# Rotational mechanics

1. Jun 19, 2009

### Urmi Roy

Hi, I've done rotational mechanics at school and there are certain things that really feel strange about it.

In Physics everything has a reason. So I want to find out the reasons to certain things.

firstly-we started off rotational mechanics by talking about angular displacement,angular velocity and angular acceleration. The vector product evidently gives us a direction for each of these quantities obeying the righthand thumb rule.

This doesn't really seem to have any physical significance at all! What does an object spinning around in a circle have to do with a direction (given by the righthand thumb rule) perpendicular to it??

Please explain this to me,I'll be greatly benefited.

2. Jun 19, 2009

### superkan619

Its all about appreciating the beauty of vector algebra.

Some simple physical situations can well explain it. For example consider two identical particles exectuing uniform circular motion with same speed (=> same radius) but in opposite directions and in parallel planes, centres facing each other. Now, if we want to add the respective angular motion parameters i.e angular velocity or angular momentum for a single particle, i.e superimposition of two angular motions for a single particle. We can intuitively say that the resulting angular velocity or angular momentum is zero. So there are two equal and opposite linear vector equivalents for each type of angular motion paramenter (any one linear vector being defined with a convention, which is choosen to be the right hand thumb rule) and it can be seen as a linear vector pointing out of the centre of the circle.

Imagine the exact situation with same directional circular motions having same particle mass and speed (=> same radius). You find reinforcement of two motions i.e the angular motion parameters become two times. This is same as doubling a linear vector for each cirular motion parameter.

The convention isn't important, you can devise your own L.H.T.R. But that needs to be obeyed in every situation. Not that you can mix both. So all are advised to use the R.H.T.R.

3. Jun 19, 2009

### Urmi Roy

So from superkan619's post, I understand that this R.H.T.R is just a convention which can be used to make all our calculations and studies of rotational motion easier and more systematic----It's just a mathematical tool ,rather than a physical one.

Is that right?

If it is, I can go one step further and ask about 'centripetal and centrifugal forces'.

Suppose we have a car moving with uniform velocity on a circular track.What provides the car this so called 'centripetal force'? I also heard that centrifugal force doesn't really exist.Please could someone elaborate?

4. Jun 20, 2009

### bp_psy

All "physical tools" are mathematical.The way in which it works is assigning physics to the mathematics.For example in your example you assigned a disk that rotates to the cross product and is perpendicular to the disc but does not accelerate it up.You may say that it does not have any physical meaning just a mathematical convention. But if you now assign the cross product to a screw it would have a meaning. The vector would point in which way would go (in or out) if you screw it in a certain direction. Certainly it still depends on the way the screw was made but now there is a 'physical' meaning atached to it . Another example when the cross product has a meaning is when describing the magnetic force . Anyway my actual point is you should not think as mathematical models as something as purely imaginary and completely separated from physics. They are sometimes almost the only thing we can understand about something.

The centripetal force is the name given to the force that makes the object move along a circular path.In the case of the car this force is the static friction between the tires and the road.In the case of ball rotating on the end of a string it is the tension. In the case of the moon rotating around the earth it is the gravitational force.
The centrifugal force is a fictitious force. It does not exist if you are working in an inertial frame. If you are in a rotational frame, such forces must be included to account for the apparent acceleration of the objects in that frame.

5. Jun 20, 2009

### superkan619

I missed out an imp point:
Angular quantities that don't depend upon type of rotation i.e they are same whether the angular motion is clockwise or anticlockwise are analogous to linear quanities which are directionless i.e scalars.

Example: Angular speed (20 rad/sec)and Linear Speed(20 m/s)
Rotational K.E(20 joules) and Linear K.E(20 joules)
work done for rotation(20 joules) and work done for translation(20 joules)

You cannot have a vector (result of the cross)popping out of the circle for the above circular entities just like you cannot have a vector along the above linear quantities, they obey simple addition and substraction.

But as soon as the circular entities have a sense of direction of rotation, they accquire vector properties and obey vector addition & substraction, the vector being equal to a cross product. eg:

Angular velocity(+20 rad/sec)and Linear velocity(+20 m/s)
Torque (-20 Nm) and Force(-20N)

Usual convention used in circular and linear systems are respectively:
+ stands for anticlockwise; + stands along +ve X-axis(or Y axis)
- stands for clockwise; - stands along -ve X-axis(or Y axis)

Just as linear vector quantities are physically meaningful so are their rotational counterparts. All these vector quantities are not just mathematical, they are physical too. So, far as the convention of +/- is concerned you can choose + as clockwise and + as -ve X or Y axis.

6. Jun 20, 2009

### Urmi Roy

Hi,
thanks for the help. There's something I would like to ask in reference to the first post.

How does it do so??

This centripetal force always acts towards the centre of the circle that the object(in concern) is moving around.Why is that?

If an aeroplane moves around in circles in mid-air,what provides it this centripetal force?

7. Jun 20, 2009

### bp_psy

Well each in situation is a little different but the main conditions for having motion around a circle or any curve are. First having a linear velocity then a force having a component perpendicular to that velocity. In the case of the plane this works something like this. First you have the plane moving in a straight line (weight equal to lift). Now you change the shape of the wing such that the lift is greater. The plane will move up and rotate slightly. Since the plane rotated, if you maintained the shape of the wing you will still have a force component perpendicular to the new velocity of the plane. The plane will therefore once again change its direction and rotate and so on until you complete a full loop.

8. Jun 20, 2009

### Cantab Morgan

Your intuition is good. Representing concepts like torque as a vector in ordinary space is an accident of our living in a three dimensional world.

Suppose we lived in "flatland" and everything was confined to the plane. A third spatial dimension had no physical meaning. Well, even in such a world, we could mathematically invent a third dimension, and represent torques and whatnot as vectors pointing into this hypothetical direction. It would have no physical reality, but all the math of rotational mechanics would still work out correctly.

(This is not a such an outlandish idea. Einstein and others showed that there are some problems in electromagnetism that can be represented more easily by adding an extra dimension. This is the celebrated theory of "hyperspace", whose name science fiction writers have seized upon and twisted to their own fantastic ends.)

However, by some happy accident, we live in a three dimensional world, not flatland, so the "hyperspace" direction in which our torques point coincides with a real spatial dimension. But this is happenstance. As I said, your intuition is good, because you are troubled by the apparent lack of physical correspondence between the "hyperspace" direction of torques, and the real physical spatial direction that it coincides with, and there isn't any.

9. Jun 20, 2009

### Bob S

Are you prepared to play ping pong in a NASA centrifuge running at 2g?

10. Jun 21, 2009

### Urmi Roy

I think I've got the centripetal force concept right,thanks to the post by bp psy and thanks for the confirmation of my intuition by Cantab Morgan.

I din't really understand the last post though,but I think I'll now go on to trouble you people about some things I'm curious about in the centre of mass and gravity concepts.

Again, these concepts (which are not really concepts as such) are tools which we use to make our studies easier,especially when the object in concern is small.

Well, why are the centres of mass and gravity different for certain objects in the first place?Since mass is what gravitational force acts upon, the point in the object where all of this force is felt should be the same as the point in which all its mass is concentrated in all cases.

11. Jun 21, 2009

### Staff: Mentor

That's true when the gravitational field is uniform, but in a non-uniform field the two centers can be different.

12. Jun 22, 2009

### Urmi Roy

Please could someone elaborate on this point about the centres of mass and gravity,I'm still having a little difficulty.

13. Jun 22, 2009

### RonL

Hi Bob,
For what it's worth, your post did not go unnoticed, I might be the most ignorant member on PF in the area of math and equations, but I'm pretty sure I can build a motion picture in my mind of exactly what you meant to imply.

Ron

14. Jun 22, 2009

### Bob S

Consider a room with a ping pong table in the center. Suppose the whole room is rotating at say 5 rpm. You and your opponent are playing ping pong. The ping pong ball always travels in a straight line because it is in an inertial reference frame, but the table is rotating under it. What are the apparent equations of motion for the ping pong ball, as observed by each player?

15. Jun 22, 2009

### D H

Staff Emeritus
The center of mass of an object is a mathematically defined construct that has nothing to do with gravity.

One definition of the center of gravity of an object is the point at which the gravitational acceleration due to some gravity field is equal to the gravitational acceleration of the object as a whole. (There are other definitions, so take care.)

The gravitational acceleration at a point x the vicinity of some point x0 can be expressed as a Taylor series in x-x0. The ith component of this Taylor series will be of the form

$$g_i(\mathbf x) = g_i(\mathbf x_0) + \sum_{j}A_{ij}(\mathbf x - \mathbf x_0)_j + \sum_{j,k}B_{ijk}(\mathbf x - \mathbf x_0)_j (\mathbf x - \mathbf x_0)_k + \cdots$$

For a uniform gravitational field, the gravitational acceleration is the same everywhere: The only term in the expansion is the constant term. The concept of center of gravity as defined above doesn't make sense for a uniform gravity field. The gravitational acceleration is the same everywhere.

Suppose x0 is the center of mass of some object. For a linear gravitational field (only the first two terms are present), the center of gravity is well defined: It is exactly equal to the center of mass. You need to go beyond the linear term to arrive at a condition where the center of mass and center of gravity do not coincide.

16. Jun 22, 2009

### RonL

It would be a little easier on me to use a round table.:uhh: I was going to raise your first example by having two players in the spin chamber, rotating in opposite directions, discounting air turbulance, how often would each player be hitting the ball he had hit on the other side of the room?

P.S. The diameters are just offset so that they can pass each other.

Guess I'm all twisted for the rest of the day

17. Jun 22, 2009

### Urmi Roy

Though this post deals with my doubt I'm afraid that I'm just an ignorant yet inquisitive high school-girl,so you can imagine the Taylor series went quite beyond my reach!

An explanation in simpler terms would be a little better please!

18. Jun 23, 2009

### Urmi Roy

19. Jun 23, 2009

### Urmi Roy

Perhaps I'll get back to the centre of mass topic later.

In rotational mechanics the most important concepts are torque and moment of inertia.

Whereas in the 'normal' force we only say that force applied is equal to the mass and the acceleration produced in the object,we define torque as being directly proportional to the force applied and the distance (of application of force) from the axis of rotation.

Its true that in practical life I find it easier to rotate something when I apply force farther away from the axis,but that's just experience isn't it?

I mean how did we come to define torque like that in the first place?

Coming to moment of inertia, it's defined as m(l squared) for a point object of mass m at l units of distance from the axis of rotation. Now that doesn't have a very simple logical explanation.Why do we define moment of inertia like that? How did scientists come to define it like that in the first place?

20. Jun 23, 2009

### D H

Staff Emeritus
If you ask your parents too many questions, they will eventually answer "because that's the way it is". Sometimes they answer this way because they want to stop the silly stream of endless questions, sometimes they answer because they themselves don't know the reason, but sometimes they answer this way because you have not grown up enough to understand. You do not have the necessary background information.

The latter is the case here. You don't know what a Taylor series is, which in turn means you do not yet know calculus. Pre-calculus physics is plagued with a whole bunch of "because that's the way it is" type answers. Once you learn calculus you will have the tools needed to take the next step and understand how these apparently ad hoc rules came to be. Until then, the only answer is "because".

Rather than asking these questions that cannot be answered without the necessary mathematical skills, you might want to think of taking a side journey into developing those skills.

21. Jun 23, 2009

### RonL

From someone that finished high school, but was a very poor student, I have spent a lifetime doing things that involve rotational mechanics and do not have a clue about how to do the math.
Whatever you do, stay the course and learn the material, try to find everyday things you do that are never given a second thought, like opening a jelly jar (sometimes you need a gripper that extends your leverage) or set up a little lab using a lawn sprinkler (the spinning kind) by changing the water flow or pressure you can change speed of the sprinkler, determine it's RPM by counting and timing each discharge as it goes around.
If you have someone that has a mechanics tool set you can use, get some different size ratchets and lever bars, then tighten some bolts and nuts until they reach a breaking point.

Your life is full of everyday things that involve rotational mechanics, if you can find some joy in the discovery of as many things as possible, I think the class will take on a new level of enthusiasm for you.

This is not much of an answer, but it might help your approach to the subject. Maybe you will forgive BobS and me, for getting a little off track to your question.

RonL

22. Jun 23, 2009

### Cantab Morgan

Wait, why are you surprised that the predictions made by Physics agree with "just experience?" If they didn't, then Physics wouldn't be worth very much, would it?

23. Jun 25, 2009

### Urmi Roy

Well,I suppose I've been asking too many questions,and I really must learn quite a lot until I make myself good enough to find the answers.

Thanks to all of you for giving me all the help in the previous posts!

24. Jun 29, 2009

### bp_psy

In introductory courses the concepts are usually simplified as much as possible. The reasons for this simplification is the lack of mathematical knowledge of the students and second the desire to go over as much material as possible in a short period of time.This approach is understandable considering that very few students will actually continue to study physics after these courses.The result is that you cant expect to get to much insight about physical concepts or understand the logic and elegance of how these concepts are expressed and derived.
In this case what I see as strange(but not really wrong) are your definitions of toque and force and the way in which you separate them as though they don't have nothing to do with each other. The general definition of a force is the time rate of change of linear momentum$$\vec{F}=dp/dt$$ where we define momentum $$\vec{p}=mv$$ which is a quantity we can measure and has very interesting properties.In the special case where mass is constant this whole thing reduces to $$\ F=ma$$ which is a very easy to understand formula.Now we have another problem, we want o describe what happens when we want to make a particle move in a circular path around a point. We know that we must apply a force .This force must have a tangential component to make it accelerate tangentially and radial (centripetal) component so as to actually follow a circular path. We also have a very nice system of describing motion in a circular path using the concepts of angular displacement, velocity and acceleration. So we want to use this system in solving the problem but we must find a way to unite all these concepts. Since there is a tangential force component there will be a rate of change of tangential momentum. $$\ dp/dt=dmv/dt$$ we know that $$\vec{v}=\omega\times r$$ and since we have 90 degree angle $$\ \textbf{v} =\omega r$$ and $$\textbf{p} =m \cdot \omega r$$.. These 'formulas' would be enough for calculating how the 1 rotating particle works but we can simplify them even more and make them work with rigid bodies.We define angular momentum as $$\vec{L}=r\times p$$ the we take the first derivative with respect to time of this new quantity $$\ dL/dt=d(r\times p)/dt$$ this gives $$\ dL/dt=(dr/dt\times p)+r\times dp/dt$$ or $$\ dL/dt=v \times p+r\times F$$ the first part is "obviously" 0 so $$\ dL/dt=r\times F$$ which we call Torque.We now should return to the moment of inertia thing.$$\vec{L}=r\times p$$ if we have a 90 degrees angle we can say $$\ L=rp$$ but in this case $$\textbf{p} =m \cdot \omega r$$ so $$\ L=r \cdot m \cdot \omega r$$ or $$\ L=mr^2 \cdot \omega$$ since we want to make this as simple as possible, we define moment of inertia as $$\ I=mr^2$$ so if I is constant is $$\ L=I\cdot \omega$$ and $$\ dL/dt=\tau= I\cdot \alpha$$.These formulas are similar to the simple force formulas but they have the same limitations, they work only in special cases. I leave it to you to work out how to apply this to rigid objects and not just 1 particle system, to figure out why the really simple formulas don't always work and why a rigid body does not have just 1 moment of inertia.If you don't understand calculus you should learn it. The amount of calculus required for this whole thing can be learned in just 3,4 weeks.

. "Physics Works!".The mathematical models of physics don't have to be easily recognizable in day to day life.Why should they?. But if you make a prediction about something using them they work. If the model is used in the correct way or if it is a correct model.The model above works very well when used correctly.[/QUOTE]

25. Jun 29, 2009

### D H

Staff Emeritus
That was a mighty long paragraph, bp_psy.

That is also a might big if. For a rigid body, the positions and orientations of the various parts of the body are constant in a reference frame fixed with respect to the body. If the masses of these components are also constant, the inertia tensor for will be constant in a such a body-fixed reference frame.

Just because the inertia tensor is constant in a body-fixed frame does not mean it is constant in all frames. In particular, if the body is rotating, the inertia tensor as expressed in inertial coordinate most likely is not constant. One solution (and this is the approach that is used almost universally) is to do the physics from the perspective of the rotating body-fixed frame.

To model translational dynamics from the perspective of a rotating frame, one must add fictitious forces to Newton's second law of motion to make the (mathematical) equations of motion accurately describe what is happening. A similar thing happens when one attempts to model rotational dynamics from the perspective of a rotating frame: One must add fictitious torques to the equations of motion. A simple version of these fictitious terms are called Euler's equations.