Luigi Fortunati <
fortunati.luigi@gmail.com> wrote:
A similar situation to my previous animation is the following: https://www.geogebra.org/m/gbqzyba2
In the inertial reference frame of the ground, when bodies are still,
there is no centripetal force and there is no centrifugal force either.
During the rotation, the blue centripetal force is activated which is
the only force acting on body A.
I would phrase this slightly differently: in the rotating state there
*must* be a "blue" centripetal force acting on each body, otherwise the
body would move in a straight line, not in a circle.
Instead, on body B (full of water but with the same overall mass as
body A), in addition to the blue force, the red force of the water
directed outwards (i.e. in the opposite direction to the centripetal
force) also acts.
The mass of the empty B must be less than the mass of A (the difference
being the mass of the water). So, the overall centripetal force acting
on the water-filled B must be sum of the centripetal force which would
be required to give the empty B that same centripetal acceleration, plus
the centripetal force which is required to give the water its centripetal acceleration. (*)
If we approximate the diameter of B as small relative to the radius
of B's orbit around the rotation axis, then the centripetal acceleration
is uniform across the diameter of B.
Let's introduce a few symbols:
M_A = mass of A
M_B = mass of the solid body B without the water
M_W = mass of the water
a = the magnitude of the centripetal acceleration
T = the tension in the string that ties each body to the rotation
axis; this tension is the same for A and B since M_A = M_B + M_W
Then by Newton's 2nd law, the net force (centripetal) acting on A
must be given by $T = M_A a$, and the net force (again centripetal)
acting on the water-filled B must be
T = (M_B + M_W) a = M_B a + M_W a (1)
which is just the statement (*) above.
Ultimately the only thing exerting forces on the water is B, i.e.,
the solid body B must exert an inward (centripetal) net force of
$M_W a$ on the water, and by Newton's 3rd law the water must exert
a corresponding (equal in magnitude, opposite in direction) centrifugal
force $M_W a$ on the solid body B.
As is often the case in Newtonian mechanics, it's instructive to draw
some free-body diagrams. I'll show these for the case where A and B
have each rotated 180 degrees, i.e., they are directly to the right
of the rotation axis, so their centripetal accelerations point left:
Free-body diagram for A:
<-------- A
There is only one force acting on A: the string tension $T$, pointing
inwards towards the rotation axis (to the left). This is shown in blue
in Luigi's animation.
Free-body diagram for the solid body B (note we are considering the water
to be a separate body):
<-------- B ----->
There are two forces acting on the solid body B (again, we are considering
the water to be a separate body):
* the string tension $T$ to the left; this is shown in blue in
Luigi's animation
* the force the water exerts on B, $M_W a$, to the right; this is
shown in red in Luigi's animation, but the animation wrongly shows
the magnitude of this force as being the same as that of the blue
force $T$; it's actually smaller than $T$
The net force on B is the algebraic sum of these two forces, $T - M_W a$
to the left. This is smaller than the net force acting on A, which is just what we expect, since A and B have the same (centripetal) acceleration $a$,
but the mass of the solid body B is less than the mass of A. In fact,
F_net = T - M_W a (2)
= (M_B + M_W) a - M_W a [by equation (1)] (3)
= M_B a (4)
And finally, here's a free-body diagram for the water:
<----- W
There is only one force acting on the water, which is exerted by B
and which points inwards towards the rotation axis (i.e., to the left).
The magnitude of this force is $M_W a$.
We can also analyze the forces between B and the water in detail, but
that takes a little bit more work (and some elementary calculus):
To keep things simple, I'll assume that the water in B rotates like
a rigid body, with the same angular frequency as B. (If the rotation
of B is uniform and lasts long enough, then the small-but-nonzero
viscosity of water will result in the water's velocity field gradually approaching this rigid-body-rotation state.)
Assuming this rigid-body rotation, let's consider what force(s) act
on each little (infinitesimal) volume element of water. In particular, consider an infintesimal volume element of mass $dm$, which is at a
radius $r$ from the rotation axis.
There are two physical processes which can exert forces on this water
element:
(a) there may be a pressure gradient in the water, and
(b) the walls of B may directly exert forces on the water immediately
adjacent to the walls
Consider process (a): if we take our infinitesimal volume element to
of water be a cylinder with flat sides of area dA facing towards & away
from the rotation axis, and to extend from radius-from-rotation-axis r
to r+dr, then we have
dm = dA dr rho (5)
where rho is the density of water (we assume rho to be constant, i.e.,
we're neglecting the compressibility of water). The net force acting
on the volume element of water due to the pressure gradient is then
the difference between the forces acting on the two opposite flat faces
of the volume element, i.e.,
dF_net = p(r+dr) dA - p(r) dA (6)
Also, by Newton's 2nd law (since this volume element of mass dm is moving
with centripetal acceleration $a$), there must be a net force acting on
the volume element of
dF_net = dm a (7)
= dA dr a [here we're using (5)] (8)
Comparing (6) and (8), we then have
dA dr rho a = (p(r+dr) dA - p(r)) dA (9)
where p = p(r) is the pressure as a function of position. Dividing
by dA, this gives
dr rho a = p(r+dr) - p(r) (10)
and hence
rho a = dp/dr (11)
and hence
p(r) = K + rho a r (12)
where $K$ is an arbitrary constant. If we think about the case where
$a = 0$, i.e., there is no rotation and hence no centripetal acceleration,
we can see that the constant $K$ is just the initial water pressure in
this non-rotating state (i.e., the water pressure in B before B started rotating).
Equation (12) tells us how the water pressure varies with position inside
B. In particular, (12) says that the water pressure is larger on the side
of B which is farther away from the rotation axis, and smaller on the side
of B which is closer to the rotation axis. (If the constant $K$ is small enough, the pressure may even be negative in some/all of the water;
this corresponds to "tension".)
Now considering process (b) above, at each point on B's inside surface
the water pressure acts perpendicular to B's inside surface, and by
Newton's 3rd law B's inside surface exerts an equal and opposite force
(also perpendicular to B's surface) on the water.
Now let's temporarily assume that $K$ is large and positive, so that
the pressure given by (12) is always positive. Then:
Since the water pressure is larger on the side of B which is farther
away from the rotation axis, where the water exerts a force on B
*away* from the rotation axis, and smaller on the side of B which is
closer to the rotation axis, where the water exerts a force on B
*towards* the rotation axis, it's easy to see that the net force of
the water on B must point *away* from the rotation axis. (We could
work this force out in detail by integrating the radial component of
the right hand side of (12) over B's inner surface.) We saw above
that this force has magnitude $M_W a$.
Similarly, since the pressure B's inside surface exerts on the water
is larger on the side of B which is farther away from the rotation axis,
where this force points *towards* the rotation axis, and smaller on
the side of B which is closer to the rotation axis, where this force
points *away* from the rotation axis, it's easy to see that the net
force of B on the water must point *towards* the rotation axis.
(Again, we could work this out in detail by integrating the radial
component of the right hand side of equation (12) over B's inner
surface.) Again, we saw above that this force has magnitude $M_W a$.
Finally, let's return to our temporary assumption that $K$ is large
and positive, so that the pressure given by (12) is always positive.
Notice that $K$ corresponds to a pressure which is constant everywhere
on B's inside surface. Thus, by symmetry, changing $K$ does *not* change
the net force the water exerts on B, or the net force B exerts on the
water. Thus, our conclusions in the previous two paragraphs about
these net forces don't depend on the value of $K$, and so remain valid
for an arbitrary $K$. That is, we can now drop our temporary assumption
that $K$ was large and positive.
--
-- "Jonathan Thornburg [remove -color to reply]" <
dr.j.thornburg@gmail-pink.com>
on the west coast of Canada
"Now back when I worked in banking, if someone went to Barclays,
pretended to be me, borrowed UKP10,000 and legged it, that was
`impersonation', and it was the bank's money that had been stolen,
not my identity. How did things change?" -- Ross Anderson
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