# The meaty part of centripetal acceleration

Discussion in 'Physics & Math' started by nicholas1M7, Jul 29, 2011.

1. ### AlexGLike nailing Jello to a treeValued Senior Member

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I'm sorry. I didn't realize that you were stupider than I thought you were.

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3. ### Oniw17ascetic, sage, diogenes, bum?Valued Senior Member

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I like it. It reminds me of the old sciforums.

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5. ### nicholas1M7BannedBanned

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Acceleration works with speed by including it.

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7. ### nicholas1M7BannedBanned

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You're wrong. Quite the opposite.

8. ### AlphaNumericFully ionizedRegistered Senior Member

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You seem capable of engaging in basic discussions, even forming coherent sentences at times, yet you make proclamations which aren't coherent. For example, you just said

That is meaningless. How does acceleration work with speed by 'including it'? Including it in what? They are related but no one would use the word 'including' when describing that relationship.

An object can experience an acceleration yet its speed remain unchanged. Can you answer my question? I asked you if you knew why a speed increase occurs if and only if $\mathbf{v}\cdot \mathbf{F} \neq 0$. Come on, you disagreed with Alex calling you stupid. What I've just asked you is high school level stuff but you've already failed to get that sort of stuff right already in this thread.

9. ### nicholas1M7BannedBanned

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Of course it is including speed. Of course we can use it to say so about the relationship between acceleration and speed as expressed in the equation. One is the other in a way.

Is it because either force or velocity has a value other than 0?

10. ### ChopsockyRegistered Member

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But what about acceleration that doesn't involve the changing of speed, but of the changing of direction?

Acceleration is simply the way we express the rate of the change in speed or direction. Often we'll say velocity, which is (as already mentioned) a vector of speed and direction. So when people say an acceleration in velocity, they're not always referring to speed.

11. ### Read-OnlyValued Senior Member

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For a self-proclaimed "genius", you certainly think and talk in a VERY childish manner! You should really remove that statement from your profile. Ugh!

One is most DEFINITELY NOT the same as the other in ANY way. Either your skill at applying the English language is very poor or your knowledge of basic physics is horrible! Which is it??

12. ### nicholas1M7BannedBanned

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You're wrong on both counts.

13. ### James RJust this guy, you know?Staff Member

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Yes. Velocity is a vector, which has magnitude and direction. If either or both of those change, you have an acceleration.

In a rotating, non-inertial reference frame, yes. In an inertial frame, no.

Acceleration is the rate of change of velocity. Speed is the magnitude of velocity.

It can. Or it can cause a decrease in speed. But it may also cause no change in speed at all - only a change in direction.

14. ### AlphaNumericFully ionizedRegistered Senior Member

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No, they aren't equivalent. If someone tells you the speed of an object over some period of time you cannot always determine the acceleration that object has been experiencing.

The example of a circular orbit proves this. An object in a circular orbit has constant speed but its velocity and the acceleration it experiences are always changing. Anyone who knows the basics of mechanics knows this.

Well done on showing you can't do high school calculus. I used the dot product, which is zero if the two forces are at right angles to one another. This is how an object can have non-zero ever changing velocity and non-zero ever changing acceleration yet its speed stay the same.

You won't understand it but here is the explicit formula for circular orbit position, velocity and acceleration :

Position : $\mathbf{x}(t) = A(\sin \omega t , \cos \omega t)$
Velocity : $\mathbf{v}(t) = \dot{\mathbf{x}}(t) = A\omega(\cos \omega t , -\sin \omega t)$
Acceleration : $\mathbf{a}(t) = \dot{\mathbf{v}}(t) = \ddot{\mathbf{x}}(t) = -A\omega^{2}(\sin \omega t , \cos \omega t)$
From which it follows that
Speed : $s = |\mathbf{v}(t)| = A\omega$
$\mathbf{v}\cdot \mathbf{a} = 0$

Despite the position, velocity and acceleration all being time dependent (they change as time progresses) the speed of the object is constant. Thus proving if you're given the speed of an object you cannot recover the acceleration uniquely.

15. ### Read-OnlyValued Senior Member

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HA!!! Your own words prove otherwise.

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16. ### nicholas1M7BannedBanned

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Okay, I'm not disagreeing with you here, I'm simply saying that acceleration includes speed for linear examples, not circular ones like you are saying.

But you are using the example of circular/non-linear acceleration. Where exactly would my statement apply? Does it only apply only to linear acceleration?

17. ### nicholas1M7BannedBanned

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Are you saying that not only a magnitude change results in acceleration, but a change in direction as well results in acceleration? So can I conclude that for at least linear examples, that acceleration includes speed?

18. ### originHeading towards oblivionValued Senior Member

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You need to go into more detail. Saying, "acceleration includes speed", is a really strange statement.

Do you mean that acceleration and speed both involve motion?

Speed is simply the motion of an object for a given time or distance/time

Uniform acceleration for a linear case is change in the speed of an object for a given time or (the change in the speed)/time.

19. ### nicholas1M7BannedBanned

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Acceleration works with speed by including it in its equation.

20. ### originHeading towards oblivionValued Senior Member

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You mean like dV/dt?

Do you mean by the comment, "acceleration works with speed" that the change in velocity is in the equation for instantaneous acceleration.

If that is what you are trying to say, well no kidding, it is trivialy obvious...

Not exactly a deeper level:shrug:

21. ### nicholas1M7BannedBanned

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Sorry for your disappointment.

22. ### originHeading towards oblivionValued Senior Member

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I'll get over it.

23. ### AlphaNumericFully ionizedRegistered Senior Member

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What's that I hear? Why, it's the sound of back peddling!