Vectors in circular motionAQA A-Level Further Maths: Revision notes

For a particle moving on a circle of radius $R$ with centre at the origin $O$, the position vector is $\mathbf{r}=R\cos\theta\,\mathbf{i}+R\sin\theta\,\mathbf{j}$, where $\theta$ is measured anticlockwise from the positive $x$-axis. At constant angular speed $\omega$ (rad s⁻¹), $\theta=\omega t$ if the particle starts at $(R,0)$: $\mathbf{r}=R\cos\omega t\,\mathbf{i}+R\sin\omega t\,\mathbf{j},\qquad |\mathbf{r}|=R.$ The period is $T=\frac{2\pi}{\omega}$. For clockwise motion starting at $(0,R)$, swap the roles: $\mathbf{r}=R\sin\omega t\,\mathbf{i}+R\cos\omega t\,\mathbf{j}$. Always check the starting position and direction by substituting $t=0$ and a small $t$.

Differentiate the position vector with respect to time: $\mathbf{v}=\frac{d\mathbf{r}}{dt}=-R\omega\sin\omega t\,\mathbf{i}+R\omega\cos\omega t\,\mathbf{j}.$ Its magnitude is $|\mathbf{v}|=R\omega\sqrt{\sin^2\omega t+\cos^2\omega t}=R\omega$, so the speed is $v=R\omega$ and is constant. Also $\mathbf{v}\cdot\mathbf{r}=0$, so the velocity is perpendicular to the radius: it acts along the tangent. The speed is constant but the velocity is not, because its direction keeps changing.

Differentiate again: $\mathbf{a}=\frac{d\mathbf{v}}{dt}=-R\omega^2\cos\omega t\,\mathbf{i}-R\omega^2\sin\omega t\,\mathbf{j}=-\omega^2\mathbf{r}.$ The acceleration is always towards the centre (opposite to $\mathbf{r}$) with constant magnitude $|\mathbf{a}|=R\omega^2=\frac{v^2}{R}=v\omega.$ The direction changes continuously, so the acceleration vector is not constant. It is perpendicular to the velocity, which is why the speed does not change.

A particle has $\mathbf{r}=5\cos2t\,\mathbf{i}+5\sin2t\,\mathbf{j}$ (metres). Find its velocity and acceleration when $t=\frac{\pi}{4}$. $\mathbf{v}=-10\sin2t\,\mathbf{i}+10\cos2t\,\mathbf{j}$. When $t=\frac{\pi}{4}$, $2t=\frac{\pi}{2}$, so $\mathbf{r}=5\mathbf{j}$ and $\mathbf{v}=-10\mathbf{i}$ m s⁻¹. $\mathbf{a}=-\omega^2\mathbf{r}=-4(5\mathbf{j})=-20\mathbf{j}$ m s⁻². Check: $\mathbf{v}\cdot\mathbf{r}=0$ and $|\mathbf{a}|=R\omega^2=20$.

By Newton's second law, the resultant force on a particle of mass $m$ is $\mathbf{F}=m\mathbf{a}=-m\omega^2\mathbf{r}$. It is directed towards the centre and has magnitude $mR\omega^2=\frac{mv^2}{R}$. This resultant force is provided by tension, friction, a normal reaction or a component of weight, depending on the context. Questions often give $\mathbf{r}$ as a function of time, ask for $\mathbf{a}$ at a chosen time and then multiply by $m$.