Vertical Circular Motion

Vertical circular motion is genuinely harder than horizontal, because the speed is no longer constant: as the particle rises, gravity does negative work and slows it; as it falls, gravity speeds it up. So the two tools must work together — conservation of energy to track the changing speed, and Newton's second law resolved radially to find the tension or reaction at each point. The headline question, and a perennial exam favourite, is: what minimum speed lets the particle complete a full circle? — and the answer famously depends on whether it is held by a string or a rod.

1 Where this sits in AQA 7367

Vertical circles are the culminating single-topic block of the Mechanics option, Paper 3 (7367/3M) circular-motion strand. Paper 3's AO1 40% / AO2 25% / AO3 35% weighting makes these multi-stage problems — combine energy, resolve radially, impose a tautness/contact condition — prime AO3 territory, with AO2 marks for the "show that $u^2 \ge 5gr$ " derivations. The prerequisites are conservation of mechanical energy, $a = v^2/r$ (circular-motion basics), and resolving forces. The crucial new idea is that the radial equation now contains a component of gravity, $mg\cos\theta$ , which varies round the circle.

2 Core theory

Forces at a general point

Take a particle of mass $m$ on a string of length $r$ , at angle $\theta$ measured from the lowest point. The two relevant directions are radial (towards the centre) and tangential:

\begin{aligned} \text{Radially (towards centre):}\quad & T - mg\cos\theta = \frac{mv^2}{r},\\ \text{Tangentially:}\quad & -mg\sin\theta = m\frac{\mathrm{d}v}{\mathrm{d}t}. \end{aligned}

The tangential equation shows gravity changing the speed; we rarely solve it directly — energy conservation does that job far more efficiently.

Energy equation — the speed at any point

With no friction or air resistance, mechanical energy is conserved. Measuring height from the lowest point, the particle at angle $\theta$ has risen $h = r - r\cos\theta = r(1 - \cos\theta)$ . If its speed is $u$ at the bottom and $v$ at angle $\theta$ ,

\tfrac12 mu^2 = \tfrac12 mv^2 + mgr(1 - \cos\theta) \;\Longrightarrow\; \boxed{\;v^2 = u^2 - 2gr(1 - \cos\theta).\;}

Tension at a general point

Substitute the energy result into the radial equation:

\begin{aligned} T &= \frac{mv^2}{r} + mg\cos\theta = \frac{m\big(u^2 - 2gr(1 - \cos\theta)\big)}{r} + mg\cos\theta\\ &= m\!\left(\frac{u^2}{r} - 2g + 2g\cos\theta + g\cos\theta\right)\\ &= \boxed{\;m\!\left(\frac{u^2}{r} + 3g\cos\theta - 2g\right).\;} \end{aligned}

Tension at the key points

\begin{aligned} \text{Bottom } (\theta = 0,\ \cos\theta = 1):\quad & T_{\text{bottom}} = m\!\left(\frac{u^2}{r} + g\right),\\ \text{Top } (\theta = \pi,\ \cos\theta = -1):\quad & T_{\text{top}} = m\!\left(\frac{u^2}{r} - 5g\right). \end{aligned}

Equivalently, at the top the radial equation reads $T_{\text{top}} + mg = \dfrac{mv_{\text{top}}^2}{r}$ (both $T$ and gravity now point inward/downward, i.e. towards the centre), so $T_{\text{top}} = \dfrac{mv_{\text{top}}^2}{r} - mg$ . The tension is greatest at the bottom and least at the top — the bottom must support the weight and provide the inward force, while at the top gravity already helps.

Condition for a complete circle — string

A string can only pull, so it stays taut iff $T \ge 0$ everywhere. The tension is smallest at the top, so the binding condition is $T_{\text{top}} \ge 0$ :

$\frac{mv_{\text{top}}^2}{r} - mg \ge 0 \;\Longrightarrow\; v_{\text{top}}^2 \ge gr.$

Now relate to the bottom: at the top $h = 2r$ , so $v_{\text{top}}^2 = u^2 - 4gr$ . Hence

$u^2 - 4gr \ge gr \;\Longrightarrow\; \boxed{\;u^2 \ge 5gr.\;}$

At the critical case $u^2 = 5gr$ : the tension at the top is exactly zero and $v_{\text{top}} = \sqrt{gr}$ — gravity alone provides the centripetal force at the very top.

Condition for a complete circle — rod or bead on a wire

A rigid rod (or a bead threaded on a smooth wire) can push as well as pull, so the tautness condition disappears. The particle completes the circle provided it merely reaches the top with $v_{\text{top}}^2 \ge 0$ :

$u^2 - 4gr \ge 0 \;\Longrightarrow\; \boxed{\;u^2 \ge 4gr.\;}$

At the critical case $u^2 = 4gr$ the particle just reaches the top with zero speed $v_{\text{top}} = 0$ ; the rod then provides an upward thrust $T_{\text{top}} = -mg$ (i.e. a push) to hold it on the circle. This string-vs-rod distinction — $5gr$ against $4gr$ — is the single most-tested discriminator in the topic.

Below the critical speed (string)

The behaviour of a particle on a string splits into three regimes according to the launch speed $u$ at the lowest point:

$u^2 \ge 5gr$ : the string stays taut all the way round — a complete circle.
$2gr \le u^2 < 5gr$ : the particle rises past the horizontal through the centre, but the string goes slack $(T = 0)$ at some point above that level (where $\cos\theta < 0$ ). From that instant the only force is gravity, so the particle becomes a projectile, following a parabola until the string snaps taut again (often with a jerk that the simple model ignores).
$u^2 < 2gr$ : the particle never reaches the level of the centre; the string stays taut throughout and the particle oscillates back and forth like a pendulum, momentarily at rest at its highest point on each side.

The two boundary speeds $u^2 = 2gr$ and $u^2 = 5gr$ are worth committing to memory, because "which regime?" is a favourite first part: compute $u^2$ , compare with $2gr$ and $5gr$ , and state the regime before doing any further work. Why is $2gr$ the lower boundary? Because reaching the level of the centre $(\theta = 90^\circ,\ h = r)$ needs $\tfrac12 mu^2 \ge mgr$ , i.e. $u^2 \ge 2gr$ ; below that the particle turns back before the string could ever go slack.

3 Worked examples with mark scheme

Worked Example 1 — minimum speed for a complete circle (string)

A particle on a string of length $0.5\ \mathrm{m}$ moves in a vertical circle. Find the minimum speed at the bottom for complete circles. (Take $g = 9.8$ .)

$u^2 \ge 5gr = 5 \times 9.8 \times 0.5 = 24.5 \;\Rightarrow\; u \ge \sqrt{24.5} = 4.95\ \mathrm{m\,s^{-1}}.$

Mark scheme: M1 quote/derive $u^2 \ge 5gr$ ; A1 $u = 4.95\ \mathrm{m\,s^{-1}}$ . (2 marks.) If the question said "rod" instead, the condition would be $u^2 \ge 4gr = 19.6$ , giving $u \ge 4.43\ \mathrm{m\,s^{-1}}$ — always check the constraint type.

Worked Example 2 — tension at the top

A particle of mass $0.5\ \mathrm{kg}$ on a string of length $2\ \mathrm{m}$ passes through the lowest point at $10\ \mathrm{m\,s^{-1}}$ . Find the tension at the top.

First the speed at the top by energy, then the radial equation there:

\begin{aligned} v_{\text{top}}^2 &= u^2 - 4gr = 100 - 4(9.8)(2) = 100 - 78.4 = 21.6,\\ T_{\text{top}} &= \frac{mv_{\text{top}}^2}{r} - mg = 0.5\!\left(\frac{21.6}{2} - 9.8\right) = 0.5(10.8 - 9.8) = 0.5\ \mathrm{N}. \end{aligned}

Mark scheme: M1 energy to find $v_{\text{top}}^2$ ; A1 $21.6$ ; M1 radial equation at the top $T = mv^2/r - mg$ ; A1 $0.5\ \mathrm{N}$ . (4 marks.) Since $T_{\text{top}} > 0$ the string is taut at the top, so the particle does complete the circle — a free check: $u^2 = 100 > 5gr = 98$ . ✓

Worked Example 3 — below critical speed: where does the string go slack?

A particle on a string of length $1\ \mathrm{m}$ has speed $5\ \mathrm{m\,s^{-1}}$ at the lowest point. Does it complete the circle? If not, find the angle (from the bottom) at which the string goes slack.

Test completion: $u^2 = 25$ , but $5gr = 5(9.8)(1) = 49$ . Since $25 < 49$ , it does not complete. The string goes slack where $T = 0$ :

\begin{aligned} 0 &= m\!\left(\frac{u^2}{r} + 3g\cos\theta - 2g\right)\\ 0 &= \frac{25}{1} + 3(9.8)\cos\theta - 2(9.8)\\ 0 &= 25 + 29.4\cos\theta - 19.6\\ \cos\theta &= \frac{-5.4}{29.4} = -0.1837 \;\Rightarrow\; \theta = 100.6^\circ \text{ from the bottom.} \end{aligned}

The height there is $h = r(1 - \cos\theta) = 1(1 + 0.1837) = 1.18\ \mathrm{m}$ above the bottom — just above the level of the centre, as expected for a string that goes slack between $2gr$ and $5gr$ . Mark scheme: M1 compare $u^2$ with $5gr$ and conclude; M1 set the general tension to zero; A1 $\cos\theta = -0.184$ ; A1 $\theta = 100.6^\circ$ . Beyond this point the particle is a projectile.

Worked Example 4 — an inverse problem from the tension

A particle of mass $0.6\ \mathrm{kg}$ is attached to a string of length $0.75\ \mathrm{m}$ and moves in a vertical circle. At the lowest point the tension is four times the particle's weight. Find the speed at the lowest point and determine whether the particle completes the circle. (Take $g = 9.8$ .)

At the lowest point $T_{\text{bottom}} = m\!\left(\dfrac{u^2}{r} + g\right)$ , and we are told $T_{\text{bottom}} = 4mg$ :

m\!\left(\frac{u^2}{r} + g\right) = 4mg \;\Rightarrow\; \frac{u^2}{r} = 3g \;\Rightarrow\; u^2 = 3gr = 3(9.8)(0.75) = 22.05 \;\Rightarrow\; u = 4.70\ \mathrm{m\,s^{-1}}.

For a complete circle a string needs $u^2 \ge 5gr = 36.75$ . Since $22.05 < 36.75$ , it does not complete — the string goes slack first (at $\cos\theta = \dfrac{2g - u^2/r}{3g} = \dfrac{2g - 3g}{3g} = -\tfrac13$ , i.e. $\theta = 109.5^\circ$ from the bottom). Mark scheme: M1 equate $T_{\text{bottom}} = 4mg$ with $m(u^2/r + g)$ ; A1 $u^2 = 3gr$ ; A1 $u = 4.70\ \mathrm{m\,s^{-1}}$ ; B1 compare with $5gr$ and conclude "does not complete". The mass cancels throughout — the ratio of tension to weight fixes $u^2/r$ regardless of $m$ , a neat point worth stating.

4 Specimen-style exam question

(Specimen-style.) A small bead of mass $0.2\ \mathrm{kg}$ is threaded on a smooth circular wire of radius $0.8\ \mathrm{m}$ fixed in a vertical plane. The bead is projected from the lowest point with speed $u$ . (a) Find the least value of $u$ for the bead to make complete revolutions. (b) With $u = 6\ \mathrm{m\,s^{-1}}$ , find the speed and the reaction of the wire on the bead at the highest point, stating its direction. (Take $g = 9.8$ .)

(a) A bead on a wire is constrained on both sides (the wire can push outward or inward), so the condition is merely that it reaches the top:

$u^2 \ge 4gr = 4(9.8)(0.8) = 31.36 \;\Rightarrow\; u \ge 5.60\ \mathrm{m\,s^{-1}}.$

(b) Energy to the top:

$v_{\text{top}}^2 = u^2 - 4gr = 36 - 31.36 = 4.64 \;\Rightarrow\; v_{\text{top}} = 2.15\ \mathrm{m\,s^{-1}}.$

Radially at the top, let $R$ be the reaction taken towards the centre (downward at the top). Then $R + mg = \dfrac{mv_{\text{top}}^2}{r}$ :

$R = \frac{mv_{\text{top}}^2}{r} - mg = 0.2\!\left(\frac{4.64}{0.8} - 9.8\right) = 0.2(5.8 - 9.8) = 0.2(-4) = -0.8\ \mathrm{N}.$

The negative sign means the reaction actually acts outward (away from the centre, i.e. upward at the top): the wire is pushing the bead up with a force of $0.8\ \mathrm{N}$ . This is precisely what a string could not do — a string would have gone slack — and is the whole point of the bead-on-wire model.

Interpretation. At $u = 6$ , $u^2 = 36 > 4gr = 31.36$ so the bead does complete the circle, but $36 < 5gr = 39.2$ , so a string would have gone slack before the top. The wire supplies the extra inward... here outward... support that keeps the bead on its circular path.

Vertical Circular Motion

Vertical Circular Motion

1 Where this sits in AQA 7367

2 Core theory

Forces at a general point

Energy equation — the speed at any point

Tension at a general point

Tension at the key points

Condition for a complete circle — string

Condition for a complete circle — rod or bead on a wire

Below the critical speed (string)

3 Worked examples with mark scheme

Worked Example 1 — minimum speed for a complete circle (string)

Worked Example 2 — tension at the top

Worked Example 3 — below critical speed: where does the string go slack?

Worked Example 4 — an inverse problem from the tension

4 Specimen-style exam question

5 Synoptic links

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