1
$\begingroup$

As I was working on a problem with a particle stuck inside a rolling drum (somewhat similar to that of a bob rotating is vertical circle), I came across a confusion : the normal force experienced by an object placed on the the ground is the result of the force it's weight exerts on the table. But what is causing the normal force on the object inside that drum?

Well the non of the components of weight of the object is exerted on the drum, so what force is the object exerting on the drum for which the drum is giving an opposite normal force to the object? Well is it the pseudo force? But if it is pseudo force, how can a force that doesn't even exist apply force in the drum?enter image description here

$\endgroup$
  • 1
    $\begingroup$ Electrostatic Coulomb force with the important ingredient of the Pauli exclusion principle that electrons cannot be in the same state. $\endgroup$ – Pieter Dec 10 '17 at 12:58
1
$\begingroup$

A normal force is an electrostatic force of repulsion exerted by a surface perpendicularly to the surface.   The force exerted upward on an object resting on a horizontal surface is only one example, but it's the most familiar one because it's the most common example used in science classes when the normal force is studied.  In order for the particle to move in a circle, there has to be a constant force on it toward the center of the circle.   This is true for anything moving in a circle, including, for instance, the moon going around the Earth.   Something has to supply that force.   In the case of the moon, it's the gravitational attraction between the Earth and the moon.   In the case of your particle, it's the surface of the barrel, and the force is a normal force. 

According to Newton's third law, the particle must exert an equal force on the barrel.   This is also a normal force.   It is a real force, not a pseudo force.   The pseudo force in this situation, the so-called centrifugal force, is the "force" which appears be pushing it away from the center (pretend that you are the particle and your eyes are closed.   You have no way of knowing that you are being forced to move in a circle.   You feel that you are being pushed against the barrel, but you are not.   You are being constantly pushed by the barrel into a circular motion instead of the straight line motion you would have without the push toward the center).

$\endgroup$
0
$\begingroup$

Usually, if a particle is on a surface there's some kind of reaction force applying that constraint. In order for the particle to stay in circular motion inside the drum, it must have some force compelling it to do so. This is due to Newton's first law, which is conveniently represented by centrifugal pseudoforce.

$\endgroup$
0
$\begingroup$

There are two possible interpretations of the question:

  1. What is the microscopic origin of the normal force?
  2. If we accept the existence of contact forces, why would the rolling drum apply a normal force (inwards)?

Since I am not at all equipped to answer the first interpretation, and because the question concerns a particular system, I will answer the second interpretation.

Brief heads up for my answer: I will proceed via logical argument, by stating my assumptions and deducing from them that a normal force must be applied by the drum. However, the assumptions I will make (and not the deduction) are the key to the answer, since the normal force will be (almost) a trivial consequence. Also, my answer is a little verbose, but I feel it is useful to examine the assumptions we make when dealing with mechanical systems.

tl;dr the drum applies a normal force to keep the particle inside.


The notion we have from studying configurations in static equilibrium, is that a normal force is applied only to oppose an applied force, to maintain equilibrium.

The scenario in the question, however, is not one of static equilibrium. This creates a question for us: should there still be a normal force exerted by the drum, even though there is no force directly opposite it?

I am assuming that the drum is a rigid constraint: it does not deform and it does not allow objects inside it to go outside by going through the walls.

I am also assuming that the drum is frictionless. More explicitly, it means that any force that the drum applies cannot have a tangential component, and that if the drum does apply a force, it has to be normal to its surface.

Let's assume then that the drum does not apply any force. If the particle, at the position $\theta$ shown, is given a velocity tangent to the drum or with a component along the outward normal, it is easy to imagine that it will, with the only force on it being gravity, leave the confines of the drum.

But since we have posited that the drum is a rigid contraint, this trajectory is a contradiction. Since by Newton's second law, the trajectory is completely by the forces acting on our particle (along with its initial position and velocity) there must be another force acting on it. Since the drum is the only other object in the system, it must be applying a force on the particle, somehow. Since it is frictionless, it must be applying a normal force.


The normal force here is an example of a constraint force. Perhaps part of your confusion is the lack of a physical reason for constraint forces, the way there is for gravity, and the way a reason can be formulated for statics (constraint forces oppose applied forces).

The reason they seem to be an ad-hoc addition to our system, just something that is there because it agrees with experiment without any fundamental theory that tells us it must be there, is because they are ad-hoc additions to our system because we have prior knowledge of our system's evolution. We know, for example, that a particle rolling on the inside of a drum does not go through the walls of the drum, and the drum does not deform (to an observable extent).

Within the framework of Newtonian mechanics, without detailed models of rigid bodies that tell us how contact forces originate from fundamental forces between elementary particles, constraint forces are simply another ingredient we must add.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.