Answer to Question #88392 in Mechanics | Relativity for Amisha

Question

The pilot of the airplane which has been diving at a speed of 540 km/h pulls out of the dive following a circular path at a constant speed.
- What is the minimum radius of the planes circular path in order that the acceleration of the pilot at the lowest point not exceed 7G?
- What is the apparent weight of an 80.0kg pilot at the lowest point of the pullout.

Answer 1

The centripetal acceleration

$$a=\frac{v^2}{R}$$

So

$$R=\frac{v^2}{a}=\frac{v^2}{7g}$$

$$=\frac{(540\times \frac{1000\:\rm{m}}{3600\:\rm{s}})^2}{7\times 9.8\:\rm{m/s^2}}=328\:\rm{m}$$

The apparent weight of an 80.0 kg pilot at the lowest point of the pullout

$$P=m(g+a)=8mg=8\times 80\:\rm{kg}\times 9.8\:\rm{m/s^2}=6272\:\rm{N}$$