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

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.

Dividing gives the design speed:

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 $r = L \sin\theta$.

DMATH_2

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):

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

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

Torque

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

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 with numbers

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.

In one sentence

Non-uniform circular motion (banked tracks, conical pendulums, vertical loops) is analysed by resolving all real forces along the radial direction so that their net inward component equals $\frac{m v^2}{r}$, with torque $\tau = r F \sin\theta$ governing any associated rotational tendency.