Question

A pilot of mass 50 kg in a jet aircraft while executing a loop-the-loop with constant speed of 250 m/s. If the radius of circle is 5 km, compute the force exerted by seat on the pilot(a) at the top of loop.(b) at the bottom of loop.(Ans : 135.0 N, 1115.0 N)

Answer 1

Let N1 be the normal force on the top and N2 be at the bottom.

N1 + mg = (mv²)/r

N2 - mg = (mv²)/r

Mv²/r = 50 × 62500 ÷ 5000 = 625 N

mg= 50 × 9.8 = 490 N

N1 = 625 - 490 = 135N

N2 = 625 + 490 = 1115N

Answer 2

Hey dear,

We'll have to consider this as vertical circular motion.

◆ Answer -

Ft = 135 N

Fb = 1115 N

◆ Explaination -

# Given -

m = 50 kg

v = 250 m/s

r = 5000 m

# Solution -

Using formulas for vertical circular motion, force exerted by seat on pilot -

(1) At the top of the loop -

Ft = mv^2/r - mg

Ft = 50×(250×250)/5000 - 50×9.8

Ft = 625 - 490

Ft = 135 N

(2) At the bottom of the loop -

Fb = mv^2/r + mg

Fb = 50×(250×250)/5000 + 50×9.8

Fb = 625 + 490

Fb = 1115 N

Hope this is helpful..