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Loops and hills

Forces and motion · level 7, lesson 5 of 6

A question to start

A buggy drives over the crest of a rounded hill. Go fast enough, and its wheels leave the ground. At what speed does that happen on Mars, and on Earth?

The idea

At the top of a crest of radius \(r\), the buggy is moving on a circle whose centre is below it. To follow the curve, it needs a downward (inward) net force \(mv^2/r\). The forces are its weight \(mg\) down and the ground's push \(N\) up:

\[mg - N = \frac{mv^2}{r} \qquad\Rightarrow\qquad N = m\left(g - \frac{v^2}{r}\right).\]

As \(v\) grows, \(N\) falls. When \(v^2/r = g\), the ground has stopped pushing, \(N = 0\): the wheels are about to leave the ground. So the buggy stays on the road only if

\[v < \sqrt{gr}.\]
mgN, smallercentre of the curve is far below

The same rule \(\sqrt{gr}\) gives the slowest speed for the top of a vertical loop: there the track pushes down and gravity pulls down, and the car needs at least \(v^2/r = g\) to keep in contact.

Crest of radius 10 m Mars Earth
Speed at which wheels lift, \(\sqrt{gr}\) 6.1 m/s, 22 km/h 9.9 m/s, 36 km/h
Ground's push at 5 m/s, as a share of weight 33% 75%

On Mars a vehicle goes light over bumps at much lower speeds. Even before it leaves the ground, the ground pushes less, so grip for steering and braking is less on every crest. Mars roads would need gentle crests or low speeds.

Worked examples

1. Liftoff on Mars. \(\sqrt{3.71 \times 10} = 6.09\) m/s.

2. Grip at 5 m/s on Mars. \(N/(mg) = 1 - v^2/(gr) = 1 - 25/37.1 = 0.33\). Only a third of the usual grip.

3. The bottom of a dip. In a dip, the centre is above: \(N - mg = mv^2/r\), so \(N = m(g + v^2/r)\). The ground pushes harder: passengers feel heavier.

Try it

  1. Derive the speed at which a vehicle leaves the top of a crest.
  2. A crest has radius 25 m. Find the liftoff speed on Mars.
  3. At the bottom of a dip of radius 10 m on Mars, at 6 m/s, what is \(N\) for a 400 kg buggy?
  4. Why is a crest more dangerous on Mars than on Earth at the same speed?
Answers
  1. At the top, \(mg - N = mv^2/r\); \(N = 0\) when \(v = \sqrt{gr}\).
  2. \(\sqrt{3.71 \times 25} = 9.6\) m/s.
  3. \(400 \times (3.71 + 3.6) = 2{,}924\) N.
  4. Weaker gravity means the ground's push, and the grip, drop to zero at a lower speed.

Common mistake

"There is a force throwing the car up off the hill." Nothing throws it up. Gravity is not strong enough to bend its path as tightly as the hill curves, so the hill falls away beneath it.

Where it leads