Theorem of Perpendicular Axis Video Lecture | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR

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in all their previous slides while give
a calculating moment of inertia you saw
that it was not a very simple and easy
process right in order to calculate the
moment of inertia we had to assume so
many things we had to do we had to do
such big calculations at all right now
in order to calculate moment of inertia
around different axis of the same body
they are there some easier methods which
were avoided right so there were two
theorems based on which we were able to
calculate the moment of inertia easily
for different objects and different
bodies so what were the two theorems the
theorem of perpendicular axis and the
theorem of parallel axis so these are
the two theorems which were extremely
helpful in calculation of moment of
inertia so let us look at each of the
theorems one by one in detail so let us
first talk about the theorem of
perpendicular axis this theorem states
that this is applicable only to planar
bottles now water cleaner bodies planar
bodies as the name suggests planar that
means plane bodies which are flat with
negligible thickness like for example
little suppose if you have a circular
plate with this much thickness so you
can consider this as a flat body or a
planar body on the other hand if you
have a circular plate with this much
thickness you cannot consider it as a
planar body so this theorem of
perpendicular axis is applicable to
planar bodies whose thickness are
negligibly small so this theorem states
that the moment of inertia of a planar
body about an axis perpendicular to its
plane is equal to the moment the sum of
the moment of inertia about two
perpendicular axis concurrent with
perpendicular axis and line
name of the bottle so this theorem
states that the moment of inertia of a
planar body about an axis perpendicular
to explain let us suppose this is some
planar body so the moment of inertia
about an axis perpendicular to explain
let us say about this axis which is
perpendicular to the plane of the body
is the sum of the moment of inertia
about two perpendicular axis concurrent
with perpendicular axis and lying in the
plane of the body so that means this is
the perpendicular axis which is this is
the axis which is perpendicular to the
plane of the body and these two are the
perpendicular axis which lie in the
plane of the body so this theorem states
that I is e is equal to I X plus I Y
that means moment of inertia about the
axis which is perpendicular to the plane
of the body is equal to the sum of the
moment of inertia of along the axis
which is perpendicular to the plane of
the body which is which are
perpendicular to this axis Z but they
lie on the plane of the body right so
here if you see this is my planar body
now x and y are the two axis which lie
in the plane of the body so moment of
inertia along the z axis I said is equal
to I X plus iy now let us look at the
proof of this theorem how do we prove
that this theorem is correct so let us
consider any clean-air body like this
let us say this is the z axis this is
the x axis and this is the y axis and
this is the origin so let us consider
any point inside this rigid body so let
us consider any point P right so how do
I calculate the moment of inertia about
the z axis so the moment of inertia will
be
summation of m-i r-i square so our I
would be nothing but the distance of
this point P from the z-axis right that
is how it will be that is suppose if
this is the y axis so the position
vector of this point P will be given in
this way so this will be equal to o P
square M I opt ed because P a denotes
its distance from y-axis P B denotes its
distance from x axis and P o denotes
this distance from z axis so I said will
be 20 summation of m-i ot square so this
will be equal to summation of m-i now
what is okey if you look at this
triangle o P is nothing but OB plus BB
right so this can be written as OB
square plus P V Square this OB can be
written as PA OB and PA are the same
because they are the opposite sides of
the parallelogram so we can write it as
e is where plus P V square so this can
be written as summation of m-i P is
where plus summation of m-i PD square
now what is summation of m-i P is where
PE is the distance from the y axis so
this becomes iy and what is this
summation of m-i PB square TB is the
distance from the x axis so this is V
this becomes I X therefore I said
becomes equal to I X plus I Y so the
moment of inertia about an axis which is
perpendicular to the plane of the planar
body is equal to the sum of the moment
of inertia about perpendicular axis
which line is the plane of the body so
let us look at and
example of this theorem of perpendicular
axis so again suppose we want to
calculate the moment of inertia of a
uniform ring about its diameter so if
you remember while we the calculating
moment of inertia's we calculated the
moment of inertia of a uniform name
about an axis passing through the center
of the Ring right there is a difference
between the D when I say about an axis
passing through the center it means this
way but when I say through the diameter
of the Ring it means this is the
diameter right so that is the difference
between the two so now let us suppose
that this is x axis and this one is the
z axis and this is the y axis right so
moment of inertia of the Ring about the
z axis is nothing but the moment of
inertia about an axis passing through
the center of the Ring so that we had
calculated in our previous slides as M
r-square now from this theorem so as for
the theorem Issac is equal to IX plus iy
now since the ring is uniform therefore
all the diameters are equal since ring
is uniform
therefore the diameters are all equal as
a result I X will be equal to I by
therefore we can say that I said is
equal to u I X or we can say I X is
equal to Isaac divided by 2 which is
equal to M R square divided by 2
therefore the moment of inertia of the
Ring about the diameter is equal to M R
square by P so that is why we told that
the value of the moment of inertia
changes with the value of the fixed
changes as the axis of rotation changes
so the moment of inertia of the ring
about an axis passing through the center
of the ring is M R square whereas the
moment of inertia
of the same ring about an axis passing
through the diameter of the ring is mr
square by two thank you please visit
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