A car of mass 1200 kg crests a hill of radius 40 m at 18 m/s. Calculate the normal force from the road on the car at the crest. [3 marks]

NSWPhysicsSyllabus dot point

Inquiry Question 2: Why do objects move in circles?

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

A focused answer to the HSC Physics Module 5 dot point on non-uniform circular motion. Banked tracks, the conical pendulum, vertical loops, the role of torque, and the worked banking-angle calculation that markers expect.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-23

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

NESA wants you to extend the centripetal-force model to situations where the path is more complicated than a flat horizontal circle: banked tracks, conical pendulums, and vertical loops. You also need to apply the torque relationship $\tau = r F \sin\theta$ to rotating mechanical systems. These are classic 4-6 mark calculation questions and almost always require a free-body diagram.

The answer

The principle is the same as for uniform circular motion: the net force toward the centre of the circular path equals $\frac{m v^2}{r}$ . What changes is the combination of real forces producing that net inward force.

Banked tracks

On a frictionless banked track angled at $\theta$ to the horizontal, the normal force $N$ acts perpendicular to the road surface. Its horizontal component provides the centripetal force; its vertical component balances gravity.

N \cos\theta = mg, \quad N \sin\theta = \frac{m v^2}{r}

Dividing gives the design speed:

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

At this speed the car needs no friction to stay on the curve. Below it, friction must act up the slope; above it, friction must act down the slope.

Conical pendulum

A mass on a string sweeps out a horizontal circle while the string traces a cone. The string makes angle $\theta$ with the vertical, length $L$ , so the radius of the circle is $r = L \sin\theta$ .

T \cos\theta = mg \quad \text{(vertical)}

T \sin\theta = \frac{m v^2}{r} \quad \text{(horizontal, centripetal)}

The speed is $v = \sqrt{r g \tan\theta}$ , identical in form to the banked-track design speed.

Vertical circle

For an object moving in a vertical circle (a ball on a string, a roller coaster loop), speed is not constant because gravity does work as the object rises and falls. At any point, the net force toward the centre still equals $\frac{m v^2}{r}$ , but the contributions of tension and gravity vary around the loop.

At the top of a vertical loop, both tension and gravity point downward (toward the centre):

T + mg = \frac{m v^2}{r}

The minimum speed for the string to stay taut (or for a passenger to stay in contact with the seat) occurs when $T = 0$ :

v_{\min} = \sqrt{g r}

At the bottom of the loop, tension points up (toward the centre) and gravity points down (away):

T - mg = \frac{m v^2}{r}

Torque

Torque is the rotational equivalent of force, the tendency of a force to cause rotation about a pivot:

\tau = r F \sin\theta

where $r$ is the distance from the pivot to the point where the force is applied, $F$ is the magnitude of the force, and $\theta$ is the angle between the force vector and the radial line. Torque is maximised when the force is perpendicular to the lever arm ( $\theta = 90°$ ). Units: newton metres (N m).

Worked example

A roller coaster loop has radius $12$ m. Find the minimum speed at the top of the loop for a $70$ kg passenger to maintain contact with the seat, and the normal force on the passenger at the bottom of the loop if the speed there is $20$ m/s.

Top of loop, minimum speed: $v_{\min} = \sqrt{g r} = \sqrt{9.8 \times 12} = 10.8$ m/s.

Bottom of loop: $N - mg = \frac{m v^2}{r}$ , so $N = m g + \frac{m v^2}{r} = 70 \times 9.8 + \frac{70 \times 20^2}{12} = 686 + 2333 = 3019$ N.

The passenger feels about $4.4$ times their normal weight at the bottom of the loop.

Try it: Banking angle calculator - solve for design angle given speed, or design speed given angle.

Common traps

Using $L$ instead of $L \sin\theta$ for the conical pendulum radius. The radius is the horizontal distance from the axis, not the string length.

Adding gravity and tension as scalars in vertical circles: Use vectors. At the top, both point down (toward centre, so add). At the bottom, tension points up (toward centre) and gravity points down (away from centre, so subtract).
Forgetting that the banked-track design speed is independent of mass: Mass cancels because both the gravitational force and the required centripetal force scale with $m$ .
Using $\tau = r F$ when the force is not perpendicular: Include $\sin\theta$ when the force is at an angle to the lever arm.
Assuming uniform speed in a vertical loop: Speed changes because gravity does work. Use conservation of energy to find the speed at different heights.

Examples in context

Example 1. Banked turn at Mount Panorama (Bathurst). Skyline-to-Esses transitions are banked at about $\theta = 8^{\circ}$ on radius $r = 90 \text{ m}$ . For no reliance on friction, $\tan\theta = v^2 / (g r)$ , so the "design speed" is $v = \sqrt{g r \tan\theta} = \sqrt{9.8 \times 90 \times \tan 8^{\circ}} = \sqrt{9.8 \times 90 \times 0.1405} = 11.1 \text{ m/s} \approx 40 \text{ km/h}$ . Racing cars take the corner faster ( $\sim 60 \text{ m/s}$ ), so the extra centripetal force is supplied by friction acting down-slope. The normal force vector tilts inward, and the component $N \sin\theta$ is what provides the inward push when friction is at zero.

Example 2. Vertical loop on the BIG6 coaster at Sydney's Luna Park. At the top of a $r = 7.0 \text{ m}$ vertical loop, the rider is upside down with gravity acting toward the centre. Newton's second law radially: $N + m g = m v^2 / r$ . For the minimum speed where the seat just exerts no force, $N = 0$ gives $v_{\min} = \sqrt{g r} = \sqrt{9.8 \times 7.0} = 8.28 \text{ m/s}$ . At the actual top-of-loop speed of $11 \text{ m/s}$ , the apparent weight for a $60 \text{ kg}$ rider is $N = m v^2 / r - m g = 60 \times 121/7.0 - 588 = 1037 - 588 = 449 \text{ N}$ , about $0.76$ g pressing them into the seat.

Try this

Q1. Distinguish between uniform and non-uniform circular motion in one sentence each, and give one physical example of each. [2 marks]

Cue. Uniform: constant speed (satellite). Non-uniform: speed varies along the circle (ball on vertical string).

Q2. A car of mass $1200 \text{ kg}$ crests a hill of radius $40 \text{ m}$ at $18 \text{ m/s}$ . Calculate the normal force from the road on the car at the crest. [3 marks]

Cue. $m g - N = m v^2 / r$ , so $N = m(g - v^2/r) = 1200 \times (9.8 - 8.1) = 2040 \text{ N}$ .

Q3. A $0.30 \text{ kg}$ ball on a $0.80 \text{ m}$ string swings in a vertical circle. (a) Find the minimum speed at the top so the string stays taut. (b) Find the string tension at the bottom when the ball's speed is $5.0 \text{ m/s}$ . (c) Explain why the speed is not constant around the loop. [2+2+2 marks]

Cue. (a) $v_{\min} = \sqrt{g L} = 2.80 \text{ m/s}$ . (b) $T = m v^2 / r + m g = 9.38 + 2.94 = 12.3 \text{ N}$ . (c) Gravity does positive/negative work as the ball descends/ascends, changing KE.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC5 marksA car of mass 1500 kg travels around a banked curve of radius 60 m. The track is banked at 20° to the horizontal. Calculate the speed at which the car can travel around the curve relying only on the normal force (no friction required).

Show worked answer →

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.

2017 HSC4 marksA conical pendulum consists of a 0.4 kg mass on a 1.5 m string moving in a horizontal circle. The string makes an angle of 30° with the vertical. Calculate the speed of the mass and the tension in the string.

Show worked answer →

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.