Question

(10%)  Problem 10:   A jet dives vertically at a speed v = 170 m/s, before pulling out of the dive along a circular arc. The pilot can survive an acceleration of a = 8.9 g. His mass is m = 81 kg.

What would be the magnitude of the normal force, in newtons, be at the top of an arc of this radius going the same velocity?

Answer 1

Find the radius of the arc from the normal force:

$$N_b=mg+m\dfrac {v^2}{R}=ma,\\\space\\ R=\dfrac{v^2}{(a-g)}.$$

At the top of the arc, the magnitude of the normal force is

$$N=m\bigg(g-\dfrac{v^2}R\bigg)=m(2g-a)=-5477\text{ N}.$$

The normal force is directed upward.