investigate and analyse theoretically and practically the uniform circular motion of an object moving in a horizontal plane and on a vertical circle, including a quantitative analysis of the forces acting at the top and bottom of the vertical circle

What this dot point is asking

VCAA wants you to analyse uniform circular motion in a horizontal plane (cars on bends, conical pendulums, balls on strings) and on a vertical circle (roller coasters, buckets of water, balls swung in a vertical loop), and to apply $F_c = \frac{mv^2}{r}$ quantitatively at the top and bottom of the vertical loop.

The answer

An object in uniform circular motion travels at constant speed $v$ around a circle of radius $r$. Speed is constant, but velocity changes direction continuously, so the object is accelerating toward the centre.

Centripetal acceleration and force

Centripetal force is not a new force. It is the net inward force supplied by gravity, friction, tension, normal force, or any combination of real forces acting toward the centre.

Period, frequency and angular velocity

The period $T$ is the time for one revolution; the frequency $f = 1/T$.

Horizontal circle: the conical pendulum

A mass on a string swung in a horizontal circle so that the string makes angle $\theta$ with the vertical. Two real forces act: tension $T$ along the string, and gravity $mg$ down. The vertical components must balance; the horizontal component of tension supplies the centripetal force.

Dividing gives $\tan\theta = \frac{v^2}{rg}$. As the speed increases, the string angle from vertical increases.

Horizontal circle: car on a flat bend

The only horizontal force is friction between tyres and road. For the bend to be completed without skidding, $\mu_s m g \geq \frac{m v^2}{r}$, so $v_{\max} = \sqrt{\mu_s g r}$.

Vertical circle: forces at the top and bottom

Consider a ball on a string of radius $r$ in a vertical circle at constant speed $v$. Gravity always acts down. Tension acts along the string toward the centre.

At the top. Both tension $T_{top}$ and gravity point down (toward the centre).

The minimum speed for the string to stay taut at the top is when $T_{top} = 0$: $v_{\min} = \sqrt{gr}$.

At the bottom. Tension $T_{bot}$ points up (toward the centre); gravity points down (away).

So tension at the bottom is greater than at the top by $2mg$.

Roller coaster at the top of a loop. Replace tension with the normal force from the rail. The cart feels lightest (sometimes momentarily weightless) at the top, when $N + mg = \frac{m v^2}{r}$.

Roller coaster at the bottom. Normal force points up; $N - mg = \frac{m v^2}{r}$, so apparent weight (felt by the rider) is maximum: $N = mg + \frac{m v^2}{r}$.

Worked example with numbers

A $1.5$ kg bucket of water is swung in a vertical circle of radius $0.80$ m. Find the minimum speed at the top of the circle so the water stays in the bucket, and the normal force from the bucket on the water at the bottom at that speed.

At the top, minimum speed is when normal force from the bucket equals zero, leaving gravity alone to supply the centripetal force on the water:

$mg = \frac{m v_{\min}^2}{r}$, so $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 0.80} = 2.8$ m/s.

At the bottom, normal force from the bucket on the water points up:

$N - mg = \frac{m v^2}{r}$, $N = m \left( g + \frac{v^2}{r} \right) = 1.5 \times \left( 9.8 + \frac{2.8^2}{0.80} \right) = 1.5 \times (9.8 + 9.8) = 29.4$ N.

Try it: Centripetal force calculator - enter mass, speed and radius and get $F_c$, $a_c$, $T$ and $\omega$.

Common traps

Treating centripetal force as a separate force on a free-body diagram. Draw the real forces (gravity, normal, friction, tension), then state that their net inward component supplies $\frac{mv^2}{r}$.

Forgetting that tension adds to gravity at the top. Both point toward the centre at the top, so $T_{top} + mg = \frac{mv^2}{r}$. At the bottom they oppose, so $T_{bot} - mg = \frac{mv^2}{r}$.

Confusing centripetal and centrifugal. Centrifugal force is fictitious; only use it in a rotating frame, which VCE does not require.

Using diameter instead of radius. $a_c = \frac{v^2}{r}$, not $\frac{v^2}{d}$.

Mixing up frequency and angular velocity. $\omega$ is in rad/s, $f$ is in Hz, related by $\omega = 2 \pi f$.

In one sentence

Uniform circular motion needs a net inward force $F_c = \frac{m v^2}{r}$ supplied by real forces (gravity, friction, tension, normal), with tension at the top of a vertical loop equal to $\frac{m v^2}{r} - mg$ and at the bottom $\frac{m v^2}{r} + mg$.