# How does an accelerating mass on a ring interact with the ring?

In outer space, imagine a mass $$m$$ on a ring with mass $$M$$. The small mass is given a velocity $$v$$ after which it moves without friction on the ring. Initially, the velocity of the center of mass (COM) of the ring is zero.

If the small mass moves along the ring, it exerts a force on the ring. This centrifugal force is directed radially outward. The ring in its turn exerts a centripetal force on the small mass. The ring starts to move because of the centrifugal force imparted on it. Because no other forces are acting on the system mass-ring, the total momentum of the system is constant.

1. Now, because the ring will gain momentum due to the centrifugal force, this means that the velocity of the moving mass must decrease, in order to keep the total momentum constant. But how can the velocity of the mass decrease if the force on it (the centripetal force, given to it by the ring) is perpendicular to its velocity?
2. It's not so hard to imagine how a ring with a rotating mass on it behaves in outer space. The mass moves with a constant velocity and lets the COM of the ring rotate. But what happens when we set the mass in motion?

@kb314 made a clear comment. When the mass is accelerated from zero to its final velocity, the reaction force of the ring contains a tangential component (it has to, due to conservation of energy(?)) which decelerates the mass. Equilibrium hasn't been reached yet. I can't see though how this tangential force is imparted on the mass. There is no friction, so...
It seems clear that you can't give the mass an instantaneous increase in speed (from zero to the final value). In the mind one can, but not in practice. In practice, the mass is given a pulsed acceleration. So, what's happening?

It can be explained because it's momentum in a specific direction that we should consider:

"Now, because the ring will gain momentum due to the centrifugal force, this means that the velocity of the moving mass must decrease, in order to keep the total momentum constant. But how can the velocity of the mass decrease if the force on it (the centripetal force, given to it by the ring) is perpendicular to its velocity?"

If we draw the ring of mass $$M$$, looking at it from above, as a circle on the paper, with the mass $$m$$ on the bottom. The ring is subject to a force that increases it's momentum down the page, but the mass $$m$$ is subject to a force that increases it's component of momentum up the page.

It's true that the force up the page doesn't change the left/right component of velocity, but the component in the up/down direction can change.

You are right that a force perpendicular to the velocity of a mass would not decrease the velocity, but as the ring is moving too, the force on the bead isn’t always perpendicular to it’s velocity and the bead doesn’t move in a perfect circle. Even without friction this could occur, as the position of the ring pulling on the bead is changing.

More detail on this below:

If we draw a circle for the ring on paper, centre O, with the bead’s starting point at the bottom B, OB is down. Then the bead goes clockwise $$45^\circ$$ around the ring to a point A.

We can draw the force on the bead, perpendicular to the ring, along AO (arrow 1, A1). We can draw the velocity of the bead (as arrow 2, A2) by starting at A and faintly drawing components 2cm left, then 2cm up, to find the end of A2. Then A2 is perpendicular to A1.

But A2 is the velocity relative to the ring. The ring has the following components of velocity, when the bead is at A. The beads left component of velocity has reduced (compared to at B), so the ring gains a left component (due to conservation of momentum, left/right). The beads component up the page has increased, so the ring has a component down the page, (conservation of momentum, up/down).

So to find the beads velocity relative to the observer (A3), we adapt A2, now starting at A, but adding components due to the rings motion, so now (for example) 1cm extra left and 1cm extra down, making A3, 3cm left and 1cm up.

We can now see that A1 is no longer perpendicular to A3 and could slow down the bead, even though the force acts towards O.

(a diagram of this is here)

• But what about energy conservation? I forget to mention this in my question. If the mass has initial velocity $v$ and the ring has initial velocity zero, how can the speed of the mass decrease when the speed of the ring increases from zero to the final value? Apr 8 '21 at 23:54
• @DescheleSchilder after the initial time instant, we agree that the speed of the bead or small mass $m$ must decrease in order to admit an increase in the speed of the ring $M$, as per the law of conservation of energy. The only way this can happen in the transient phase of motion, is if forces internal to the system of the two masses transfer energy between the two. Indeed, the contact reaction force acted by $M$ on $m$ is not along the radius with the non-normal component caused by the accelerated ring pushing into the bead of small mass as it accelerate. Apr 9 '21 at 2:01
• For the question “how can the velocity of the mass decrease if the force on it (the centripetal force, given to it by the ring) is perpendicular to its velocity”…as the ring is moving too, the force on the bead isn’t always perpendicular to it’s velocity and the bead doesn’t move in a perfect circle. Even without friction this could occur, as the position of the ring pulling on the bead is changing. Apr 9 '21 at 8:54
• it's been added to the answer Apr 9 '21 at 8:56
• @kb314 Thanks, maybe it can be understood like this…Part1: If we draw a circle for the ring on paper centre O with the bead’s starting point at the bottom B, OB is down. If the bead has gone clockwise $45^\circ$ around the ring to A. We can draw the force on the bead, perpendicular to the ring, along AO (arrow 1, A1). We can draw the velocity of the bead (as arrow 2, A2) by starting at A and faintly drawing components 2cm left, then 2cm to find the end of A2. Then A2 is perpendicular to A1. Apr 10 '21 at 9:51

When we reach a stationary state, with the mass $$m$$ moving with a speed $$v$$ along the ring, it is true that the pair of forces on the contact mass and ring are radial.

But in the initial condition, that is for $$t=0$$, when mass $$m$$ starts to move tangentially, the contact force is zero. There is no centrifugal or centripetal force.

If we consider the initial velocity along the axis $$y$$ ($$0), for any $$t = \epsilon > 0$$ there is a horizontal component and a vertical component of the normal force. The vertical component is the force responsible for the vertical deceleration of the mass $$m$$. And the vertical force of the mass $$m$$ on the ring accelerates it in the $$y$$ direction.

After a transient time of exchanging momentum, the COM of the system ring and mass will be moving vertically so that $$mv_{my} = (M+m)v_{COM}$$, where $$0 is the speed of the COM of the system along the $$y$$ axis.

• This is answer contains a good qualitative analysis. However, it seems clear that the speed of the COM of the system is given by the linear momentum conservation equation $mv_{my}=(M+m)v_{COM}$ at all time instants and not just at equilibrium. Apr 9 '21 at 2:02
• @kb314 yes, in the case there is no acceleration step, and $v_{my}$ is constant since $t=0$. Apr 9 '21 at 14:10

This answer is arrived at as a summary of the discussion in the comments to the answer by @JohnHunter.

Please see this schematic which shows a (different) mechanism where a frictionless bead rotates on a ring which is horizontal to the ground. The schematic contains the free body diagram (FBD) of forces acting on the bead at the time instant at which a jet (such as a miniscule firecracker quality rocket) applies an impulse force on the bead. The resulting reaction force (recall that the phenomenon is devoid of friction) acted by the ring on the bead is clearly not radial since the reaction to the impulse will cause a non-radial reaction component besides the instantaneous centripetal force component. The situation considered is the converse of the situation in the OP which was analyzed in a similar fashion in the answer by @JohnHunter.

The reaction forces that the answer by @JohnHunter alludes to and to which my subsequent comments ascribe the non-radial reaction force components, are indeed somewhat friction-like. This is because such contact forces are not perpendicular to the contact surface, which is the behavior of friction forces as well. The distinction between the non-radial contact force and the friction force acted by a surface on another surface can be made by observing the cause of the forces. Non-radial contact reaction forces are caused by local mechanical stresses resulting from the pushing or pulling i.e. contraction or extension of the surfaces induced on each other while friction contact reaction forces (static or kinetic) are cause by the relative motion between the surfaces. Indeed, although the static friction does not require relative motion between the surfaces to occur, impending relative motion or abrasion between the surfaces is the cause resulting in the genesis of the static friction between them.

This post elicits a great example illustrating the fact that directions of contact reaction forces may not always be perpendicular to the perceived tangent to the surface. Such an assumption therefore needs to be treated with caution.