Circular motion question

Question

A 76-kg pilot at an airshow performs a loop de loop with his plane. At the bottom of the 52-m radius loop, the plane is moving at 48 m/s. Determine the net force acting upon the pilot.

Answer 1

Let

m = 76 kg be the mass of the pilot,

r = 52 m be the radius of the loop,

v = 48 m/s be the plane's speed,

g = 9.81 m/s² be gravitational acceleration.

There are three forces action on the pilot.

1) Downward force of gravity of magnitude

mg

2) Reaction to the force of gravity by the seat acting upward, also of magnitude

mg

3) Centrital force provided by the seat directed towards the center of the loop of magnitude

mv²/r

The net force is the vector sum of all three.

The first two cancel each other.

So the net force is equal the centripetal force

mv²/r = 76×48²/52 = 3367 N