We were
discussing the concept of area moment of inertia, Moment
of inertia of a circular section , Area
moment of inertia for rectangular section and also we have also seen the Basic
concept of radius of gyration with the help of our previous posts.

Now we are
going further to start our discussion to understand the basic concept of mass
moment of inertia with the help of this post.

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**Let us first see here the basic concept of mass moment of inertia**

Mass
moment of inertia is basically defined as the sum of second moment of mass of
individual sections about an axis.

Let us see
the following figure which indicates one lamina with entire mass M. Let us
assume that lamina, displayed here, is made with number of small elemental
masses m

_{1}, m_{2}, m_{3}, m_{4}…etc.
As
we have considered above that the total mass of the lamina is M, now we need to
determine here the mass moment of inertia for this lamina.

Where,

x

_{1}= Distance of the C.G of the mass m_{1}from OY axis
x

_{2}= Distance of the C.G of the mass m_{2}from OY axis
x

_{3}= Distance of the C.G of the mass m_{3}from OY axis
x

_{4}= Distance of the C.G of the mass m_{4}from OY axis
Similarly,
we will have

y

_{1}= Distance of the C.G of the mass m_{1}from OX axis
y

_{2}= Distance of the C.G of the mass m_{2}from OX axis
y

_{3}= Distance of the C.G of the mass m_{3}from OX axis
y

_{4}= Distance of the C.G of the mass m_{4}from OX axis
Total mass
of the lamina = Sum of all small elemental masses

A = m

_{1 }+ m_{2 }+ m_{3 }+ m_{4 }+ ----------
Moment of
mass about the OY axis will be determined by multiplying the mass with the
perpendicular distance between the C.G of mass and axis OY.

Moment of
mass about the OY axis = M .X

Let us
determine the moments of all small elemental masses about the OY axis and we
will have following equation.

Moments of
all small elemental masses about the OY axis = m

_{1}.x_{1 }+ m_{2}.x_{2}+ m_{3}.x_{3}+ m_{4}.x_{4 }+…….
Above
equation will be termed as the first moment of mass about the OY axis and it is
used to determine the centroid or centre of gravity of the mass of the lamina.

If
the first moment of mass will again multiplied with the vertical distance
between the C.G of the mass and axis OY, then we will have second moment of
mass i.e. M.X

^{2}.By taking the definition of mass moment of inertia in to consideration, we can write the equation for the sum of second moment of mass of all small elemental masses about the OY axis and we will have following equation.

Second
moments of all small elemental masses about the OY axis = m

_{1}.x_{1}^{2}_{ }+ m_{2}.x_{2}^{2}+ m_{3}.x_{3}^{2}+ m_{4}.x_{4}^{2}_{ }+...
(I

_{m})_{yy}= m_{1}.x_{1}^{2}_{ }+ m_{2}.x_{2}^{2}+ m_{3}.x_{3}^{2}+ m_{4}.x_{4}^{2}_{ }+…….
(I

_{m})_{yy}= Σ m.x^{2}^{}

Similarly,
we will determine the sum of second moment of mass of all small elemental
masses about the OX axis and we will have following equation.

(I

_{m})_{xx}= m_{1}.y_{1}^{2}_{ }+ m_{2}.y_{2}^{2}+ m_{3}.y_{3}^{2}+ m_{4}.y_{4}^{2}_{ }+…….
(I

_{m})_{xx}= Σ m.y^{2}^{}

Therefore,
we can say here that mass moment of inertia about an axis could be calculated
by taking the product of mass and the square of the perpendicular distance
between the center of gravity of the mass and that axis.

Therefore,
we have following formula for determining the mass moment of inertia around the
X axis and Y axis and it is as mentioned here.

###
*I*_{xx} =
Σ m.y^{2}*I*_{yy} =
Σ m.x^{2}

*I*

_{xx}= Σ m.y^{2}*I*

_{yy}= Σ m.x^{2}
Do you
have any suggestions or any amendment required in this post? Please write in
comment box.

###
**Reference:**

Strength
of material, By R. K. Bansal

Image
Courtesy: Google

We will
see another important topic i.e. moment of inertia of area under curve, in the
category of strength of material, in our next post.

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