Normal force at the top of a vertical loop -- Circular Motion Dynamics

  • #1
Idontknowhatimdoing
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Homework Statement:
I know that when an object moves around a loop, there is a normal force and the force of gravity acting on the object. I am also aware that the normal force is what causes the centripetal acceleration and that at the top of the loop, both the force of gravity and normal force point downward.

My question, then, is what causes this normal force at the very top of a loop if the object is moving faster than the minimum speed required to clear the loop? From what I've learned so far, normal force usually acts as a response to some other force (it's a type of supportive force). What exactly is the normal force at the top of the loop supporting? Is there some other way of thinking about normal force that explains why it appears in this situation?
Relevant Equations:
centripetal force = m * centripetal acceleration --> Fc = m * (v^2)/r
From the equation for centripetal force, I can see that the centripetal force is proportional to v^2. Does this have something to do with why there is a normal force at the top? Does the velocity of the object require there to be a normal force? If so, why is that the case?
 

Answers and Replies

  • #2
kuruman
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I am assuming that the mass is moving on the inside of the loop as opposed to the outside. What "causes this normal force at the very top of a loop if the object is moving faster than the minimum speed required to clear the loop" is the presence of the loop. If the loop were not there, the mass would be subject only to force of gravity and would follow a parabolic path of larger radius of curvature than the radius of the loop. In other words, the loop constrains the mass to follow a tighter circular path. According to Newton's first law the mass will travel in a straight line. Gravity alone modifies that to a parabolic path. Gravity plus normal force modify it the path further to a circle. Because the force of gravity is constant and the same everywhere, the normal force adjusts itself to whatever is necessary for the mass to stay on the circle at every point along the path.
 
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  • #3
DrJohn
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For an aircraft looping, the wings at the top of the loop are now upside down and still generating a force called lift as they fly through the air. They start with enough speed to be sure this will be true and not stall upside down or in the vertical up stage which is very much worse than when at the top of the loop - I have heard people scream when they get that bit wrong while going up.

The lift is at (roughly but definitely not absolutely) 90 degrees pointing up through the wing from the under surface to the upper surface of the wing (when you look at the aircraft on the ground) . But upside down, the lift force is now pointing down. So even with no physical constraint (as is the case when a ball or rollercoaster goes round a loop) an aircraft will fly a loop, if structurally strong enough and permitted to carry out aerobatics.

So all the way round the loop, there is a force at roughly 90 degrees to the direction the aircraft is pointing at any given instant, which constrains the flight path into a circle within the aircraft's frame of reference (the air it is moving through). And that lift is your normal force.

If it's windy a ground observer will see a distorted circular flight path, and in aerobatic competitions, this will loose points against a perfect maneuver. So these pilots, knowing the wind direction before takeoff, adjust how hard they pull back at any given point in the loop to give the illusion that the loop was a perfect circle. Normal pilots just accept that it doesn't look perfect from the ground and enjoy themselves.

Hope I've phrased this clearly enough - familiarity can lead to careless phrases that could be misinterpreted.

Specialist aerobatic aircraft can carry out an outside loop, because their wings have a nearly symmetric cross section and are fitted are almost zero angle of incidence to the fuselage. Normal aircraft can't safely do outside loops, so don't try it.
 
  • #4
Lnewqban
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Welcome @Idontknowhatimdoing !
What the first Newton’s law said about a straight line?
If at any point, the loop suddenly disappeared, how the trajectory would change and why?

P.D.- I now see that @kuruman explained this above, sorry.

Please, see:
https://www.physicsclassroom.com/class/circles/Lesson-2/Amusement-Park-Physics

Yes, the normal force depends on the velocity of the object.
But that normal force is a reaction to the centrifugal force, which is a ficticious force induced by the circular movement.
For the object to press on the loop at its top (not falling from it), its velocity must be such that the centrifugal force (always pointing out the loop) is greater than the gravity force at the top angular position.


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  • #5
haruspex
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what causes this normal force at the very top of a loop if the object is moving faster than the minimum speed required to clear the loop?
The normal force that one body exerts on an adjacent one is the minimum force necessary to prevent interpenetration. Necessarily, that is at right angles to the contact surface; hence the epithet "normal".
 
  • #6
Idontknowhatimdoing
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But that normal force is a reaction to the centrifugal force, which is a fictitious force induced by the circular movement.
This makes sense, but the fact that the normal force can be a reaction to a fictitious force just doesn't sit right with me. Does this mean that you don't need a real force for normal force to respond?
 
  • #7
haruspex
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This makes sense, but the fact that the normal force can be a reaction to a fictitious force just doesn't sit right with me. Does this mean that you don't need a real force for normal force to respond?
You have a choice of any number of reference frames.

If you choose the rotating frame centred always on the object then to correct for its acceleration you have to introduce a centrifugal force on every object in the frame. Within that frame, it is just like a real force; and in that frame the object is not accelerating so something has to balance the centrifugal force.

If you choose an inertial frame then the object has an acceleration normal to its velocity so the net force must have a component normal to the velocity. We call this component the centripetal force, but that is all it is: it is not an additional applied force.

Neither of these views give a causal explanation, so do not appear to answer your original question. The view I gave in post #5 is causal.
 
  • #8
DrJohn
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This makes sense, but the fact that the normal force can be a reaction to a fictitious force just doesn't sit right with me. Does this mean that you don't need a real force for normal force to respond?
Think of spinning a ball around on the end of a bit of string. What happens if you let go of the string? (Newton's First Law)
The string is providing a centripetal force to pull the ball on the end around the corner at every instance. You, the string holder, are providing that force though the string. Forget centrifugal farce. Oops. A real typo, but I am leaving it in place ;-)
If you use a very heavy ball, say 50kg and try to spin it around fast in circles around your head with a thin string, the string will probably break as it can't be pulled hard enough by you before it breaks.

Now let's look a physical loop runways.

Imagine a small steel ball, say 5cm in diameter, that you fire around the inside of a strong metal loop and at a speed high enough to ensure a number of loops, five or ten. Also imagine a thin paper loop where you fire a table tennis ball around, and it successfully performs several loops. Your loops are converting the balls' KE into centripetal force, until the KE drops to too low a value and enough energy simply can't be extracted..

Now imagine you use the steel ball and fire it around your thin paper loop at the same initial velocity as before - the steel ball will burst through your thin paper loop. The paper can't hold back against the momentum and convert enough of the steel ball's kinetic energy into a centripetal force and force it round the track. And as we know, if we fire the steel ball round the metal loop with insufficient energy, it might only manage one loop or even half a loop, then fall back. You can think of these solid physical loops extracting KE from the ball and converting that into the centripetal force. The loop is not reacting to an imaginary force, the loop is converting some of the KE into a real force.

Your science teacher can probably set up an experiment with markers on the strong loop and a timer visible, video the loops, and show that the speed decays faster than if you rolled the same ball at the same initial speed along a long metal surface, to show that the energy loss is greater than that due to friction and the ball doesn't travel as far when looping (you can use the radius of your metal loop to get the distance travelled). That excess energy loss is your centripetal force. You'd use a ramp of a height about say 50% higher than the loop diameter in each experiment to fire the ball into the loop, high enough that at a loop is guarenteed of course, and that the ball has the same KE at the instance it starts to go round the loop as when you use the ramp to roll it along the metal surface.

Hope I've phrased this correctly, if not I am sure some will correct things for me. (I'm currently unwell and mistakes might slip in. I've re-read it several times and corrected several already but just now my brain hurts.)
 
  • #9
Lnewqban
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This makes sense, but the fact that the normal force can be a reaction to a fictitious force just doesn't sit right with me. Does this mean that you don't need a real force for normal force to respond?
I am unable to explain it better than post #7 does.
Imagine that the loop has a huge radius and the imaginary centrifugal force is like artificial gravity.
The, flip that mental picture 180 degrees, and you will have a situation somehow similar to a heavy block resting on a surface: normal force points upwards.
 

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