Physics: forces on occupant of aircraft flying in a vertical loop?

Question

If anyone could help me out with the following physics question, I would appreciate it greatly:

"An aircraft is traveling in a vertical loop of diameter 3000m. The pilot has a mass of 90kg. At the bottom of the loop, the seat exerts 2550N on the man.

a) At what speed is the plane flying?
b)What force does the seat exert on the man when he is upside down at the top of the loop?
c) What is the minimum speed that the plane can fly at so that the man will feel "weightless" at the top of the loop? (ie when will the seat not exert force on him?)"

If you could possibly give me full working out, that would be great. I especially need help with knowing when to add and subtract any force vectors, such as force of gravity and centrifugal force.

Thanks!

Answer 1

a) At the bottom, the force of gravity (Fg) is pointing down, the normal force (Fn, the force the seat exerts on the pilot) is pointing up, and there is a net centripedal force, Fc, up. So:

Fc = Fn - Fg, knowing Fn=2550 N, Fg = (90kg)(9.8m/(s^2)) = 882 N, so Fc = 1668 N.

Now remember that Fc = (M(v^2))/R, where M is the mass of the pilot (90kg), v is the speed (the unknown), and R is the radius of the loop (3000m/2 = 1500m). So using some simple algebra, you get a v of roughly 167 m/s.

b) Fg is still down, but Fn and Fc are now down too. Remember that Fc is a NET FORCE, and always points towards the center of circular motion. Fc = Fg + Fn. Assuming speed hasn't changed, Fc is still 1668 N and Fg is still 882 N. So Fn = 786 N. This should make sense if you've been on a roller coaster: you feel a lot of weight at the bottom of a loop, but almost weightless at the top. That's because Fn is greater at the bottom and less on the top.

c) Same as previous problem, only this time Fn = 0 N (the seat exerts no force on the pilot), and now you must again find velocity. So Fc = Fg:

Fc = ((90kg)(v^2))/(1500m) = (90kg)(9.8m/(s^2)) = Fg

Again, simple algebra yields a velocity of roughly 121 m/s.

Hope this makes sense.

EDIT: I assume the speed is the same at the bottom and top of the loop because during circular motion, it is commonly assumed that one travels at a constant speed. This question is obviously for homework from a basic physics course. Applying conservation of energy would complicate this problem too much for this level, as this is an ideal scenario and not a real world problem. Even if kinetic energy is lost to potential, you might as well account for the plane's engine keeping it at a constant speed too, and air resistance for that matter.

Answer 2

a) At what speed is the plane flying?

The seat's force comes from the man's weight W = mg = 90*10 ~ 900 kg-m/sec^2 (aka Newton) and the A/C's cetripetal force C = mV^2/R. Thus fotal force F = W + C and C = F - W = 2550 N - 900 N ~ 1650 = mV^2/R; so that V = sqrt(1650R/m) and R = 3000/2 = 1500 meters and m = 90 kg. HOWEVER...this is only the velocity of the A/C at the bottom of the loop.

b)What force does the seat exert on the man when he is upside down at the top of the loop?

The equations here are similar as for a: F = W - C, but the weight and centripetal forces are acting in opposite, not the same, direction. Also note C = mv^2/R; where v < V because kinetic energy is converted to potential energy PE = mgh as the plane climbs from the bottom of the loop. In this case we note that TE(0) = KE(0) = 1/2 mV^2 = mgh + 1/2 mv^2 = PE(h) + KE(h) = TE(h); where h is the height (diameter) of the loop, and and PE and KE are the potential and kinetic energy of the A/C at the top of the loop (h = 3000m). Solve for v. 1/2 V^2 = gh + 1/2 v^2; so that, 1/2 V^2 - gh = 1/2 v^2 and v^2 = 2[1/2 V^2 - gh].

Solve for v^2 and plug into F = W - C = mg - mv^2/R to find the net force on the seat. You can do the math. [Note, I'm using g ~ 10 m/sec^2 for easier math; use 9.81 when you do it.]

c) What is the minimum speed that the plane can fly at so that the man will feel "weightless" at the top of the loop? (ie when will the seat not exert force on him?)"

Set W = C; so that F = W - C = 0, which is weightlessness. Thus mg = mw^2/R and w^2 = Rg so that w = sqrt(Rg); where w is the minimum velocity needed to fly weightless upside down at the top of the loop. You can do the math.

But what about the physics? It's important to realize that the velocity at the bottom of the loop (V) will be greater than that at the top of that loop (v). That stems from the conservation of energy. As the A/C climbs out of the bottom, it bleeds off airspeed and consequent kinetic energy as the aircraft's potential energy increases. And that's how the total energy of the aircraft is conserved from bottom to top.

PS: The assumption that the airspeed is the same at the bottom and the top of the loop is erroneous. A proper vertical loop has vastly different airspeeds at those two points. And by invoking the conservation of energy law, both airspeeds can be calculated.

Answer 3

The helicopter pictured is the MBB Bo 105CBS belonging to crimson Bull u . s .. right this is an outline from Wikipedia: The 4-blade inflexible important rotor, a worldwide first, with fiberglass blades ensures intense maneuverability. A Bo 105CBS used for promotional purposes via crimson Bull u . s . is completely aerobatic, appearing loops, rolls, Immelmanns and different maneuvers generally regarded as for fastened-wing airplane in basic terms.