# Rotational mechanics

1. Jun 27, 2009

### Urmi Roy

I've just got some small problems this time.
The first being,do the particles at the axis of a rotating body really stand still??Don't they spin about themselves??

2. Jun 27, 2009

### ham

I would think of it this way...If you think that a particle at the axis of rotation has a radius (dimension), then there would be rotation. However, since the axis has no dimension or radius, it is just a line, then basically there is no partical there hence no rotation. If there were rotation, the radius would be infinately small and of no consequence.

3. Jun 28, 2009

### Urmi Roy

I think that means that this notion of the axis being completely still stems from the fact that we consider the particle to be infinitesimally small,which allows us to consider the axis as sttionary. Right?

Another thing that I was wondering about is that-I found that all the quantities that we discuss in rotational mechanics,can be obtained if we do a Vector product with the analogous linear quantities,like angular momentum is 'p (cross product) r' and then torque is ' F (cross product) r' , etc.

Is that just a coincidence, or does it have something to say about the physics of rotational motion?

4. Jun 29, 2009

### Urmi Roy

Recently I observed something really strange. When I was toying around with my yoyo,the yoyo stopped at the end of the string and it started spinning around itself.

Simultaneously,the turns in the string seemed to be moving upward towards my hand.

I was wondering,is this what we mean by vectorial direction?

It might sound crazy,but I think that it kinda makes sense. I mean angular velocity has a direction,which is perpendicular to the plane of rotation--so it was in the yoyo!

5. Jun 30, 2009

### Urmi Roy

6. Jun 30, 2009

### gmax137

UrmiRoy - in the forum view (when you can see all the threads by title/subject) you can also see how many people have viewed the thread (right now this one has 140+ views). So you see, most people who view the thread do not post to it. Maybe they are too busy, maybe they think they have nothing useful to add. This is typical of the PF forum, fortunately. It is fortunate because it means most of the posts that people DO make are worthwhile. In other words, your patience will be rewarded.

7. Jul 1, 2009

### Urmi Roy

I'm really sorry for being too impatient!!

8. Jul 1, 2009

### jimcharoen

Ha haaa.. this is really a philosophy question. Indeed, this is a question of either discrete or continuity. If you believe in Democritus (atom = undevidable) then there must be a very very small particle at the center of the object which stands still. If you, on the other hand, believe in continuity, there is no such particle i.e. even if the particle is extremely small, it is still devidable.

Cheers,
Charoen Peetiya

9. Jul 1, 2009

### BobbyBear

When you are studying rotational mechanics you are studying it within the branch of continuum mechanics in general, in which you're modeling matter as a continuum, and you can talk of there being matter at each point in space, and you can talk of velocity at each point, and two points can be infinitely near each other and their respective velocities will differ by an infinitesimal amount, and, and . . . and you can use calculus as your mathematical language to describe the phenomena you are studying. Remember though that this is just a model of reality, for matter is not continuous at all, but the model serves us well enough to describe and predict the behaviour concerning the motion of objects, as long as we don't get too close to the object so that we see that it is really not continuous at all (ie. when the continuum hypothesis is breached). I'm just trying to say, don't eat your head over what happens at a point (the axis of rotation), because it is a mathematical abstraction of reality - and for our particular interests, we don't really care what reality looks like, as long as with our model we can extract conclusions and make predictions about the particular aspect of reality that we are interested in.

I have no idea what really happens at the axis of rotation as far as the matter contained in it is concerned. Firstly, are you sure the axis would be occupied by matter at all (since in reality we know that only a very tiny percentage of the space even within an atom is really occupied by matter). And if there was matter exactly at the axis, would it remain there? aren't molecules always vibrating (if they are above the absolute zero temperature)? so wouldn't the centrifugal force, which would be very large near the axis, force them outward? Or maybe they'd get squished? I really don't know . . . but this is not what rotational mechanics is concerned about.

10. Jul 2, 2009

### Urmi Roy

I get the point--the concept of a stationary axis is,number one an assumption,and number two,not that improtant when it comes to real life situations,thanks for the clarification!

11. Jul 2, 2009

### BobbyBear

The assumption is an assumption about reality, in our case, about matter being continuous. For the disciplines of continuum mechanics, what's really happening at a microcopic level is not important (cannot be predicted). Just like singularities that arise in our mathematical description don't really exist, and we cannot predict with our model what is really happening at a singular point in space (eg, shockwaves are mathematical discontinuities, but in reality all magnitudes vary continuously, however large their gradients may be). And that our model is not valid for those points, doesn't necessarily mean that for other purposes it's not important to understand some aspect of it - but we need to study it using different models of reality.

About post 3, whether the equations for angular motion can be derived from those of linear motion by simply cross multiplying with the position vector . . . well, for indeformable bodies, I don't know what to say to you, perhaps so. But not in general: take for example, the equations of conservation of linear momentum and conservation of angular momentum. They are independent principles: one cannot be deduced from the other, however tempting it may be to try and do so (unless additional hypotheses are made - but actually these hypotheses are consequences of both principles having to hold). Remember that a principle is a statement that assumes a certain truth - and as such, no principle can be deduced from another.

I'm not sure I understand post 4. A vector quantity is simply one that is described by a direction as well as a magnitude. The direction of the angular velocity vector doesn't have a physical significance as far as I know, except that it is useful (for mathematical purposes) to define the angular velocity as a vectorial quantity whose cross product with the position vector equals the linear velocity vector.
I'm not sure what you mean by the angular velocity direction being in the yoyo.

12. Jul 2, 2009

### Urmi Roy

Well, actually the problem with me is that though I find every concept and theorem in linear mechanics very logical and easy to understand with each of them related to one another,even if they can't be deduced from eachother,I can't find that lucidity in my concepts of rotational mechanics.

I have a strong intuition that every part of mechanics-be it linear or rotational,must be correlated,must be explanable on the basis of eachother-but when it comes to some parts of rotational mechanics,I just can't find any explanation--these expressions of torque etc. etc. seem to have sprung out of nowhere!

Newton devised his second law of motion,F=ma, after observing the world around him,and having found a deep logic behind it--where does that apply to those expressions of rotational mechanics,with all that vector product and mathematical assumptions?

Actually I wasn't saying that there was an "angular velocity direction in the yoyo," I meant that the direction that we would assign the angular velocity in the yoyo ( by the right hand thumb rule) would be pointing out of the plane of the yoyo,which is found to be same as the direction that the yoyo's string was seen to be coiling up (like when we twist a piece of string,we see sort of a movement up the string).
However,I understand that it was only my imagination,since,as BobbyBear said,the direction assigned to angular motion is only a mathematical assumption.

13. Jul 5, 2009

### Urmi Roy

Recently I got into a tangle while doing rolling without slipping and rolling with slipping.

Well,in one of my text books it says that rolling without slipping involves rolling friction which provides the requisite amount of torque to stop a wheel from slipping and in another,it sates absolutely nothing about rolling friction and just states about static and dynamic friction (and that too very ambiguously).

How does a tyre slip with its linear velocity faster than its angular velocity would ideally allow it anyway? Can't really imagine that happenning.

I would be grateful if someone please gives me a description of how the forces that act and their individual roles on a wheel during rolling without slipping and rolling with slipping.

14. Jul 5, 2009

### D H

Staff Emeritus
If you have seen people laying rubber with a car, that is rolling with slipping. Step on the gas hard enough (assuming an adequately powered vehicle) and the torque applied by the engine will exceed the maximum torque possible from rolling friction. Result: The tire rotates faster than a rolling tire and leaves a strip of rubber on the road due to the higher sliding friction.

15. Jul 5, 2009

### sganesh88

DH, Urmi is bothered about linear velocity being greater than "what the angular velocity would allow"- meaning R X W(omega). I had a similar pre-occupation before. I thought that if a rolling wheel loses contact with the ground and gets effused into space, it will abruptly come to a stop because the liaison, friction, that "should" continuously act to equate V and "R X W" isn't present anymore.
The thing is that the wheel's linear velocity and angular velocity are actually independent. You can have any pair of values for these two quantities if friction doesn't interfere(so i was partly right in my misconception).
A wheel kept on a frictionless incline will just slide without any angular velocity. An automobile with its engine revved to even 12000 rpm on a frictionless surface will stay put with an enormous W (measured at its wheels) but zero linear velocity. As friction does its duty of preventing relative motion, it acts in a "less villainous" way when it comes to wheels. By modifying V and W such that V=R X W, friction stops functioning because in such a configuration, there is no point on the wheel that tries to slide or slip past the surface(observe a cycloid curve). Slipping and sliding arises when the max friction force possible isn't quite enough to bring about the synchronization. So in DH's example, as the accelerator pedal is pressed hard, wheels start rotating at an insane speed and friction is at a loss to sync V and W according to the above mentioned relation that quickly resulting in slipping. Same applies to sliding during hard braking.

Ya. Rotational mechanics doesn't operate on a different set of laws. The same three laws suffice. Conservation of angular momentum can't arise if conservation of momentum is false.

16. Jul 6, 2009

### Urmi Roy

I think this explanation and the examples provided were excellent,thanks for the clarification!!

I read recently that the law of conservation of angular momentum can be derived directly (with quite meddlesome maths) from the expression of the law of conservation of momentum and also that both can be derived from something called the Noether's Theorem.

I was wondering if you all could perhaps throw some light on this issue and ofcourse I would love to recieve more information relating to 'rotational mechanics' from all of you experts,especially pertaining to the last few posts in this thread(e.g. posts three and four perhaps).

17. Jul 6, 2009

### D H

Staff Emeritus
If you read that conservation of angular momentum derives from conservation of momentum, it is wrong. The two are fundamentally different. Conservation of momentum implies the weak form of Newton's third law. The strong form is needed for conservation of angular momentum. Alternatively, one could just use Noether's Theorem, but that is a topic of rather advanced (upper level undergraduate / lower level graduate) physics.

18. Jul 7, 2009

### Urmi Roy

I'm sorry for making a blunder,I just remembered that it wasn't the conservation of momentum that the conservation of angular momentum is derived from,it was Newton's third law (every action has equal and opposite reaction,).

Since the two laws (conservation of angular momentum and conservation of momentum ) are analogues, they also can be derived from the same law-Newton's third law---that's what I read.

19. Jul 7, 2009

### Urmi Roy

By the way, I've been pondering about what sganesh88 said about the rolling concept--well there are 2 things I want to confirm-

1. rolling friction seems to have no role here,it was probably just an alternate explanation in one of my books (H.C Verma).

2. When the rolling does not have any slipping, the frictional froces don't work,because they don't need to.

Last edited: Jul 7, 2009
20. Jul 7, 2009

### BobbyBear

Rolling Friction

Lol for a minute I imagined you rolling with and without slipping on your living room carpet:P
Urmi, you are very sweet when you get all excited about physics :)

I think your queries on this have been addressed to, although you did ask about rolling friction. I'd just like to give my opinion that rolling friction is not some special or seperate form of friction at all - I'd be more happy simply calling it 'rolling resistance'. Basically, it's the consideration of the distribution of forces involved within the contact area of the bodies, which is in reality a finite area, as all bodies are deformable. From these considerations one can conclude for example that a minimum torque is needed to set the wheel in rolling motion (due to the distribution of forces in general, not necessarily due to a dissipative force) - however, there'd be no way to explain this minimum torque if you are considering the bodies to be non-deformable. Can anyone please confirm that if we're considering the wheel to be perfectly rigid, we'd not be able to talk of a 'rolling friction force' associated with a 'rolling friction coefficient'?

Urmi, I think you should concentrate first on understanding how the sliding friction force (the only kind of friction force I'd consider, ie. proportional to the coefficient of friction times the normal force and pointing in the direction so as to oppose impending relative movement) plays a part in producing or balancing the torques acting upon the wheel. Consider different scenarios: a wheel rolling down a hill (there is a gravitational force-component acting upon the centre of mass causing it to accelerate), a wheel rolling up a hill, a wheel connected to a shaft which is exerting a torque that tends to increase the angular velocity of the wheel, or a torque that is opposing the angular velocity of the wheel . . . in each of these cases, which must be the direction of the friction force? Consider what the movement would be like if there was no friction (completely slippery surface) and what role the friction force is playing in the movement of the wheel. It's always helpful to draw a free body diagram of the wheel with all the forces acting upon it, and simply apply Newton's second law for linear and rotational motion.

I'd also like you to consider, if you haven't already, that frictional force does no work when a body rolls without slipping. Do you see this clearly? (again, this assuming that there is no deformation).

Last edited: Jul 7, 2009
21. Jul 7, 2009

### BobbyBear

Um, I think I'm getting into a little tangle myself:P

Consdier an indeformable wheel rolling down an inclined plane. The wheel is subject to the force of gravity (Fg), which we can separate into a component normal to the plane and a component tangent to the plane, acting at the centre of mass of the wheel, say FN and FT. If there was no friction at all, then the wheel would not spin but slide downward with a constant linear acceleration given by Newton's law (a=FT/m). But if friction is not zero it will set the wheel in angular motion. Let's suppose that the maximum static (sliding) friction is greater than FT so that there is no slipping. Then the friction force at the point of contact, f, will be in the direction opposite to FT and equal to FT. So the net torque upon the wheel will be T=FT*R, and the angular acceleration of the wheel ( $$\alpha$$ ) will be given by Newton's law: $$T=\alpha *I.$$

http://img150.imageshack.us/img150/2525/planef.gif [Broken]

Obviously the angular velocity of the wheel is increasing, but since linear and angular quantities are related by $$v= R *\omega$$, the wheel has a linear acceleration too, or rather each point of the wheel - in particular, the centre of mass of the wheel has a linear acceleration given by $$a=r*\alpha$$ . . . no? If my calculations are correct,
I=3/2 * m * R2 (with respect to the instantaneous centre of rotation) and the linear acceleration of the centre of mass is therefore 2/3 * F/m.

But if we apply Newton's second law to the wheel however it tells us that since the net force upon the wheel is zero, the acceleration of its centre of mass is zero.

Um, where am I going wrong?

Last edited by a moderator: May 4, 2017
22. Jul 7, 2009

### BobbyBear

This is really beyond me, but does Noether's Theorem have something to say about the stress tensor being symmetric?

23. Jul 7, 2009

### BobbyBear

I think it's because the relative velocity at the point of contact is zero, thus the friction force is not doing work.

24. Jul 7, 2009

### Urmi Roy

Re: Rolling Friction

This is quite a good way of putting it.

This is what I was confused about--we're doing rotational mechanics of rigid bodies,so ideally there should not be any such deformation considered.

25. Jul 7, 2009

### Urmi Roy

In rolling without slipping v=R*w should be valid, but as we know the tyre may also slip,in which case this relation will not be valid, so we can't just assume that v=R*w is valid (and then explain the why there is no net acceleration of the tyre on this basis),----That's the whole point ,since I'm asking about what it is that makes it valid in the first place, for the special case of rolling without slipping,(in terms of forces acting therein).

Whereas the net linear force is zero,the net torque is not zero.

BobbyBear stated that the net torque about the centre of mass of the wheel is T=(alpha)*I,due to the friction--the question is, which force balances this torque due to friction in order to have zero angular acceleration of the tyre?

Last edited: Jul 8, 2009