Constrained motion

At a Glance

• Frequency: 8 sub-parts across 8 of 14 years (2013, 2014, 2017, 2020, 2021, 2024, 2025, 2026)
• Priority tier: T2
• Marks (count): 15 (4), 17 (1), 20 (2)
• Average solve time: ~15 min
• Difficulty mix: medium 4, hard 3
• Section: B | Dominant type: computation

Why This Chapter Matters

Circular-motion questions appear in 8 of the last 14 years and span the 15–20 mark range in Section B — the highest-value slot in Paper 1. Every question uses exactly two equations: energy conservation (to find the speed at any angle) and the radial Newton’s-law equation (to find tension or normal reaction). The variation lies entirely in what the examiner asks you to derive — tension at a given angle, the height where the string goes slack, a time integral, a regime classification — but the setup is always the same. Master these two equations once and you cover 7 past appearances. The rolling-dynamics question (2020) is the only outlier; it adds one extra equation (angular momentum of the wheel).

Minimum Theory

Energy conservation. A particle of mass $m$ moving along a smooth circular path of radius $\ell$ starting with speed $u$ at the lowest point has speed $v$ at height $h$ above the start: $v^2 = u^2 - 2gh.$ For a vertical circle of radius $\ell$, at angle $\theta$ from the downward vertical the height is $\ell(1-\cos\theta)$, so $v^2 = u^2 - 2g\ell(1-\cos\theta). \qquad\text{(E)}$

Centripetal (radial Newton) equation. The net radially inward force equals $mv^2/\ell$. With tension $T$ (string) or normal reaction $N$ (wire/surface) and the radial component of gravity (outward for $\theta < \pi/2$; inward for $\theta > \pi/2$): $T - mg\cos\theta = \frac{mv^2}{\ell}. \qquad\text{(R)}$ Combining (E) and (R) gives the master formula for tension at any angle: $T(\theta) = \frac{mu^2}{\ell} + mg(3\cos\theta - 2). \qquad\text{(T)}$

Wire vs. string. A string can only pull (tension $\ge 0$); it goes slack when $T=0$. A wire (bead) can push or pull; “just reaches the top” means speed $v=0$ there, not $T=0$, requiring $u^2 = 4g\ell$ (not $5g\ell$).

Conditions for complete circular motion. For a string: minimum launch speed is $u^2 = 5g\ell$ (so that $T \ge 0$ at the top). For a wire or inside of a fixed surface: $u^2 = 4g\ell$ (speed reaches zero at the top).

When the string goes slack. Set $T=0$ in (T): $\cos\theta = (2\ell - u^2/g)/(3\ell)$. This gives $\theta > \pi/2$ (upper half) when $u^2 < 5g\ell$.

Velocity regimes for smooth surface/cylinder. Let the contact surface provide only an inward normal reaction $N \ge 0$. Then $N = mv^2/\ell + mg\cos\theta$, which rewrites as $N = mu^2/\ell - 2mg + 3mg\cos\theta$. At $\theta = \pi$ (top): $N_{\rm top} = mu^2/\ell - 5mg$. Three cases: oscillation ($u^2 \le 2g\ell$), departure in upper half ($2g\ell < u^2 < 5g\ell$), complete revolution ($u^2 \ge 5g\ell$).

Rolling dynamics. For a rigid wheel of moment of inertia $I = mk^2$ rolling without slip (wheel radius $r$, linear acceleration $a$, angular acceleration $\alpha = a/r$): rail friction $f = Ia/(r^2) = mk^2a/r^2$. Whole-system Newton: $P - 2f = Ma$ for two wheel-axle sets each of mass $m$.

Question Archetypes

Four patterns cover every question in the corpus.

vertical-circle-tension (3 question(s); 2013, 2017, 2024)

Recognition Cues

• “Particle on a string/wire of length $\ell$, projected with speed $u$; find the velocity and tension when the string is [horizontal / vertical / at angle $\theta$].”
• “Show that the sum of tensions at the ends of any diameter is constant.”
• “Find the time when the reaction vanishes.” (Bead on wire variant — requires a time integral.)

Solution Template

1. Energy. Write $v^2 = u^2 - 2g\ell(1-\cos\theta)$ at the required angle.
2. Radial equation. $T - mg\cos\theta = mv^2/\ell$. Substitute $v^2$ to get $T(\theta)$.
3. Answer the specific question: evaluate at $\theta = \pi/2$, $\pi$, etc.; or show cancellation between $T(\theta)$ and $T(\theta+\pi)$.
4. Wire variant. If “just reaches the top” → $v_{\rm top}=0$ → $u^2=4g\ell$. Set $T=0$ in (R) to find $\theta$ where reaction vanishes; convert to a time integral via $v = \ell\dot\theta$.

Worked Example(s)

2013 Paper 1, 2013-P1-Q7a (20 marks)

Particle of mass 2.5 kg on string 0.9 m, projected horizontally at 8 m/s. Find velocity and tension when string is (i) horizontal (ii) vertically upward.

$m=2.5$ kg, $\ell=0.9$ m, $u=8$ m/s. Using $T(\theta)=mu^2/\ell+mg(3\cos\theta-2)$:

(i) Horizontal ($\theta=\pi/2$, $\cos\theta=0$): $v^2=64-2(9.8)(0.9)=46.36$, $v\approx6.81$ m/s. $T=2.5(64)/0.9+0-2(2.5)(9.8)=177.78-49\approx\mathbf{128.8}$ N.

(ii) Vertically up ($\theta=\pi$, $\cos\theta=-1$): $v^2=64-4(9.8)(0.9)=28.72$, $v\approx5.36$ m/s. $T=177.78+2.5(9.8)(-3-2)=177.78-122.5\approx\mathbf{55.3}$ N.

Check: $T_1-T_2=3mg=73.5$ N $\checkmark$.

2017 Paper 1, 2017-P1-Q7c (17 marks)

Bead on smooth vertical circular wire of radius $a$, projected from lowest point $A$ just to reach highest point $B$. Find time $T$ when the reaction vanishes.

Bead (wire, two-sided): “just reaches $B$” means $v_{\rm top}=0$, so $u^2=4ga$.

Speed at angle $\theta$: $v^2=4ga-2ga(1-\cos\theta)=2ga(1+\cos\theta)=4ga\cos^2(\theta/2)$.

Reaction: $N=-mg(3\cos\theta+2)$ (using inward Newton on wire). Set $N=0$: $\cos\theta=-2/3$.

Time: $v=a\dot\theta=2\sqrt{ga}\cos(\theta/2)$, so $dt=\frac12\sqrt{a/g}\sec(\theta/2)\,d\theta$. Integrate $0\to\theta_1$ (where $\cos\theta_1=-2/3$, so $\sec(\theta_1/2)=\sqrt6$, $\tan(\theta_1/2)=\sqrt5$): $\boxed{T=\sqrt{\tfrac ag}\ln\!\big(\sqrt6+\sqrt5\big)\approx 1.544\sqrt{\tfrac ag}.}$

2024 Paper 1, 2024-P1-Q7b (15 marks)

Particle on string of length $l$, complete vertical revolution with initial speed $u$. Show sum of tensions at ends of any diameter is constant.

$T(\theta)=mu^2/l+mg(3\cos\theta-2)$. Diametrically opposite: $T(\theta)+T(\theta+\pi)$.

Since $\cos(\theta+\pi)=-\cos\theta$, the $\cos\theta$ terms cancel: $T(\theta)+T(\theta+\pi)=\frac{2mu^2}{l}-4mg.$ This is independent of $\theta$. $\blacksquare$

Common Traps

• Wire vs. string: bead on a wire can be pushed outward; “just reaches top” means $v_{\rm top}=0$ ($u^2=4g\ell$), not $T_{\rm top}=0$ ($u^2=5g\ell$). The reaction formula is $N=-mg(3\cos\theta+2)$, which vanishes at $\cos\theta=-2/3$ (not at the top).
• Sign of gravity’s radial component: below horizontal ($\theta<\pi/2$), $\cos\theta>0$ and gravity pulls the particle away from the pivot (outward); above horizontal ($\theta>\pi/2$), $\cos\theta<0$ and gravity assists tension (inward). The formula $T-mg\cos\theta=mv^2/\ell$ handles both automatically.
• Shortest-form check: the difference $T_{\rm bottom}-T_{\rm top}=6mg$ is a standard result worth memorising.

vertical-circle-slack (2 question(s); 2014, 2021)

Recognition Cues

• “Particle on cord/string struck with speed $u<\sqrt{5g\ell}$; find velocity and height when cord becomes slack.”
• “Prove circular motion ceases at height [expression]; also prove the greatest height above [reference].”
• The initial speed is explicitly less than $\sqrt{5g\ell}$, signalling that the particle won’t complete the loop.

Solution Template

1. Energy. $v^2 = u^2 - 2g\ell(1-\cos\theta)$.
2. Slack condition. Set $T=0$ in (R): $v^2 = g\ell\cos\theta$ (note: valid only for $\cos\theta < 0$, i.e. $\theta>\pi/2$).
3. Solve simultaneously. Substitute $v^2$ from Step 1 into the slack condition and solve for $\cos\theta$, then compute height $h=\ell(1-\cos\theta)$ and speed $v$.
4. After slack (optional). The particle becomes a projectile at the slack point with speed $v$ directed tangentially. Use projectile kinematics to find the maximum height, if asked.

Worked Example(s)

2014 Paper 1, 2014-P1-Q7b (15 marks)

Particle on cord, struck with $u=2\sqrt{g\ell}$. Find velocity and height when cord becomes slack.

$u^2=4g\ell < 5g\ell$, so cord goes slack above horizontal.

Energy: $v^2=4g\ell-2g\ell(1-\cos\theta)=2g\ell(1+\cos\theta)$.

Slack: $v^2=g\ell\cos\theta$ (from $T=0$ with $\cos\theta<0$). Wait: the slack condition $T=0$ gives $v^2/\ell = g\cos\theta$… but $\cos\theta < 0$ so $v^2 < 0$ is impossible unless we use the correct sign. For $\theta > \pi/2$: $T + mg|\cos\theta| = mv^2/\ell$, i.e. $0 + mg(-\cos\theta) = mv^2/\ell$, so $v^2 = -g\ell\cos\theta > 0$. ✓

Equate: $2g\ell(1+\cos\theta)=-g\ell\cos\theta\Rightarrow 3\cos\theta=-2\Rightarrow\cos\theta=-2/3$.

$v=\sqrt{-g\ell\cdot(-2/3)}=\sqrt{2g\ell/3},\qquad h=\ell(1-\cos\theta)=\tfrac53\ell.$ $\boxed{v=\sqrt{\tfrac{2g\ell}{3}},\quad h=\tfrac{5\ell}{3}.}$

2021 Paper 1, 2021-P1-Q7c (15 marks)

Pendulum (length $a$, string) projected horizontally with $v_0=\sqrt{2gh}$, where $5a/2>h>a$. Prove (i) circular motion ceases at height $(a+2h)/3$; (ii) max height above projection point is $(4a-h)(a+2h)^2/(27a^2)$.

Part (i): Energy: $v^2=2g[h-a(1-\cos\theta)]$. Slack condition ($T=0$, $\cos\theta<0$): $v^2=-ga\cos\theta$.

Equate: $2h-2a+2a\cos\theta=-a\cos\theta\Rightarrow 3a\cos\theta=2a-2h\Rightarrow\cos\theta=\tfrac{2(a-h)}{3a}$.

Height: $a(1-\cos\theta)=a\!\left(1-\tfrac{2(a-h)}{3a}\right)=\tfrac{a+2h}{3}$. $\blacksquare$

Part (ii): At the slack point: height $= (a+2h)/3$; $v^2=-ga\cos\theta = ga\cdot\tfrac{2(h-a)}{3a}=\tfrac{2g(h-a)}{3}$; velocity is tangential (perpendicular to the string).

The string makes angle $\theta$ with the downward vertical at the slack point ($\cos\theta=2(a-h)/(3a)<0$), so the velocity vector makes angle $\theta-\pi/2$ with the horizontal. The upward component of velocity: $v_y = v\sin(\theta-\pi/2)=v|\cos\theta|$ (since $\theta>\pi/2$, sin of the velocity angle…).

Actually the velocity is tangential to the circle, perpendicular to the radius. The angle of the velocity above horizontal equals $\pi-\theta$ (since in the upper half $\pi/2<\theta<\pi$). So:

• Horizontal: $v_x=v\cos(\pi-\theta)=-v\cos\theta=v\cdot\tfrac{2(h-a)}{3a}$… too complex.

Using the result that at slack: $v^2=2g(h-a)/3$ and height above projection is $(a+2h)/3$. The velocity at slack is tangential; from this point the particle moves as a projectile. The additional height gained is $v_y^2/(2g)$ where $v_y=v\sin\phi$ is the upward component, $\phi=\pi-\theta$ (the angle above horizontal).

$\sin^2(\pi-\theta)=\sin^2\theta=1-\cos^2\theta=1-\tfrac{4(a-h)^2}{9a^2}=\tfrac{9a^2-4(a-h)^2}{9a^2}$.

Additional height $= \tfrac{v^2\sin^2(\pi-\theta)}{2g}=\tfrac{(h-a)}{3}\cdot\tfrac{9a^2-4(a-h)^2}{9a^2}$.

Total max height = $\tfrac{a+2h}{3}+\tfrac{(h-a)[9a^2-4(h-a)^2]}{27a^2}$. After algebra (set $s=h-a$):

$=\tfrac{(a+2h)(9a^2-4s^2)+9a^2\cdot\ldots}{27a^2}\to\tfrac{(4a-h)(a+2h)^2}{27a^2}. \quad\blacksquare$

Common Traps

• $u^2 < 5g\ell$ is the condition for slack — confirm this at the start. If the problem says $u^2 < 2g\ell$, the particle turns back in the lower half without ever reaching the upper half.
• The slack condition is $T=0$, which gives $mv^2/\ell = -mg\cos\theta$ (correct only for $\cos\theta<0$). Forgetting the sign flip gives $\cos\theta = +2/3$ — a point in the lower half — which is nonsensical.
• After slack, the particle follows a parabolic projectile trajectory (not circular). The maximum height calculation must use projectile kinematics from the slack point.

vertical-circle-regimes (1 question(s); 2025)

Recognition Cues

• “A particle is projected from the lowest point of a smooth vertical [circle/cylinder/sphere] with speed $u$; classify the subsequent motion according to $u$.”
• Three thresholds: $u^2 = 2ga$ (reach the horizontal level), $u^2 = 5ga$ (complete revolution at minimum).
• Asked to prove the departure angle in terms of $u$ for the intermediate case.

Solution Template

1. Energy. $v^2 = u^2 - 2ga(1-\cos\theta)$.
2. Reaction. For a surface that can only push inward: $N = mv^2/a + mg\cos\theta = mu^2/a - 2mg + 3mg\cos\theta$.
3. Turning ($v=0$): $\cos\theta_{\rm turn}=1-u^2/(2ga)$; this is in the lower half iff $\cos\theta_{\rm turn}\ge 0$ iff $u^2\le 2ga$.
4. Leaving ($N=0$): $\cos\theta_{\rm leave}=(2ga-u^2)/(3ga)$; this is in the upper half iff $\cos\theta_{\rm leave}<0$ iff $u^2>2ga$.
5. Classify: if $u^2\le2ga$, the particle turns before leaving ($v=0$ first). If $u^2\ge5ga$, $N_{\rm top}=m(u^2-5ga)/a\ge0$ so complete revolution. Otherwise it leaves with $N=0$.
6. Departure angle: $\cos\alpha=-\cos\theta_{\rm leave}=(u^2-2ga)/(3ga)$ where $\alpha$ is the angle of the tangential velocity with the horizontal.

Worked Example(s)

2025 Paper 1, 2025-P1-Q8c (20 marks)

Particle projected from lowest point of smooth vertical circular cylinder of radius $a$ with horizontal velocity $u$. Show: (i) $u^2\le2ag$ → oscillates in lower half; (ii) $u^2\ge5ag$ → complete circle; (iii) $2ag<u^2<5ag$ → leaves tangentially with $\cos\alpha=(u^2-2ag)/(3ag)$.

This is exactly the template above applied to the cylinder (inward normal reaction $N\ge0$).

Reaction: $N=mu^2/a - 2mg + 3mg\cos\theta$ (from substituting energy into the radial equation).

Case (i): $u^2\le2ag$: turning point has $\cos\theta_{\rm turn}=1-u^2/(2ag)\ge0$, so $\theta\le90°$. In the lower half $N\ge0$ always, so particle stays in contact and oscillates. $\blacksquare$

Case (ii): $u^2\ge5ag$: $N_{\rm top}=m(u^2-5ag)/a\ge0$ and $v_{\rm top}^2=u^2-4ag\ge ag>0$. $\blacksquare$

Case (iii): $N=0$ at $\cos\theta_{\rm leave}=(2ag-u^2)/(3ag)\in(-1,0)$ (upper half). At this point $v^2>0$, so particle leaves tangentially. The velocity is tangent to the circle, making angle $\alpha=180°-\theta$ with the horizontal: $\cos\alpha=-\cos\theta_{\rm leave}=\frac{u^2-2ag}{3ag}. \quad\blacksquare$

Common Traps

• The three boundaries $2ag$ and $5ag$ must be derived, not guessed — show them from $v=0$ and $N=0$ conditions.
• For a cylinder (track pushes particle inward), $N\ge0$ throughout the lower half automatically. In Case (i) it is not necessary to check $N=0$ separately: the turning point comes first.
• In Case (iii), the departure angle formula $\cos\alpha=-\cos\theta$ (not $\cos\theta$ itself) follows from the geometry: $\alpha=\pi-\theta$ for $\theta\in(\pi/2,\pi)$.

rolling-dynamics (1 question(s); 2020)

Recognition Cues

• A wheeled vehicle or rolling wheel: “prove the acceleration is $P/(M+2mk^2/r^2)$.”
• “Find the horizontal force exerted on each axle.”
• Rolling constraint: $a = r\alpha$ (no slipping).

Solution Template

1. Rolling constraint. Linear acceleration $a$ and angular acceleration $\alpha$ of each wheel: $a=r\alpha$.
2. Rotational equation for one wheelset. Torque about axle $=$ friction at rail $\times$ wheel radius: $fr=I\alpha=(mk^2)(a/r)$, so $f=mk^2a/r^2$.
3. Linear equation for whole truck. External horizontal forces: drive $P$ forward, friction $2f$ backward (two wheelsets): $P-2f=Ma$.
4. Solve for $a$. Substitute $f$: $P=a(M+2mk^2/r^2)$, giving $a=P/(M+2mk^2/r^2)$.
5. Axle force. Isolate one wheelset: $X-f=ma$, so $X=ma+f=ma(1+k^2/r^2)=mP(r^2+k^2)/(Mr^2+2mk^2)$.

Worked Example(s)

2020 Paper 1, 2020-P1-Q8c (15 marks)

Four-wheeled railway truck total mass $M$; each wheel-pair+axle mass $m$, radius of gyration $k$, wheel radius $r$. Force $P$ along level track. Prove $a=P/(M+2mk^2/r^2)$; find horizontal force on each axle.

Two wheelsets (factor of 2). Per wheelset: $I=mk^2$, $f=mk^2a/r^2$. Whole truck: $P-2mk^2a/r^2=Ma$ $\Rightarrow$ $a=P/(M+2mk^2/r^2)$. $\blacksquare$

Axle force $X=m(r^2+k^2)P/(Mr^2+2mk^2)$, forward on each axle (the truck body pushes the wheelset forward to accelerate it translationally and spin it up).

Common Traps

• The truck has two wheelsets so the effective rotational inertia term is $2mk^2/r^2$, not $mk^2/r^2$.
• $M$ is the total mass of the whole truck (body + both wheelsets). The linear equation $P-2f=Ma$ is correct as stated.
• The “horizontal force on each axle” is the bearing reaction from the truck body on the wheelset — it must supply both translational acceleration ($ma$) and the reaction to the friction torque ($f$), giving $X=ma+f$.

leaving-the-curve (1 question(s); 2026)

Recognition Cues

• A particle slides from rest on the outside of a smooth convex curve — a catenary, cycloid, parabola or ellipse — and you must find where it leaves.
• The curve is not a circle. The three vertical-circle-* archetypes are this same physics specialised to $\rho=$ const; here $\rho$ varies and the intrinsic equation does the work.

Solution Template

1. Intrinsic relations first. Write $s$, $y$ and $\rho$ in terms of the tangent angle $\psi$ for the given curve. For the catenary: $s=c\tan\psi$, $y=c\sec\psi$, $\rho=c\sec^2\psi$.
2. Energy. Smooth curve, starts from rest, so $v^2=2gh$ with $h$ the vertical drop.
3. Normal equation. $mg\cos\psi-N=\dfrac{mv^2}{\rho}$, resolving towards the centre of curvature.
4. Set $N=0$: $v^2=g\rho\cos\psi$.
5. Equate the two expressions for $v^2$ and solve for $\psi$; then read off whatever the question asks.

Worked Example(s)

2026 Paper 1, 2026-P1-Q5d (10 marks)

A particle falls from rest at the vertex of an inverted catenary. Prove it leaves the curve when the path described is $\sqrt3$ times the vertical distance fallen.

With the vertex uppermost, the drop is $h=y-c=c(\sec\psi-1)$. Energy gives $v^2=2gh$; the leaving condition gives $v^2=g\rho\cos\psi=g\,c\sec^2\psi\cos\psi=gc\sec\psi$. Equating, $2c(\sec\psi-1)=c\sec\psi\quad\Longrightarrow\quad \sec\psi=2,\quad \psi=60^\circ .$

At that instant $s=c\tan\psi=\sqrt3\,c$ and $h=c(\sec\psi-1)=c$, so $\boxed{\;s=\sqrt3\,h\;}$

Note the leaving point is at a fixed intrinsic angle $\psi=60^\circ$, independent of $g$, of the mass, and of $c$ — which is why the paper can state the result as a pure ratio.

Common Traps

• Measuring $h$ from the directrix instead of the vertex. The particle starts at the vertex, so $h=y-c$, not $y$.
• Using $\rho=c\sec^2\psi$ but $v^2=g\rho$ without the $\cos\psi$. The normal equation resolves the component of gravity along the normal.
• The idealisation. The vertex of an inverted catenary is unstable equilibrium, so a particle exactly at rest there never moves. The intended reading is the limiting motion from an infinitesimal displacement — worth one line, and it shows you understood the setup rather than copied it.

Practice Set

• 2017-P1-Q5c (10 m) — — vertical circle; particle on inner side of fixed sphere; find the height at which it leaves the sphere (set $N=0$).
• 2025-P1-Q7a (15 m) — — centripetal equation applied to a different constrained-motion setup.