8
   

Can an object be accelerating and yet -not- moving?

 
 
dalehileman
 
  1  
Reply Tue 26 Feb, 2013 01:28 pm
@contrex,
Quote:
I never said that nor withdrew it either.
Yes, no, sorry, Con, it was Drew:

Quote:
There is a point in time when your speed is still zero, but your acceleration is some amount greater than zero.
contrex
 
  1  
Reply Tue 26 Feb, 2013 01:36 pm
@dalehileman,
dalehileman wrote:

Quote:
I never said that nor withdrew it either.
Yes, no, sorry, Con, it was Drew:


So he with-Drew it?
dalehileman
 
  1  
Reply Tue 26 Feb, 2013 02:09 pm
@contrex,
Con for the seventh time--by actual count--you've again made my entire morning if not day
contrex
 
  1  
Reply Tue 26 Feb, 2013 02:16 pm
@dalehileman,
dalehileman wrote:

Con for the seventh time--by actual count--you've again made my entire morning if not day


It's nice of you to say so. I have to say I thought this thread was getting a bit over serious.
dalehileman
 
  1  
Reply Tue 26 Feb, 2013 02:19 pm
@contrex,
Quote:
a bit over serious
….to say the very least
0 Replies
 
MattDavis
 
  1  
Reply Tue 26 Feb, 2013 04:26 pm
@dalehileman,
Not to get all "deconstructionist" or anything but it will depend on what is a "practical situation". Wink
This question was originally posted (I think) to test understanding from the Newtonian mechanics frame of looking at reality. I doubt there is any regard given to Plank's constant in terms of spacial quantization. For most "practical" purposes the calculus answer is pretty damn good. Very Happy
On a quantum level maybe not so much, but this doesn't make much difference on the scale of balls thrown in the air.
contrex
 
  1  
Reply Tue 26 Feb, 2013 04:40 pm
@MattDavis,
MattDavis wrote:
Plank's constant


Is that a wooden person?
MattDavis
 
  1  
Reply Tue 26 Feb, 2013 04:47 pm
@contrex,
contrex wrote:
Is that a wooden person?

Haha... yes it probably is, but it is also this....
Quote:
a physical constant that is the quantum of action in quantum mechanics.

http://en.wikipedia.org/wiki/Planck_constant Very Happy
dalehileman
 
  1  
Reply Tue 26 Feb, 2013 06:37 pm
@MattDavis,
Quote:
…...will depend on what is a "practical situation".
Again forgive me Matt if sometimes I'm not xtal clear

I see the ball looked at in two different ways, yielding two different answers to the OP: The theoretical in which there's no stopping of the ball and the practical where the peak really does flatten a bit

..the difference between just thinking about it and throwing an actual ball up into the air

But Matt I don't see how Newtonian mechanics, Plank's constant, spacial quantization, the calculus, or quantum level have any pertinence whatever to this difference

If you're trying to impress me with your erudition, then you've succeeded admirably
MattDavis
 
  1  
Reply Tue 26 Feb, 2013 06:47 pm
@dalehileman,
Nothing to forgive Very Happy
I understand where you're coming from (I think).
If you were asked for an answer to a homework problem in a classical mechanics physics class, the answer would be that:
"Yes. An object (macroscopic) can be both still (velocity=0) and accelerating (velocity in the process of change)."
dalehileman
 
  1  
Reply Tue 26 Feb, 2013 07:39 pm
@MattDavis,
Quote:
Nothing to forgive
Why thank you Matt. From some responses I'd assume I was upsetting nearly everyone

Quote:
I understand where you're coming from (I think).
I'm not sure I do any more


Quote:
…..the answer would be that:
"…...can be both still (velocity=0) and accelerating……."
Oh yes yes, I do agree

Not to backtrack however, there's one interesting aspect of the whole thing that still intrigues me. I understand that the peak is round but unlike the situation of the automobile starting, for that instant of zero duration the peak is flat, suggesting, just like the different kinds of infinity, a different kind of zero from the typical situation where one straight line crosses another

…...and yes I realize you can restore the curvature if you also stretch out the vertical. Still it's flat for zero seconds, stopped

Intriguing
engineer
 
  1  
Reply Tue 26 Feb, 2013 07:55 pm
@dalehileman,
You are way over-thinking this. The position of the ball forms a parabola. It does up, it comes down. The velocity of the ball linearly drops from X upward to X downward. If the entire flight takes four seconds, at one second, the velocity is X/2. At three seconds it is -X/2. At two seconds it is zero. When the velocity goes from positive to negative, the ball stops moving momentarily. The acceleration never changes. It is always negative (never zero, never changing) and is solely due to the force of gravity.
MattDavis
 
  1  
Reply Tue 26 Feb, 2013 08:03 pm
@dalehileman,
Perhaps this will help:
It is for an actual "ideal" pendulum.
Notice first the velocity arrow does shrink to zero at the peaks.
Notice next that the acceleration arrows are never absent (at the peaks), though they change direction.


If you want to expand on this even more:
Look into the concept of jerk.
I remember in high school doing some "work" on this regarding pendulums.
http://en.wikipedia.org/wiki/Jerk_%28physics%29
Boomslang
 
  1  
Reply Tue 26 Feb, 2013 09:05 pm
@MattDavis,
There is nothing more to look at, in regards to jerk, for understanding this idea. Jerk is not involved with the momentary instance where velocity is zero while acceleration is nonzero. I suspect bringing the topic of jerk into this discussion will confuse those who don't understand this simple concept yet, even more.
georgeob1
 
  1  
Reply Tue 26 Feb, 2013 09:09 pm
@Boomslang,
Perhaps so, but we may find some of our intuitive jerks have a natural affinity for it.
0 Replies
 
MattDavis
 
  1  
Reply Tue 26 Feb, 2013 09:11 pm
@Boomslang,
Thanks.
I agree it may be a digression, but I think that jerk is a natural extension of the theme: position, velocity, acceleration...jerk.
Jerk does pertain to a pendulum's movement and that was the "idea" I was referencing.
In my defense:
The topic has already been down the long and winding path of digression.
Boomslang
 
  1  
Reply Tue 26 Feb, 2013 09:34 pm
@MattDavis,
MattDavis wrote:
Jerk does pertain to a pendulum's movement.

The acceleration is constant for a simple pendulum. Throughout the bob's journey, the acceleration is the same.
a = -g
When the acceleration is constant, jerk doesn't come into play.
MattDavis
 
  1  
Reply Tue 26 Feb, 2013 09:35 pm
@Boomslang,
Incorrect.
Note that acceleration is a vector not a simple magnitude.
If you would like you can watch the video, notice the changing "a" vector.
Boomslang
 
  1  
Reply Tue 26 Feb, 2013 10:14 pm
@MattDavis,
There is a tangential vector and a perpendicular vector. It may not fall at a constant acceleration, but the resultant vectors at any point adds up to the value of gravity. Thus we can calculate gravity from a simple pendulum. Here is a nice diagram that illustrates the resultant vectors by itself very well.
http://upload.wikimedia.org/wikipedia/commons/2/21/Pendulum_animation.gif
Tension may change, causing the net forces to change, however the magnitude of the resultant acceleration vectors, or the sum of the two vectors, always is a constant, or g.
MattDavis
 
  1  
Reply Tue 26 Feb, 2013 10:31 pm
@Boomslang,
The bob on the end of the pendulum.
It's movement can be described with a change in position over time (velocity).
It's movement can be described with a change in velocity over time (acceleration).
It's movement can be described with a change in acceleration over time (jerk).

With respect to a "stationary" reference frame
(as in the way you can sit and look at it, as if it is right in front of you, like on a computer screen):
The bob's position is never constant.
The bob's velocity is never constant.
The bob's acceleration is never constant.
---------------------------------------------------------------------------------
I don't understand your question.
Which two vectors are you asking to add together?
The tangential acceleration vector and the perpendicular acceleration vector?
The vectors in your video are a velocity vector and an acceleration vector.
Are you asking what will happen if you add those together?
 

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