Vertical circular motionEdexcel A-Level Further Maths: Revision notes

In vertical circular motion the speed changes, because gravity does work as the particle rises and falls. The acceleration therefore has two components:

• radial (towards the centre): $\frac{v^2}{r}=r\omega^2$,
• tangential (along the path): $\frac{dv}{dt}=r\frac{d\omega}{dt}$. Resolve the real forces along the radius and along the tangent. The radial resultant equals $\frac{mv^2}{r}$ and the tangential resultant equals $m\frac{dv}{dt}$. Example: a bead is on a smooth wire with $OB$ horizontal. The weight is tangential, so the tangential acceleration is $g$ (downwards) and the wire's normal reaction alone provides $\frac{mv^2}{r}$.

The tension in a string, or the normal reaction of a smooth surface or wire, is always perpendicular to the motion and does no work. Only gravity does work, so mechanical energy is conserved: $\tfrac12mv^2+mgh=\text{constant}.$ Take the lowest point $A$ with speed $u$. When $P$ has risen a height $h$, $v^2=u^2-2gh$. At angle $\theta$ from the downward vertical, $h=r(1-\cos\theta)$. At the horizontal level of $O$, $h=r$; at the top, $h=2r$. Example: $u=12$, $r=2$, $g=9.8$. At the top $v^2=144-4(9.8)(2)=65.6$, so $v=8.10$ m s⁻¹.

Write Newton's second law along the radius, towards the centre. Remember that the weight has a radial component $mg\cos\theta$ at angle $\theta$ from the downward vertical.

• At the lowest point: $T-mg=\frac{mv^2}{r}$.
• At the highest point: $T+mg=\frac{mv^2}{r}$.
• At angle $\theta$ from the downward vertical: $T-mg\cos\theta=\frac{mv^2}{r}$. Combine this with the energy equation to find $T$. Example ($u=12$, $r=2$, $m=0.5$): at the top $T=0.5\left(\frac{65.6}{2}-9.8\right)=11.5$ N, and at the bottom $T=0.5\left(\frac{144}{2}+9.8\right)=40.9$ N. The tension is greatest at the bottom.

A particle on a string, or on the inside of a surface, stays on its circle only if $T\ge0$ (or $R\ge0$). The critical point is the top, where $T+mg=\frac{mv^2}{r}$ with $T=0$ gives $v^2\ge gr$. Using energy from the bottom, $u^2-4gr\ge gr$, so $u^2\ge5gr.$ A bead on a wire or rod can be pushed or pulled by the wire, so it never goes slack. It completes the circle if it just reaches the top with $v\ge0$: $u^2\ge4gr.$ Example: bead with $r=0.9$ m needs $u\ge\sqrt{4\times9.8\times0.9}=5.94$ m s⁻¹, but a particle on a string of the same length needs $u\ge6.64$ m s⁻¹.

If $u^2\le2gr$ the particle does not rise above the level of $O$. Its speed falls to zero first, the tension never reaches zero, and it oscillates like a pendulum. If $2gr<u^2<5gr$ (string or inside surface) the particle rises above the horizontal level of $O$ but cannot complete the circle. It leaves the circle at the angle where $T=0$ (or $R=0$), then moves as a projectile. Method for an angle $\alpha$ above the horizontal: (1) energy: $v^2=u^2-2gr(1+\sin\alpha)$; (2) radial equation with $T=0$: $mg\sin\alpha=\frac{mv^2}{r}$, so $v^2=gr\sin\alpha$; (3) equate and solve for $\sin\alpha$. Example: $u=4$, $r=0.6$ gives $5.88\sin\alpha=16-11.76-11.76\sin\alpha$, so $\sin\alpha=\frac{4.24}{17.64}$ and $\alpha=13.9^\circ$.