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.

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 This 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