Linear Acceleration

Rigid-body Subjected to Accelerations 75
Basic Equation for Rigid-Body Subjected to Acceleration
Although we have proved that the pressure at a given point has the
same magnitude in all directions, and thus it is a scalar functions
question, we are now faced with an equally important question- how
does the pressure in a fluid in which there are no shearing stresses vary
from point to point?. To answer this question, we consider a small
element of fluid removed from some arbitrary position within the mass of
fluid of interest as illustrated in Fig. (1). There are two types of forces
acting on this element:
 Surface forces (due to the pressure), and
 Body force (equal to the weight of the element)
x
y
z
z
y
x
g 




z
x
2
y
y
p
p 









 



j
i
k
y
x
2
z
z
p
p 









 



y
x
2
z
z
p
p 









 



z
x
2
y
y
p
p 









 



x

y

z

Fig. (1) – The Surface and Body Forces acting on a Differential Fluid
Element in Vertical and Horizontal Directions
If we let the pressure at the center of the element be designated as p,
then the average pressure on the various faces can be expressed in

terms of P and its derivation, as shown in Fig. 1 (the surface forces in
the x-direction are not shown). The resultant surface force in the y-
direction is
z
x
2
y
y
p
p
z
x
2
y
y
p
p
Fy 








 














 






or
z
y
x
y
p
Fy 








Similarly, for the x- and z-directions the resultant surface forces are:
z
y
x
x
p
FX 







 and z
y
x
z
p
FZ 








The resultant surface force acting on the element can be expressed as:
k
F
j
F
i
F
F z
y
x
s










or
z
y
x
k
z
p
j
y
p
i
x
p
Fs 























The pressure gradient in vector form
where i

, j

, and k

are the unit vectors along the coordinate axes.
The group of terms in parentheses can be written as:
p
k
z
p
j
y
p
i
x
p









 
The symbol  is the gradient or “del” vector operation. Thus, the
resultant surface force per unit volume can be expressed as
p
z
y
x
Fs








Since the z axis is vertical, the weight of the element is
z
y
x
g
k
W 








where the negative sign indicates that the force due to the weight is
downward (in the negative z direction).
Newton’s second law, applied to the fluid element, can be expressed as
a
m
F 




where:
F : represents the resultant force acting on the element,
a : is the acceleration of the element, and
m
 : is the element mass, which can be written as z
y
x
g 



a
m
k
W
F
F s 









or
a
)
z
y
x
(
k
)
z
y
x
(
g
)
z
y
x
(
p 

















and cancelling z
y
x 

 , the general equation of motion for a fluid that
acts as a rigid body (no shearing stresses) is determined to be
a
k
g
p 






……………….. (1)
this can be expressed in scalar form in the three orthogonal directions,
as
 x
a
x
p





, y
a
y
p





and )
a
g
(
z
p
z






…………….. (2)

Application (i) – Rigid-Body at Rest
For fluid at rest or moving on a straight path at constant velocity, all
components of acceleration are zero, and the relation in Eq. (2) reduces
to
0
x
p



, 0
y
p



, and g
z
p





(fluid at rest) ……. (3)
which confirm that, in fluid at rest, the pressure remains constant in
horizontal direction (p is independent of x and y) and varies only in the
vertical direction as a result of gravity.
Application (ii) – Rigid-Body in Free Fall
A freely falling body accelerates under the influence of gravity, Fig. (2).
When the air resistance is negligible, the acceleration of the body equals
the gravitational acceleration and accelerate in any horizontal direction is
zero. Therefore,
0
a
a y
x 
 , and g
az 
 (fluid accelerates downward) ……. (4)
Then, Eq. (2) reduces to
0
z
p
y
p
x
p









, (p = constant)……. (5)
When the direction of motion is reversed and the fluid is forced to
accelerate vertically with g
az 
 by placing the fluid container in an
elevator, the pressure gradient in the z-direction is g
2
z
p 



 .

g
az 

h
Z
g
az 
h
Z
1
p 1
p
1
2 p
p  h
g
2
p
p 1
2 


Density  Density 
Free Fall
of a Liquid az = - g
Upward Acceleration
of a Liquid az = + g
Free Fall of a Fluid Body
Fig. ( ) - The Effect of Acceleration on the Pressure of a Liquid
During Free Fall and Upward Acceleration
Fig. (2) – The Effect of Acceleration on the Pressure of a
Fluid during Free Fall and Upward Acceleration.
Therefore, the pressure difference across a fluid layer now doubles
relative to the stationary fluid case.
az  0 az  0
 g h  g h
h
Fig. ( ) - Variation of the Magnitude of Pressure with the Variation of
az Variation.
Fig. (3) – Variation of the Magnitude of Pressure with the Variation of az
.
Application (iii) – Rigid-Body Acceleration
Many fluids such as water, gasoline and milk are transported in tankers.
In an accelerated tanker, the fluid rushes to the tank, and some initial
splashing occurs but then a new free surface (usually horizontal) is

formed, each fluid particle assumes the same acceleration, and the
entire fluid moves like a rigid body. Rigid body-motion of a fluid also
occurs when the fluid is contained in a tank rotates about an axis.
We consider an open container of a liquid that is translating along a
straight path with a constant acceleration “a” as illustrated in Fig (4).
Since 0
ax  it follows from Eqs .(2) that the pressure gradient in the x-
direction is zero ( 0
x
p 

 ) . In the y- and z-directions
y
a
y
p





, and )
g
a
(
z
p
z 





……. (6)
The change in pressure between two closely spaced points located at y,
z and dy
y  , dz
z  can be expressed as
dz
z
p
dy
y
p
dp





 ……..…. (7)
or in terms of the results from Eqs.(6) and (7)
dz
)
g
a
(
dy
a
dp z
y 




 ……..…. (8)
Along a line of constant pressure, dp = 0, and therefore from Eq. (8) it
follows that the slope of this line is given by the relation





 tan
g
a
a
y
d
z
d
z
x
……..…. (9)

g
aZ 
y
a
.
const
p1
.
const
p2
.
const
p3
Free surface slope = dz/dy
y
x
z
y
a
z
a a
Fig. (4) – Lines of Constant Pressure (which are the Projections of
the Surfaces of Constant Pressure on the y - z plane) in a
Linearly Accelerating Liquid, and the Vertical Rise.
Application (iv) –Rigid-Body Rotation
As a second case, consider rotation of the fluid about the z axis without
any translation as sketched in Fig. (5). We know from experience that
when a glass filled with water is rotated about its axis, the fluid is forced
outward as a result of the so-called centrifugal force, and the free
surface of the liquid becomes concave. This is known as the forced
vortex motion.
Consider a vertical cylindrical container partially filled with a liquid. The
container is now rotated about its axis at a constant angular velocity of
, as shown in Fig. (5). After initial transients, the liquid will move as a
rigid-body together with the container.

2
r r
a 

Still Water Level

r
Axis of rotation
R
Fig. ( ) – Rigid-body Rotation of a Liquid in a Tank.
z
Fig. (5) – Rigid-body Rotation of a Liquid in a Tank.
This problem is best analyzed in cylindrical coordinates (r, , z), with z
taken along the centerline of the container directed from the bottom
towards the free surface, since the shape of the container is a cylinder,
and the fluid particles undergo a circular motion.
 The centripetal acceleration of a fluid particle rotating with a
constant angular velocity  at a distance r from the axis of
rotation is 2
r  and is directed radially towards the axis of
rotation (negative r- direction). That is, 2
r r
a 


 There is a symmetry about the z-axis, which is the axis of
rotation, and thus there is no  dependence.
 Then p = p ( r, z ) and 0
a 
 . Also, since there is no motion in
the z-direction, 0
az 

Then the equation of motion for rotating fluids (Eq. ) reduce to
2
r
r
p





, 0
p




, g
z
p





……..…. (10)
Then the total difference of )
,
r
(
p
p 
 , which is
dz
z
p
dr
r
p
dp 




















 ……..…. (11)
becomes,
dz
g
dr
r
dp 2





 ……..…. (12)
Along a surface of constant pressure, such as the free surface, dp = 0,
so that from Eq. (12)
g
2
r
r
d
z
d 2



r
.
const
p1
.
const
p2
.
const
p3
z
.
const
p4
C
o
n
s
t
a
n
t
p
r
e
s
s
u
r
e
l
i
n
e
s
g
2
/
r2
2

Axis of rotation
2
2
max r
g
2
z 



Fig. (6) – Surfaces of Constant Pressure in a Rotating Liquid

Integration of this result gives the equation for the surface of constant
pressure as
t
tan
cons
g
2
r
z
2
2


 ……..…. (13)
This equation reveals that these surfaces of constant pressure are
parabolic, as illustrated in Fig. (6).
Integration of Eq. (12) yields


 




 z
d
g
dr
r
p
d 2
or
t
tan
cons
z
g
2
r
p
2
2





 ……..…. (14)
where the constant of integration can be expressed in terms of a
specified pressure at some arbitrary point ro, zo. This result shows that
the pressure varies with the distance from the axis of rotation, but at a
fixed radius, the pressure varies hydrostatically in the vertical direction
as shown in Fig. (6).
Since the volume of a parabolic is one-half the base area times its
height, as shown in Fig. (7), the still-water level is exactly halfway
between the height and low points of the free surface. The center of the
fluid drops an amount g
4
R
2
h 2
2

 and the edges raise an equal
amount.

h
R
2
Volume 2


g
2
R
h
2
2


2
h
2
h
Still Water Level

R
Axis of rotation
Fig. (7) – Determining the free-surface Position for Rotation
of a Cylinder of Fluid about its Central Axis.

Fluid Masses subjected to Accelerations
ax
az
z
x
dz
dx
g
z
P
Fz 












 
dz
dx
x
P
Fx 












Fig. ( )-
Applying Newton's second law to the fluid element of Fig. (1) which is
being accelerated in such a way that its components of acceleration are
x
a and z
a . The summation of force components on such an element is
dz
dx
x
P
Fx 













…………..(1)
dz
dx
g
y
P
Fy 















With the mass of the element equal to ( dz
dx
 ) the component forms of
Newton's second law may be written:

dz
dx
a
dz
dx
x
P
x













………………………………..…..(2)
dz
dx
a
dz
dx
g
y
P
y














Eq. (2) in component form, based on rectangular coordinates with the
positive z axis being vertically upward, can be expressed as
x
a
x
P





……………………………………………………………..(3)
)
g
a
(
z
P
z 





These equations characterize the pressure variation through an
accelerated rigid-body of fluid and with them specific application may be
studied.
The change in pressure between two closely spaced points located at x,
z and x+dx, z+dz can be expressed as
dz
y
P
dx
x
P
dp





 ………….. (4)
and substituting the above expression for x
P 
 and z
P 
 , we
obtain
dz
)
g
a
(
dx
a
dp z
x 




 ………….. (5)
However, along a line of constant pressure dP = 0 and hence, for such a
line

)
g
a
(
a
x
d
z
d
z
x


 ………….. (6)
a- For constant Linear Acceleration with ax = 0
Here a container of liquid is accelerated vertically upward,
0
x
P 

 , and with no change of pressure with x, eq. (5) becomes
)
g
a
(
z
P
z 





………….. (7)
Eq. (7) shows that the pressure will vary linearly with depth, but the
variation is due to the combined effects of gravity and the
externally induced acceleration, )
a
g
( z

 rather than simply
specific weight g
 .
B- For constant Linear Acceleration with ay = 0
Here the slope free surface (which is a line of constant pressure dP = 0)
is given by Eq.( )
ax
Az+g
z
x
dx
dz


Along a free surface the pressure is constant, so that for the accelerating
mass shown in Fig. ( ) the free surface will be inclined as 0
ax  . In
addition, all lines of constant pressure will be parallel to the free surface
as illustrated.
 Free surface of slope = dz/dx
P1
P2
P3
Constant
pressure
lines
ax
az
Fig. ( ) – Linear Acceleration with a Free Surface.

Linear Acceleration

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