Circular motion
Study circular motion with vectors for Year 12 Specialist Mathematics in Queensland (QCAA). Starting from the position vector r(t), you differentiate to reach a velocity that is tangent to the circle and an acceleration that points straight back to the centre.
You will learn to find the speed, the centripetal acceleration and the period of a rotating particle, and to convert a vector path into its Cartesian equation — a key application of vector calculus to motion in a plane.
Theory
Circular motion in Year 12 Specialist Mathematics (QCAA, Queensland) describes a particle moving on a circle using the position vector \(\mathbf{r}(t)=R\cos\omega t\,\mathbf{i}+R\sin\omega t\,\mathbf{j}\). Differentiating gives a velocity that is tangent to the circle with speed \(R\omega\), and an acceleration of magnitude \(R\omega^2\) that points back to the centre. This page shows how to find speed, acceleration, period and the Cartesian path.
In uniform circular motion a particle travels a circle of radius \(R\) at a constant angular speed \(\omega\) (radians per second). Taking the centre as the origin, its position is the vector function \(\mathbf{r}(t)=R\cos\omega t\,\mathbf{i}+R\sin\omega t\,\mathbf{j}\), so the particle sweeps out an angle \(\omega t\) in time \(t\).
Differentiating \(\mathbf{r}(t)\) with respect to time gives the velocity \(\mathbf{v}=\dot{\mathbf{r}}=-R\omega\sin\omega t\,\mathbf{i}+R\omega\cos\omega t\,\mathbf{j}\). This vector is always tangent to the circle (perpendicular to \(\mathbf{r}\)), and its magnitude is the constant speed \(|\mathbf{v}|=R\omega\).
Differentiating again gives the acceleration \(\mathbf{a}=\ddot{\mathbf{r}}=-\omega^2\mathbf{r}\). Because it is a negative multiple of \(\mathbf{r}\), the acceleration points from the particle straight back to the centre — this is the centripetal acceleration, with magnitude \(|\mathbf{a}|=R\omega^2=\dfrac{|\mathbf{v}|^2}{R}\).
The motion repeats every period \(T=\dfrac{2\pi}{\omega}\). If the centre is the point \((a,b)\) instead of the origin, the position becomes \(\mathbf{r}(t)=(a+R\cos\omega t)\,\mathbf{i}+(b+R\sin\omega t)\,\mathbf{j}\), and eliminating \(t\) with \(\cos^2+\sin^2=1\) gives the Cartesian path \((x-a)^2+(y-b)^2=R^2\).
For \(\mathbf{r}(t)=R\cos\omega t\,\mathbf{i}+R\sin\omega t\,\mathbf{j}\), differentiate componentwise to get the velocity and the acceleration:
The speed and the magnitude of the centripetal acceleration are:
The period (time for one revolution) and the Cartesian path about a centre \((a,b)\) are:
Analysing a circular motion
- Read off the radius \(R\) and the angular speed \(\omega\) from \(\mathbf{r}(t)=R\cos\omega t\,\mathbf{i}+R\sin\omega t\,\mathbf{j}\) (or from a period, since \(\omega=\dfrac{2\pi}{T}\)).
- Differentiate componentwise for \(\mathbf{v}=\dot{\mathbf{r}}\) and \(\mathbf{a}=\ddot{\mathbf{r}}\), or use the shortcut \(\mathbf{a}=-\omega^2\mathbf{r}\).
- Apply the magnitudes: speed \(|\mathbf{v}|=R\omega\), centripetal acceleration \(|\mathbf{a}|=R\omega^2\), period \(T=\dfrac{2\pi}{\omega}\).
- Substitute a value of \(t\) for a vector at an instant, or eliminate \(t\) with \(\cos^2+\sin^2=1\) for the Cartesian path.
Read off the radius and angular speed, then apply the magnitude formulas:
| \(R\) | \(=\) | \(5,\quad \omega=3\) |
| \(|\mathbf{v}|\) | \(=\) | \(R\omega\) |
| \(=\) | \(5 \times 3 = 15\text{ m/s}\) |
Centripetal acceleration magnitude \(R\omega^2\):
| \(|\mathbf{a}|\) | \(=\) | \(R\omega^2\) |
| \(=\) | \(5 \times 3^2\) | |
| \(=\) | \(45\text{ m/s}^2\) |
Period from the angular speed:
| \(T\) | \(=\) | \(\dfrac{2\pi}{\omega}\) |
| \(=\) | \(\dfrac{2\pi}{3}\text{ s}\) |
\(|\mathbf{v}|=15\text{ m/s}\), \(|\mathbf{a}|=45\text{ m/s}^2\), \(T=\dfrac{2\pi}{3}\text{ s}\).
Differentiate componentwise for the velocity:
| \(\mathbf{v}\) | \(=\) | \(\dot{\mathbf{r}}\) |
| \(=\) | \(-12\sin(2t)\,\mathbf{i}+12\cos(2t)\,\mathbf{j}\) |
Substitute \(t=\dfrac{\pi}{4}\), so \(2t=\dfrac{\pi}{2}\):
| \(\mathbf{v}\!\left(\dfrac{\pi}{4}\right)\) | \(=\) | \(-12\sin\dfrac{\pi}{2}\,\mathbf{i}+12\cos\dfrac{\pi}{2}\,\mathbf{j}\) |
| \(=\) | \(-12(1)\,\mathbf{i}+12(0)\,\mathbf{j}\) | |
| \(=\) | \(-12\,\mathbf{i}\) |
Use \(\mathbf{a}=-\omega^2\mathbf{r}\) with \(\omega=2\) and \(\mathbf{r}\!\left(\dfrac{\pi}{4}\right)=6\,\mathbf{j}\):
| \(\mathbf{a}\) | \(=\) | \(-4\,\mathbf{r}\) |
| \(\mathbf{a}\!\left(\dfrac{\pi}{4}\right)\) | \(=\) | \(-4(6\,\mathbf{j})\) |
| \(=\) | \(-24\,\mathbf{j}\) |
\(\mathbf{v}=-12\,\mathbf{i}\text{ m/s}\) and \(\mathbf{a}=-24\,\mathbf{j}\text{ m/s}^2\).
Angular speed from the period \(\omega=\dfrac{2\pi}{T}\):
| \(T\) | \(=\) | \(2\text{ s}\) |
| \(\omega\) | \(=\) | \(\dfrac{2\pi}{2}=\pi\text{ rad/s}\) |
Speed \(|\mathbf{v}|=R\omega\):
| \(|\mathbf{v}|\) | \(=\) | \(0.5 \times \pi\) |
| \(=\) | \(\dfrac{\pi}{2}\text{ m/s}\) |
Centripetal acceleration \(|\mathbf{a}|=R\omega^2\):
| \(|\mathbf{a}|\) | \(=\) | \(0.5 \times \pi^2\) |
| \(=\) | \(\dfrac{\pi^2}{2}\text{ m/s}^2\) |
\(\omega=\pi\text{ rad/s}\), \(|\mathbf{v}|=\dfrac{\pi}{2}\text{ m/s}\), \(|\mathbf{a}|=\dfrac{\pi^2}{2}\text{ m/s}^2\).
The oscillating part has radius \(R=2\) and \(\omega=4\), so:
| \(|\mathbf{v}|\) | \(=\) | \(R\omega = 2 \times 4 = 8\text{ m/s}\) |
| \(|\mathbf{a}|\) | \(=\) | \(R\omega^2 = 2 \times 4^2 = 32\text{ m/s}^2\) |
The centre is the constant part of \(\mathbf{r}\):
| \(\text{centre}\) | \(=\) | \((3,\ 1)\) |
Eliminate \(t\) using \(\cos^2+\sin^2=1\):
| \(\dfrac{x-3}{2}\) | \(=\) | \(\cos(4t),\quad \dfrac{y-1}{2}=\sin(4t)\) |
| \((x-3)^2+(y-1)^2\) | \(=\) | \(4\) |
\(|\mathbf{v}|=8\text{ m/s}\), \(|\mathbf{a}|=32\text{ m/s}^2\), centre \((3,1)\), path \((x-3)^2+(y-1)^2=4\).
Common pitfalls
Frequently asked questions
How do you find the speed of a particle in circular motion?
Differentiate the position vector to get the velocity; for \(\mathbf{r}(t)=R\cos\omega t\,\mathbf{i}+R\sin\omega t\,\mathbf{j}\) the speed is the constant \(|\mathbf{v}|=R\omega\), the radius times the angular speed.
Why does the acceleration point to the centre?
Differentiating twice gives \(\mathbf{a}=-\omega^2\mathbf{r}\), a negative scalar times \(\mathbf{r}\). A negative multiple reverses the direction, so \(\mathbf{a}\) points from the particle back to the centre. This is the centripetal acceleration.
What is the magnitude of the centripetal acceleration?
It is \(|\mathbf{a}|=R\omega^2\), which can also be written as \(\dfrac{|\mathbf{v}|^2}{R}\) using the speed \(|\mathbf{v}|=R\omega\).
How do you find the period of circular motion?
The period is the time for one full revolution, \(T=\dfrac{2\pi}{\omega}\). Rearranged, the angular speed is \(\omega=\dfrac{2\pi}{T}\).
How do you get the Cartesian equation of a circular path?
Write \(\cos\omega t\) and \(\sin\omega t\) in terms of \(x\) and \(y\), then use \(\cos^2+\sin^2=1\). For a centre \((a,b)\) and radius \(R\) this gives \((x-a)^2+(y-b)^2=R^2\).
Why is the velocity perpendicular to the position vector?
Their dot product \(\mathbf{v}\cdot\mathbf{r}\) works out to zero at every instant, and a zero dot product means the two vectors are at right angles — the velocity is tangent to the circle.