model the force vectors acting on an object on a banked track moving in uniform circular motion in a horizontal plane and identify the design speed at which friction is not required to keep the object on the track

What this dot point is asking

VCAA wants you to draw the free-body diagram of an object on a banked track in uniform circular motion, resolve the forces, and find the design speed at which the normal force alone supplies the centripetal force and friction is not required.

The answer

A banked track is a curve whose surface is tilted (inward edge lower than the outer edge) so that the normal force from the surface points partly toward the centre of the circle. The horizontal component of this tilted normal force can supply some or all of the centripetal force needed to keep an object moving on the curve.

Free-body diagram on a frictionless banked track

Consider a car of mass $m$ on a banked curve of radius $r$ at angle $\theta$ to the horizontal, moving at speed $v$. Two forces act:

• Gravity $mg$, vertically down.
• Normal force $N$, perpendicular to the road surface.

If friction is zero, the only horizontal force is the horizontal component of $N$. Resolve along horizontal (toward the centre) and vertical axes:

Vertical (no vertical acceleration):

Horizontal (centripetal):

The design speed

Dividing the horizontal equation by the vertical equation:

This is the design speed condition. At this speed, the horizontal component of the normal force exactly supplies the centripetal force, and no friction is needed.

Above and below the design speed

At the design speed

Friction is zero. The road feels smooth; passengers feel pushed into their seats but not sideways.

Below the design speed

The horizontal component of the normal force is more than needed for the (smaller) required centripetal force. The car tends to slide down the bank (inward); friction must act up the bank to prevent it.

Above the design speed

The horizontal component of the normal force is not enough for the (larger) required centripetal force. The car tends to slide up the bank (outward); friction must act down the bank to keep the car on the curve.

Why engineers bank curves

Highway off-ramps, velodromes, race tracks and railway curves are banked so that vehicles can take the curve safely at the typical traffic speed without relying on friction. This reduces tyre wear, lowers the risk of skidding on wet roads, and is more comfortable for passengers because no sideways force is felt.

Examples in context

Example 1. Calder Park Thunderdome banking. Calder Park's Thunderdome speedway near Melbourne has turns banked at $24^\circ$ with radius $244$ m. The design speed at which no friction is needed is $v_d = \sqrt{rg\tan\theta} = \sqrt{244 \times 9.8 \times \tan 24^\circ} = \sqrt{244 \times 9.8 \times 0.445} = \sqrt{1064} = 32.6$ m s$^{-1}$, or $117$ km h$^{-1}$. Above this speed, friction must point down the banking to prevent the car sliding outward; below it, friction points up the banking to prevent the car sliding inward. NASCAR-era racers exceeded the design speed by $50\%$ and relied heavily on tyre friction supplementing the banking.

Example 2. Eastern Freeway tunnel access banking. The CityLink eastern off-ramp to Burnley Tunnel has a curve of radius $80$ m banked at $6^\circ$, designed for a $60$ km h$^{-1}$ ($16.7$ m s$^{-1}$) approach. The frictionless design speed is $v_d = \sqrt{80 \times 9.8 \times \tan 6^\circ} = \sqrt{82.4} = 9.07$ m s$^{-1}$ ($32.6$ km h$^{-1}$), well below the actual sign-posted speed. So drivers always rely on tyre friction up the bank. For wet weather, the available friction coefficient drops from $\mu = 0.7$ to $0.4$, reducing maximum speed via $v^2 = rg(\tan\theta + \mu)/(1 - \mu \tan\theta)$ from $74$ km h$^{-1}$ down to $57$ km h$^{-1}$, justifying the sign-posted reduction.

Try this

Q1. State the formula for the design speed of a banked track at which no friction is required. [2 marks]

• Cue. $v_d = \sqrt{rg\tan\theta}$.

Q2. A car drives around a banked curve of radius $50$ m banked at $15^\circ$. Calculate the design speed in m s$^{-1}$ and in km h$^{-1}$. [3 marks]

• Cue. $v = \sqrt{50 \times 9.8 \times \tan 15^\circ} = \sqrt{131.3} = 11.5$ m s$^{-1}$ = $41.3$ km h$^{-1}$.

Q3. Refer to Calder Park Thunderdome. (a) Calculate the design speed for $r = 244$ m and $\theta = 24^\circ$. (b) Determine the direction of friction if a car travels at $40$ m s$^{-1}$. (c) Explain why higher banking angles reduce wear on tyres. [3+2+2 marks]

• Cue. (a) $32.6$ m s$^{-1}$. (b) Friction acts down the slope (toward centre of curve). (c) Banking supplies more of the centripetal force, reducing tyre demand.