Driving A Car Over a Circular Hill

Here's a textbook physics question about circular motion:

A car drives over a hill with a circular vertical profile. At what speed will the car lift off?

Finding the minimum liftoff speed at the top of the hill is easy enough. Here's the free body diagram:

Figure 1

Two forces act on the car: weight, \(mg\), pulls downward, and a normal force, \(N\), pushes upward. If the car stays on the hill, then \(N\) must be less than \(mg\), since the net force \(F_\text{net} = mg - N\) must point downward to hold the car on a circular path around the hill's curve. As usual, the centripetal force required for the car's circular motion is \(F_\text{c} = mv^2/r\), where \(v\) is the car's speed and \(r\) is the hill's radius of curvature. This centripetal force must be provided entirely by \(F_\text{net}\):
\[\begin{equation}
\frac{mv^2}{r} = mg - N. \label{eq1}
\end{equation}
\]As soon as the car leaves the ground, the normal force will have nothing to act on, so it will vanish. As a result, setting \(N = 0\) gives the threshold speed at which the car lifts off:
\[\begin{align}
\frac{mv^2}{r} &= mg - 0 \nonumber \\
v &= \sqrt{gr}
\end{align}\]
The car will lift off if its speed at the top of the hill exceeds \(\sqrt{gr}\).

That was easy because we limited our analysis to the top of the hill. But the question didn't ask specifically about just the top of the hill. If the car begins its climb at some initial speed, would it still lift off right at the crest of the hill? Or would it fly out of control immediately? Or maybe it would lift off somewhere on the other side of the hill?

Depending on your hobbies, you might not have any intuition about how cars behave on circular hills with red and blue lights strobing in the bullet-cracked rear-view, your buddy firing back with his good arm while shrieking about churches in his native tongue. So let's analyze a more general question:

A car coasts up a circular hill with initial speed \(v_0\). Where on the hill does the car lift off, and for what initial speed?

Figure 2. Notes from a speed bump designer's last day on the job.

The car's position on the hill is specified by its angle from vertical, \(\theta\). The angle subtended by the hill from vertical is \(\theta_0\). The hill's height is \(h_0 = r - r\cos\theta_0\).

Some Restrictions

• \(\theta_0 > 0\), otherwise there would be no hill.
• Let's stick to the regime \(\theta_0 < \pi/2\), otherwise the "hill" would really be a terrain bubble with walls steeper than vertical.
• We'll require \(\theta \leq \theta_0\). If not, then the car is beyond the circular extent of the hill, which doesn't make sense.

One Way to Proceed: Energy Conservation

The car rolls up the hill without applying power to its wheels, so its mechanical energy (kinetic + gravitational potential) is conserved. (Another plausible model would be to hit the gas to keep the car moving at constant speed, but that turns out not to be as interesting.) Initially, the car has total energy \(mv_0^2/2\). After climbing through a height \(h\), the car will slow to speed \(v < v_0\), having gained gravitational potential energy \(mgh\). Energy conservation gives the car's speed in terms of height:
\[\begin{align}
\frac{1}{2}mv_0^2 + 0 &= \frac{1}{2}mv^2 + mgh \nonumber \\
v^2 &= v_0^2 - 2gh \label{asdfjkl}
\end{align}\]

Getting Over the Hill

Before trying to find the liftoff speed, let's find the minimum initial speed required to get over the hill at all. Equating initial kinetic energy with potential energy at the crest of the hill gives \(mv_0^2/2 = mgh_0\), or \(v_0 = \sqrt{2gh_0}\). Substituting \(h_0 = r - r\cos \theta_0\) gives \(v_0 = \sqrt{2gr(1 - \cos \theta_0)}\) as the threshold initial speed for cresting the hill. If the car starts up the hill slower than this, it will roll to a stop partway up and roll back down the same side.

It seems reasonable that if the car can't even get over the hill, it wouldn't have enough speed to fly off the road. We'll look at this again toward the end.

The Liftoff Criterion

The car will stay on the hill if the radial component of its weight, \(mg \cos \theta\), minus the opposing normal force, \(N\), is strong enough to provide the centripetal force required for circular motion. This gives something very close to Eq. (\(\ref{eq1}\)):
\[\begin{align}
m\frac{v^2}{r} &= mg\cos \theta - N.
\end{align}\]
As before, setting \(N = 0\) gives the speed above which the car will lift off:
\[\begin{align}
v^2 &= gr\cos \theta. \nonumber
\end{align}\]
The car will lift off if its speed exceeds this threshold, corresponding to the inequality
\[\begin{align}
v^2 > gr\cos \theta. \nonumber
\end{align}\]
Substituting \(v^2 = v_0^2 - 2gh\) from Eq. (\(\ref{asdfjkl}\)) gives
\[\begin{align}
v_0^2 - 2gh &> gr \cos \theta. \label{someeq}
 \end{align}\]
Finally, inserting \(h = r\cos \theta - r\cos \theta_0\) gives
\[\begin{align}
v_0^2 - 2gr(\cos \theta - \cos \theta_0) &> gr \cos \theta \nonumber \\
v_0^2 + 2gr\cos \theta_0 &> 3gr \cos \theta \nonumber \\
\frac{v_0^2}{3gr} + \frac{2}{3} \cos \theta_0 &> \cos \theta \label{coastsol}
\end{align}\]
The car will leave the ground whenever this liftoff criterion is met. Note that everything on the left-hand side is constant.

A Closer Look at the Left Side of Equation (\(\ref{coastsol}\))

Let's start with the cosine term on the left side of the liftoff criterion. Since \(\cos \theta_0\) is bounded above by 1, \(0 < \cos \theta_0 < 1\), so \(0 < \frac{2}{3} \cos \theta_0 < \frac{2}{3}.\) Nothing surprising here. For any sizable hill, this term will be somewhat less than \(\frac{2}{3}\).

The second term on the left side of Eq. (\(\ref{coastsol}\)), \(v_0^2/(3gr)\), is never less than zero, and can be made arbitrarily large by choosing a high enough \(v_0\). If \(v_0\) is zero, the car obviously shouldn't lift off. Sure enough, setting \(v_0 = 0\) in Eq. (\(\ref{coastsol}\)) makes the liftoff criterion \(\frac{2}{3}\cos\theta_0 > \cos\theta\). In other words, \(\cos\theta_0\) must be so large that a mere \(2/3\) of it is enough to exceed \(\cos\theta\). Obviously, \(\theta = \theta_0\) won't work, since the inequality goes the wrong way: \(\frac{2}{3}\cos\theta < \cos\theta\) for \(0 < \theta < \pi/2\). We'll definitely need the strict inequality \(\theta_0 < \theta\) to make the car to lift off when \(v_0 = 0\). But we've already noted the restriction that \(\theta_0 ≥ \theta\) for any hill, so the liftoff criterion is never satisfied for \(v_0 = 0\).

Reassuringly, then, Eq. (\(\ref{coastsol}\)) makes sense in the zero-speed limit: given zero initial speed, the car will never lift off (unless \(\theta\) and \(\theta_0\) both exceed \(90°\), and the car just falls off). Furthermore, choosing a large enough \(v_0\) will ensure that the inequality (\(\ref{coastsol}\)) is satisfied for any value of \(\theta\), meaning that if the car begins with a high enough speed, it will leave the hill immediately.

The Minimum Initial Speed Required for Liftoff

Given \(g\), \(r\), and \(\theta_0\), we can choose a particular initial speed \(v_0\) such that \(v_0^2/(3gr) = \frac{1}{3}\cos \theta_0\). The special \(v_0\) value that makes this true is
\[\begin{equation}
v_0 = \sqrt{gr\cos\theta_0}. \label{liftoffspeed} \end{equation}\]
At this special speed, the liftoff criterion, Eq. (\(\ref{coastsol}\)), becomes
\[\begin{align}
\frac{1}{3}\cos \theta_0 + \frac{2}{3}\cos \theta_0 = \cos \theta_0 &> \cos \theta \nonumber \\
\implies \theta_0 &< \theta
\end{align} \]
As noted above, this inequality is never true, since it implies that the car's position has exceeded the maximum angular extent of the hill. Thus, the car will remain grounded if its initial speed is lower than \(\sqrt{gr\cos\theta_0}\).

If the car starts up the hill at a speed greater than \(\sqrt{gr\cos\theta_0}\), then \(v_0^2/(3gr) > \frac{1}{3}\cos \theta_0\), and the liftoff criterion becomes \(A \cos \theta_0 > \cos \theta\), where \(A > 1\). Even if \(\cos \theta_0\) is smaller than \(\cos \theta\), we can still satisfy \(A \cos \theta_0 > \cos \theta\) with a big enough \(A\). In particular, whenever \(A\) is a tiny bit bigger than 1, then \(\theta_0 = \theta\) is always enough to satisfy the liftoff criterion. But \(\theta_0 = \theta\) describes the car's initial position. Thus, if \(v_0 > \sqrt{gr\cos\theta_0}\), then the car flies off the surface of the hill right away!

In conclusion, there is no way to send a car coasting up a hill so that it lifts off at the top of the hill, or anywhere else, for that matter. The car will either stay on the hill for its entire motion, or it will lift off immediately (and may crash back onto the hill a short time later).

Incidentally, recall that the minimum speed needed to get over the hill is \(v_0 = \sqrt{2gr(1-\cos\theta_0)}\). This will be lower than the minimum lift-off speed \(\sqrt{gr\cos\theta_0}\) if
\[\begin{align}
2 - 2\cos\theta_0 &< \cos\theta_0 \nonumber \\
2 &< 3\cos\theta_0 \nonumber \\
\theta_0 &< \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2° \nonumber \\

\end{align}\]
So, for a small to sizable hill with half-angle \(\theta_0 < 48.2°\), the car can roll over smoothly to the other side if it starts with speed \(\sqrt{2gr(1-\cos\theta_0)} < v_0 < \sqrt{gr\cos\theta_0}\). For a steeper hill with \(\theta_0 > 48.2°\), the inequality reverses to \(\sqrt{2gr(1-\cos\theta_0)} > \sqrt{gr\cos\theta_0}\), meaning the liftoff speed is less than the speed required to get over the hill. On such a steep hill, either the car stops and rolls back down, or it starts with enough speed that it lifts off immediately; there's no way to roll it to the other side without catching air.

Some Actual Numbers

For a concrete example, let's choose \(\theta_0 = \pi/4 = 45°\), so \(\cos\theta_0 = 1/\sqrt{2}\). Let's give the car an initial speed below the liftoff speed given by Eq. (\(\ref{liftoffspeed}\)) of \(v_0^2 = gr\cos\theta_0\). Let's go with \(v_0^2 = \frac{1}{2}gr\cos\theta_0 = gr/(2\sqrt{2})\). The car will lift off as soon as \(\theta\) satisfies
\[\begin{align}
\cos \theta &< \frac{2}{3} \cos \left( \frac{\pi}{4}\right) + \frac{gr/(2\sqrt{2})}{3gr} \nonumber \\ &= \frac{2}{3\sqrt{2}} + \frac{1}{6\sqrt{2}} \nonumber \\ &= \frac{4}{6\sqrt{2}} + \frac{1}{6\sqrt{2}} = \frac{5}{6\sqrt{2}}\nonumber \\ \theta &> \cos^{-1}\left(\frac{5}{6\sqrt{2}}\right) = 53.9°
\end{align}\]
The car would lift off if \(\theta\) could exceed 53.9°, but it can't, since the hill's half-angle is only 45°.

Now, what if the initial speed slightly exceeds the lift-off threshold? Let's set \(v_0^2 = 1.1 gr\cos\theta_0\) \(= 1.1 gr/\sqrt{2}\). The liftoff criterion in this case is
\[\begin{align}
\cos \theta &< \frac{2}{3} \cos \left(\frac{\pi}{4}\right) + \frac{1.1gr/\sqrt{2}}{3gr} \nonumber \\ &= \frac{2}{3\sqrt{2}} + \frac{1.1}{3\sqrt{2}} \nonumber \\ \theta &> 43.1°
\end{align}\]
Given this initial speed, the car will take off whenever it's more than 43.1° from vertical. This first occurs at the car's initial position, where \(\theta = 45°\), so the car lifts off immediately.