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What does rest mean?

by tony873004
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tony873004
#1
Aug26-12, 02:53 PM
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If I throw a ball straight up, at its highest point, is it momentarily at rest? Its velocity is momentarily 0, but it is still accelerating.
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phinds
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Aug26-12, 03:02 PM
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Quote Quote by tony873004 View Post
If I throw a ball straight up, at its highest point, is it momentarily at rest?
Yes, "at rest" means "not moving in the chosen frame of reference". In the frame of reference of a spot on the moon, for example, it is never at rest throughout your example, but I'm assuming you mean relative to YOU as the frame of reference.
ModusPwnd
#3
Aug26-12, 03:02 PM
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I would consider rest to be v=0, regardless of what a or any other higher derivative is. I think this is pretty standard.

russ_watters
#4
Aug26-12, 03:02 PM
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What does rest mean?

I always took it to mean not moving, so the object can still be accelerating.
mrspeedybob
#5
Aug26-12, 03:53 PM
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Quote Quote by russ_watters View Post
I always took it to mean not moving, so the object can still be accelerating.
Agreed. This is not even limited to the given example of a ball at the top of it's trajectory. In that example the ball is at rest for an instant of zero duration as it reverses direction. If I am in a free falling elevator and have a ball in my hand the ball is at rest in my frame of reference for period of non-zero duration (until we hit the ground). An observer on the ground sees the ball and I accelerating for the duration of the experiment.
tony873004
#6
Aug26-12, 04:09 PM
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But does "not moving" mean v=0? Rather, could moving mean that you are not where you were some amount of time ago? In other words, in the following formula, does Δy = 0 represent not moving, or does v0 = 0 mean not moving?

[tex]\Delta y = v_{0}t + 0.5 at^{2}[/tex]

Any non-0 number for t will result in an accelerating object to have a non-zero Δy, even if v0 = 0.

My curiosity is from a problem where 5 free-body diagrams are shown: A, B, C, D, E.
It asks "Which of the following could represent an object at rest?"
If we take v0 = 0 to be rest, then the answer is "all of them".
If we take Δy = 0 to be rest, then the answer is only the ones with balanced forces.
Zula110100100
#7
Aug26-12, 04:32 PM
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Isn't v[sub]0[\sub] the initial velocity, and (at) from the .5at^2 is the v(current), since the velocity only starts at 0 of course there is a change in y. In what situation can you have Δy!=0 and v = 0? I would consider v = Δy/t (if we are talking that one dimension only)
tony873004
#8
Aug26-12, 04:39 PM
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Rather than considering t being the time I threw the ball straight up, I was thinking of initial time being the moment of its highest point. So v0 = 0 and t is actually Δt, some non-zero amount of time after the instant of the highest point.
Zula110100100
#9
Aug26-12, 06:38 PM
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Meh, I am confused...say t=0 is the time it's at the top, at that instant v0 = 0. So if we are talking a non-zero Δt, then for that time it accelerated a, so v = a(Δt)... Moving means exactly not being where you were some time ago, which in a 1d world, means exactly having a Δy in some Δt.... v = Δy/Δt... so far as I know having a velocity is the only way to accomplish a change in y...

To the main point, at rest means v = 0, it means that the distance between the object at rest and the reference point is constant
A.T.
#10
Aug27-12, 05:57 AM
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Quote Quote by tony873004 View Post
My curiosity is from a problem where 5 free-body diagrams are shown: A, B, C, D, E.
It asks "Which of the following could represent an object at rest?"
If we take v0 = 0 to be rest, then the answer is "all of them".
If we take Δy = 0 to be rest, then the answer is only the ones with balanced forces.
Yes, in problem statements like this the term "being at rest" is sometimes used for "remaining at rest for a non-zero time period". It's a bit ambiguous and you have to get from the context what is meant.
Naty1
#11
Aug27-12, 06:41 AM
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If I throw a ball straight up, at its highest point, is it momentarily at rest? Its velocity is momentarily 0, but it is still accelerating.

yes,no, wait......., no...er...... who knows??

One could argue the average acceleration Δv/Δt is not zero while dv/dt is zero....oh well,
let's just say 'words are often ambiguous'...

My old Halliday and Resnick does not even have 'rest' or 'at rest' in the index.
K^2
#12
Aug27-12, 07:57 AM
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Quote Quote by Naty1 View Post
yes,no, wait......., no...er...... who knows??

One could argue the average acceleration Δv/Δt is not zero while dv/dt is zero....oh well,
let's just say 'words are often ambiguous'...
dv/dt is non-zero. There is absolutely nothing ambiguous about it. The object undergoes constant acceleration. That means dv/dt=-g at all times, regardless of whether v is zero or not.
HallsofIvy
#13
Aug27-12, 08:01 AM
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Thanks
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"At rest" means that, at least momentarily, the velocity is 0. That has nothing at all to do with the acceleration.
genericusrnme
#14
Aug27-12, 08:06 AM
P: 615
Rest means v=0 in some chosen frame of reference, regardless of acceleration, jerk or whatever other higher derivatives you want to concider.
You could define 'super rest' or something, if you wanted, which would mean v=0 and a=0 but I'm guessing from it's lack of useage that it's not that useful of a concept.
I think the problem here is that you are doing physics with no calculus, go look up some khan academy videos and learn aboud calculus and then stationary points.
azizlwl
#15
Aug27-12, 08:32 PM
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Very good information.
Since there is no icon that I click that I can follow this tread, then i have to make a reply.

Since there is no Δy in Δt for v=0 but for acceleration Δy is not zero in (Δt)2 at the top the flight.
genericusrnme
#16
Aug28-12, 07:01 AM
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Quote Quote by azizlwl View Post
Very good information.
Since there is no icon that I click that I can follow this tread, then i have to make a reply.
There is, at the top there is a thread tools drop down, then in that there is the option to 'subscribe to thread'
azizlwl
#17
Aug28-12, 07:49 AM
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Quote Quote by genericusrnme View Post
There is, at the top there is a thread tools drop down, then in that there is the option to 'subscribe to thread'
Thanks so much.


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