# A question about angular motion

1. Nov 14, 2013

### PhysicsKid0123

If a bead slides frictionlessly along a wire on a horizontal table. The wire is in the shape of a spiral, and the bead is given an initial push at the outer side of the spiral, so that it slides along in an "orbit" of decreasing radius.Is the bead going faster as it moves inward? Is its energy conserved? If not, how is it acquiring energy? Is the angular momentum of the bead conserved? If not what torque is acting on it?

I would say that the bead goes faster because it getting closer to the radius and that energy is not conserved because it gain energy as it gets closer to the center. Is the right? I'm not if angular momentum is conserved, or if there is torque... Can someone please elaborate?

2. Nov 14, 2013

### vtahmoorian

a bead slides frictionlessly along a wire on a horizontal table?! is that what you mentioned? if it is so, then how will the energy be wasted and how is it not conserved?
On the other hand , if you have taken it as an isolated system what is the origin of the gained energy? seems the only exchange of energy is gravitational potential energy to kinetic, right?

3. Nov 14, 2013

### PhysicsKid0123

No, there is no gravity here. The spiral is on a flat plane. I'm trying to figure out what the bead experiences as it travels through the spiral when it is given a push.

4. Nov 14, 2013

### PhysicsKid0123

I don't see why there would be any torque either... I'm not sure if energy is conserved or not. I'm also not sure if momentum is conserved either. I don't think neither are conserved and my reasons are because it gains energy as it spirals towards the center of the spiral so energy is different before and after. Also, I don't think momentum is conserved because there is a centripetal force towards the center. So since there a force momentum is not constant...

5. Nov 14, 2013

### voko

The energy and angular momentum questions can be answered by considering the forces acting on the bead.

What forces are present?

As regards energy, do these forces do any work?

Once you know what happens with kinetic energy, you can then deduce what happens with angular momentum.

6. Nov 14, 2013

### PhysicsKid0123

The only forces is the centripetal force caused by the centripetal acceleration, and the work done is the product of distance and the force or how far the force was applied in terms of distance. So work is done... Right? So since there is positive work according to the work-energy theorem there is kinetic energy, which are both equal. Still don't know if momentum or energy is conserved, however. I don't think they are but that is just intuition. I really don't know why.

7. Nov 14, 2013

### mic*

PhysicsKid0123,
I'm not sure if part of your question is uncertainty about whether the bead would actually gain speed (angular velocity), but if it is, here is a somewhat analogous scenario you might have encountered

http://www.flickr.com/groups/swingball/

When the ball on a string gets to the end of the rotating coil, it has no option but to start wrapping itself around the pole. The behaviour definitely changes

8. Nov 14, 2013

### vtahmoorian

Conservation energy says :" when there are not non conservative forces working on the system, mechanical energy which is the sum of kinetic and potential energies should stays unchanged"
your first push gives the bead momentum, which is the root of its kinetic energy as well. so there is a bead which has an initial kinetic energy, if there is no non conservative forces , it will keeps that amount of energy, but the radius of the spiral is becoming smaller as it moves forward, therefore its angular momentum will change and consequently the bead will take torque.
there is no energy gaining here! how do you think now?

9. Nov 14, 2013

### voko

This is completely wrong.

Force is not caused by acceleration.

Acceleration is caused by force.

There is no centripetal acceleration here, because motion is not circular, but spiral, so the acceleration is not always directed toward the same center, hence it is not centripetal.

This is partially correct. But it misses the other part: that the mutual direction of the force and displacement is factored into this product. How?

This is not a guesswork matter. Carefully consider the force, its direction, the direction of the bead's motion, and you will have the answer.

10. Nov 14, 2013

### haruspex

I have to object to that. Centripetal acceleration is the component of acceleration orthogonal to the velocity. All of the acceleration here is centripetal. The difference from a circular path is that here the acceleration increases.
Vectorially, $\vec{v}.\dot{\vec{v}} = 0$, which has the solution $\vec{v}^2 = constant$. But this does not prescribe the path. It applies to any smooth bead on any smooth shape of wire where there are no external forces.

11. Nov 14, 2013

### mic*

I would agree with haruspex. If the orthogonal force toward the centre of the decaying spiral is not centripetal force, how would you describe it Voko?

12. Nov 15, 2013

### voko

According to Newton, who I believe coined the term, "a centripetal force is that by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre". I interpret "towards a point as to a centre" as a way of saying "towards a fixed point", not towards a point that depends on the current position or the velocity of the body acted upon by the force. If this is not a fixed point, then the whole definition becomes tautological.

The acceleration component that is orthogonal to velocity is the normal acceleration. Normal acceleration happens to be centripetal in circular motion.

I disagree with the definitions that relate centripetal acceleration with the "center of curvature". They make "centripetal" synonymous with "normal"; that creates confusion and does not seem to have any merit whatsoever because the term "normal" is widely understood and used.

13. Nov 15, 2013

### haruspex

OK, but that is only an interpretation, and it is certainly not clear from the quote.
According to http://en.wikipedia.org/wiki/Centripetal_force, it is directed toward "the instantaneous center of curvature of the path". OTOH http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration specifically reserves centripetal for the circular case.
http://theory.uwinnipeg.ca/physics/circ/node6.html defines Centripetal acceleration as the rate of change of tangential velocity (which is wrong, but I think the intent is clear).
http://www.britannica.com/EBchecked/topic/102869/centripetal-acceleration gives only the circular path, as does http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html.
http://www.britannica.com/EBchecked/topic/102869/centripetal-acceleration only mentions elliptical paths (!)
http://omega.albany.edu:8008/calc3/curves3-dir/lecture.html regards normal and centripetal as equivalent
http://physics.tutorvista.com/motion/tangential-acceleration.html says "The radial component of acceleration is due to the centripetal force which is acting towards the center of the curved path. "
http://en.wikipedia.org/wiki/Mechan...ictitious_forces_in_a_local_coordinate_system tells me "in an inertial frame of reference one can identify the centripetal and tangential forces"
So on that more-or-less random sample, it's 6:3 in favour of the more general definition.

To me it would seem strange to have a special term for the circular case.

14. Nov 15, 2013

### voko

Centripetal is not necessarily only for the circular case. It is for any attractive central force. That is a fairly broad class of motions.

Newton introduced the concept in De motu corporum in gyrum, where he studied motion under a central force, where the motions under study were circular, elliptical and parabolic. In all the cases it was about a fixed center.

It may be that the confusion is produced at high-school levels of physics teaching, where the only really examined curvilinear motion is circular, and where normal and centripetal are identical. Those notions then get stuck and "logically developed" as the level of instruction increases, but the logical development that keeps normal = centripetal makes the latter acquire a meaning different from the original. I think the original meaning is good and useful, and we have the term "normal" for the other meaning.

Another concept that adds to the confusion is radial, which sometimes is equivalent to centripetal or normal.

A different yet related set of confusion includes tangential, transverse and azimuthal :)

15. Nov 15, 2013

### haruspex

That's a fair position, but not one reference I found laid it out that way. Most never mention "radial" or "normal". Clearly some scope for standardisation. Is there an international body that ordains such things?

16. Nov 15, 2013

### voko

You are right, haruspex. And wrong, too. I have flipped through a few books (Symon, Goldstein, Kleppner) and am surprised they do not contain any systematic exposure of the nomenclature. Neither in the way I have it, nor in the alternative view. Or at least their indices all point at all the wrong places. They just mention those names, almost in passing, as if the reader is supposed to know what they mean.

This is rather strange because the decomposition of velocity and acceleration in curvilinear motion is such an important topic.