Investigate uniform circular motion, including the centripetal acceleration $a = v^2 / r$ and the net force required to maintain circular motion ($F_c = m v^2 / r$)

What this dot point is asking

VCAA wants you to recognise uniform circular motion as motion in a circle at constant speed, to derive the centripetal acceleration from the changing velocity vector, and to identify the source of the centripetal force in named physical situations.

Why circular motion requires acceleration

In uniform circular motion the speed is constant but the velocity vector changes direction continuously. By Newton's second law, this changing velocity requires a net force directed toward the centre of the circle.

The acceleration produced is the centripetal acceleration:

directed toward the centre. Magnitude only; the direction continuously changes.

Centripetal force

Newton's second law applied to uniform circular motion:

This is not a separate kind of force. It is whatever real force (or net force) happens to be directed toward the centre. Common sources:

Period, frequency, angular speed

Period $T$: time for one revolution. Frequency $f = 1/T$. Angular speed $\omega = 2\pi/T = 2\pi f$.

Speed: $v = 2\pi r / T = \omega r$.

Substituting: $a_c = \omega^2 r$ (equivalent form for centripetal acceleration).

Worked example (banked curve)

A car of mass $m$ rounds a banked curve of radius $r$ and bank angle $\theta$, designed so that no friction is needed. Find the design speed.

Free-body diagram: weight $mg$ down, normal force $N$ perpendicular to the road surface.

Horizontal: $N \sin\theta = m v^2 / r$.

Vertical: $N \cos\theta = mg$.

Divide: $\tan\theta = v^2 / (rg)$.

Design speed: $v = \sqrt{rg \tan\theta}$.

A banked curve of radius $50$ m at $20°$ has design speed $\sqrt{50 \cdot 9.8 \cdot \tan 20°} = \sqrt{50 \cdot 9.8 \cdot 0.364} = \sqrt{178.4} = 13.4$ m s$^{-1}$.

Common traps

Treating centripetal force as a "new" force

It is the net inward force from the actual forces acting (tension, friction, gravity, normal).

Pointing centripetal force outward

Always toward the centre. The "centrifugal" effect is a perceived inertia, not a force in an inertial frame.

Forgetting that speed is constant

$a_c$ has constant magnitude but changing direction. KE is constant in uniform circular motion; net work done by centripetal force is zero (force perpendicular to velocity).

Using $a_c = v^2/r$ when motion is not uniform

If speed is also changing, there is a tangential acceleration component in addition. Year 11 problems use uniform circular motion.

In one sentence

Uniform circular motion has constant speed but changing direction, requiring a centripetal acceleration $a_c = v^2/r$ directed toward the centre and a net force $F_c = mv^2/r$ supplied by whatever real force (tension, friction, gravity, normal component) acts inward.

Examples in context

Example 1. Mt Buller chairlift wire-rope turnaround. A Mt Buller resort chairlift wire-rope passes around a $3.0$ m radius bullwheel at the top station, moving cable at $5.0$ m s$^{-1}$. Centripetal acceleration at the rim is $a = v^2/r = 25/3 = 8.33$ m s$^{-2}$, slightly less than $g$. Each $25$ kg passenger chair travelling along the cable, on entry to the wheel, experiences a centripetal force $F = mv^2/r = 25 \times 8.33 = 208$ N pulling it toward the wheel centre, supplied by tension in the cable. Engineers use this force to specify the wheel bearings and the cable tensile rating, ensuring no slippage at peak winter load.

Example 2. Bathurst 1000 banked turn at Reid Park. A Supercar entering Reid Park at Mt Panorama travels at $200$ km h$^{-1}$ ($55.6$ m s$^{-1}$) around a corner of effective radius $50$ m. Centripetal acceleration is $a = v^2/r = 55.6^2/50 = 61.8$ m s$^{-2}$, or $6.3g$. For a $1300$ kg car, the centripetal force required is $F = mv^2/r = 1300 \times 61.8 = 8.0 \times 10^4$ N. This force is supplied by friction between tyre and tarmac plus the horizontal component of the normal force on banking. If friction alone can only supply $4g \times m = 5.1 \times 10^4$ N, the bank must contribute the remaining $2.9 \times 10^4$ N.

Try this

Q1. Define centripetal acceleration and state its direction. [2 marks]

• Cue. Acceleration of an object in uniform circular motion, of magnitude $v^2/r$, directed toward the centre of the circle.

Q2. A $500$ kg car travels at $20$ m s$^{-1}$ around a flat curve of radius $40$ m. Calculate (a) the centripetal acceleration, and (b) the friction force needed. [4 marks]

• Cue. (a) $v^2/r = 400/40 = 10$ m s$^{-2}$. (b) $F = ma = 5000$ N.

Q3. Refer to the Bathurst Supercar example. (a) Calculate the centripetal acceleration in units of $g$. (b) Determine the required centripetal force for a $1300$ kg car. (c) Explain how track banking reduces tyre wear. [2+2+2 marks]

• Cue. (a) $61.8/9.8 = 6.3g$. (b) $8.0 \times 10^4$ N. (c) Banking provides part of the centripetal force from normal-force component, reducing reliance on tyre friction.