Question

A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a constant rate. Know that the plane has a speed of 550 km/h at point A (starting point of loop), and the pilot experiences weightlessness at point C (ending point of loop) i.e., the normal force from the seat bottom is zero.
Determine (a) the deceleration of the plane, and (b) the force exerted on her by the seat of the trainer when the trainer is at point B (midway of the loop).

Answer 1

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a)

At the point C (top position of the loop), the pilot feel weightless, so the normal reaction exerted by the seat is zero:

N = 0

Therefore, the equation of the forces at position C is:

$mg=m\frac{v^2}{r}$

where the term on the left is the weight of the pilot and the term on the right is the centripetal force, and where:

$g=9.8 m/s^2$ is the acceleration due to gravity

$v$ is the velocity of the jet at the top

$r=1200 m$ is the radius of the loop

Solving for v,

$v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

So, this is the velocity of the jet at position C.

The velocity at position A (bottom) is

$v_A=550 km/h =152.8 m/s$

The distance covered by the jet is the length of a semi-circumference of radius r, so

$s=\pi r=\pi(1200)=3770 m$

Since the deceleration of the plane is constant, we can find it by using the following suvat equation:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$

b)

The force exerted on the pilot by the seat is equal to the normal force.

At point B (half of the loop), we have:

- The normal force exerted by the seat, N, acting towards the center of the loop

- There are no other forces acting  towards the center of the loop, so N must be equal to the centripetal force:

$N=m\frac{v_B^2}{r}$ (1)

where $v_B$ is the velocity at position B.

To find the velocity at position B, we notice that the distance covered by the jet between position A and position B is a quarter of a circle:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

Since we know the deceleration, we can use the suvat equation to find the velocity at point B:

$v_B^2-v_A^2=2as\\v_B=\sqrt{v_A^2+2as}=\sqrt{152.8^2+2(-1.54)(1885)}=132.4 m/s$

Therefore, we can now use eq.(1) to find the normal force exerted by the seat on the pilot at point B:

$N=(55)\frac{(132.4)^2}{1200}=803.4 N$