Circular Motion Applications

This final lesson welds the whole circular-motion toolkit onto the hardest synoptic problems on Paper 3: the loop-the-loop, particles on the inside and outside of curved surfaces, cars cresting hills, water in a swung bucket, and hybrids that fold in elastic springs. There is no new theory — only the disciplined three-step method energy → radial Newton → contact/tautness condition applied to richer geometry. The reward is genuine: these are the questions that separate the strong Further-Maths candidate from the merely competent.

1 Where this sits in AQA 7367

These are the synthesis problems of the Mechanics option, Paper 3 (7367/3M) circular-motion strand. They are the purest AO3 in the option — Paper 3's AO1 40% / AO2 25% / AO3 35% weighting is felt most here, because a single question may chain energy conservation, a radial equation, a contact condition, and sometimes a projectile or spring sub-part. The prerequisites are everything from the previous three lessons: $a = v^2/r$ , energy conservation, the $N = 0$ / $T = 0$ critical idea, and Hooke's law for the spring hybrids.

2 Core theory — the master method

Every problem here is solved by the same three steps:

\begin{aligned} &\textbf{(1) Energy: } \quad \tfrac12 mu^2 + mgh_{\text{start}} = \tfrac12 mv^2 + mgh \;\Rightarrow\; v^2 \text{ at the point of interest.}\\ &\textbf{(2) Newton radially: } \quad (\text{resultant towards centre}) = \frac{mv^2}{r}.\\ &\textbf{(3) Critical condition: } \quad N = 0 \text{ (loses contact)} \;\text{or}\; T = 0 \text{ (goes slack)}. \end{aligned}

Loop-the-loop — minimum release height

A particle slides from rest at height $h$ down a smooth track into a vertical circular loop of radius $r$ . At the top of the loop the track pushes inward (downward), so radially $N + mg = \dfrac{mv_{\text{top}}^2}{r}$ . Contact needs $N \ge 0$ , so the critical case is $N = 0,\ v_{\text{top}}^2 = gr$ . Energy from the start (height $h$ ) to the top (height $2r$ ):

$mgh = \tfrac12 mv_{\text{top}}^2 + mg(2r) \;\Rightarrow\; gh = \tfrac12 gr + 2gr = \tfrac52 gr \;\Rightarrow\; \boxed{\;h = \tfrac52 r.\;}$

The particle on the inside of a loop is exactly like a particle on a string: it needs $v_{\text{top}}^2 \ge gr$ (equivalently $u^2 \ge 5gr$ at the bottom).

Particle on the inside of a smooth sphere/bowl

Let $\theta$ be measured from the downward vertical through the centre, so height above the lowest point is $r(1 - \cos\theta)$ . Released from rest at the rim $(\theta = 90^\circ)$ , energy gives

$\tfrac12 mv^2 = mg\big[r - r(1-\cos\theta)\big] = mgr\cos\theta \;\Rightarrow\; v^2 = 2gr\cos\theta.$

Radially the surface pushes inward (towards the centre, which is above the particle in the lower bowl), so $N - mg\cos\theta = \dfrac{mv^2}{r}$ :

$N = mg\cos\theta + \frac{m(2gr\cos\theta)}{r} = 3mg\cos\theta.$

Since $N = 3mg\cos\theta \ge 0$ for all $0 \le \theta \le 90^\circ$ , the particle never leaves the inside of the bowl — it simply oscillates. (Contrast the outside of a sphere below.)

Particle on the outside of a smooth sphere

The classic. A particle starts at rest at the top of a smooth sphere of radius $r$ and slides down the outside; $\theta$ is measured from the top. Energy from the top:

$v^2 = 2gr(1 - \cos\theta).$

Now the centre is below the surface, so the inward (centripetal) direction is towards the centre and the reaction points outward: $mg\cos\theta - N = \dfrac{mv^2}{r}$ . Hence

$N = mg\cos\theta - \frac{m\cdot 2gr(1-\cos\theta)}{r} = mg(3\cos\theta - 2).$

The particle leaves the surface when $N = 0$ :

$3\cos\theta - 2 = 0 \;\Rightarrow\; \boxed{\;\cos\theta = \tfrac23\;} \;\Rightarrow\; \theta \approx 48.2^\circ.$

Strikingly, this leaving angle is independent of $g$ , $r$ and the mass — a universal result for a particle starting from rest at the top.

Car over a hill, water in a bucket

A car cresting a circular hill of radius $r$ : at the top the road pushes up (outward) and gravity down (inward), so $mg - N = \dfrac{mv^2}{r}$ . Contact needs $N \ge 0$ , giving a maximum speed $v_{\max} = \sqrt{gr}$ . Water in a bucket swung in a vertical circle stays in provided the bucket need not "pull" it inward at the top, i.e. $\dfrac{v^2}{r} \ge g$ , so the minimum top speed is $v_{\min} = \sqrt{gr}$ — the same critical $\sqrt{gr}$ , but a maximum in one case and a minimum in the other.

3 Worked examples with mark scheme

Worked Example 1 — loop-the-loop minimum height

A bead slides from rest down a smooth track into a vertical loop of radius $0.3\ \mathrm{m}$ . Find the minimum starting height for it to complete the loop.

$h = \tfrac52 r = \tfrac52 (0.3) = 0.75\ \mathrm{m}.$

Mark scheme: M1 critical condition $v_{\text{top}}^2 = gr$ at the top; M1 energy from $h$ to height $2r$ ; A1 $h = 0.75\ \mathrm{m}$ . (3 marks.) The $\tfrac52 r$ result is worth knowing, but a "show that" wants the energy line written out.

Worked Example 2 — does it complete the loop?

A particle enters the bottom of a smooth vertical loop of radius $0.5\ \mathrm{m}$ at $4\ \mathrm{m\,s^{-1}}$ . Does it complete the loop?

$v_{\text{top}}^2 = u^2 - 4gr = 16 - 4(9.8)(0.5) = 16 - 19.6 = -3.6 < 0.$

A negative $v_{\text{top}}^2$ is impossible, so the particle never reaches the top — it does not complete the loop. Its greatest height comes from $\tfrac12 mu^2 = mgh_{\max}$ : $h_{\max} = \dfrac{u^2}{2g} = \dfrac{16}{19.6} = 0.816\ \mathrm{m}$ , which is below the top at $2r = 1\ \mathrm{m}$ — consistent. Mark scheme: M1 $v_{\text{top}}^2 = u^2 - 4gr$ ; A1 $-3.6$ , conclude "does not complete"; B1 support via $h_{\max} < 2r$ . The negative-square argument is the cleanest justification.

Worked Example 3 — particle leaving a smooth sphere

A particle rests at the top of a smooth sphere of radius $2\ \mathrm{m}$ and is slightly displaced. Find the angle at which it leaves the surface, and its speed and height (above the centre) at that instant.

Leaving condition (from rest at the top): $\cos\theta = \tfrac23$ , so $\theta = \arccos\tfrac23 \approx 48.2^\circ$ . Then

\begin{aligned} v^2 &= 2gr(1 - \cos\theta) = 2(9.8)(2)\big(1 - \tfrac23\big) = 39.2 \times \tfrac13 = 13.07 \;\Rightarrow\; v = 3.61\ \mathrm{m\,s^{-1}},\\ h &= r\cos\theta = 2 \times \tfrac23 = \tfrac43 \approx 1.33\ \mathrm{m} \text{ above the centre.} \end{aligned}

Mark scheme: M1 $N = mg(3\cos\theta - 2)$ , set $N=0$ ; A1 $\cos\theta = \tfrac23$ ; M1 energy for $v^2$ ; A1 $v = 3.61\ \mathrm{m\,s^{-1}}$ ; A1 $h = \tfrac43\ \mathrm{m}$ . (5 marks.) The leaving angle is the same for any sphere — only $v$ and $h$ depend on $r$ .

Worked Example 4 — circular motion on a spring

A particle of mass $0.2\ \mathrm{kg}$ is attached to a spring (natural length $0.5\ \mathrm{m}$ , stiffness $k = 40\ \mathrm{N\,m^{-1}}$ ) fixed at $O$ on a smooth horizontal table. It moves in a horizontal circle of radius $0.8\ \mathrm{m}$ . Find the angular speed.

The spring is stretched by $x = 0.8 - 0.5 = 0.3\ \mathrm{m}$ , so its tension $T = kx = 40(0.3) = 12\ \mathrm{N}$ provides the centripetal force:

$T = mr\omega^2 \;\Rightarrow\; 12 = 0.2(0.8)\omega^2 \;\Rightarrow\; \omega^2 = \frac{12}{0.16} = 75 \;\Rightarrow\; \omega = \sqrt{75} = 5\sqrt3 \approx 8.66\ \mathrm{rad\,s^{-1}}.$

Mark scheme: M1 extension and $T = kx$ ; A1 $T = 12\ \mathrm{N}$ ; M1 $T = mr\omega^2$ ; A1 $\omega = 5\sqrt3 \approx 8.66\ \mathrm{rad\,s^{-1}}$ . The radius here is the stretched radius $0.8\ \mathrm{m}$ , not the natural length — a classic trap.

Worked Example 5 — car over a hill

A car of mass $1000\ \mathrm{kg}$ drives over a hill modelled as a circular arc of radius $50\ \mathrm{m}$ . Find the maximum speed at the crest for the car to stay in contact with the road.

At the crest, radially (inward is downward): $mg - N = \dfrac{mv^2}{r}$ . Contact needs $N \ge 0$ :

$mg \ge \frac{mv^2}{r} \;\Rightarrow\; v^2 \le gr = 9.8 \times 50 = 490 \;\Rightarrow\; v \le 22.1\ \mathrm{m\,s^{-1}}.$

Mark scheme: M1 radial equation at the crest; M1 set $N \ge 0$ ; A1 $v_{\max} = 22.1\ \mathrm{m\,s^{-1}}$ (about $79.6\ \mathrm{km\,h^{-1}}$ ). The mass cancels — the limit is independent of how heavy the car is. Above this speed the car leaves the road at the crest.

Worked Example 6 — water in a swung bucket

A bucket containing $1.5\ \mathrm{kg}$ of water is swung in a vertical circle of radius $0.8\ \mathrm{m}$ . Find (a) the minimum speed at the top for the water to stay in, (b) the minimum speed at the lowest point for a complete swing, and (c) the force the bucket base exerts on the water at the lowest point in that case. (Take $g = 9.8$ .)

(a) At the top the water stays in provided the bucket base need not pull it inward, i.e. gravity alone can supply at most the centripetal force: $\dfrac{v_{\text{top}}^2}{r} \ge g$ , so

$v_{\text{top,min}} = \sqrt{gr} = \sqrt{9.8 \times 0.8} = \sqrt{7.84} = 2.8\ \mathrm{m\,s^{-1}}.$

(b) Treating the water like a particle on a "string" (the bucket base can push but not pull), a complete swing needs $u^2 \ge 5gr$ :

$u_{\min} = \sqrt{5gr} = \sqrt{5 \times 9.8 \times 0.8} = \sqrt{39.2} = 6.26\ \mathrm{m\,s^{-1}}.$

(c) At the lowest point the base pushes up (inward, towards the centre above), so $N - mg = \dfrac{mu^2}{r}$ . With $u^2 = 5gr$ , $\dfrac{u^2}{r} = 5g$ , so

$N = m(g + 5g) = 6mg = 6 \times 1.5 \times 9.8 = 88.2\ \mathrm{N}.$

Mark scheme: M1 $v^2/r \ge g$ at the top; A1 $2.8\ \mathrm{m\,s^{-1}}$ ; M1 $u^2 \ge 5gr$ ; A1 $6.26\ \mathrm{m\,s^{-1}}$ ; M1 radial equation at the bottom $N - mg = mu^2/r$ ; A1 $88.2\ \mathrm{N}$ . The base force at the bottom is $6mg$ — six times the water's weight — which is why a fast vertical swing feels so heavy at the lowest point and why the bucket must be sturdy.

4 Specimen-style exam question

(Specimen-style.) A smooth solid hemisphere of radius $0.9\ \mathrm{m}$ rests with its flat face on a horizontal table, curved side up. A small particle of mass $0.3\ \mathrm{kg}$ is placed at the top and given a tiny horizontal nudge, then slides down the outside. (a) Find the angle to the upward vertical at which the particle leaves the surface. (b) Find its speed at that instant. (c) Find the height above the table at which it leaves the surface. (Take $g = 9.8$ .)

(a) For a particle from rest at the top of a smooth sphere/hemisphere, $N = mg(3\cos\theta - 2)$ . It leaves when $N = 0$ :

$\cos\theta = \tfrac23 \;\Rightarrow\; \theta = \arccos\tfrac23 = 48.19^\circ.$

(b) Energy from the top:

$v^2 = 2gr(1 - \cos\theta) = 2(9.8)(0.9)\big(1 - \tfrac23\big) = 17.64 \times \tfrac13 = 5.88 \;\Rightarrow\; v = 2.42\ \mathrm{m\,s^{-1}}.$

(c) The height above the centre (the flat face, which sits on the table) is $r\cos\theta = 0.9 \times \tfrac23 = 0.6\ \mathrm{m}$ . Because the flat face rests on the table, the centre is at table level, so the particle leaves at $0.6\ \mathrm{m}$ above the table.

Why the leaving angle is universal. Setting $N = mg(3\cos\theta - 2) = 0$ gives $\cos\theta = \tfrac23$ regardless of $m$ , $g$ or $r$ : the geometry of "gravity's radial component vs the centripetal demand" balances at the same angle on every smooth sphere. Only the speed and height there scale with $r$ . Stating this universality explicitly is exactly the kind of insight that earns the AO2 communication marks.

Circular Motion Applications

Circular Motion Applications

1 Where this sits in AQA 7367

2 Core theory — the master method

Loop-the-loop — minimum release height

Particle on the inside of a smooth sphere/bowl

Particle on the outside of a smooth sphere

Car over a hill, water in a bucket

3 Worked examples with mark scheme

Worked Example 1 — loop-the-loop minimum height

Worked Example 2 — does it complete the loop?

Worked Example 3 — particle leaving a smooth sphere

Worked Example 4 — circular motion on a spring

Worked Example 5 — car over a hill

Worked Example 6 — water in a swung bucket

4 Specimen-style exam question

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