Lesson 9 · 35 min

Choosing a System: Engineering Applications

You now have two complete toolkits. This lesson is about judgement: recognizing which one a problem is asking for, setting it up cleanly, and using the answers to make engineering decisions about roads, rides and aircraft.

Learning objectives

Which system?

Both systems always give the same vectors, so a "wrong" choice is never incorrect, only longer. Let the information in the problem decide:

Choosing a coordinate system
The problem gives or asks for…UseBecause
\(x(t)\) and \(y(t)\), or \(a_x\) and \(a_y\) separately\(x\)–\(y\)differentiate or integrate each coordinate on its own
Gravity only (a projectile)\(x\)–\(y\)\(a_x = 0\), \(a_y = -g\): two independent, simple motions
A path \(y = f(x)\) with \(v_x\) or \(v_y\) known (guides, slots, cams)\(x\)–\(y\) with the chain rule\(v_y = f'v_x\), \(a_y = f''v_x^2 + f'a_x\)
A known path and the speed along it (vehicles, trains, rides)\(n\)–\(t\)\(a_t = \dot v\) and \(a_n = v^2/\rho\) come straight from the data
Circular motion with \(\omega\), \(\alpha\)\(n\)–\(t\)\(a_t = \alpha r\), \(a_n = \omega^2 r\)
"How sharply does the path bend?", "is it speeding up?", "normal force / friction needed"\(n\)–\(t\), often converted from \(x\)–\(y\)\(\rho\), \(a_t\) and \(a_n\) are path quantities (Lesson 8)

A procedure that always works

  1. Sketch the path, the particle and the axes or the \(\et\), \(\en\) directions at the instant of interest.
  2. Choose the system with the table above, and write down what is known in that system (with signs).
  3. Relate the unknowns with the kinematic equations: differentiate, integrate, or use \(a_t\), \(a_n\), \(\rho\).
  4. Convert if the question asks for the other system (Lesson 8).
  5. Check: units, signs, \(|\avec|\) the same in both systems, and whether the size makes physical sense (compare with \(g = 9.81\ \text{m/s}^2\)).

Roads: curves, crests and sags

Road designers limit the normal acceleration so that drivers stay comfortable and tyres keep their grip. On a flat curve at steady speed, \(a_n = v^2/\rho\) must stay below a chosen limit \(a_{n,\max}\), which sets the smallest radius allowed for a given design speed:

\[ \rho_\text{min} = \frac{v^2}{a_{n,\max}} \]

Example 9.1 — Designing a highway curve

A highway is designed for \(100\ \text{km/h}\) with a comfort limit of \(0.15g\) on the sideways acceleration. (a) Find the minimum curve radius. (b) A car on that curve at the design speed brakes at \(3\ \text{m/s}^2\). What is its total acceleration?

Show solution

(a) \(v = 100/3.6 = 27.78\ \text{m/s}\) and \(a_{n,\max} = 0.15(9.81) = 1.472\ \text{m/s}^2\):

\[ \rho_\text{min} = \frac{27.78^2}{1.472} = 524.4\ \text{m} \]

(b) \(a_t = -3\ \text{m/s}^2\) and \(a_n = 1.472\ \text{m/s}^2\):

\[ |\avec| = \sqrt{3^2 + 1.472^2} = 3.341\ \text{m/s}^2 \]

Real highways also bank (superelevate) their curves so that part of \(m v^2/\rho\) comes from the normal force of the road; that is a Newton's-law refinement of the same kinematics.

Over a crest, \(\en\) points down. Gravity can supply at most \(g\) of downward acceleration, so if \(v^2/\rho\) exceeds \(g\) the road would have to pull the car down, which it cannot do: the wheels lift off. The limiting speed is

Speed limit over a crest of radius \(\rho\)

\[ \frac{v^2}{\rho} \le g \quad\Rightarrow\quad v_\text{max} = \sqrt{g\rho} \]

In a sag, \(\en\) points up, the road pushes harder than usual, and occupants feel heavier: there is no speed at which contact is lost, only a comfort limit.

Figure 9.1 A car at constant speed over a hilly track \(y = 6\cos(2\pi x/60)\) (metres). The crest radius is \(\rho = 15.2\ \text{m}\), so the car stays in contact only while \(v \le \sqrt{g\rho} = 12.2\ \text{m/s}\). Raise the speed and watch the status box at the crest. With constant speed \(a_t = 0\): the acceleration is all normal, \(v^2/\rho\), largest where the track bends most sharply.

Example 9.2 — Airtime over a hill

A rural road goes over a hump whose crest has a radius of curvature of \(30\ \text{m}\). Above what speed do the wheels of a car leave the road at the crest?

Show solution
\[ v_\text{max} = \sqrt{g\rho} = \sqrt{9.81(30)} = 17.16\ \text{m/s} \approx 61.8\ \text{km/h} \]

Notice that the answer does not depend on the car's mass: it is pure kinematics. This is why "humpback" bridges carry low speed limits.

Rides and aircraft: \(g\)-loads

Pilots and ride designers describe normal accelerations in multiples of \(g\). At the bottom of a pull-out from a dive, the aircraft is on a curve with the center of curvature above it, \(a_n = v^2/\rho\) points up, and the pilot is pressed into the seat.

Example 9.3 — Pulling out of a dive

A jet at \(150\ \text{m/s}\) pulls out of a dive along a vertical arc. The pilot's normal acceleration must not exceed \(5g\). What is the smallest radius of the pull-out?

Show solution
\[ \rho_\text{min} = \frac{v^2}{a_{n,\max}} = \frac{150^2}{5(9.81)} = 458.7\ \text{m} \]

Doubling the speed would need four times the radius: fast aircraft need a lot of sky to turn.

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Key takeaways