Circular motion in a vertical planeAQA A-Level Further Maths: Revision notes

In a vertical circle the weight does work as the particle rises or falls, so the speed is not constant. The motion is no longer uniform circular motion, but at every position the component of the resultant force towards the centre still equals $\frac{mv^2}{r}$. Let $\theta$ be the angle between the radius to the particle and the downward vertical. The height above the lowest point is $h=r(1-\cos\theta)$. The particle may be on a string, on the inside of a track or sphere, or threaded on a wire or rod. A string and a track can only pull or push inwards; a wire or rod can provide a reaction in either direction.

On a smooth path only gravity does work, so mechanical energy is conserved. With speed $u$ at the lowest point, $\frac12mv^2=\frac12mu^2-mgr(1-\cos\theta)\quad\Rightarrow\quad v^2=u^2-2gr(1-\cos\theta).$ At the top ($\theta=180^\circ$) the particle has risen $2r$, so $v_{\text{top}}^2=u^2-4gr$. Tension and normal reaction do no work, because they act perpendicular to the motion.

Resolve along the radius towards the centre. With $T$ the tension (or $R$ the reaction towards $O$): $T-mg\cos\theta=\frac{mv^2}{r}.$ At the bottom, $T=mg+\frac{mu^2}{r}$, which is the greatest tension. At the top, $T+mg=\frac{mv^2}{r}$, so $T=\frac{mv^2}{r}-mg$. At the horizontal level the weight is tangential and $T=\frac{mv^2}{r}$. The tension decreases steadily as the particle rises.

For a string or the inside of a track, the string goes slack (or contact is lost) if $T$ or $R$ reaches zero. The most dangerous point is the top, where $T=\frac{mv^2}{r}-mg\ge0$, so $v_{\text{top}}^2\ge gr$. Combining with $v_{\text{top}}^2=u^2-4gr$: $u^2\ge5gr.$ For a wire or rod the support can push, so the only condition is $v_{\text{top}}\ge0$, giving $u^2\ge4gr$. Example: $m=0.5$ kg, $L=0.9$ m, $u=7$ m s⁻¹, so $5gL=44.1<49$. Then $v_{\text{top}}^2=49-4(9.8)(0.9)=13.72$, $T_{\text{top}}=\frac{0.5\times13.72}{0.9}-4.9=2.72$ N, and $T_{\text{bottom}}=4.9+\frac{0.5\times49}{0.9}=32.1$ N.

If $u^2\le2gr$ the particle rises no higher than the level of $O$ and oscillates, with $T\ge0$ throughout; its greatest height above the lowest point is $\frac{u^2}{2g}$. If $2gr<u^2<5gr$ the string goes slack above the horizontal and the particle then moves as a projectile. Setting $T=0$ at angle $\phi$ above the horizontal gives $v^2=gr\sin\phi$, and energy gives $v^2=u^2-2gr(1+\sin\phi)$, so $\sin\phi=\frac{u^2-2gr}{3gr}.$ For $u^2=3gr$, $\sin\phi=\frac13$. A wire or rod never goes slack, so the bead either oscillates or completes the circle.