Calculate Moment of Inertia : Sphere - System of Particles & Rotational Motion Video Lecture - Class 11

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particles and rotational motion part 23
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don't move your arm exam so let us try
to calculate the moment of inertia of a
uniform hollow sphere about its diameter
so in this case we have drawn this fear
trying to show that the inside of the
sphere is empty so it is follow from
inside it is just that the outer surface
which is present here so let us say that
the mass of the sphere is Capital m and
the radius of the sphere is capital I
okay now let us assume any point p on
the surface of this sphere so the
coordinate of this point will be given
as its iy + Z I right so let us assume
that the center of the sphere is
considered as the origin so what will be
the coordinate of the center it will be
0 0 0 right and the position vector of
this point P is given as X I Y I Z I
right now let us suppose if I draw a
perpendicular to the axis of the sphere
and say I join it to point 2 so looking
at this diagram what do we see we can
say that P Q square is equal to O P
Square minus o Q square as per
Pythagoras theorem now what does Okies
where open square is nothing but X I
square plus y I squared plus Z I square
this - Oh Q square what will be oq
square it is nothing but its I square
because the y&z components will be 0 so
this X is quite an X I square will get
cancelled so this will be equal to Y I
square plus Z I square right so
therefore we can see that moment of
inertia about x axis so this is my x
axis so moment of inertia about x axis
say I denote it by eye X so this I H
will be equal to summation of m-i r-i
I square plus Z I square right because
the distance from the x axis is given by
P Q square right now similarly we can
write I Y will be equal to summation of
m-i Z I square plus X I square and in a
very similar fashion we can write I side
is equal to summation of m-i X I square
plus y square so therefore the total
moment of inertia of a hollow sphere
will be equal to I X plus iy plus I said
so this can be written as summation of
2m I X square plus y square plus Z I
square so this can be written as
summation of two and I X I square plus y
square del Z del square is nothing but
an I square so this is equal to 2m R
square so therefore the moment of
inertia of the sphere will be equal to I
X plus I Y plus Isaac divided by 3 so
this will be equal to 2m R square
divided by 3
so therefore the moment of inertia of a
hollow sphere is given by 2 by 3 M R
square right so here what did we do but
if it right I will be equal to this
divided by 3 by so that's because I X is
the moment of inertia about x axis
similarly this iy is the moment of
inertia about y axis and similarly about
z axis so the mean of the moment of
inertia about all the three axis will
give the net moment of inertia so meanly
the ih+
iy + n Z divided by 3 so here I have
calculated I x + iy + I said which is
equal to this therefore I is equal to 2
by 3 M r-square
so in so let us look at the last
scenario that is the scenario of a
uniforms on its sphere about its
diameter so whenever we have a solid
sphere or a solid cylinder we again
follow the same approach of considering
a small element so let us suppose that
the mass of the sphere is Capital m and
the radius of the sphere is capital I
okay so let us consider a small element
so let us consider a small element if
this is the center of the sphere we
consider a small element of thickness D
at a distance X from the center this is
at a distance X and this thickness is DX
so consider a small element of thickness
DX at a distance X from the center right
now in this case also we will calculate
the mass per unit volume of the sphere
so what will be mass per unit volume M
divided by 4 by 3 PI R cube so therefore
we can write the mass of mass of the
small element is equal to DM which will
be equal to mass per unit volume
multiplied by the volume of this small
element so what will be the volume of
the small element it will be area of the
small element which is given as 4 PI R
square that is 4 PI X square into the
thickness and the thick
vx because this small element is also a
small portion of the sphere and the
surface area of a sphere is 4 Pie X
square so we wrote it as 4 PI X square
into DX so this 4 & 4 will get cancelled
so from this we get and as T M divided
by r cube into X square DX therefore the
moment of inertia associated with this
small element will be DM into X square
therefore di will be equal to 3 M
divided by a cube into X to the power 4
DX therefore the total moment of inertia
of the solid sphere will be equal to
integration of e I that is equal to
integration of 3 m by r cube into X to
the power 4 DX so the integration will
be from 0 to I so this is equal to 3 m
divided by r cube into X to the power 5
divided by 5 from 0 to R so this is
equal to GM R square divided by 5 now we
have to calculate the moment of inertia
about the diameter right we want to
calculate the moment of inertia about
the diameter so what would be the moment
of inertia about the diameter the total
moment of inertia is equal to the mean
of the moment of inertia about all the
three diameters right so the moment of
inertia about one diameter will be equal
to 2/3 of the total moment of inertia so
this will be equal to 2 by 3 into G M R
square divided by 5 so T 3 will get
cancelled so this will be equal to 2 by
5 em R square so this is the moment of
inertia of a solid sphere thank you
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