A pilot's body can withstand a maximum acceleration of…

A pilot's body can withstand a maximum acceleration of $a_{\max} = 4.6\,g$. What is the smallest radius of curvature of the plane's path when pulling out of a dive, if the plane flies at a constant speed of $v = 200\ \mathrm{km/h}$?

Explanation of Method

Smallest radius of a dive pull-out.

Step 1: Only normal acceleration

At constant speed there is no tangential acceleration; on a curve of radius $R$ the acceleration is $a_n = v^2/R$. It must not exceed $4.6\,g$: $\frac{v^2}{R} \le 4.6\,g \;\Longrightarrow\; R \ge \frac{v^2}{4.6\,g}$.

Step 2: Numbers

\[ v = \frac{200}{3.6} \approx 55.56\ \mathrm{m/s}, \]

\[ 4.6\,g = 45.13\ \mathrm{m/s^2}, \]

\[ R_{\min} = \frac{55.56^2}{45.13} = \frac{3086.4}{45.13}. \]

\[R_{\min} \approx 68.4\ \mathrm{m}\]

$R_{\min} \approx 68\ \mathrm{m}$ $v = 200\ \mathrm{km/h}$ $a_n = \tfrac{v^2}{R} \le 4.6g$ $\text{centre}$

Pulling out of a dive, the plane moves on a curve of radius $R$. At constant speed its acceleration points to the centre of the curve; the tighter the curve, the larger it is, so there is a smallest radius the pilot can survive.

Result

\[\boxed{\,R_{\min} \approx 68.4\ \mathrm{m}\,}\]

Remark

Note. Doubling the speed would require a four times larger radius ($R \propto v^2$). At the bottom of the pull-out the pilot also feels his normal weight, so the load he feels there is $a_n + g$; with that stricter reading, $R_{\min} = v^2/(3.6\,g) \approx 87\ \mathrm{m}$. The problem as stated uses the kinematic acceleration only.