Calculating Loop Radius for Force of 5x Pilot Weight on Plane

[Sirsh]

A plane flying at 720 kilometres per hour loops the loop. What must the radius of the loop be so that the maximum force of the pilot on the seat is 5 times her weight?

Just wondering how if I am doing this correctly (:

r = ? v = 200m/s g = 9.8m/s^2 m = ?

So, at the top of the loop Reaction force = 0 therefore

R = 0, mv^2/r = mg

v = $$\sqrt{rg}$$, r = v^2/g

Therefore; Radius = 200^2/9.8 = 4.081km.

After finding this how would i find the force exerted on pilot?

Thanks.

 

[PhanthomJay]

A plane flying at 720 kilometres per hour loops the loop. What must the radius of the loop be so that the maximum force of the pilot on the seat is 5 times her weight?

Just wondering how if I am doing this correctly (:

r = ? v = 200m/s g = 9.8m/s^2 m = ?

So, at the top of the loop Reaction force = 0 therefore

R = 0, mv^2/r = mg

v = $$\sqrt{rg}$$, r = v^2/g

Therefore; Radius = 200^2/9.8 = 4.081km.

After finding this how would i find the force exerted on pilot?

Thanks.

This problem is not worded clearly. Presumably, the plane starts the loop and maintains constant speed while in the loop (it must adjust it's engine thrust to do so). The normal force is not necessarily zero at the top of the loop. But it is given that the normal force is 5mg as it enters the loop (presumably from the bottom). Regardless, set the normal force equal to 5mg at the bottom, draw the free body diagram, and use Newton 2 to solve for r.

 

[Sirsh]

okay ill give that a go, thanks alot

 

[Sirsh]

Also, does this mean that 5g's will be affecting the pilot?

 

[xcvxcvvc]

how do we know the loop's orientation? is it a circle parallel to the ground, perpendicular to the ground, or some other orientation?

 

[Sirsh]

It's a vertical circle

 

[inutard]

If it is a vertical loop, the maxima of normal force occurs at the bottom of the loop (as the normal force has to counteract gravitational force). A free-body diagram should show you the exact proportions (to g) of these forces. After this, one could just use the simple formula of a = v^2/r to calculate for the radius (Assuming velocity is constant of course!)

 

[Sirsh]

Hey inutard, Thanks for the information on the max Nforce, didnt know that! Okay, because the question states that the max force is 5times the pilots mass i would have thought that it means that the normal force is 5 times greater, and what this means to ME is that it is 5g's. therefore when using the formula to figure out the radius.. a = v^2/r . a would just = 5*9.8m/s^2? i did the calculations with this and it seems logical. what do you think mate?

 

[PhanthomJay]

Also, does this mean that 5g's will be affecting the pilot?

The term 5g's means that that the force exerted on the pilot (by the seat) is 5 times his weight. The pilot's acceleration , however, is not 5g. His acceleration is dependent on the net force acting on him. There is another force acting on him besides the seat force. Please identify it and use Newton 2 to solve for r.

 

[Sirsh]

The term 5g's means that that the force exerted on the pilot (by the seat) is 5 times his weight.

That's what i meant sorry, wouldn't that be like: F = 5*(mg) ? but then 2nd law says that if the force is 5 times as much, the acceleration will be directly proportional, therefore 5 times as great. So i thought that becase mv²/r = 5*mg and mass cancles out, so, v²/r = 5*g therefore r = v²/5*g ? or r = v²/5*a which i'd think because a is inrespect to gravity which is 9.81m/s^2.

Im not sure if I am right, correction would be apprechiated if i am.

 

[holezch]

This problem is not worded clearly. Presumably, the plane starts the loop and maintains constant speed while in the loop (it must adjust it's engine thrust to do so). The normal force is not necessarily zero at the top of the loop. But it is given that the normal force is 5mg as it enters the loop (presumably from the bottom). Regardless, set the normal force equal to 5mg at the bottom, draw the free body diagram, and use Newton 2 to solve for r.

how is it given that the normal force is given to be 5mg at the bottom ? :S is the max. normal force going to be when he is at the bottom? Then, wouldn't the kind of loop the pilot will go through be arbitrary? thanks

 

[cepheid]

how is it given that the normal force is given to be 5mg at the bottom ? :S is the max. normal force going to be when he is at the bottom? Then, wouldn't the kind of loop the pilot will go through be arbitrary? thanks

holezch,

Any time the centripetal force has an upward component, the normal force felt by the pilot (from the seat pushing upward on him) will be larger than his weight. Since, at the bottom of the loop, the centripetal force points entirely upward, it is at this point that the pilot will "feel" heaviest. Therefore, by saying that the normal force should never exceed 5mg, the problem does indeed specify that this is what the force should be at the bottom of the loop.

As to your second question, on the contrary, this condition does constrain the geometry of the loop very precisely. By requiring that the normal force at the bottom of the loop doesn't exceed a certain amount, an upper limit is being placed on the magnitude of the centripetal force. Therefore, since m and v are both set to specific values as well, this places a lower limit on the radius, r, of the loop. If the pilot tries to make a tighter turn than that, he'll experience more than 5g's.

 

[holezch]

holezch,

Any time the centripetal force has an upward component, the normal force felt by the pilot (from the seat pushing upward on him) will be larger than his weight. Since, at the bottom of the loop, the centripetal force points entirely upward, it is at this point that the pilot will "feel" heaviest. Therefore, by saying that the normal force should never exceed 5mg, the problem does indeed specify that this is what the force should be at the bottom of the loop.

As to your second question, on the contrary, this condition does constrain the geometry of the loop very precisely. By requiring that the normal force at the bottom of the loop doesn't exceed a certain amount, an upper limit is being placed on the magnitude of the centripetal force. Therefore, since m and v are both set to specific values as well, this places a lower limit on the radius, r, of the loop. If the pilot tries to make a tighter turn than that, he'll experience more than 5g's.

thanks a lot, so ultimately, the net force is centripetal. and that means it may be expressed as m(v^2/r)... then, at the bottom, we have to know all the forces (the net force) and equate it to m(v^2/r).. how should I find all the forces?
thanks

 

[cepheid]

thanks a lot, so ultimately, the net force is centripetal. and that means it may be expressed as m(v^2/r)... then, at the bottom, we have to know all the forces (the net force) and equate it to m(v^2/r).. how should I find all the forces?
thanks

Er, well, I don't think the net force is purely centripetal. There are two forces acting on the pilot. One is the centripetal (a.k.a. normal) force, and the other is gravity.

 

[holezch]

hm, okay, I assumed that the 720 km/h was a constant speed throughout the looping.. if it were , then the net force would be centripetal.. since that normal force IS our centripetal force, can we just say 5mg = m(v^2/r) and solve for r? thanks

 

[PhanthomJay]

hm, okay, I assumed that the 720 km/h was a constant speed throughout the looping.. if it were , then the net force would be centripetal.. since that normal force IS our centripetal force, can we just say 5mg = m(v^2/r) and solve for r? thanks

No. Newton 2 tells us that F_net =ma, not F_n =ma.

 

[holezch]

okay, so we know that there's the normal force and gravity, that means ma - mg = m(v^2/r) = 5mg? but I'm confused about a couple of things.. isn't the normal force and the weight in equilibrium? the plane would be "stationary" at the bottom of the loop, floating in the air.. thanks

 

[PhanthomJay]

okay, so we know that there's the normal force and gravity,

Yes!

that means ma - mg = m(v^2/r) = 5mg?

what law is this?
F_net =ma,where F_net is _____ and 'a' is the centripetal acceleration of the plane caused by it's curved motion.

but I'm confused about a couple of things.. isn't the normal force and the weight in equilibrium?

the plane can't be in equilibrium if it is accelerating inwardly

the plane would be "stationary" at the bottom of the loop, floating in the air.. thanks

It has no vertical velocity at that instant,but since it's velocity is changing direction with time, it must be accelerating, inwardly, toward the center of the loop...

 

[inutard]

isn't the normal force and the weight in equilibrium? the plane would be "stationary" at the bottom of the loop, floating in the air.. thanks

aha! but that is the crux of the matter! the plane is not stationary in an inertial frame of reference. there is an upward NET force on the pilot (which, in the modern sense, is taken to be THE centripetal force). How can the plane travel in a loop otherwise? despite the normal force, there's always going to be an mg ("weight") force downwards. so the net force (or THE centripetal force) is 5mg - mg = 4mg! Of course, in some senses, the normal force is ONE centripetal force, but note that the formula v^2/r = a only applies to THE centripetal force (aka the net force.)

 

[holezch]

okay, I thought earlier that the net force was centripetal, isn't that true? since in the end, the acceleration from the net force is calculated by v^2/r ? then, Fnet = -mg + normal force? and the normal force is 5mg? then we solve for r

 

[holezch]

aha! but that is the crux of the matter! the plane is not stationary in an inertial frame of reference. there is an upward NET force on the pilot (which, in the modern sense, is taken to be THE centripetal force). How can the plane travel in a loop otherwise? despite the normal force, there's always going to be an mg ("weight") force downwards. so the net force (or THE centripetal force) is 5mg - mg = 4mg! Of course, in some senses, the normal force is ONE centripetal force, but note that the formula v^2/r = a only applies to THE centripetal force (aka the net force.)

thanks a lot!

 

[Sirsh]

Holezch, this is my question.. lol

 

[_wolfgang_]

What if you had to find the force at the top of the loop would you use 6mg+(mv^2/r)?? because the weight force is acting downwards now?? Sorry Sirsh its everyones question now!

 

[Sirsh]

6mg = mv^2/r don't you mean?

 

[_wolfgang_]

hmm yeah i think i may have answered my own question since it is at the top of the loop now we add weight force to the centripetal force which is in turn 4mg so it would be 5mg then??

 

[Sirsh]

pretty sure fc is negative also R is negative but W is positive. so... -Fc = -R + W or.. sign conversion or wtv makes it Fc = R - W i think..

 

[PhanthomJay]

This is getting a bit mixed up, partly because the problem statement is a lttle unclear. It says nothing about constant speed , nor even if it's in a vertical circle with constant radius. So one can only make assumptions as to what is happening at the top of the loop. But anyway, I haven't yet seen anyone calculate the radius of the loop at the bottom of the path, nor the centripetal acceleration at the bottom of the loop,nor the net centripetal force at the bottom. Takers, anyone? Only after solving that can you address what might be happening at the top, with assumptions of constant speed and constant radius. The normal force at the bottom is 5mg (given); it is less than that at the top.

 

[Sirsh]

Fc = R - W

Therefore mv^2r/r = 5mg - mg

So, mv^2/r = 4mg, mass cancles.

v^2/r = 4g, r = v^2/4g

so (800/3.6)^2 / (4*9.8) = r , so r = ~1260m. ??

 

[Sirsh]

pretty sure fc is negative also R is negative but W is positive. so... -Fc = -R + W or.. sign conversion or wtv makes it Fc = R - W i think..

my bad _wolfgang_

thats for bottom ;D

 

[PhanthomJay]

Fc = R - W

Therefore mv^2r/r = 5mg - mg

So, mv^2/r = 4mg, mass cancles.

v^2/r = 4g, r = v^2/4g

so (800/3.6)^2 / (4*9.8) = r , so r = ~1260m. ??

You are correct that the net centripetal force at the bottom is 4mg, and the centripetal acceleration at the at the bottom is 4g. Now we already established (given) that v = 720 km/hr = 200m/s. So what's that 800/3.6 for the v value?

 

[Sirsh]

Oops. there is an identical question to this, forgot different Velocity. so we would substitute 200ms into it instead of (800/3.6) so we'd end up getting.. ~1020m :)

 

[PhanthomJay]

Oops. there is an identical question to this, forgot different Velocity. so we would substitute 200ms into it instead of (800/3.6) so we'd end up getting.. ~1020m :)

Now you've got it! I'll leave it up you as an academic exercise if you what to assume something to see what may be happening at the top of the loop. Note that in more typical loop the loop problems, like yoyos in a vertcal circle , or loop the loop roller coasters , the speed is far from constant, so don't take this problem as a typical example of vertcal circular motion problems.