Investigate the relationship between the forces acting on objects in non-uniform circular motion (banked tracks, conical pendulums, vertical circles) and apply the relationship tau = r F sin theta for torque

On a banked track with no friction, the horizontal component of the normal force provides the centripetal force. The vertical component balances gravity.

Vertical: $N \cos\theta = mg$.
Horizontal: $N \sin\theta = \frac{m v^2}{r}$.

Dividing the second by the first:

$\tan\theta = \frac{v^2}{r g}$.

Solving for $v$:

$v = \sqrt{r g \tan\theta} = \sqrt{60 \times 9.8 \times \tan 20°}$
$v = \sqrt{60 \times 9.8 \times 0.364}$
$v = \sqrt{214} = 14.6$ m/s.

Markers reward the resolved force diagram, the explicit statement that no friction is required at this "design speed," and the substitution with correct units. Mass cancels out, so the answer is independent of vehicle mass.

Resolve the tension into vertical and horizontal components. Vertical balances gravity; horizontal provides centripetal force.

Vertical: $T \cos\theta = mg$, so $T = \frac{mg}{\cos\theta} = \frac{0.4 \times 9.8}{\cos 30°} = \frac{3.92}{0.866} = 4.53$ N.

Horizontal: $T \sin\theta = \frac{m v^2}{r}$, where $r = L \sin\theta = 1.5 \times \sin 30° = 0.75$ m.

$v^2 = \frac{r T \sin\theta}{m} = \frac{0.75 \times 4.53 \times 0.5}{0.4} = 4.25$.

$v = \sqrt{4.25} = 2.06$ m/s.

Markers reward a clear force diagram showing tension resolved into components, the relationship $r = L \sin\theta$ (not $L$), and correct units.

Uniform circular motion has constant speed around the circle (e.g. a satellite in a stable orbit, or a car on a flat, level roundabout at constant speed).

Non-uniform circular motion has a speed that changes as the object moves around the circle, because a force such as gravity does work on it (e.g. a ball on a string swinging in a vertical circle, which speeds up as it falls and slows as it rises).

Marks: one for a correct distinguishing statement (constant vs changing speed, with the reason - work done by a force), one for one correct example of each.

At the top of a hill the centre of the circular path is below the car, so weight acts toward the centre and the normal force acts away from the centre. Newton's second law radially:

$mg - N = \dfrac{mv^2}{r}$, so $N = m\left(g - \dfrac{v^2}{r}\right)$.

$N = 1200\left(9.8 - \dfrac{18^2}{40}\right) = 1200(9.8 - 8.1) = 1200 \times 1.7 = 2.0 \times 10^3\ \text{N}$.

Marks: one for the correctly signed radial equation ($mg - N$, not $N - mg$), one for the substitution, one for the final answer to two significant figures with the unit newton.

Radius

The radius is the horizontal distance from the axis, not the string length: $r = L\sin\theta = 1.2 \times \sin 25^{\circ} = 1.2 \times 0.4226 = 0.507\ \text{m}$.

Speed

$v = \sqrt{rg\tan\theta} = \sqrt{0.507 \times 9.8 \times \tan 25^{\circ}} = \sqrt{0.507 \times 9.8 \times 0.4663} = \sqrt{2.317} = 1.52\ \text{m/s}$.

Period

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 0.507}{1.52} = 2.1\ \text{s}$.

Marks: one for $r = L\sin\theta = 0.507\ \text{m}$, one for $v = \sqrt{rg\tan\theta} = 1.52\ \text{m/s}$, one for $T = 2\pi r/v = 2.1\ \text{s}$ with the unit.

(a) From $\tan\theta = \dfrac{v^2}{rg}$, for a fixed radius $r$ the quantities $g$ and $r$ are constants, so $\tan\theta$ is directly proportional to $v^2$. A directly proportional relationship graphs as a straight line through the origin.

(b) Using the points $(100\ \text{m}^2\text{s}^{-2},\ 0.204)$ and $(400\ \text{m}^2\text{s}^{-2},\ 0.816)$:

$\text{gradient} = \dfrac{0.816 - 0.204}{400 - 100} = \dfrac{0.612}{300} = 2.04 \times 10^{-3}\ \text{s}^2\text{m}^{-1}$.

(c) The gradient of $\tan\theta$ against $v^2$ equals $\dfrac{1}{rg}$, so $r = \dfrac{1}{g \times \text{gradient}} = \dfrac{1}{9.8 \times 2.04 \times 10^{-3}} = 50\ \text{m}$.

Marks: one for the proportionality argument (constants $r$, $g$; line through origin), one for a correctly computed gradient with correct units, one for equating the gradient to $1/(rg)$, one for $r = 50\ \text{m}$.

(a) At the minimum speed the tension is zero and gravity alone supplies the centripetal force: $mg = \dfrac{mv_{\min}^2}{r}$, so $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 0.80} = 2.8\ \text{m/s}$.

(b) At the bottom, tension acts up (toward the centre) and gravity acts down (away from the centre): $T - mg = \dfrac{mv^2}{r}$, so

$T = mg + \dfrac{mv^2}{r} = (0.30)(9.8) + \dfrac{(0.30)(5.0^2)}{0.80} = 2.94 + 9.375 = 12\ \text{N}$.

(c) Gravity has a component along the direction of motion everywhere except at the very top and bottom of the circle, so it does positive work as the ball descends (speeding it up) and negative work as the ball ascends (slowing it down), changing its kinetic energy and hence its speed.

Marks: one for $v_{\min} = \sqrt{gr} = 2.8\ \text{m/s}$, one for the correctly signed bottom equation $T - mg = mv^2/r$, one for $T = 12\ \text{N}$ with the unit, one for the work-done-by-gravity explanation.

$\tau = rF\sin\theta = (0.25)(40)\sin 30^{\circ} = (0.25)(40)(0.5) = 5.0\ \text{N m}$.

To increase the torque without changing $F$, the person should either apply the force further from the pivot (increase $r$, e.g. use a longer spanner) or apply it closer to $90^{\circ}$ to the handle (increase $\sin\theta$ toward its maximum value of $1$).

Marks: one for the correct formula with substitution, one for $\tau = 5.0\ \text{N m}$ with the unit, one for a correct method to increase torque (longer lever arm or force closer to perpendicular).

Band-6 plan. (1) Derive the no-friction "design" banking angle from the resolved forces. (2) Compute it for the given data. (3) Explain what happens to vehicles above/below design speed on that bank, and on a shallower bank. (4) Weigh the two options against a real spread of speeds and reach a justified verdict.

Model answer. On a frictionless banked curve, the normal force $N$ is the only force apart from gravity. Resolving vertically and horizontally, $N\cos\theta = mg$ and $N\sin\theta = mv^2/r$. Dividing gives the design condition $\tan\theta = \dfrac{v^2}{rg}$, independent of mass.

For $v = 25\ \text{m/s}$ and $r = 80\ \text{m}$: $\tan\theta = \dfrac{25^2}{80 \times 9.8} = \dfrac{625}{784} = 0.797$, so $\theta = \tan^{-1}(0.797) = 38.6^{\circ}$.

If the engineer banks the curve at this steep $38.6^{\circ}$, a car travelling at exactly $25\ \text{m/s}$ needs no friction at all. However, real traffic travels over a range of speeds. A car travelling slower than $25\ \text{m/s}$ has too little required centripetal force for that steep angle, so the excess component of $N\sin\theta$ would push it down and inward - it must rely on friction acting up the slope to avoid sliding down into the bank. A car travelling faster needs friction acting down the slope to avoid sliding up and off the road. A very steep bank is also dangerous for slow-moving or stationary vehicles (e.g. in traffic), which can slide down the slope under gravity alone if friction is insufficient.

A shallower bank, designed with a lower nominal speed and relying on some friction across the whole range of realistic speeds, keeps $\tan\theta$ smaller and the road closer to level, so vehicles travelling well below the design speed are not put at risk of sliding down a steep slope while stationary or slow, and the required friction at both high and low speeds stays within the typical range tyres can supply in wet or dry conditions.

On balance, a road carrying a real distribution of speeds (not a single design speed) is safer with a shallower bank supplemented by friction across the operating range, rather than a steep zero-friction bank optimised for one speed only, because the steep design fails badly (relying entirely on friction, at an unfavourable angle) for any vehicle far from the design speed.

Marker's note: the top band derives $\tan\theta = v^2/(rg)$ from resolved forces (not quoted from memory), correctly computes $\theta = 38.6^{\circ}$, explains the direction friction must act both above and below the design speed, and reaches an explicit, reasoned safety verdict for a range of speeds rather than just describing the single design-speed case.

Band-6 plan. (1) Set up the radial equation at the top for a general normal force. (2) Explain the sign/direction constraint that a one-sided rail imposes (push only) versus a locked track (push or pull). (3) Derive the minimum speed condition for Design A. (4) Explain why Design B has no equivalent minimum-speed constraint from the track. (5) Reach an assessment of the two designs.

Model answer. At the top of a vertical loop of radius $r$, both the track's normal force $N$ and gravity act toward the centre (downward), so Newton's second law radially gives $N + mg = \dfrac{mv^2}{r}$, hence $N = \dfrac{mv^2}{r} - mg$.

In Design A, the car rests on a rail below it, so the rail can only push the car away from itself (here, downward, toward the centre) - it cannot pull. This means $N$ must be greater than or equal to zero. If the required $N$ from the equation above becomes negative (i.e. if $v$ is too small), the physical rail simply cannot supply a negative (pulling) force, so the car loses contact with the rail and leaves the circular path, following a projectile path instead. The minimum speed for the car to maintain circular contact is where $N = 0$: $mg = \dfrac{mv_{\min}^2}{r}$, giving $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 10} = 9.9\ \text{m/s}$. Below this speed, Design A is unsafe at the top of the loop.

In Design B, the car is mechanically locked to an overhead track (e.g. by wheels above and below the rail), so the track can supply either a positive (pushing) or a negative (pulling) normal force as needed. There is no physical requirement that $N \geq 0$: even at very low speed, or $v = 0$, the track can supply enough upward (away-from-centre) restraining force to hold the car in place, so there is no minimum-speed condition analogous to Design A's. The radial equation still holds, but $N$ is simply solved for and can be negative (a pull) without any contradiction.

Assessing the two designs: Design A is only safe at or above $v_{\min} = 9.9\ \text{m/s}$ at the top of the loop, so the ride must guarantee sufficient speed there (e.g. via a sufficiently high entry point and energy conservation), and any energy loss to friction or air resistance is a real safety risk if it drops the car below $v_{\min}$. Design B removes this specific failure mode because the locked track can restrain the car at any speed, including momentarily near zero, making it the more robust design against speed shortfalls, though it introduces its own engineering demands (the track and wheel assembly must be strong enough to supply large pulling forces).

Marker's note: the top band derives the radial equation for both designs from the same physics, correctly identifies that the constraint $N \geq 0$ is a property of a one-sided rail (not a universal law of circular motion), computes $v_{\min} = 9.9\ \text{m/s}$ for Design A, and explains specifically why Design B has no equivalent constraint before reaching a comparative safety judgement. A response that only calculates $v_{\min}$ without explaining why Design B differs caps in the middle band.