Horizontal circular motionEdexcel International A Level Further Maths: Revision notes

A particle moving in a horizontal circle of radius $r$ at constant speed $v$ has angular speed $\omega=\frac{v}{r}$, so $v=r\omega$. One revolution takes the period $T=\frac{2\pi}{\omega}=\frac{2\pi r}{v}$. The velocity changes direction, so the particle accelerates even at constant speed. The acceleration is directed towards the centre and has magnitude $a=\frac{v^2}{r}=r\omega^2.$ By Newton's second law, the resultant force towards the centre is $F=\frac{mv^2}{r}=mr\omega^2$. There is no acceleration in the vertical direction, so vertical forces balance.

1. Draw a diagram with every real force on the particle and mark the radius $r$ of the horizontal circle.
2. Resolve vertically: the forces balance, because the particle stays at the same height.
3. Resolve horizontally towards the centre: the resultant equals $\frac{mv^2}{r}$ or $mr\omega^2$.
4. Solve the two equations together, often by dividing to eliminate $T$ or $R$. Check whether the question gives $v$ or $\omega$ and choose the form of the acceleration to match. The radius of the circle is not always the length of the string.

A particle on a string of length $L$ moves in a horizontal circle with the string at angle $\theta$ to the vertical. The radius is $r=L\sin\theta$. Vertically: $T\cos\theta=mg$. Horizontally: $T\sin\theta=mr\omega^2$. Substituting $r=L\sin\theta$ gives $T=mL\omega^2$, and so $\omega^2=\frac{g}{L\cos\theta}$, which does not depend on the mass. Example: $m=0.3$ kg, $L=0.8$ m, $\theta=30^\circ$. $r=0.8\sin30^\circ=0.4$ m. $T=\frac{0.3\times9.8}{\cos30^\circ}=3.39$ N. $\omega^2=\frac{9.8}{0.8\cos30^\circ}=14.1$, so $\omega=3.76$ rad s$^{-1}$ and the period is $\frac{2\pi}{\omega}=1.67$ s. A faster rotation needs a larger $\theta$, since $\cos\theta$ must fall.

On a smooth horizontal table the only horizontal force on the particle is the tension $T$ in the string, and it provides the centripetal force. For an elastic string of natural length $l$ and modulus $\lambda$, Hooke's law gives $T=\frac{\lambda x}{l}$ where $x$ is the extension. The radius is the stretched length, $r=l+x$. Example: $m=0.2$ kg, $l=0.5$ m, $\lambda=24$ N, $\omega=6$ rad s$^{-1}$. Then $\frac{24(r-0.5)}{0.5}=0.2r\times36$, so $48r-24=7.2r$ and $r=0.588$ m. The particle must reach the correct $r$ for the speed: a larger $\omega$ needs a larger tension and so a larger extension.

A vehicle on a track banked at angle $\alpha$ has a normal reaction $R$ perpendicular to the surface, so $R\sin\alpha$ is horizontal and acts towards the centre. For a smooth surface: $R\cos\alpha=mg$ and $R\sin\alpha=\frac{mv^2}{r}$, giving the design speed $v^2=rg\tan\alpha$. Example: $r=50$ m, $\alpha=20^\circ$: $v=\sqrt{50\times9.8\times\tan20^\circ}=13.4$ m s$^{-1}$. Above that speed the vehicle tends to slide up the slope, so friction acts down the slope; below it, the vehicle tends to slide down and friction acts up. At the point of slipping $F=\mu R$. Resolve horizontally and vertically with $F$ included and solve for $\mu$ or the maximum speed.

Turntable or flat road: friction alone provides the centripetal force, so $F=mr\omega^2\le\mu mg$. A coin at $r=0.15$ m with $\mu=0.4$ stays put while $\omega\le\sqrt{\frac{\mu g}{r}}=5.11$ rad s$^{-1}$; the mass cancels. Smooth cone or bowl: the reaction is perpendicular to the surface, so resolve it into vertical and horizontal parts exactly as for a banked surface; in a bowl, find $r$ from the geometry. Aircraft banking: the lift is perpendicular to the wings and acts like $R$ on a bank. In every case find the radius of the horizontal circle from the geometry before anything else.