Numerical Torque & Angular Momentum (Part - 14) - Rotational Motion, Physics, Class 11 Video Lecture

hello friends this video on system of
particles and rotational motion part 14
is brought to you by exam comm no move
you from exam let us now look at the
fourth problem it says find the
components along the XYZ axis of the
angular momentum L of a particle whose
position vector is ad with components X
Y Z and momentum is P with components P
X py and PZ show that if the particle
moves only in the XY plane the angular
momentum has only a set component so let
us look at the first part first so first
we have to calculate the components of
angular momentum that means we have to
calculate n X n violent ends it so what
do we know about angular momentum we
know that angular momentum is the cross
product of radius vector and linear
momentum now we can also write angular
momentum in terms of its components and
x and y and L is it in this way that is
Alex I cap l YJ cap plus NZ k cap right
also in determinant form we can write
this as a chain killing this will be
what is our R is nothing but X what how
do we write our ad is equal to X I cap
plus YJ cap plus Z K cap similarly how
do I write P we write it as P X I cap
plus P Y J cap + pz k cap right this is
how we write R and P so we can write n
as ijk here R cross P so for I this
would be X Y Z for T this will be px py
PZ right so from this we will get ypz
minus zpy into i cap plus z PX minus xpz
into j
Plus X P Y minus y DX in Topeka
so you would have studied all these
things in a case of your mattresses and
returning on site so now if you compare
this with equation one so what you get
for I cap you have LX here and for I cap
you have this term here therefore you
can see that comparing with equation one
what do we get we get LX is equal to Y
is Z minus Z P Y we get L y is equal to
Z px minus xpz and we get L Z is equal
to X P Y minus y P X so that's how we
solved the first part of the problem
first part asked us to calculate the
components along x y&z axis of angular
momentum n so to calculate M X ly and LZ
now the second part see is that if the
particle moves only the XY plane the
angular momentum has only a Z component
so that means the particle moves only in
XY plane the movement of the particle is
denoted by the position vector that
means the particle has an X component it
has a Y component but it does not have a
Z component note it says if that is the
case in that case the angular momentum
will have only the Z component so let us
try to look into that so we see that the
particle moves in X Y K right therefore
you have Z is equal to 0 right now if Z
is equal to 0 now what do you do in this
expression let me call this expression
as equation future in this entire
expression if you put Z is equal to 0
what happens not only Z but if your Z
becomes 0 that means your piece it also
becomes 0 because when there is no
movement along the Z plane then from the
ending the momentum come along the Z
plane okay so when if you put Z is equal
to 0 and PZ is equal to 0 in the
expressions we obtain so fact so in this
if you put Z equal to 0 P is equal to 0
again here you put Z is equal to 0 means
it is equal to 0 so you see that LX
comes out to be 0 ly come out to be 0
but NZ is the only one which will not be
0 because it neither has said nor has PZ
so therefore you see that LX is equal to
0 ly is also equal to 0 but L Z is equal
to X P Y minus y e X so that means when
it moves in XY plane
it only has a nonzero component of
angular momentum along the z axis the
problem says with two particles each of
mass m and speed V travel in opposite
directions along parallel lines
separated by a distance D so that means
we have two particles let us say this is
particle a and this is another particle
B which are traveling in opposite
direction that means if this particle is
traveling in this direction then this
particle is traveling in this direction
and on parallel lines separated by a
distance D that means this distance
between the two parallel lines is d so
that the vector angular momentum of the
two particle system is the same whatever
be the point about which the angular
momentum is taken so we have to
calculate the vector angular momentum of
this two particle system so let us
calculate the angular momentum about up
the point a itself so angular momentum
about the point a so L a will be equal
to what is angular momentum angular
momentum is given as the cross product
of the position vector
and linear momentum and what is linear
momentum it is nothing but mass into
velocity right so then while talk about
the angular momentum about point F we
will consider the mass for what we first
we will consider the angular momentum of
the object a plus the angular momentum
of the object B right so what is the
angular momentum of the object a so what
will be our in this case R will be equal
to 0 because we are considering this
point itself right because right now we
are considering this point a as the
origin so R will be equal to 0 therefore
this will be equal to 0 into NV + for
the particle B what will the I R will be
nothing but D because D is the distance
of the point of the object B from the
origin which is now assumed to be at a
so this will be equal to D into mass
into velocity so what is the magnitude
which we get we get it as MV d now let
us calculate the angular momentum about
point D so angular momentum about point
B will be equal to again F a plus L V
now in this case what will be lb l a l e
is will be the ad will be equal to d in
this case it will be B into NV plus 0
into NV because in this case for the
position vector of the point B from the
origin which is again B is 0 so here
also we get the angular momentum as mg T
now let us try to calculate the angular
momentum at any arbitrary point let us
say at this point C because the question
says that the vector angular momentum
will be the same whatever be the point
about which the angular momentum is
taken so we have taken first at Point a
now then we have taken at point B and we
found that they were same now let us
take at any some other point say Point C
so let us calculate the angular momentum
about Point C so this is
again be equal to the angular momentum
of particle a plus angular moment of
particle B so what does the distance of
this particle a it is D by 2 so this
will be D by 2 into NV again for
particle B also if this distance is D by
2 right so this will be D by 2 into NV
so here also we get the value as MV D so
therefore we see that the value of
angular momentum remains the same
whatever be the point about which the
angular momentum is taken right so with
this I hope that you got the concepts
clearly about angular velocity and sugar
momentum torque and angular acceleration
and you also understood how do we apply
them in solving problems thank you
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