Circular motion — derive centripetal acceleration a = v²/r

Question

A particle moves in a circle of radius $r$ with constant speed $v$ . Derive the expression for its centripetal acceleration $a = v^2/r$ from first principles.

(NCERT Class 11, Chapter 4 — frequently asked in boards and JEE Main)

Solution — Step by Step

Set up position vectors at two instants

Consider a particle moving along a circle of radius $r$ . At time $t$ , let the particle be at point P, and at time $t + \Delta t$ , at point Q. The angle swept is $\Delta\theta$ .

The position vectors are:

$\vec{r_1} = r\cos\theta\,\hat{i} + r\sin\theta\,\hat{j}$

$\vec{r_2} = r\cos(\theta + \Delta\theta)\,\hat{i} + r\sin(\theta + \Delta\theta)\,\hat{j}$

Find the change in velocity

Since speed is constant, the velocity at P is tangent to the circle with magnitude $v$ . At Q, the velocity has the same magnitude but a different direction — it has rotated by $\Delta\theta$ .

The velocity vectors:

$\vec{v_1} = v(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j})$

$\vec{v_2} = v(-\sin(\theta + \Delta\theta)\,\hat{i} + \cos(\theta + \Delta\theta)\,\hat{j})$

The magnitude of $\Delta\vec{v}$ for small $\Delta\theta$ : $|\Delta\vec{v}| = v \cdot \Delta\theta$ .

Compute acceleration as Δv/Δt in the limit

$a = \lim_{\Delta t \to 0} \frac{|\Delta\vec{v}|}{\Delta t} = \lim_{\Delta t \to 0} \frac{v \cdot \Delta\theta}{\Delta t} = v \cdot \omega$

Since $\omega = v/r$ :

$\boxed{a = v \cdot \frac{v}{r} = \frac{v^2}{r}}$

The direction of $\Delta\vec{v}$ (and hence acceleration) points radially inward — toward the centre. That is why we call it centripetal (centre-seeking) acceleration.

Why This Works

Even though the speed is constant, the velocity is continuously changing direction. Any change in velocity — whether in magnitude or direction — requires acceleration. In uniform circular motion, the acceleration does zero work (it's perpendicular to displacement) but it constantly redirects the velocity vector to keep the particle on its circular path.

The key insight: the velocity vector traces out its own circle (of radius $v$ ) in velocity space. The rate at which this "velocity circle" is traversed gives the acceleration: $a = v\omega = v^2/r$ .

We can also write this as $a = \omega^2 r$ or $a = 4\pi^2 r / T^2$ , depending on what's given.

Alternative Method — Using Calculus Directly

Write position as $\vec{r} = r\cos(\omega t)\,\hat{i} + r\sin(\omega t)\,\hat{j}$ .

Differentiate once for velocity:

$\vec{v} = -r\omega\sin(\omega t)\,\hat{i} + r\omega\cos(\omega t)\,\hat{j}$

Differentiate again for acceleration:

$\vec{a} = -r\omega^2\cos(\omega t)\,\hat{i} - r\omega^2\sin(\omega t)\,\hat{j} = -\omega^2 \vec{r}$

Magnitude: $a = \omega^2 r = v^2/r$ . The negative sign confirms it points toward the centre (opposite to $\vec{r}$ ).

💡 Expert Tip

In JEE, if a question gives angular velocity $\omega$ , use $a = \omega^2 r$ . If it gives linear speed $v$ , use $a = v^2/r$ . If it gives time period $T$ , use $a = 4\pi^2 r/T^2$ . Pick the formula that avoids extra conversions — saves precious seconds.

Common Mistake

⚠️ Common Mistake

Students sometimes claim "centripetal acceleration is zero because speed is constant." Speed being constant does NOT mean acceleration is zero — acceleration is the rate of change of velocity (a vector), not speed (a scalar). In circular motion, direction changes constantly, so there is always acceleration directed toward the centre.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Calculus Directly

Common Mistake

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