Uniform circular motion problems

AI Thread Summary
The discussion revolves around solving uniform circular motion problems involving a pilot experiencing an upward acceleration of 8.5g. The key calculations include determining the force exerted by the airplane seat on the pilot and the radius of the circular arc. The centripetal force is derived from both the pilot's mass and gravitational force, leading to a total force calculation of approximately 4659.75 N. For the radius, using the formula r = v²/a, the correct radius is found to be around 110 meters. The conversation emphasizes the importance of understanding the relationship between centripetal acceleration, gravitational force, and the forces acting on the pilot.
smally
Messages
9
Reaction score
0
[SOLVED] Uniform circular motion problems

Homework Statement


A pilot of mass 50kg comes out of a vertical dive in a circular arc such that his upward acceleration is 8.5g

Homework Equations


a) What is the magnitude of the force exerted by the airplane seat on the pilot at the bottom of the arc?
b) If the speed of the plane is 345 km/h, what is the radius of the circular arc?

The Attempt at a Solution


I've done F = ma, so 50 x 8.5 = 425, which is massively as the answer is 4659.75N

At the bottom of the ark the upward acceleration is the same as the magnitude of its centripetal acceleration (I think) therefore r = v^{2} / a, which comes to 14002.94, which is also massively wrong as the answer provided is 110m

I tried alsorts of other ways too but they never match the answers provided, what am I missing.
 
Physics news on Phys.org
You're not too far off

So draw the plane and all the forces acting on the poor pilot(who is pulling nearly 9 Gs and probably about to black out or rip his plane apart)

The centripetal FORCE acting on him is his mass times the centripetal acceleration(which is given in the problem as you correctly saw). What is the CAUSE of the centripetal force? ie what is pushing on him that hard? The seat. So the centripetal force is part of the answer. Is that all?

Also just good 'ol gravity is acting on him, pushing him straight down. That means the seat is exerting an equal normal force

The answer will be Fcent+Fnormal=Fouch
 
Ok so I've now done:
50 x 9.81 x 8.5 = 4169.25
50 x 9.81 = 490.5
4169.25 + 490.5 = 4659.75

but I'm a bit confused as to why I times gravity for when working out Fcent.

I also assumed that the answer to a, would help me with b; but I'm still getting the answer wrong
 
g represents an acceleration, it's in units of acceleration. A plain old unitless number(8.5) times g is also going to be an acceleration

It is customary to say the acceleration in terms of g, they could say "oh it's 83.3 m/s^2" or they could say "oh it's 8.5 times the acceleration caused by gravity" which immediately tells you he's being pushed into his seat with 8.5 times more force than normal, and pilots normally have a good sense of what the human body can stand(even trained fighter pilots don't pull more than 7 or 8 gs for just seconds)and their plane can stand

For b)You found the centripetal acceleration as PART of a(it wasn't the final answer!), and you're given v, and you know r=v^2/a
 
Oh right, the "he's being pushed into his seat 8.5 times more force than normal" got lightbulb above my head appear then thanks.

and for b)
r = 95.83^2 / (9.81 x 8.5)
r = 110.13

correct, thank you
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top