Banked Curves & Conical Pendulums

A curve you can take on ice

Watch a bobsled run, or the corner of a NASCAR oval, and you notice the track is not flat. It leans. The outside edge is lifted, sometimes by a few degrees, sometimes by more than thirty, and the sleds and cars carve through the corner far faster than anything could manage on level ground.

A flat road turns a car using friction between the tyres and the tarmac. Wet the road, ice it, or wear the tyres smooth, and the turn stops working. A banked road turns the car using the road's own push: the force that would be there even if the surface were polished glass. Get the angle right for the speed and you can round the corner with no friction whatsoever. That is a strange claim, so this lesson earns it.

The only new decision is which axes to use

From the last lesson you have the one fact this whole topic rests on: a body moving in a circle of radius $r$ at speed $v$ has acceleration of magnitude

$a = \frac{v^2}{r}$

pointed at the centre of the circle. Not along the velocity. Not along the surface. At the centre. For a car on a banked road the centre of the circle is off to the side at the same height as the car, so that acceleration is horizontal.

Now list the forces. The road pushes on the tyres perpendicular to the road surface: that is the normal force $N$, where "normal" is the geometry word for perpendicular. Gravity pulls down with $mg$. Leave friction out for now.

Here is the move that decides everything. When you did blocks on ramps you tilted your axes to lie along the slope, because the acceleration was along the slope. That reason is gone. Here the acceleration is horizontal, so use horizontal and vertical axes and let $N$ be the vector you break apart.

$N$ is perpendicular to a surface tilted by $\theta$ from the horizontal, so $N$ itself is tilted by $\theta$ from the vertical. Its vertical component is $N\cos\theta$ and its horizontal component is $N\sin\theta$, pointing toward the inside of the curve. Vertically nothing accelerates, so the vertical forces cancel. Horizontally, the only horizontal force is that component of $N$, and it must supply the whole $mv^2/r$:

$N\cos\theta = mg, \qquad N\sin\theta = \frac{mv^2}{r}$

The design speed

You have two equations and two unknowns you did not ask for, $N$ and $m$. Divide the second by the first and both disappear at once, because $N$ appears in each and so does $m$:

There is no mass in it. A motorcycle and a loaded truck behave identically on the same bank, which is why engineers can post one number for everyone. And for a given track (given $\theta$ and $r$) exactly one speed satisfies it:

$v_{\text{design}} = \sqrt{rg\tan\theta}$

This is the design speed. At that speed and only at that speed, a frictionless banked curve holds you. Go faster and $N\sin\theta$ is too small to bend your path tightly enough, so you drift outward and up the bank. Go slower and $N\sin\theta$ is more inward force than you need, so you slide down the bank toward the centre. On ice, both of those happen. On tarmac, friction covers the gap, and the next section is about how much gap it can cover.

Worked example 1. A curve of radius $80\;\mathrm{m}$ is to be banked so a car at $20\;\mathrm{m/s}$ needs no friction. Find the angle.

Write down the relation before anything else, because it is the only one you have: $\tan\theta = v^2/(rg)$. Then

$\tan\theta = \frac{(20)^2}{(80)(9.8)} = \frac{400}{784} = 0.5102$

$\theta = \arctan(0.5102) = 27.0^\circ$

Sanity check: a higher design speed would grow the numerator and so the angle. Faster corners need steeper banks, which matches every racetrack you have seen.

Friction as the correction term

Real roads have grip, so a real banked curve works over a range of speeds.

Friction acts along the surface, up the slope or down it, depending on which way you would otherwise slide. Above the design speed you tend to ride up the bank, so static friction acts down the slope. Below it you tend to slide down, so friction acts up the slope. Either way the most it can give is $f = \mu_s N$.

Take the fast case. Let up-the-slope be the positive direction along the surface, and set up the two components as before, with $f$ pointing down the slope:

$\text{vertical:}\quad N\cos\theta - f\sin\theta = mg$ $\text{horizontal:}\quad N\sin\theta + f\cos\theta = \frac{mv^2}{r}$

Check each term against the picture. $f$ points down the slope, which is down and outward, so it has a downward vertical part $f\sin\theta$ (hence the minus sign when it joins $N\cos\theta$ on the left) and an inward horizontal part $f\cos\theta$ (hence the plus). At the maximum speed the tyres are on the verge of sliding, so $f = \mu_s N$. Substitute that and divide the second equation by the first; $N$ cancels again, and so does $m$:

$\frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta} = \frac{v^2}{rg}$

Divide top and bottom by $\cos\theta$ to make it readable:

$v_{\max} = \sqrt{rg\,\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}}$

The slow case is the same calculation with $f$ flipped, which flips the sign of $\mu_s$ in both places:

$v_{\min} = \sqrt{rg\,\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}}$

Both formulas collapse to $\sqrt{rg\tan\theta}$ when $\mu_s = 0$, as they must. And if $\mu_s \ge \tan\theta$ the quantity under the root in $v_{\min}$ goes negative, which is the algebra's way of saying there is no minimum: friction alone can hold the car at rest on that slope, so you can crawl round at any speed you like, even zero.

Worked example 2. A track of radius $90\;\mathrm{m}$ is banked at $12^\circ$ with $\mu_s = 0.50$. What is the maximum speed?

Get the pieces first. $\tan 12^\circ = 0.2126$, and $rg = (90)(9.8) = 882$.

$v_{\max} = \sqrt{(882)(0.7973)} = \sqrt{703.2} = 26.5\;\mathrm{m/s}$

Worth comparing: the design speed here is $\sqrt{882 \times 0.2126} = 13.7\; \mathrm{m/s}$. Friction nearly doubles what the bank alone can do.

Worked example 3. A $1200\;\mathrm{kg}$ car creeps at $10\;\mathrm{m/s}$ around a curve of radius $70\;\mathrm{m}$ banked at $18^\circ$. How large is the friction force on it, and which way does it point?

This one is different in kind. Nothing is on the verge of sliding, so $f \ne \mu_s N$: the friction is whatever it needs to be, and you have to solve for it. First decide the direction. The design speed is $\sqrt{(70)(9.8)\tan 18^\circ} = \sqrt{686 \times 0.3249} = 14.9\;\mathrm{m/s}$. The car is slower than that, so it tends to slide down the bank and friction points up the slope.

Here the tilted axes finally pay off, because they let you get $f$ in one line without ever finding $N$. Take the axis pointing up the slope. Along that axis, gravity contributes $-mg\sin\theta$, friction contributes $+f$, and $N$ contributes nothing. The acceleration is horizontal with magnitude $v^2/r$ pointing inward, and its component along the up-slope direction is $-(v^2/r)\cos\theta$, since inward is partly down the slope. So

$f - mg\sin\theta = -m\frac{v^2}{r}\cos\theta$

$f = mg\sin\theta - m\frac{v^2}{r}\cos\theta$

Numbers. $\sin 18^\circ = 0.3090$, $\cos 18^\circ = 0.9511$, and $v^2/r = 100/70 = 1.429\;\mathrm{m/s^2}$.

$f = (1200)(9.8)(0.3090) - (1200)(1.429)(0.9511) = 3634 - 1630 = 2004\;\mathrm{N}$

Positive, so it does point up the slope, confirming the guess. And it is well under $\mu_s N$ for any ordinary road, so the car holds.

The conical pendulum is the same problem

Tie a ball to a string, whirl it so it travels a horizontal circle while the string sweeps out a cone at angle $\theta$ from the vertical. Two forces act: gravity $mg$ down, and tension $T$ along the string: one vertical, one tilted by $\theta$ from the vertical. It is the banked curve's free body diagram with $T$ written where $N$ was, so the equations copy across with no new thinking:

$T\cos\theta = mg, \qquad T\sin\theta = \frac{mv^2}{r}$

with one extra piece of geometry: the radius is not given to you, it is set by the string. If the string has length $L$, the ball is a horizontal distance $r = L\sin\theta$ from the axis, and hangs a vertical distance $h = L\cos\theta$ below the pivot.

Dividing gives $\tan\theta = v^2/(rg)$ again, and substituting $r = L\sin\theta$ lets you solve for the speed:

$v^2 = rg\tan\theta = gL\sin\theta\tan\theta$

The tension comes straight from the vertical equation alone: $T = mg/\cos\theta$. That grows without limit as $\theta$ approaches $90^\circ$, which is the precise reason you can never whirl a string fully horizontal. A horizontal string has no upward component at all, so nothing holds the weight up.

The prettiest result is the period. Substitute $v = 2\pi r / T_{\text{per}}$ into $\tan\theta = v^2/(rg)$:

$\tan\theta = \frac{4\pi^2 r^2}{T_{\text{per}}^2 \, r g} = \frac{4\pi^2 r}{T_{\text{per}}^2 g}$

Now put in $r = L\sin\theta$ and use $\tan\theta = \sin\theta/\cos\theta$. The $\sin\theta$ cancels off both sides:

$T_{\text{per}} = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{h}{g}}$

The period depends only on $h$, the height of the pivot above the circle. Not on the mass, not on the radius, not on the speed. Two pendulums of the same length hanging at the same angle, one with a marble and one with a brick, come round together.

Worked example 4. A $0.60\;\mathrm{kg}$ ball on a $1.2\;\mathrm{m}$ string is swung as a conical pendulum. The string snaps at $20\;\mathrm{N}$. What is the fastest the ball can go?

You are not given the angle, which is the whole difficulty. But tension and angle are locked together by the vertical equation, so the breaking tension fixes the angle:

$\cos\theta = \frac{mg}{T} = \frac{(0.60)(9.8)}{20} = \frac{5.88}{20} = 0.2940$

so $\theta = 72.9^\circ$ and $\sin\theta = 0.9558$. The radius follows:

$r = L\sin\theta = (1.2)(0.9558) = 1.147\;\mathrm{m}$

Now the horizontal equation, which you have not used yet:

$\frac{mv^2}{r} = T\sin\theta = (20)(0.9558) = 19.12\;\mathrm{N}$

The pattern is worth naming: one equation gave the angle, geometry gave the radius, the other equation gave the speed. Almost every problem in this topic is that three-step chain in some order.

A puck in a funnel, an aeroplane in a turn

The reason this topic is worth a lesson of its own is that the same two-equation skeleton keeps reappearing in disguise.

A puck sliding on the smooth inside of a funnel has exactly two forces on it: its weight, and a normal force from the wall pointing up and inward. That is the banked curve with the road replaced by the funnel wall, so $v = \sqrt{rg\tan\theta}$ with $\theta$ the tilt of the wall.

An aeroplane in a level turn banks its wings by $\theta$. The lift force is perpendicular to the wings, so it tilts by $\theta$ from the vertical. Two forces: lift and weight. Same equations, with $L$ in place of $N$. For a plane at $150\;\mathrm{m/s}$ banked at $30^\circ$,

which is why airliners need a great deal of sky to turn around in.

A bead on the inside of a smooth cone, a motorcyclist leaning into a corner: all the same. Two forces, one vertical, one tilted, and a horizontal acceleration.

How to recognise this kind of problem

Reach for the banked-curve method when all of these are true:

• Something moves in a horizontal circle at constant speed. Horizontal is the key word. A ball on a vertical circle is a different lesson, because there the acceleration is not horizontal.
• Exactly one force is tilted out of the vertical: a normal force from a sloped surface, a string tension, an aerodynamic lift. Everything else is gravity.
• The problem hands you an angle, or asks for one.

When you see that, write these three lines on blank paper before you do anything else:

1. Vertical: the tilted force's vertical component equals $mg$.
2. Horizontal: its horizontal component equals $mv^2/r$.
3. Geometry: what is $r$? Given directly on a road, but $r = L\sin\theta$ on a string, and it may need a sketch inside a cone.

Then decide which of the four quantities $\theta$, $v$, $r$, $T$ or $N$ you were given and which you want, and divide or substitute accordingly. If friction is mentioned, ask first whether the problem says "maximum", "minimum" or "on the verge": those mean $f = \mu_s N$ and you use the $v_{\max}$ or $v_{\min}$ formula. If it gives you an actual speed and asks for the friction force, friction is an unknown you solve for, as in worked example 3.

Everything here came from one idea: a force tilted out of the vertical does two jobs at once, holding the weight up and bending the path. In the next lesson that tilted force is replaced by gravity itself, and the circle becomes an orbit, but the two equations you write down first will look very familiar.