Circular motion involves a particle moving along the circumference of a circle. In VCE Specialist Mathematics, this is primarily analyzed using vector calculus in two dimensions.
The position of a particle moving in a circle of radius $r$ centered at the origin $(0,0)$ at time $t$ can be expressed by the vector function:
$$\mathbf{r}(t) = r\cos(\theta(t))\mathbf{i} + r\sin(\theta(t))\mathbf{j}$$
For uniform circular motion (constant speed), the angle $\theta$ changes at a constant rate such that $\theta = \omega t$, where $\omega$ is the angular velocity.
$$\mathbf{r}(t) = r\cos(\omega t)\mathbf{i} + r\sin(\omega t)\mathbf{j}$$
KEY TAKEAWAY: In circular motion centered at the origin, the position vector $\mathbf{r}(t)$ is always equal to the radius of the circle, and the acceleration vector $\mathbf{a}(t)$ is always directed toward the origin (opposite to $\mathbf{r}(t)$).
Angular velocity is the rate of change of the angle $\theta$ with respect to time.
$$\omega = \frac{d\theta}{dt}$$
EXAM TIP: Always ensure your calculator is in Radian mode. VCE Specialist Mathematics questions involving calculus and circular motion almost exclusively use radians for angular measurements.
Using vector differentiation, we can derive the velocity and acceleration for uniform circular motion.
$$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = -r\omega\sin(\omega t)\mathbf{i} + r\omega\cos(\omega t)\mathbf{j}$$
* Speed: $|\mathbf{v}(t)| = \sqrt{(-r\omega\sin\omega t)^2 + (r\omega\cos\omega t)^2} = r\omega$.
* Direction: The velocity vector is always tangential to the path and perpendicular to the position vector ($\mathbf{r} \cdot \mathbf{v} = 0$).
$$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = -r\omega^2\cos(\omega t)\mathbf{i} - r\omega^2\sin(\omega t)\mathbf{j}$$
This can be rewritten in terms of the position vector:
$$\mathbf{a}(t) = -\omega^2 \mathbf{r}(t)$$
COMMON MISTAKE: Students often forget that while the speed ($v$) is constant in uniform circular motion, the velocity ($\mathbf{v}$) is NOT constant because its direction is constantly changing. Therefore, there is always a non-zero acceleration.
According to Newton’s Second Law ($\mathbf{F}_{net} = m\mathbf{a}$), a net force must act on an object to maintain circular motion. This net force is called the centripetal force.
$$F_c = m a = \frac{mv^2}{r} = mr\omega^2$$
| Feature | Formula (Angular) | Formula (Linear) |
|---|---|---|
| Speed | $v = r\omega$ | $v = \frac{2\pi r}{T}$ |
| Acceleration | $a = r\omega^2$ | $a = \frac{v^2}{r}$ |
| Force | $F = mr\omega^2$ | $F = \frac{mv^2}{r}$ |
VCAA FOCUS: Questions often ask to “Show that the acceleration is always perpendicular to the velocity.” To do this, calculate the dot product $\mathbf{v}(t) \cdot \mathbf{a}(t)$ and show that it equals zero.
In uniform circular motion:
* $\mathbf{r}(t)$ and $\mathbf{v}(t)$ are perpendicular ($\mathbf{r} \cdot \mathbf{v} = 0$).
* $\mathbf{v}(t)$ and $\mathbf{a}(t)$ are perpendicular ($\mathbf{v} \cdot \mathbf{a} = 0$).
* $\mathbf{a}(t)$ and $\mathbf{r}(t)$ are parallel but opposite in direction ($\mathbf{a} = -k\mathbf{r}$).
If the circle is centered at $(x_0, y_0)$, the position vector is:
$$\mathbf{r}(t) = (x_0 + r\cos(\omega t))\mathbf{i} + (y_0 + r\sin(\omega t))\mathbf{j}$$
In this case, $\mathbf{a}(t) = -\omega^2 (\mathbf{r}(t) - \mathbf{r}_c)$, where $\mathbf{r}_c$ is the position vector of the center.
STUDY HINT: Practice converting between parametric equations and vector equations. A path given by $x = 3\cos(2t)$ and $y = 3\sin(2t)$ is simply a circle of radius 3 with an angular velocity $\omega = 2$ rad/s.
