Luigi Fortunati <
fortunati.luigi@gmail.com> wrote:
[[previous animation of finger pushing trolley, which pushes on
a spring, compressing the spring]]
So, the animation I made is not good and I changed it like that <https://www.geogebra.org/m/an3veznf>
eliminating the trolley.
Now, the tip of the finger acts directly against the tip of the spring
(which reacts).
Your description suggests that the trolley begins at rest and
instantaneously starts moving when the animation starts. That
(an instantaneous jump in velocity) can't happen.
You're right.
The change of speed from zero to v cannot be instantaneous and, then,
let's say that it happens in a hundredth of a second (any other time
interval is fine too).
I would like to make the animation (necessarily in slow motion) with
this initial acceleration but I need confirmation: in this time interval
is it correct that the blue force is greater than the red one?
There are 3 forces of particular interest here:
A: The finger exerts a rightward force on the point P which is at the
end of the spring.
B: The Newton's-3rd-law reaction to force A is that the end of the
spring exerts a leftward force on the hand. By Newton's 3rd law,
this force B has the same magnitude as force A.
C: If the spring is compressed, the body of the spring exerts a leftward
force on the end of the spring (the point P). Assuming that the spring
obeys Hooke's law, this force C is proportional to the distance the
spring is compressed. Notice that this means that force C depends on
the distance the spring is compressed, but NOT directly on the force A.
The net force on the end of the spring (point P) is the signed sum of A
and C. That is, since A acts to the right and C acts to the left, the
net force to the right acting on P is A - C. This means that by Newton's
2nd law,
(rightward acceleration of P) = (A - C)/M
where M is the "effective mass" of the spring (not all of the spring's
mass moves with the same acceleration, so the "effective mass" is going
to be something like 1/2 the actual mass of the spring).
Initially the spring isn't compressed, so C = 0. Therefore, to
initially accelerate the end of the spring (point P) to the right, A
must be positive. As the spring begins to compress, C will increase,
and in order to continue accelerating point P to the right, A - C must
remain positive.
Once the end of the spring (point P) has reached a suitable velocity,
A should be decreased to equal C, so that A - C = 0. To maintain P at
a constant velocity, as the spring compresses and C increases, A should
be increased so that A continues to equal C.
Thus a free-body diagram of the end of the spring (point P) at the
initial time (just after the force A is "turned on" would look like this:
P -----------> (right arrow = A, force C = 0)
Somewhat later, but still during the initial acceleration phase, the
free-body diagram might look like this (I've chosen A to increase so
that A - C is a constant during the acceleration phase, giving a constant acceleration of point P):
<---- P ---------------> (right arrow = A, left arrow = C)
During the later constant-velocity phase, the free-body diagram might
look like this:
<------- P -------> (right arrow = A, left arrow = C)
For your animation, the motion of the end of the spring (point P) should
be governed by Newton's 2nd law, i.e., at each time step in your animation,
C = determined by how far the spring is compressed
A = you get to choose this to give whatever acceleration profile
for P you want
(rightward acceleration of P) = (A - C)/M
Note that B doesn't appear in the dynamics of point P because it doesn't
act on point P. Rather, B acts on the finger, and we can assume that
the arm/hand exerts whatever forces on the finger are needed so that the finger's motion is whatever we desire.
As to your question (about equality or inequality of the red & blue
forces in your animation), I'm a bit confused by the arrows in your
animation: Your animation shows a red arrow labelled "Finger" pointing
to the right, but the legend below the animation shows "Finger action"
with a blue arrow pointing to the right. And the animation shows a
blue arrow labelled "Spring" pointing to the left, but the legend shows
"Spring reaction" with a red arrow pointing to the left.
What we can say for sure is that B = A (these are a Newton's-3rd-law action-reaction pair), while C varies with the compression of the spring. Remember, A and C both act on the end of the spring (point P), whereas
B acts on the finger.
For a future version of your animation, it might be useful to use 3
different colors for the 3 forces I've described as A, B, and C. It
might also be useful to show a free-body diagram of the end of the spring (point P) at each time step.
--
-- "Jonathan Thornburg [remove -color to reply]" <
dr.j.thornburg@gmail-pink.com>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently on the west coast of Canada
"Why would we install sewers in London? Everyone keeps getting cholera
again and again so there's obviously no reason to install sewers. We
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