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.

Worked example with numbers

A racing track has a bend of radius $50$ m banked at $25°$. Find the design speed.

$v = \sqrt{r g \tan\theta} = \sqrt{50 \times 9.8 \times \tan 25°} = \sqrt{50 \times 9.8 \times 0.4663} = \sqrt{228.5} = 15.1$ m/s.

That is about $54$ km/h.

Try it: Banking angle calculator - enter radius, speed and angle and get the design speed and required banking.

Common traps

Forgetting that the normal force changes magnitude on a banked track. $N$ is not equal to $mg$ when $\theta > 0$; it is $N = \frac{mg}{\cos\theta}$, which is greater than $mg$.

Resolving forces along the slope instead of horizontally and vertically. The centripetal acceleration is horizontal (toward the centre of the circle), so resolve along horizontal and vertical axes, not along and perpendicular to the slope.

Including friction in the design-speed calculation. The whole point of the design speed is that friction equals zero.

Confusing the direction of the friction force. Below the design speed, friction acts up the slope; above, it acts down. The car tendency is opposite to the friction.

Quoting an angle that depends on mass. $\tan\theta = \frac{v^2}{rg}$ has no mass term; the design speed is the same for a motorbike and a truck.

In one sentence

On a banked track, the horizontal component of the normal force supplies the centripetal force, and the design speed at which no friction is needed is given by $\tan\theta = \frac{v^2}{rg}$, so $v_{design} = \sqrt{rg\tan\theta}$.