a) [tex]-1.54 m/s^2[/tex]
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:
[tex]mg=m\frac{v^2}{r}[/tex]
where the term on the left is the weight of the pilot and the term on the right is the centripetal force, and where:
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]v[/tex] is the velocity of the jet at the top
[tex]r=1200 m[/tex] is the radius of the loop
Solving for v,
[tex]v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s[/tex]
So, this is the velocity of the jet at position C.
The velocity at position A (bottom) is
[tex]v_A=550 km/h =152.8 m/s[/tex]
The distance covered by the jet is the length of a semi-circumference of radius r, so
[tex]s=\pi r=\pi(1200)=3770 m[/tex]
Since the deceleration of the plane is constant, we can find it by using the following suvat equation:
[tex]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[/tex]
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:
[tex]N=m\frac{v_B^2}{r}[/tex] (1)
where [tex]v_B[/tex] 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:
[tex]s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m[/tex]
Since we know the deceleration, we can use the suvat equation to find the velocity at point B:
[tex]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[/tex]
Therefore, we can now use eq.(1) to find the normal force exerted by the seat on the pilot at point B:
[tex]N=(55)\frac{(132.4)^2}{1200}=803.4 N[/tex]
