Banked tracks and vertical loops

In 5D, the centripetal force was provided by a single obvious force. Here it comes from a combination of forces, so the skill is resolving them. Always start with a force diagram, and remember that the centripetal force is horizontal (towards the centre of a horizontal circle), not down the slope.

Cars on flat curves

On a flat road, the only horizontal force on a car is sideways friction from the road on the tyres. The vertical forces balance ($N = mg$), and friction provides the centripetal force:

$f = \dfrac{mv^2}{r}$

Friction can only rise to a maximum of $\mu N = \mu mg$, where $\mu$ is the coefficient of friction. So the fastest safe speed is when $\mu mg = mv^2/r$:

$v_{\max} = \sqrt{\mu g r}$

Mass cancels: a truck and a hatchback have the same limit on the same bend (if their tyres grip equally). On a dry road ($\mu \approx 0.7$) with a 50 m radius, $v_{\max} = \sqrt{0.7 \times 9.8 \times 50} = 18.5$ m/s, about 67 km/h. On a wet road ($\mu \approx 0.4$) it drops to 14.0 m/s (50 km/h), and on ice almost nothing is safe.

Banked curves

On a road tilted at angle $\theta$, the normal force is perpendicular to the road surface, so it tilts towards the centre of the curve. Its horizontal component $N\sin\theta$ can provide the centripetal force with no friction at all. At this design speed:

• vertically, forces balance: $N\cos\theta = mg$
• horizontally, the net force is centripetal: $N\sin\theta = \dfrac{mv^2}{r}$

Dividing the second equation by the first:

$\tan\theta = \dfrac{v^2}{rg}$

Again, mass cancels. Every vehicle can take the curve at the same design speed without relying on friction.

Faster or slower than the design speed

Real cars don't all travel at the design speed. Friction takes up the difference:

• Faster than the design speed, the car needs more centripetal force than $N\sin\theta$ provides. The car tends to slide up and out, so friction acts down the slope, adding an inward component.
• Slower than the design speed, $N\sin\theta$ is more than needed. The car tends to slide down the bank, so friction acts up the slope.

If the friction needed is more than the road can supply, the car slides. That's why a steeply banked track can be dangerous for a slow car on ice: it slides down into the inside of the curve.

The conical pendulum

A mass on a string swung in a horizontal circle, with the string at angle $\theta$ to the vertical, is a conical pendulum. The forces are tension and weight. Vertically $T\cos\theta = mg$; horizontally $T\sin\theta = mv^2/r$. Dividing gives $\tan\theta = v^2/rg$, exactly the banked-track result, because the geometry is the same: a single tilted force supports the weight and supplies the centripetal force. A chair-swing ride at a fair works this way; the chairs swing out further as the ride speeds up.

Vertical loops

In a vertical circle the speed changes (the rider slows going up), so it isn't uniform circular motion, but at the top and bottom we can still apply $F_{\text{net}} = mv^2/r$ towards the centre.

At the bottom, the centre is above. The seat pushes up with $N$ and weight acts down:

$N - mg = \dfrac{mv^2}{r} \quad\Rightarrow\quad N = mg + \dfrac{mv^2}{r}$

The seat pushes harder than your weight, so you feel heavy.

At the top, the centre is below. Both the seat (now above you) and gravity push down:

$N + mg = \dfrac{mv^2}{r} \quad\Rightarrow\quad N = \dfrac{mv^2}{r} - mg$

If the rider goes slowly enough that $N$ would be negative, they fall out of the circle (which is why roller coasters lock riders in and keep wheels on both sides of the track). The minimum speed at the top is when $N = 0$, so gravity alone provides the centripetal force:

$mg = \dfrac{mv^2}{r} \quad\Rightarrow\quad v_{\min} = \sqrt{gr}$

That 5 g is uncomfortable and dangerous, so modern loops aren't circles. They are tall teardrop shapes (clothoid loops) with a tight radius at the top, where the rider is slow, and a wide radius at the bottom, where the rider is fast. That keeps $v^2/r$, and so the forces, more even all the way round.