triangles. Set the magnitude of the velocity change $|\Delta\vec v|$ against the chord: $\frac{|\Delta \vec v|}{v} = \frac{|\text{AB}|}{r} = \frac{r\,\Delta\theta}{r}=\Delta\theta.$ So $|\Delta \vec v| = v\,\Delta\theta.$ Why this step? Corresponding sides of similar triangles are proportional. The "short side over long side" ratio is the same in both triangles.

Step 4 — Take the limit. Acceleration magnitude: $a_c=\lim_{\Delta t\to 0}\frac{|\Delta \vec v|}{\Delta t}=\lim_{\Delta t\to 0}\frac{v\,\Delta\theta}{\Delta t}=v\,\frac{d\theta}{dt}=v\,\omega.$ Since $v=\omega r \Rightarrow \omega = v/r$: $\boxed{a_c=v\omega=\frac{v^2}{r}=\omega^2 r}$

Step 5 — Direction. As $\Delta t\to 0$, $\Delta\theta\to 0$ and $\Delta\vec v$ becomes perpendicular to $\vec v$, pointing toward the centre. Hence "centripetal."

Method 2 — Calculus (position vector)

Let the position be $\vec r(t) = r\big(\cos\omega t,\; \sin\omega t\big),$ constant $r$, constant angular speed $\omega$.

Why this form? Any point on the circle of radius $r$ has these coordinates; $\omega t$ is the angle swept.

Differentiate once (velocity): $\vec v=\dot{\vec r}= r\omega\big(-\sin\omega t,\;\cos\omega t\big),\qquad |\vec v|=r\omega=v.$

Differentiate again (acceleration): $\vec a=\dot{\vec v}= -r\omega^2\big(\cos\omega t,\;\sin\omega t\big) = -\omega^2\,\vec r.$

Why this step? The second derivative of $\cos$ brings down two factors of $\omega$ and a minus sign.

The factor $-\vec r$ shows $\vec a$ points opposite to the radius vector, i.e. straight to the centre, with magnitude $|\vec a| = \omega^2 r = \frac{v^2}{r}. \checkmark$

Both methods agree — that is your confidence check.

Worked examples

Forecast-then-Verify

Common mistakes (Steel-man + fix)

Flashcards

Connections

• Newton's Second Law — $F_c = ma_c$ ties acceleration to the real inward forces.
• Newton's First Law (Inertia) — why the object would otherwise go straight.
• Angular velocity and period — source of $\omega = 2\pi/T$ and $v=\omega r$.
• Banking of roads — application: normal force component as $F_c$.
• Gravitation — orbital motion — gravity as the centripetal force for orbits.
• Vectors — derivative of a unit vector — the calculus method's backbone.
• Centrifugal force (non-inertial frames) — the "fictitious" counterpart.

Concept Map

Hinglish (regional understanding)