action reaction forces: what is the reaction to tension?

Ken G

I would have agreed at any time, including right from the start.I still say that an action/reaction pair is always the same type, and I see no contradictions to this widely held view. I don't even see why you seem to see a contradiction. I see nothing "special" about the pair at the interface that distinguishes it from the pairs throughout the rope, which is exactly why they are all regarded as "tension."

Ken G

No, I'd say you have it exactly.
Yes, picturing this is actually one of the values of the concept of "centrifugal force" (which is still a useful concept you see invoked all the time). The centrifugal force is what we call a "coordinate force" (which is a better term than "fictitious force", which is somewhat denigrating!), which means it is a term that acts as though it were a force, but in a noninertial reference frame.

It stems from our choice of a rotating coordinate system, and is there because we need to include it to get coordinate velocities that don't change with time in the absence of any real forces. By "coordinate velocity", I mean rate of change of a distance component in our rotating coordinate system, which is not the same as a real velocity (which is a vector and is coordinate independent). For example, the rock on the string has a coordinate velocity of zero in the rotating coordinates-- its coordinate is staying the same, so it is not "moving" at all in those coordinates.

But you well know that to have a rotating coordinate velocity stay zero, it must have a real net force on it, called the centripetal force. The point of the centrifugal force is to be able to imagine we have a force balance, since the coordinate velocity is not changing, so that there's no acceleration in the coordinate velocity components. So if we manually tack on the centrifugal force in our rotating coordinates, and include it in F in F=ma, we can treat a as the coordinate acceleration, and not worry that having zero coordinate acceleration does not mean the object isn't really accelerating. In short, centrifugal force is a way to pretend that F=ma pertains to coordinate acceleration a, whereas if we limit F to real forces, then F=ma only pertains to the coordinate-free vector acceleration (which you might call the real acceleration).

The reason I'm resuscitating the centrifugal force in this way is that it is very useful for understanding the behavior of the spinning rock, which indeed does seem mysterious seen from our inertial perspective. Once the rock reaches its steady-state revolution, we can enter the rotating frame, and draw an effective free-body diagram, tacking on the centrifugal force, and use it to completely understand why the rock does what it does, exactly as if the rock was not revolving at all, and had a real force on it that is the same as the centrifugal force. It's a way to make F=ma work the same in a rotating frame as it does in a non-rotating frame, so if you can understand a stationary rock's behavior in a non-rotating frame with a real force on it, you can understand its behavior in the rotating frame with a centrifugal force on it, as the two situations are equivalent.

Andrew Mason

The reaction to the centripetal force on the object is the object's force on the earth. Both forces are communicated between the object and the earth via the rope. Since the rope is considered massless its function is only to conduct the forces between the earth and the object.

In any description of the forces involved in the motion of macroscopic bodies you will have forces which result in acceleration of macroscopic bodies and tensions within those bodies and the objects that connect them. To analyse them, you have to separate the tensions from the forces that create accelerations.

A better example would be two masses tethered by a rope of negligible mass and each rotating about the centre of mass of the two-body system. There are two accelerating bodies and equal and opposite forces between them. You do not have to worry about the tension as a separate force but merely as the means by which the rotating masses interact.

AM

fisico30

Hello Andrew,
thanks for your input. I think I follow your reasoning: the pivotal point is connected to the rope which is connected to the rotating object. The rope (massless) exerts a force (centripetal) on the object. The object, since the rope is massless, acts on the Earth, which is connected to the pivotal point.....

Surely, if a person holds the rope, the person is connected to Earth (by standing on it), so eventually everything that is not flying is connected to Earth.

How do you explain that the object, with its sideway motion, is able to cause that force?
I tend to think that a force is caused by an object when it tries to move in a certain direction (push or pull) such that the resulting force is along the direction of motion of the object. In the case of the rotating object, the object moves along the tangent to the circle while the reaction force is along the radial direction...

thanks,
fisico30

Andrew Mason

Newton's third law.
That is why we study Newton's laws of motion. Newton showed that the net force on a given body is always in the direction of the change in motion. The direction of the change in motion is always toward the centre of rotation.

AM