Lesson 7 · 35 min

Banked Curves, Crests and Loops

Tilt a curve and the road itself pushes the car toward the center, so it needs less friction, or none at all, to turn. Curve the road up or down instead, and the normal force changes: a dip presses you into your seat, a crest lifts you out of it, and a fast enough crest launches you. Each case is \(\sum F_n = mv^2/\rho\) with a well-drawn free-body diagram.

Learning objectives

The banked curve

A curve banked at angle \(\theta\) tilts the road's normal force toward the center. Before you open Figure 7.1, make a prediction.

Take \(n\) horizontal, toward the center, and \(b\) vertical. With no friction, the free-body diagram has only the weight and the normal force:

\[ \textstyle\sum F_b = 0{:}\quad N\cos\theta - mg = 0 \qquad\qquad \sum F_n = m\dfrac{v^2}{\rho}{:}\quad N\sin\theta = m\frac{v^2}{\rho} \]

Design speed of a banked curve

\[ \tan\theta = \frac{v_0^2}{g\rho} \qquad\Rightarrow\qquad v_0 = \sqrt{g\rho\tan\theta} \]

At \(v_0\) the car needs no friction at all: it would take the curve on ice. The mass cancels again.

At any other speed, friction makes up the difference. Faster than \(v_0\), the car tends to slide up and out, and friction acts down the bank; slower, it tends to slide down and in, and friction acts up the bank. With \(F_f\) taken positive down the bank,

\[ N = m\left(g\cos\theta + \frac{v^2}{\rho}\sin\theta\right) \qquad\qquad F_f = m\left(\frac{v^2}{\rho}\cos\theta - g\sin\theta\right) \]

and the car holds the road while \(|F_f| \le \mu_s N\). Setting \(|F_f| = \mu_s N\) gives the limits:

\[ v_{\max} = \sqrt{g\rho\,\frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta}} \qquad\qquad v_{\min} = \sqrt{g\rho\,\frac{\sin\theta - \mu_s\cos\theta}{\cos\theta + \mu_s\sin\theta}} \]

If \(\tan\theta \le \mu_s\), there is no lower limit: the car could stop on the bank without sliding down.

Figure 7.1 A car rounding a curve of radius \(120\ \text{m}\) banked at \(\theta\). Top: the curve seen from above at a slant, with the road drawn wider than it is. Press Play: the car's velocity \(\colV{\vvec}\) runs along the road, its acceleration \(\colMA{\avec}\) points horizontally at the center \(O\), and friction \(\colF{F_f}\) on its tires acts along the bank, up or down. Bottom: the car seen from behind, with the center to the left. The free-body diagram shows the weight \(\colW{W}\), the normal force \(N\), perpendicular to the road, and friction along it; the kinetic diagram shows \(\colMA{mv^2/\rho}\), horizontal. Below the design speed friction points up the bank, above it down the bank, and at \(v_0\) it vanishes. Outside \(v_{\min}\) and \(v_{\max}\) friction cannot hold the car: too fast, it slides up and off the bank; too slow, it slides down onto the infield.

Example 7.1 — Designing a banked curve

A highway curve of radius \(120\ \text{m}\) is banked at \(15^\circ\). (a) What is its design speed? (b) With \(\mu_s = 0.3\), what is the fastest a car can take it?

Show solution
\[ \text{(a)}\quad v_0 = \sqrt{9.81(120)\tan 15^\circ} = 17.8\ \text{m/s} = 64\ \text{km/h} \] \[ \text{(b)}\quad v_{\max} = \sqrt{9.81(120)\frac{\sin 15^\circ + 0.3\cos 15^\circ}{\cos 15^\circ - 0.3\sin 15^\circ}} = \sqrt{1177\,\frac{0.5486}{0.8883}} = 27.0\ \text{m/s} = 97\ \text{km/h} \]

There is no lower limit: \(\tan 15^\circ = 0.27 < 0.3\), so a stopped car stays put. On a flat curve the same friction would allow only \(\sqrt{0.3(9.81)(120)} = 18.8\ \text{m/s}\).

Over a crest, through a dip

On a road that curves in the vertical plane, the normal direction is vertical at the top of a crest or the bottom of a dip. At a crest the center of curvature is below, so the acceleration \(v^2/\rho\) points down; in a dip it points up.

\[ \begin{aligned} \text{Crest:}\quad & mg - N = m\frac{v^2}{\rho} \qquad && \Rightarrow\quad N = m\left(g - \frac{v^2}{\rho}\right) \\[4pt] \text{Dip:}\quad & N - mg = m\frac{v^2}{\rho} \qquad && \Rightarrow\quad N = m\left(g + \frac{v^2}{\rho}\right) \end{aligned} \]

Over a crest the road pushes up with less than the weight, and at \(v = \sqrt{g\rho}\) it stops pushing at all: \(N = 0\), and the car is about to leave the road. Any faster and gravity alone cannot bend the path enough, so the car flies.

Figure 7.2 A car at the top of a crest and another at the bottom of a dip, both of radius \(\rho\), driving to the right at the same speed. The road is drawn to scale, so the radius slider reshapes it: a small \(\rho\) gives a sharp hill and a deep dip, a large \(\rho\) a gentle swell. The dashed circles of radius \(\rho\) fit the road under each car; their centers, marked when they fall inside the figure, are the centers of curvature. Each car's velocity \(\colV{\vvec}\) is horizontal there, and its acceleration \(\colMA{v^2/\rho}\) points toward its center of curvature: down at the crest, up in the dip. The weight \(\colW{W}\) ends at each car's center of gravity, and the road's push on the two tires is drawn as one normal force \(N\), up into the car under its center of gravity, to scale. At the crest the normal force shrinks as the speed grows and reaches zero at \(\sqrt{g\rho}\); in the dip it grows. (The cars are drawn about twice their size.)

Example 7.2 — A crest and a dive

(a) A \(1200\ \text{kg}\) car goes over a crest of radius \(50\ \text{m}\) at \(72\ \text{km/h}\). Find the normal force, and the speed at which the car would leave the road. (b) A \(75\ \text{kg}\) pilot pulls out of a dive at \(100\ \text{m/s}\) on a circle of radius \(500\ \text{m}\). How hard does the seat push at the bottom?

Show solution

(a) \(v = 20\ \text{m/s}\), so \(v^2/\rho = 8\ \text{m/s}^2\):

\[ N = 1200(9.81 - 8) = 2172\ \text{N} \qquad v_{\max} = \sqrt{9.81(50)} = 22.1\ \text{m/s} = 80\ \text{km/h} \]

(b) \(v^2/\rho = 100^2/500 = 20\ \text{m/s}^2\):

\[ N = 75(9.81 + 20) = 2236\ \text{N} = 3.04\,mg \]

The pilot feels "three g": the seat pushes with three times their weight.

Around a vertical loop

A car running on the inside of a vertical loop of radius \(r\) is pressed against the track by \(N\), which points toward the center. At the top, the weight points toward the center too:

\[ \text{Top:}\quad N + mg = m\frac{v^2}{r} \]

The track can only push, \(N \ge 0\), so the car stays on only if \(v^2/r \ge g\) at the top: \(v_{\text{top}} \ge \sqrt{gr}\). A ball whirled on a cord in a vertical circle obeys the same equation with the tension \(T\) in place of \(N\), since a cord can only pull.

The speed is not constant around the loop. Along the path the tangential equation, with the weight's component \(-mg\sin\phi\) at angle \(\phi\) from the bottom, gives the change in speed exactly as in Lesson 5:

\[ m\,v\frac{dv}{ds} = -mg\sin\phi, \quad ds = r\,d\phi \quad\Rightarrow\quad v^2 = v_b^2 - 2gr(1 - \cos\phi) \]

Making it around the loop

\[ v_{\text{top}}^2 = v_b^2 - 4gr \ge gr \quad\Rightarrow\quad v_b \ge \sqrt{5gr} \]

\(v_b\) is the speed at the bottom. The Work and Energy module gets the same relation between \(v_b\) and \(v_{\text{top}}\) in one line.

Figure 7.3 A small car, drawn as a block, enters a smooth vertical loop of radius \(1\ \text{m}\) at the speed \(v_b\) you choose. The track's push \(N\) ends on the car, from the track's side; the weight \(\colW{W}\) ends at the center of gravity. Below \(v_b = \sqrt{5gr} = 7.0\ \text{m/s}\), \(N\) falls to zero before the top, and the car leaves the track and falls inside the loop.

Example 7.3 — Upside down on a roller coaster

A roller coaster car goes over the top of a loop of radius \(10\ \text{m}\), upside down, at \(12\ \text{m/s}\). How hard does the seat push on a \(70\ \text{kg}\) rider? What is the least speed at the bottom that gets a car over the top?

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At the top both \(N\) and the weight point down, toward the center:

\[ N + 70(9.81) = 70\,\frac{12^2}{10} \quad\Rightarrow\quad N = 70(14.4 - 9.81) = 321\ \text{N} = 0.47\,mg \] \[ v_b \ge \sqrt{5(9.81)(10)} = 22.1\ \text{m/s} \]

Real coasters use loops that are tighter at the top than at the bottom, to keep the riders' loads lower at the bottom.

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Key takeaways