Understanding the Minimum Speed to Keep Carriage on Tracks in a Loop

In summary, the normal force must always act towards the centre of the circle loop in order for the rail to exert a pushing force on the carriage and keep it on the trajectory. In Q1 and Q2, the reaction force must be greater than or equal to zero for the carriage to maintain a circular trajectory. This is because in the limiting case of minimum speed, the centripetal force must be equal to or greater than the weight of the carriage to keep it on the track. However, it is assumed that the normal force is zero in this case, which raises the question of whether the passenger and seat would be in physical contact. This is because if the normal force is zero, everything in contact with the track would be in free
  • #1
User1265
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1
Homework Statement
Carriage of rollercoaser about to enter a vertical loop of diameter 17m. The carriage is initially at rest delta h above the bottom of the loop. Such that the passenger remains in contact with their seat, calculate the minimum speed of the car at the top of the loop.
Relevant Equations
mg
= mv^2/R
I recognise that the normal force must alwayss act towards the centre of the circle loop, as the rail always has to be exertign a pushing force on the car/carriage in order for it to follow the trajectoryof the loop. However , I cannot understand why, the reaction force has to be greater than or equal to zero in Q1 and Q2.
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My solution :

F net = mv2/R = Fnormal + mg

where m - mass of whole carriage (inc. seat + passenger)

then Fnormal greater than or equal to 0
--> for, mv2/R to be greater than mg , to allow a weight of value mg going round the top of the tracks.

Otheriwse

If mv^2/R <mg - hoz velocity at the top of track will be too small to continue on the track and the carriage will fall off initally a projectile manner
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I understand that for the whole carriage ( including the passenger and seat) to stay on the track, then the centripetal force in the limiting case of minumum speed, would have to be at least equal to the weight of the whole carriage in order to produce a great enough speed that will keep the trajectory (of that weight) at that instant - thus staying on the tracks at that instance of the carriage at the top of the loop. But this is assuming the only force acting in radial plane is weight , as normal force is assumed to be zero in the limiting case.

But Q1) is why can the normal force be assumed to be zero in the case of slowest speed to keep a circular trajectory?

With a zero Normal force, would passenger and seat not be in physical contact? So wouldn't the minimum speed really need to be slightly greater than Fn =0 such to keep physical contact between seat and passenger, or is it that lack of physical contact is negligble , and thus even with Fn=0 can the smallests safe, velocity be produced in reality.

Q2)
Also,
Surely if the seat exerted an upwards force on the passenger, to somehow stick onto the seat, isn't the condition that the passenger remains in contact with their seat is still satisified , regardless of whether the centripetal force (net force radially) is too small (<weight whole carriage) to produce a sufficient velocity to keep the whole carriage on the track at the top of the loop, and falling off.

So surely the minium speed at the top can be zero, and the seat and the passenger can both fall at the same rate, where lack of physical contact is again neglible.
 
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  • #2
User1265 said:
With a zero Normal force, would passenger and seat not be in physical contact?
They would not. Everything that is in contact with the tract will be instantaneously in free fall. Does that bother you? For the same reason that the carriage will continue on after the normal force is instantaneously zero, the passenger's rear end will continue on. Of course we are assuming that carriage and passenger are point masses, i.e. at the same radius from the center of the circle.

Rethink question 2 in view of the above.
User1265 said:
So surely the minium speed at the top can be zero, and the seat and the passenger can both fall at the same rate, where lack of physical contact is again neglible.
Not so "surely". If the speed at the top were (even instantaneously) zero it's as if the carriage and the passenger are released from rest. They would drop straight down through the center of the circle.
 
  • #3
kuruman said:
They would not. Everything that is in contact with the tract will be instantaneously in free fall. Does that bother you? For the same reason that the carriage will continue on after the normal force is instantaneously zero, the passenger's rear end will continue on.

Thanks for the reply. I just don't understand how that means a carriage in contact with the tract will be instanteously in free fall, as surely if it is in physically in contact with the tract, there must be a force exerted by the tract, as the normal force is a contact force, so the net acceleration of the carriage is not g.
 
  • #4
User1265 said:
Thanks for the reply. I just don't understand how that means a carriage in contact with the tract will be instanteously in free fall, as surely if it is in physically in contact with the tract, there must be a force exerted by the tract, as the normal force is a contact force, so the net acceleration of the carriage is not g.
One has to think of the situation as being a limiting case. You must think of the carriage as being on the ##verge## of losing contact. So basically the normal force is zero but just at the very top.
 
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  • #5
You can think of it as writing the equations for the constrained circular motion, and then finding the critical values for when the constraints can no longer be satisfied.

So at an arbitrary point on the loop, with ##\theta## measured from the downward vertical, you might say$$N - mg\cos{\theta} = \frac{mv^{2}}{r}$$We also impose the constraint that ##N\geq 0##, if we assume the rollercoaster is not bound to the track! Of course, if we had a situation like a bead on a wire, we might indeed permit negative values of ##N## because the normal force could act in both directions! But I'm assuming that isn't the case here.

As the rollercoaster goes up the loop, the normal force will decrease and decrease until eventually it will reach zero. That is, the radial component of the weight force becomes sufficiently large to fulfil the entire requirement for the centripetal force. If it carried on for an infinitesimally small amount of time, ##N## would become negative (as if to pull the rollercoaster back toward the track) - but that's not allowed! The result is that the motion must become unconstrained.

You want this to happen at the top of the loop, so you can set ##(N, \theta) = (0, \pi)## in the above equation.
 
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  • #6
User1265 said:
Thanks for the reply. I just don't understand how that means a carriage in contact with the tract will be instanteously in free fall, as surely if it is in physically in contact with the tract, there must be a force exerted by the tract, as the normal force is a contact force, so the net acceleration of the carriage is not g.
You release a ball from rest at 10 cm below a ceiling. What is the normal force exerted by the ceiling on the ball?
Answer: Zero because the ceiling exerts no force on the ball and, by Newton's 3rd, the ball exerts no force on the ceiling. In other words, if the ceiling cannot tell the ball (through the normal force) that it is not there, the ball cannot tell the ceiling (again through the normal force) that it is not there. How would that change if the ball were held just below the ceiling without exerting a force on it? No change.

Lesson to be learned: If the force (normal or otherwise) between two objects is zero, as far as one object is concerned, the other might as well not exist because it has no way to communicate its presence.
 
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Related to Understanding the Minimum Speed to Keep Carriage on Tracks in a Loop

1. What is the minimum speed required to keep a carriage on tracks in a loop?

The minimum speed required to keep a carriage on tracks in a loop depends on various factors such as the radius of the loop, the weight of the carriage, and the friction between the wheels and the tracks. However, in general, the minimum speed is calculated using the formula v = √(rg), where v is the minimum speed, r is the radius of the loop, and g is the acceleration due to gravity.

2. Why is minimum speed important in a loop track?

Minimum speed is important in a loop track because it ensures that the carriage stays on the track without derailing. If the carriage goes too slow, it may not have enough centripetal force to maintain its circular motion and will fall off the track. On the other hand, if the carriage goes too fast, it may experience excessive centrifugal force, causing it to derail.

3. How does the weight of the carriage affect the minimum speed in a loop track?

The weight of the carriage affects the minimum speed in a loop track because it determines the amount of centripetal force needed to keep the carriage on the track. A heavier carriage will require a higher minimum speed to generate enough centripetal force, while a lighter carriage may require a lower minimum speed.

4. Can the minimum speed be different for different types of trains?

Yes, the minimum speed can be different for different types of trains. This is because different trains have different weights, sizes, and designs, which can affect their centripetal force and minimum speed in a loop track.

5. How does friction play a role in determining the minimum speed in a loop track?

Friction plays a crucial role in determining the minimum speed in a loop track. Friction between the wheels of the carriage and the tracks provides the necessary centripetal force to keep the carriage on the track. Without enough friction, the carriage may not have enough centripetal force and will derail, even at high speeds.

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