Reaction force at the top of a rollercoaster loop

AI Thread Summary
In a rollercoaster loop, the centripetal force required for the cart's circular motion is a combination of the cart's weight and the reaction force from the track. The discussion highlights confusion about the origin of the reaction force, emphasizing that the cart exerts an equal and opposite force on the rails due to its acceleration towards the center of the loop. It clarifies that the cart's desire to move in a straight line is countered by the rails, which apply a force to change its direction. The forces act simultaneously, with no cause-effect relationship, challenging the traditional interpretation of Newton's Third Law. Understanding this dynamic is crucial for grasping the physics of rollercoaster loops.
dainckp
Messages
1
Reaction score
0
I understand that in a rollercoaster loop, two forces combine to make the centripetal force required to keep the cart moving in a circular path - the cart's weight, and the reaction force of the track against the cart (except in the special case where the cart's weight alone is sufficient for the centripetal force)

What I don't understand is, why is there a reaction force at all? If the rails are pushing back on the cart, by Newton's Third Law the cart must be in the first place exerting an equal and opposite force on the rails, but I can't see where such a force would come from. The cart's velocity is perpendicular to the rail, and the acceleration is directed towards the centre of the loop, and I was taught there is no such thing as an outward centrifugal force, so why does the cart push on the rails in the first place?

A reaction force yes, but a reaction to what?
 
Physics news on Phys.org
Welcome to PF!

The force is from (or, rather, causes) part of the acceleration: f=ma. Gravity causes the rest.
 
If the rails are pushing back on the cart, by Newton's Third Law the cart must be in the first place exerting an equal and opposite force on the rails, but I can't see where such a force would come from.

The cart wants to go in a straight line but the rails force the cart to move in a curve. eg The rails apply a force on the cart.
 
dainckp said:
If the rails are pushing back on the cart, by Newton's Third Law the cart must be in the first place...
No, not "in the first place". The two equal and opposite forces act simultaneously, and are on the same footing. There is no cause-effect relationship between them, contrary to what the unfortunate naming "action & reaction" suggests.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top