Moment of Momentum Theorem Video Lecture | Fluid Mechanics for Mechanical Engineering

the moment of momentum equation uses
cross products and integrals over
surfaces and volumes and can be a little
bit difficult to follow here's a little
bit simpler explanation of what's going
on if we've got covers the same physics
if we've got a control volume like this
defined by this outer surface this
dotted line is the control surface the
outer boundary of the control volume
then the difference between stuff coming
in and stuff going out has to be
balanced the momentum coming in and the
momentum going out has to be balanced by
applied forces so if something comes in
going slowly and goes out going quickly
there must have been a force applied to
accelerate it in vector terms that force
is the difference between the momentum
it has when it's coming out and the
momentum it had when it was coming in a
force vector a momentum vector defined
by the direction of the momentum the
velocity of the mass that's coming out
and the velocity of the mass that's
coming in if we separate that into two
components in a two dimensional
representation with x and y-axes then
the sum of the forces in the x-direction
the components in the X Direction times
the X Direction vector I plus the sum of
the forces in the Y Direction times the
Y Direction vector J will give us the
total force vector that was applied
similarly we can keep track of the
components of the momentum out and the
momentum in first the momentum out mass
flow times velocity component in the X
Direction times the X Direction vector I
out plus mass flow times velocity in the
Y direction ties the Y Direction vector
again for summed over all of the
elements where there's a note flow then
minus the inflow in the same way taking
into account the components when we
separate that into two separate
equations for the separate components
some of the forces in the X Direction
equal to m dot u out minus m dot u in
summation over all all of the inflows no
clothes say man why now we can do
exactly the same kind of analysis with
moments these momentum flows in and out
act just like forces they produce
resulting effects on the on the control
volume so if we take the sum of the
moments around the origin that's around
this location here and it could be
anywhere because we can pick our
coordinate system wherever we want it
some of the moments around the origin
will be equal to the sum of the forces
their magnitudes and the radius
perpendicular locating that force
relative to the origin so to make that
fairly clear if the origin is here and
this is force f1 then it's an offset
distance r1 perpendicular to the origin
so the important distance here is how
far separated the line of action of the
forces from a parallel line running
through the origin so there's our one
perpendicular times f1 the magnitude of
F 1 gives us a moment acting in a
counterclockwise direction if we apply
the right-hand rule which says if I take
my right hand like this thumb up
align my hand with the radius vector and
my fingers with the force vector if it
lines up so that it's going in the same
direction as my fingers then that's
positive that's a counterclockwise
moment if I lined it up and the force is
going in the opposite direction that
would be a clockwise moment so sum of
all the forces times the perpendicular
radii so there's f1 here's f5 and it's
perpendicular radius r5 and it's
actually acting in the clockwise
direction if I look at my right hand
rule line my hand up with the radius
vector my fingers are pointing in the
opposite direction from the force
so f1 are one perpendicular winds up
giving me a positive contribution
f-five are five perpendicular winds up
giving me a negative contribution so
this is generating a moment
counterclockwise
f5 a moment counterclockwise over on the
other side we wind up with exactly the
same sort of thing we had before m dot
times V is the momentum and it's
equivalent to a force effect on the
control volume multiply that times the
perpendicular radius to where the flow
went in or went out and balance between
what went out and what went in so if
things go out with more momentum or
further away from the origin than they
came in then they'll be in that moment
applied so when we look at m dot V
perpendicular sorry magnitude of the
velocity times R perpendicular let's
look at M - V - it should generate a
moment exactly the same sense as f1 so
r1 perpendicular and r2 perpendicular
are the same m dot two times magnitude
of V - the f1 force will offset the
thrust applied by m2 v2 so it's over
here on the opposite side right hand
rule again lining my hand up with the R
to measure the radius perpendicular out
to this line of action my fingers line
up with M 2 so it's positive I get
positive M 2 times the magnitude of the
V 2 vector times R 2 perpendicular each
of these are positive quantities so it
makes a net positive contribution on
this side and over on this side that's a
clockwise thrust tending to work against
the f1 force that we had over here and
the moment it generated so there's the
outflow now for the incoming momentum we
get a negative sign here because n is
the opposite of a
and then inside here we have to take the
sum of m dot V / V magnitude times R
perpendicular for all the inflows so
here's our little right hand rule
mnemonic if we look at M 3 m dot 3 V 3
are perpendicular three which direction
is that going to be in if we look at m
dot 3 it lines up exactly with the
origin so our perpendicular is 0
it's neither positive or negative so our
perpendicular is equal to 0 that term
makes no contribution to the moments
even though it did make a contribution
to the forces applied to the control
volume then we have this flow here going
into the control volume m dot 4v4 so
we've got m dot 4 magnitude of V 4 times
R perpendicular for there is the art
perpendicular from a line that runs
through the origin parallel to the flow
R perpendicular 4 times m dot 4
magnitude of V 4 and if we line up with
the right hand rule hand in the
direction of the radius and fingers are
in the direction of the flow so that
winds up being positive there by the
right hand rule but when we take it into
account it winds up making a negative
contribution overall as it's coming in
instead of going out so we need to keep
this negative sign to account for
inflows versus outflows we need to make
sure all of these signs here work out
according to the right-hand rule to sort
out whether we're looking at clockwise
or counterclockwise to get the signs
right and we'll get a balance like this
which gives us some information about
the moments now if we took that same
system and looked at the cross product
of the radius vector and the force
vector that's F 1 R 1 perpendicular if
we integrated over the
control surface of the entire system the
inverses out sign would be taken care of
by vom which is positive for a velocity
vector going out of the control volume
and negative for a velocity vector going
into the control volume so this is the
mass flow out through a small element of
area and this is the perpendicular
radius times the velocity vector
magnitude with its sign adjusted to
account for whether it's clockwise or
counterclockwise all taken care of by
the cross-product here and the
cross-product here so this is Munson's
equation 542 for a steady situation and
it handles the signs automatically but
mathematically it's a little more
complex