An aeroplane moves at 64 m/s in a vertical loop of radius 120 m, as shown in figure. Calculate the force of the plane’s seat on 72 kg pilot while passing through the bottom part of the loop.

Explanation

Two forces acts on the pilot his downward weigth force w and the upward force of the aeroplane’s seat `F_("seat")`. Because the pilot moves in a circular path, these forces along the radial direction must, according to Newton’s seconds law ( `sum`F = ma), equal the pilot’s mass times his centripetal acceleration, where `a_(c ) = v_(2)`/r. We find that `sum`F
(in radial direction) = `F_("seat") – w = (mv^(2))/(r )`
Remember that force pointing towards the center of the circle (`F_("seat")`) are positive & those pointing away from the center (w) are negative. Substituting `omega` = mg and rearranging, we find that the force of the aeroplane seat on the pilot is
`F_("seat") = m ((v^(2))/(r ) + g) = 72kg [ (64 (m//s)^(2))/(120m) + 9.8 m//s^(2) ]`
= 72kg (34.1 m/`s^(2)` + 9.8 m/`s^(2)` ) = 3160.8 N
The pilot in this example feels very heavy. To keep him in the circular path, the seat must push the pilot upwards with a force of 3160 N, 4.5 times his normal weight. He experiences an acceleration of 4.5 g, that is, 4.5 times the acceleration of gravity.