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Loops the loop! need help

  • Thread starter Sirsh
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  • #1
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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 im 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 = [tex]\sqrt{rg}[/tex], 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.
 

Answers and Replies

  • #2
PhanthomJay
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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 im 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 = [tex]\sqrt{rg}[/tex], 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.
 
  • #3
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okay ill give that a go, thanks alot
 
  • #4
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Also, does this mean that 5g's will be affecting the pilot?
 
  • #5
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how do we know the loop's orientation? is it a circle parallel to the ground, perpendicular to the ground, or some other orientation?
 
  • #6
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It's a vertical circle
 
  • #7
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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!)
 
  • #8
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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?
 
  • #9
PhanthomJay
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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.
 
  • #10
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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 im right, correction would be apprechiated if i am.
 
  • #11
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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
 
  • #12
cepheid
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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.
 
  • #13
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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
 
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  • #14
cepheid
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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.
 
  • #15
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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
 
  • #16
PhanthomJay
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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.
 
  • #17
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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
 
  • #18
PhanthomJay
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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...
 
  • #19
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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, theres 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.)
 
  • #20
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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
 
  • #21
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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, theres 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!
 
  • #22
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Holezch, this is my question.. lol
 
  • #23
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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!
 
  • #24
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6mg = mv^2/r dont you mean?
 
  • #25
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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??
 
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