Do these equations belong to kinematics or dynamics? Or both, maybe?

[Femme_physics]

I'm not sure how to prepare myself to the new subject, because I'm still not clear whether it's "kinematics" or "dynamics". What is the actual difference between the two? Can I see a difference with examples? (and no, wiki didn't clear it up for me)

I did attach the formulas that we'll probably be using (got it out of the tests solution manual), I'm just not sure which subject do they belong to out of the two.

 

[Doc Al]

I'm not sure how to prepare myself to the new subject, because I'm still not clear whether it's "kinematics" or "dynamics". What is the actual difference between the two? Can I see a difference with examples?

Kinematics is the description of motion without regard for the forces involved in creating the motion (if any). Dynamics deals with the forces acting on things and the resulting accelerations they produce (think Newton's 2nd law). Many problems involve both kinematics and dynamics.

Those equations you posted look mainly kinematic.

 

[Femme_physics]

Thank you, I'll go practice me some kinematics then :)

 

[Cleonis]

I'm still not clear whether it's "kinematics" or "dynamics".

Different people are using the words 'kinematics' and 'dynamics' in different ways, drawing different distinctions.

Engineers
Engineers will probably favor distinction between motion of moving parts of machinery, with gears and so on, and on the other hand free motion of loose objects.

For instance, how do you steer a robot arm? Let's say the robot arm must pick up boxes from one surface, and place it on another. The robot arm will have several hinges. For each spot on the working surface the robot's hinges must move to a particular angle. That sort of thing deals with motion, but not with force; you just want to work out how the hinges must move.

When an object is falling you calculate its velocity by working out the gravitational force and the frictional force. That's dynamics.
Physicists
But now the following case: two marbles, equal weight, are moving towards each other, with the same velocity, say, 1 meter per second. They collide and bounce back again, once again each moving 1 meter per second.

Question: what will happen if a moving marble hits a stationary marble? Let's say the moving marble moves with 2 meters per second. Well, that is really the same case as the one above, but from a different perspective. All of the velocity of the moving marble will transfer to the other one.

This is thinking about motion, and predicting how objects will move in such and such circumstances, but you don't look at the actual forces that are involved. Probably physicists wil favor the above as the meaning of 'kinematics'.In general:
Distinguishing between kinematics and dynamics is not necessary, particularly not when first learning about physics concepts. There is no clearcut distinction anyway.

On the sheet with expressions you show most contain 'g', which usually stands for the acceleration from gravity. If those expressions are used to handle cases of falling under the influence of gravity then the expressions are used for dynamics.

So the expressions themselves aren't necessarily 'kinematics' or 'dynamics', it depends on the case that they are used for.

 

[Femme_physics]

I see. Thanks for the comprehensive reply :) I decided to start crunching on kinematics first having seen the exercises in the manual. Although I might be jumping the gun and should wait for my lecturer to start us nice and easy.

 

[Studiot]

What is the actual difference between the two? Can I see a difference with examples? (and no, wiki didn't clear it up for me)

This is a difference that will probably never be cleared up since many people use the terms interchangeably. Even dictionaries disagree.

One guiding thought may be that dynamics deals with both cause and effect and the realtionship between, whereas kinetics, kinematics etc deals only with effects.

So most would have that dymanics deals with both the forces causing motion and the motion itself, whereas kinetics deals solely with the motion.

This is all nice and pat but then some clown called a certain type of energy kinetic energy.

More widely the terms are used to describe change, not necessarily involving force at all.

So in Chemistry we have reaction dynamics = reaction kinetics

In Control Engineering we have system dynamics which deals with the response of a system to a control mechanism

In Social Science we have social dynamics which deals with interpersonal relationships and their development and the effect on a overall population.

and so on.

Shalom

 

[Femme_physics]

Thanks studiot. (I somehow missed your latest reply)

I'm back home exercising after our first kinematics class. I see there are some things that are unclear to me.

In this equation,

http://img193.imageshack.us/img193/7749/thisform.jpg

Now I know that Vi stands for initial velocity, and that Vf stands for final velocity (not in this equation). But when I saw Vo I'm confused. As well as Xo. What does it mean? I'm only aware of initial (i), final (f), average (line on top) and difference of (delta)

 

[I like Serena]

Thanks studiot. (I somehow missed your latest reply)

I'm back home exercising after our first kinematics class. I see there are some things that are unclear to me.

In this equation,

http://img193.imageshack.us/img193/7749/thisform.jpg

Now I know that Vi stands for initial velocity, and that Vf stands for final velocity (not in this equation). But when I saw Vo I'm confused. As well as Xo. What does it mean? I'm only aware of initial (i), final (f), average (line on top) and difference of (delta)

V₀ is the initial velocity, or more precisely the velocity at time t=0.

x₀ is the initial x, that is, x at time t=0.
If you fill in t=0 in the equation, you'll see :)

 

[Femme_physics]

Hmm...

So
Xi = Xo
And
Vi = Vo

Just different ways of writing it?

 

[I like Serena]

Hmm...

So
Xi = Xo
And
Vi = Vo

Just different ways of writing it?

Yep!

 

[Femme_physics]

Thanks for clearing it up!

 

[Studiot]

I don't recommend using V_i as V(initial) since i is often used as a 'dummy' counter or index variable, both in mathematical expresions and computer programming.

eg

Sum = \sum\limits_{i = 1}^{i = n} {{P_i}}

or

for I=32 to 90
Print P(I)
next I

 

[Femme_physics]

Dooly noted! But what does "o" stand for? "original"?

 

[I like Serena]

Dooly noted! But what does "o" stand for? "original"?

It's not an "o". It's a zero. For t is zero seconds. :)

 

[Femme_physics]

I see. Math is certainly a language. Thank you, native speakers ;)

 

[Femme_physics]

In these formulas, does "V" mean "final velocity"?

http://img861.imageshack.us/i/formulaskin.jpg/

 

[Doc Al]

In these formulas, does "V" mean "final velocity"?

Yes. V is the velocity at time t (the 'final' velocity); V₀ is the velocity at time 0 (the 'initial' velocity).

 

[I like Serena]

In these formulas, does "V" mean "final velocity"?

http://img861.imageshack.us/i/formulaskin.jpg/

Yes, "V" means "final velocity".

More precisely "v" is the velocity after time "t" seconds.
Since you will use a time "t" that represents the time between the initial situation and the final situation, "v" will be the velocity in the final situation.

 

[Femme_physics]

Thanks Doc! This is like translating synonyms. I need a math thesaurus!

Edit: Thanks ILS too!

 

[Studiot]

Actually v on its own is a variable.

The value of v at any time, t , is given by the formula concerned.

v₀ is a fixed value (ie a constant).

So for instance the equation v = v₀ + at
should be read

At any time t, the instantaneously velocity, v, is given by adding the acceleration times the time to the initial velocity.

 

[rock.freak667]

Yes it does.

 

[Femme_physics]

So far I've been only using these two equations. I've been wondering if these are the only equations I need for acceleration/position/velocity problems.http://img831.imageshack.us/img831/4514/using.jpg ----------------------------------------------------------------------------------------------------I noticed that adding this equation below is completely useless since it's basically the same equation as the one above without acceleration! http://img715.imageshack.us/img715/30/noneedex.jpg

----------------------------------------------------------------------------------------------------

I've only been curious of the need to use these two equations below. Should I forgo putting them in my formulas page? They're only confusing me since I've been able to solve 3 exercises ignoring them.

http://img831.imageshack.us/img831/4797/notusing.jpg

 

[Studiot]

I'm not sure what x₀ is for, you will never need it if you start at x=0 at t=0

You have 5 quantities to play with.

Time
Distance
Initial speed at time t= 0
Speed at time t
Acceleration

You have to know at least 3 of these which means your require at most three equations.

Most courses use the following (in your notation)

v² = (v₀)² + 2ax

v = v₀ +at

x = v₀t + 1/2 a t²

Edit - see post27
Using these three you can take the three given quantities to calculate all possibilites.

As a matter of interest I found the most efficient method for exam purposes was to make the above list of quantities and equations and then 'fill in the gaps in the quantities list' by calculation with the equations.

Notice I said speed, not velocity.
The problem with using x for distance comes when you do projectile motion and need to resolve velocities in horiz and vertical directions.

go well

 

[Femme_physics]

With respect to this equation


EQUATION#1:

v² = (v₀)² + 2ax


It appears to me that I can derive anything I want by using the two equations you mentioned with it.

EQUATION#2:
v = v₀ +at

EQUATION#3:
x = v₀t + 1/2 a t²

So what can equation#1 give me that equations #2 and #3 can't?

 

[Doc Al]

So what can equation#1 give me that equations #2 and #3 can't?

It just saves you time for some problems.

 

[Femme_physics]

I see, I guess I'll see how it goes when I solve more problems and try to use it. Thanks :)

 

[Studiot]

Oops I guess I'm just having a bad hair day, you need to be told three quantities not 2.



I can't put it better than the sadly now defunct English Universities Press.

 

[Femme_physics]

Wow, that's a lot of info to shut me up :)

Appreciated.

 

[Femme_physics]

http://img269.imageshack.us/img269/5428/tryingtound.jpg

In this equation, where do I plug in the ²? Because, if I plug it after typing in the cosine function in my calculator without providing an angle, I'm getting an error! I just want to make sure, do I put the ² AFTER plugging the angle.

 

[BruceW]

it means:
( \cos( \alpha ) )^2
The notation is difficult to get used to at first

 

[BruceW]

(you can write it either way, they both mean the same thing, but physicists often write it the way in your textbook, because its easier)
lazy so-and-so's...

 

[Femme_physics]

it means:
( \cos( \alpha ) )^2
The notation is difficult to get used to at first

Thanks :) (missed that reply)

Another question with respect to these formulas.

Sometimes I see "g" with a minus before it. Sometimes I see "g" without a minus before it. If I decide to plug "g" as a minus-- where there is already a minus it's supposed to cancel-out, and where there is no minus it's actually plugged as a minus-- according to the formula. Or, do I always plug "g" without the sign, and then it accepts whatever sign the formula gives it?

 

[I like Serena]

Sometimes I see "g" with a minus before it. Sometimes I see "g" without a minus before it. If I decide to plug "g" as a minus-- where there is already a minus it's supposed to cancel-out, and where there is no minus it's actually plugged as a minus-- according to the formula. Or, do I always plug "g" without the sign, and then it accepts whatever sign the formula gives it?

Always accept the sign the formula gives for "g".

It should help if you start understanding the meaning of "g" and in particular in which direction it "works".

 

[Femme_physics]

Thanks :) Great, will do!

 

[rock.freak667]

Thanks :) (missed that reply)

Another question with respect to these formulas.

Sometimes I see "g" with a minus before it. Sometimes I see "g" without a minus before it. If I decide to plug "g" as a minus-- where there is already a minus it's supposed to cancel-out, and where there is no minus it's actually plugged as a minus-- according to the formula. Or, do I always plug "g" without the sign, and then it accepts whatever sign the formula gives it?

It depends on the convention used. For example if you have an object falling under gravity, so you only need to consider up and down motion, so the particle is moving down, you can take down as positive and use g = + 9.81 m/s².

If you take up as positive then g acts downwards and thus g = -9.81 m/s².

In cases where the formula is like v=u-gt, then down is negative and the minus sign covers for that, so you'd just put in g= 9.81 into the equation.

 

[I like Serena]

It depends on the convention used. For example if you have an object falling under gravity, so you only need to consider up and down motion, so the particle is moving down, you can take down as positive and use g = + 9.81 m/s².

If you take up as positive then g acts downwards and thus g = -9.81 m/s².

In cases where the formula is like v=u-gt, then down is negative and the minus sign covers for that, so you'd just put in g= 9.81 into the equation.

I'm sorry, I'm going to have to disagree here.

"g" is a physical constant and never has a sign. It is just g = 9.81 m/s².

This is opposed to usage of for instance a generic acceleration "a" of which the direction is unspecified, which might for instance be -g or +g.

If you look at the formulas in the first post, you'll see any sign of "g" is definitely accounted for! ;)

 

[Femme_physics]

I thought rock.freak's reply had somewhat contradicted yours, but yours did make more sense to me. I appreciate anyone's feedback, regardless! :)

 

[rock.freak667]

I'm sorry, I'm going to have to disagree here.

"g" is a physical constant and never has a sign. It is just g = 9.81 m/s².

This is opposed to usage of for instance a generic acceleration "a" of which the direction is unspecified, which might for instance be -g or +g.

If you look at the formulas in the first post, you'll see any sign of "g" is definitely accounted for! ;)

My post was supposed to make the point that your sign for 'g' would depend on what direction you use as positive, not that 'g' will be -9.81 m/s² regardless of direction chosen.

 

[Doc Al]

My post was supposed to make the point that your sign for 'g' would depend on what direction you use as positive, not that 'g' will be -9.81 m/s² regardless of direction chosen.

The point that I Like Serena was making is that while the acceleration can be +g or -g depending on your sign convention, g itself is just the magnitude of the acceleration due to gravity. It's always positive (unsigned, actually).

 

[BruceW]

"g" is a physical constant and never has a sign. It is just g = 9.81 m/s2

This is true, but if she gets an exam that explicitly gives g = -9.81 m/s^2, and then this equation (for example):
depth = \frac{gt^2}{2}
Then she should use the given value for g.
In other words, the notation which the exam gives is the most important.
So here, the examiner is looking for a negative value for depth, but if she used g as positive, then she'd get an incorrect positive value for depth. (Incorrect in the sense that it doesn't conform to the examiner's notation).

 

[Femme_physics]

Eep! I hope the system would do its best to not confuse me!

Thanks for the heads up, and the input! :)

 

[nrqed]

Eep! I hope the system would do its best to not confuse me!

Thanks for the heads up, and the input! :)

Bonjour!

J'espère que la préparation pour tes examens se passe bien!



Some people do use g to be -9.81 m/s^2 although they are the exceptions.
But I can tell from your formula sheet that your class is using g = 9.81 m/s^2.



So if your teacher and textbook are consistent, you should be using a positive value all the time, with no exception.

Now, if sometimes your teacher or your book chooses to put the positive y-axis pointing down, all the signs of the velocities (V, Vi) as well as the acceleration will flip signs in the equations, but g will remain positive. Hopefully your teacher does not do that as it is always confusing to students. I avoid doing this in my classes.

Bonne chance dans tes études!