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 const

Question

3 months ago

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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 const

ant 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).

Physics

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-03-20T03:47:08+03:00

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

$mg=m\frac{v^2}{r}$ where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top $g=9.8 m/s^2$

= loop radius

$v$ By solving for v,

$r=1200 m$

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from $v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

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

Given the plane's deceleration is consistent, we can obtain it using the following equation:

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

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$ - The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B: $N=m\frac{v_B^2}{r}$

$v_B$

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

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