Angular speed and radial accelerationEdexcel International A Level Further Maths: Revision notes

A particle moving in a circle of radius $r$ has position given by the angle $\theta$ (in radians) from a fixed line. The angular speed is $\omega=\frac{d\theta}{dt}$, measured in rad s$^{-1}$. For motion at constant angular speed, $\theta=\omega t$ and the particle completes a revolution (angle $2\pi$) in time $T=\frac{2\pi}{\omega}$, the period. To convert from revolutions per minute: $\omega=\frac{2\pi n}{60}$ where $n$ is the number of revolutions per minute. For example $30$ rev min$^{-1}$ gives $\omega=\pi$ rad s$^{-1}$.

The arc length is $s=r\theta$, so the speed along the circle is $v=r\omega$. The velocity is always tangential, perpendicular to the radius, so its direction changes even when the speed is constant. All parts of a rotating rigid object, or particles on one string in a line, have the same $\omega$ but different speeds, since $v=r\omega$ increases with $r$.

Because the direction of the velocity changes, a particle moving in a circle at constant speed is accelerating, with acceleration directed towards the centre of the circle (the radial direction). Its magnitude is $a=r\omega^2=\frac{v^2}{r}.$ The two forms are equivalent because $v=r\omega$. Use $r\omega^2$ when the angular speed is known and $\frac{v^2}{r}$ when the speed is known. At constant speed there is no acceleration along the tangent.

By Newton's second law, the resultant force towards the centre is $F=mr\omega^2=\frac{mv^2}{r}$. This is not a new force: it is the resultant of the real forces, such as tension in a string. For a particle on a smooth horizontal table attached to a fixed point by a string, the tension is the only horizontal force, so $T=mr\omega^2$. Do not draw a 'centrifugal' force on a diagram.

Particles $A$ ($0.3$ kg) at $0.4$ m and $B$ ($0.2$ kg) at $0.9$ m from $O$ rotate with $\omega=5$ rad s$^{-1}$ on a smooth table. For $B$: $T_{AB}=0.2(0.9)(25)=4.5$ N. For $A$: $T_{OA}-T_{AB}=0.3(0.4)(25)=3$, so $T_{OA}=7.5$ N. The inner string supports both particles, so it has the larger tension. If strings break at $12$ N, $OA$ goes first when $0.30\omega^2=12$, i.e. $\omega=6.32$ rad s$^{-1}$.