Motion in a vertical circleEdexcel International A Level Further Maths: Revision notes

In a vertical circle the speed is not constant: the particle slows as it rises and speeds up as it falls. The tension (string) or normal reaction (smooth surface, smooth wire) acts perpendicular to the motion and does no work, so mechanical energy is conserved: $\tfrac12mu^2=\tfrac12mv^2+mgh,$ where $h$ is the height gained from the starting point. For a circle of radius $r$ starting at the lowest point, the height of the particle when $OP$ makes an angle $\theta$ with the downward vertical is $h=r(1-\cos\theta)$. At the top, $h=2r$. Example: $u=7$ m s$^{-1}$, $r=0.8$ m: at the top $v^2=49-4(9.8)(0.8)=17.64$, so $v=4.2$ m s$^{-1}$.

Newton's second law along the radius, towards $O$, gives the force that the string or surface provides. With $\theta$ measured from the downward vertical: $T-mg\cos\theta=\frac{mv^2}{r}.$ At the lowest point, $T=mg+\frac{mu^2}{r}$. Level with $O$, the weight is tangential so $T=\frac{mv^2}{r}$. At the highest point, $T+mg=\frac{mv^2}{r}$. The tension changes continuously and is greatest at the lowest point. Always take the positive direction towards the centre, and make sure each force is resolved along the radius.

A string can only pull, so $T\ge0$ throughout. The tension is smallest at the top, so the condition for a complete circle is $T\ge0$ there: $v_{top}^2\ge gr.$ Combined with energy, $u^2=v_{top}^2+4gr$, giving $u^2\ge5gr.$ Example: $r=0.6$ m: $u\ge\sqrt{5\times9.8\times0.6}=5.42$ m s$^{-1}$. The tension at the top is zero when $u$ is exactly this value.

If $u^2$ is between $2gr$ and $5gr$, the particle rises above $O$ but the string goes slack at an angle $\phi$ above the horizontal. Example: $r=0.8$ m, $u=5$ m s$^{-1}$. Height above the start is $0.8(1+\sin\phi)$, so $v^2=25-15.68(1+\sin\phi)=9.32-15.68\sin\phi$. At slack, $T=0$ and the radial equation gives $mg\sin\phi=\frac{mv^2}{r}$, so $7.84\sin\phi=9.32-15.68\sin\phi$ and $\sin\phi=0.396$, $\phi=23.3^\circ$. If $u^2\le2gr$, the particle does not rise above $O$ and oscillates, because the tension stays positive.

A bead on a wire, or a particle in a smooth tube, can be pushed or pulled by the reaction $R$, so $R$ may act towards or away from $O$ and there is no tension condition. The bead reaches the top if $v\ge0$ there, which needs only $u^2\ge4gr.$ To find the reaction, take $R$ towards $O$ and use the radial equation: a negative answer means the force acts away from $O$. Example: $u=5$, $r=0.5$, $m=0.1$ kg: at the top $v^2=5.4$ and $0.98+R=\frac{0.1\times5.4}{0.5}$, so $R=0.1$ N towards $O$.

Inside a smooth hemispherical bowl or sphere, the particle stays in contact while $R\ge0$, the same condition as for a string, so the circle conditions above apply. Outside a sphere the reaction pushes away from the surface, so towards the centre we have $mg\cos\theta-R=\frac{mv^2}{r}$ with $\theta$ from the upward vertical. The particle leaves the surface when $R=0$. Example: starting from rest at the top of a sphere of radius $a$: $v^2=2ga(1-\cos\theta)$ and $mg\cos\theta=\frac{mv^2}{a}$, so $\cos\theta=2(1-\cos\theta)$, $\cos\theta=\frac23$ and $\theta=48.2^\circ$.