Question

A fighter plane is pulling out for a dive at speed of 900 km/hr. assuming its path to be vertical circle of radius 2000 m and its mass to be 16000 kg, find the force exerted by the air on it at the lowest point. take g = 9.8 m/s2

Answer 1

For the lowest point of the vertical circular motion we can write the net force on the fighter jet as

$F_{net} = F_{air} - mg$

now this net force on the fighter jet is towards the center of vertical circle

so this is known as centripetal force

$F_{net} = m*a_c$

$F_{air} - mg = m*\frac{v^2}{R}$

now the force due to air is given by

$F_{air} = mg + \frac{mv^2}{R}$

given that

m = 16000 kg

R = 2000 m

g = 9.8 m/s^2

v = 900 km/h = 250 m/s

now plug in all above values

$F_{air} = 16000*9.8 + \frac{16000*250^2}{2000}$

$F_{air} = 656800 N$

so the force due to air on the aircraft will be 656800 N towards the centre of the loop