Horizontal circular motionEdexcel A-Level Further Maths: Revision notes

A particle moving in a circle of radius $r$ turns through an angle $\theta$ (in radians) in time $t$. Its angular speed is $\omega=\frac{d\theta}{dt}$, measured in rad s⁻¹. For uniform circular motion $\omega$ is constant and the speed along the circle is constant. The arc length is $r\theta$, so the linear speed is $v=r\omega.$ One revolution is $2\pi$ radians, so the period (time for one revolution) is $T=\frac{2\pi}{\omega}=\frac{2\pi r}{v}$. Convert revolutions per minute to rad s⁻¹ by multiplying by $\frac{2\pi}{60}$. For example 120 rev min⁻¹ is $4\pi$ rad s⁻¹. Example: a particle moves at $\omega=6$ rad s⁻¹ on a circle of radius 0.5 m. Then $v=0.5\times6=3$ m s⁻¹ and $T=\frac{2\pi}{6}=1.05$ s.

Even at constant speed the velocity changes direction, so the particle accelerates. The acceleration is directed towards the centre of the circle (radial) and has magnitude $a=r\omega^2=\frac{v^2}{r}.$ The two forms are equivalent because $v=r\omega$. Use $r\omega^2$ when $\omega$ is given and $\frac{v^2}{r}$ when the speed is given. By Newton's second law the resultant force towards the centre is $F=mr\omega^2=\frac{mv^2}{r}$. This is not an extra force: it is the resultant of the real forces (tension, normal reaction, friction, weight components) in the radial direction. Example: a 0.4 kg particle with $r=0.5$ m and $\omega=6$ rad s⁻¹ has $a=0.5\times36=18$ m s⁻² and a resultant inward force of $7.2$ N.

A particle on a string of length $l$ moves in a horizontal circle, the string making a constant angle $\theta$ with the vertical. The radius is $r=l\sin\theta$. Only the tension $T$ and the weight act. Vertically (no acceleration): $T\cos\theta=mg$. Horizontally (towards the centre): $T\sin\theta=mr\omega^2=ml\sin\theta\,\omega^2$. The second equation gives $T=ml\omega^2$, and combining the two gives $\omega^2=\frac{g}{l\cos\theta},\qquad \tan\theta=\frac{r\omega^2}{g}.$ Example: $m=0.5$ kg, $l=1.2$ m, $\theta=60^\circ$. Then $T=\frac{0.5\times9.8}{\cos60^\circ}=9.8$ N, $\omega^2=\frac{9.8}{1.2\cos60^\circ}=16.3$, so $\omega=4.04$ rad s⁻¹ and the period is $\frac{2\pi}{\omega}=1.55$ s.

If the particle is attached to an elastic string and moves on a smooth horizontal table, the tension is the only horizontal force. Hooke's law gives $T=\frac{\lambda x}{l}$, where $\lambda$ is the modulus of elasticity, $l$ the natural length and $x$ the extension. The radius of the circle is the stretched length, $r=l+x$. Then $T=\frac{mv^2}{r}=mr\omega^2$. Example: $m=0.3$ kg, $l=0.8$ m, $\lambda=24$ N, $r=1.0$ m. The extension is $0.2$ m, so $T=\frac{24\times0.2}{0.8}=6$ N, then $6=\frac{0.3v^2}{1.0}$ gives $v=4.47$ m s⁻¹. Because $r$ depends on $\omega$, the tension grows as the particle goes faster: $\frac{\lambda(r-l)}{l}=mr\omega^2$ links $r$ and $\omega$. A breaking tension therefore gives a maximum $\omega$.

On a surface banked at angle $\theta$ the normal reaction $R$ is perpendicular to the surface. Its horizontal component $R\sin\theta$ supplies the inward force. With no sideways friction: vertically $R\cos\theta=mg$ and horizontally $R\sin\theta=\frac{mv^2}{r}$, so $\tan\theta=\frac{v^2}{rg}.$ Example: $r=40$ m, $\theta=20^\circ$ gives $v^2=392\tan20^\circ$, so $v=11.9$ m s⁻¹, and $R=\frac{90\times9.8}{\cos20^\circ}=939$ N for a 90 kg rider. On a flat bend, friction alone supplies the inward force: $F=\frac{mv^2}{r}$ with $F\le\mu R=\mu mg$, so no slipping needs $\mu\ge\frac{v^2}{rg}$. On a rough bank, include friction along the slope as well. At the limit of sliding use $F=\mu R$, with friction acting down the slope if the vehicle is about to slide up, and up the slope if it is about to slide down.