Circular Motion Mathematical Derivations

Overview

This note gives the compact vector-based derivation behind the standard uniform-circular-motion results:

• $v=r\omega$
• $a_c=r{\omega }^2$
• $a_c=\frac{v^2}{r}$

At H2 level, you usually apply these results directly. The derivation is still useful because it shows clearly why the acceleration is inward even when the speed is constant.

Definition

In uniform circular motion, the object moves around a circle with constant speed but continuously changing velocity direction. The acceleration is therefore centripetal: it points towards the centre of the circle.

Key Representations

The derivation below uses component equations for position, velocity and acceleration. This makes the vector direction of centripetal acceleration explicit rather than treating $a_c=r{\omega }^2$ as a memorised scalar formula.

Position Model

For a particle moving anticlockwise in a circle of radius $r$ in the $xy$-plane, let

$$\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}$$

with

$$x(t)=r\cos (\omega t),y(t)=r\sin (\omega t)$$

So

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

The magnitude of the position vector remains constant:

$$|\vec{r}|=r$$

Velocity Components

Differentiate $\vec{r}(t)$ with respect to time:

$$\vec{v}(t)=\frac{d\vec{r}}{dt}$$

Hence

$$v_x(t)=-r\omega \sin (\omega t),v_y(t)=r\omega \cos (\omega t)$$

so

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

The speed is

$$|\vec{v}|=\sqrt{v_x^2+v_y^2}=\sqrt{r^2{\omega }^2{\sin }^2(\omega t)+r^2{\omega }^2{\cos }^2(\omega t)}=r\omega$$

Therefore:

$$v=r\omega$$

This also shows that $\vec{v}$ is always tangent to the path and perpendicular to $\vec{r}$.

Acceleration Components

Differentiate $\vec{v}(t)$ with respect to time:

$$\vec{a}(t)=\frac{d\vec{v}}{dt}$$

Hence

$$a_x(t)=-r{\omega }^2\cos (\omega t),a_y(t)=-r{\omega }^2\sin (\omega t)$$

so

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

Factor out the position-vector form:

$$\vec{a}(t)=-{\omega }^2\vec{r}(t)$$

This is the clearest vector statement of centripetal acceleration:

• $\vec{a}$ is proportional to $\vec{r}$
• the minus sign shows it points opposite to the outward radius vector
• therefore $\vec{a}$ points toward the centre

Magnitude of Centripetal Acceleration

The magnitude is

$$|\vec{a}|=\sqrt{a_x^2+a_y^2}=\sqrt{r^2{\omega }^4{\cos }^2(\omega t)+r^2{\omega }^4{\sin }^2(\omega t)}=r{\omega }^2$$

Therefore:

$$a_c=r{\omega }^2$$

Using $v=r\omega$:

$$a_c=r{\omega }^2=r{\left(\frac{v}{r}\right)}^2=\frac{v^2}{r}$$

So the standard centripetal-acceleration results are

$${\vec{a}}_c=-{\omega }^2\vec{r}$$ $$a_c=r{\omega }^2=\frac{v^2}{r}$$

Why It Matters

This derivation explains two important ideas:

1. Constant speed does not imply zero acceleration.
2. In uniform circular motion, the acceleration changes the direction of $\vec{v}$ rather than its magnitude.

That is why the centripetal force:

• is required continuously to maintain the circular path
• points inward
• does no work in uniform circular motion because it is perpendicular to $\vec{v}$

Links

• Related: circular motion
• Related: circular motion core concepts
• Related: centripetal acceleration and force
• Related: Vectors
• Misconception: vector direction consistency