Angular speed and circular motionAQA A-Level Further Maths: Revision notes

A particle moving in a circle at constant speed has a velocity that is always tangential, so its direction is changing and it is accelerating even though its speed is constant. The acceleration is directed towards the centre of the circle, and since the speed is constant it has no component along the motion. A resultant force towards the centre is needed to maintain the motion, with $F=ma$ applied along the radius. Angles are measured in radians throughout: one revolution is $2\pi$ radians, and an arc of radius $r$ subtending $\theta$ radians has length $r\theta$.

The angular speed $\omega$ is the rate at which the radius line turns, measured in radians per second: $\omega=\frac{\theta}{t}\quad(\text{constant speed}).$ The period (time for one revolution) is $T=\frac{2\pi}{\omega}$. To convert revolutions per minute to rad s⁻¹ multiply by $\frac{2\pi}{60}$. Example: 12 rev/min gives $12\times\frac{2\pi}{60}=1.26\text{ rad s}^{-1}$. To convert back, $\omega$ rad s⁻¹ is $\frac{60\omega}{2\pi}$ rev/min: $5\text{ rad s}^{-1}$ is $47.7$ rev/min.

In time $t$ the particle travels an arc $r\theta$, so its speed is $v=r\omega.$ Example: a rider 4.5 m from the axis with $\omega=1.257\text{ rad s}^{-1}$ has speed $v=5.65\text{ m s}^{-1}$. All points on a rotating body share the same $\omega$, but the speed grows with the distance from the axis. On a turntable at 30 rev/min, $\omega=\pi$, so $v=0.4\pi=1.26\text{ m s}^{-1}$ at 0.4 m and $v=0.9\pi=2.83\text{ m s}^{-1}$ at 0.9 m.

The magnitude of the acceleration is $a=r\omega^2=\frac{v^2}{r},$ directed towards the centre. Using $v=r\omega$, the two forms are equivalent. Choose the one that matches the data. Example: $v=3\text{ m s}^{-1}$, $r=0.6$ m: $a=\frac{9}{0.6}=15\text{ m s}^{-2}$. Example: $r=4.5$ m, $\omega=1.257$: $a=4.5\times1.58=7.11\text{ m s}^{-2}$. The resultant force towards the centre is $F=ma$. For a 900 kg car at $15\text{ m s}^{-1}$ on a bend of radius 50 m, $a=4.5\text{ m s}^{-2}$ and $F=4050$ N.

Time to turn through an angle $\theta$ at angular speed $\omega$: $t=\frac{\theta}{\omega}$. A quarter turn is $\frac{\pi}{2}$ radians, so on a bend with $\omega=0.3\text{ rad s}^{-1}$, $t=\frac{\pi/2}{0.3}=5.24$ s. For objects on the same turntable, $\omega$ is shared, so the one at the larger radius needs the larger acceleration $r\omega^2$ and is the first to slip when friction can supply a limited value. If friction can give at most $12\text{ m s}^{-2}$ and $r=0.9$ m, then $\omega^2=\frac{12}{0.9}$, $\omega=3.65\text{ rad s}^{-1}=34.9$ rev/min.