An airplane flying at v = 600 km/h makes a vertical loop of…

An airplane flying at $v = 600\ \mathrm{km/h}$ makes a vertical loop of radius $R = 600\ \mathrm{m}$. With what force does the pilot, of mass $m = 80\ \mathrm{kg}$, press on the seat at the highest and lowest points?

Explanation of Method

Seat force on a pilot in a vertical loop.

Step 1: Centripetal force needed

\[ v = \frac{600}{3.6} = 166.7\ \mathrm{m/s}, \]

\[ \frac{mv^2}{R} = \frac{80 \cdot 166.7^2}{600} \approx 3704\ \mathrm{N}, \]

\[ mg = 784.8\ \mathrm{N}. \]

Step 2: Highest point

The pilot is upside down; both the seat's push $N$ and gravity point toward the centre: $N + mg = \frac{mv^2}{R} \;\Longrightarrow\; N = 3704 - 785$.

\[N_{\text{top}} \approx 2919\ \mathrm{N}\]

Step 3: Lowest point

\[ N - mg = \frac{mv^2}{R} \;\Longrightarrow\; N = 3704 + 785. \]

\[N_{\text{bottom}} \approx 4489\ \mathrm{N}\]

By Newton's third law, the pilot presses on the seat with these forces.

$N$ $mg$ $N$ $mg$ $\text{top}$ $\text{bottom}$ $\text{top: } N = \tfrac{mv^2}{R} - mg \approx 2.92\ \mathrm{kN}$ $\text{bottom: } N = \tfrac{mv^2}{R} + mg \approx 4.49\ \mathrm{kN}$ $\tfrac{mv^2}{R} = 3.70\ \mathrm{kN}$

Forces on the pilot (to scale, blue = weight, red = seat). Both point toward the centre at the top; at the bottom the seat must also cancel the weight. At the top the pilot is upside down but still pressed into the seat.

Result

\[\boxed{\,N_{\text{top}} \approx 2919\ \mathrm{N}\,}\]

\[\boxed{\,N_{\text{bottom}} \approx 4489\ \mathrm{N}\,}\]

Remark

Check. The original results are correct. In terms of his normal weight (785 N), the pilot feels about 3.7 "g" at the top and 5.7 "g" at the bottom.