Basic Mechanical Engineering topic - moment of inertia of plane figures along with examples

1. MOMENT OF
INERTIA OF PLANE
FIGURES
Submitted to :
Dr.Biranchi Narayana Padhi

2. MOMENT OF INERTIA
Moment of inertia is the name given to rotational inertia, the rotational analog of mass for linear
motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia
must be specified with respect to a chosen axis of rotation. For a point mass, the moment of inertia is
just the mass times the square of perpendicular distance to the rotation axis, I = mr2. That point mass
relationship becomes the basis for all other moments of inertia since any object can be built up from a
collection of point masses.

3. PARALLEL AXIS THEOREM(contd.)
Derivation
We may assume, without loss of generality, that in a Cartesian coordinate system the perpendicular distance
Between the axes lies along the x-axis and that the center of mass lies at the origin. The moment of inertia
relative to the z-axis is
Icm=∫(x2+y2)dm
The moment of inertia relative to the axis z′, which is a perpendicular distance D along the x-axis from the
centre of mass, is
I=∫((x+d)2+y2)dm
Expanding the brackets yields
I=∫(x2+y2)dm + d2∫dm
The first term is Icm and the second term becomes mD2. The integral in the final term is a multiple of the x-coordinate
of the center of mass – which is zero since the center of mass lies at the origin. So, the equation becomes:

4. PERPENDICULAR AXIS THEOREM
For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two
perpendicular axes through the same point in the plane of the object. The utility of this theorem goes beyond that of calculating
moments of strictly planar objects. It is a valuable tool in the building up of the moments of inertia of three dimensional objects such
as cylinders by breaking them up into planar disks and summing the moments of inertia of the composite disks.

5. PRODUCT OF INERTIA
The product of inertia of area A relative to the indicated XY rectangular axes is IXY = ∫ xy dA . The
product of inertia of the mass contained in volume V relative to the XY axes is IXY = ∫ xyρ dV—
similarly for IYZ and IZX.
Relative to principal axes of inertia, the product of inertia of a figure is zero. If a figure is mirror
symmetrical about a YZ plane, IZX = IXY = 0.
•Products of inertia of a body are measures of symmetry .
•If a particular plane is a plane of symmetry, then the products of inertia associated with any axis perpendicular to that plane are zero.
•For example, consider a thin laminate. The mid-plane of the laminate lies in the XY-plane so that half its thickness is above the plane
and half is below.Hence,the XY-plane is a plane of symmetry and
Ixy=Iyz=0
•Bodies of revolution have two planes of symmetry. For
the configuration shown, the XZ and YZ planes are
planes of symmetry. Hence, all products of inertia are
zero about the X, Y, and Z axes.
•Products of inertia are found either by measurement or calculation. Calculations are based on direct integration or on the “body build-up” technique. In the body build-up technique,
products of inertia of simple shapes are added (or subtracted) to estimate the products of inertia of a composite shape.
•The products of inertia of simple shapes (about their individual mass centers) are found in standard inertia tables. These values are transferred to axes through the composite mass
center using the Parallel Axes Theorem for Products of Inertia.

6. MOMENT OF INERTIA OF RECTANGULAR PLATE
y’
z
z’
x’
Using mass distribution,
The moment of inertia about XX’=Ixx’=mb2/12
The moment of inertia about YY’=Iyy’=ma2/12
Once we have Ixx’ and Iyy’ we can apply perpendicular axis theorem to get the
moment of inertia along ZZ’,i.e.
Izz’= Ixx’+ Iyy’
=(a2+b2)m/12
To get the moment of inertia along I’,we have to use parallel axis theorem,
I’=Ixx’+mb2/4
=mb2/3.
I’
‘m’ is the mass of the plate

7. MOMENT OF INERTIA OF CIRCULAR PLATE
In the figure, we can see a uniform thin disk with radius r rotating about a Z-axis passing through the centre.
As we have a thin disk, the mass is distributed all over the x and y plane. Then, we move on to establishing the relation for surface mass
density (σ) where it is defined as or said to be the mass per unit surface area. Since the disk is uniform, therefore, the surface mass
density will also be constant where;
dm=σ(dA).
Now it is time for the simplification of the area where it can be assumed the area to be made of a collection of rings that are mostly thin
in nature. The thin rings are said to be the mass increment (dm) of radius r which are at equal distance from the axis. The small area (dA)
of every ring is further expressed by the length (2πr) times the small width of the rings (dr.) It is given as;
A = πr2, dA = d(πr2) = πdr2 = 2rdr
Now, we add all the rings from a radius range of 0 to R to get the full area of the disk. The radius range that is given is the value that is
used in the integration of dr. If we put all these together then we get;
I = O∫R r2σ(πr)dr
I = 2 π σO∫R r3dr
I = 2 πσ r4 / 4 |o
R
I = 2 π (m / π r2 )(R4 / 4)
I = ½ mR2
(Using perpendicular axis theorem,the moment of inertia along its diameter comes out to be mR2/4.)

8. MOMENT OF INERTIA OF RING
First we’ll calculate the moment of inertia along AA’.
We will assume the mass of the ring to be M and radius be R.
Now we need to cut an elemental ring (dx) at the circumference of the ring. Hence, the
mass (dm) of the elemental ring will be;
dm = m / 2πR dx
Next, we calculate IAA’ = (dm) R2
Substituting the values, we get;
IAA’ = m / 2πR dx R2
Using integration;
IAA’ = m / 2π o∫2πR dx
IAA’ = mR / 2π 2πR
IAA’ = mR2
Using perpendicular axis theorem the moment of inertia along its diameter comes out to be
mR2/2.

9. MOMENT OF INERTIA OF ELLIPSE(PLANAR)
The first step involves determining the parametric equation of an ellipse.
Then, calculating the area (we have to remember that r will be integrated from
0 to semi-major axis a)
Area=pie*ab.
Calculating moment of inertia
In this case, the moment of inertia formula will be;
I = ρ ∫ (x2 + y2) dA
I = ρ o∫a
o∫2π λ r3 (cos2 θ + λ2 sin2 θ drdθ
I = ρλ a4 / 4 o∫2π (cos2 θ + λ2 sin θ 2 dθ
By symmetry we get;
I = ρλ a4 ½ 2 o∫2π (cos2 θ + λ2 sin2 θ dθ
I = ρλ a4 ½ [B(½ , 3/2) + λ2b (3/2, ½ ) ]
B = beta function.
Now if we make use of the beta-gamma relation we get;
I = ρ λab / 4 (a2 + b2)
I = M (a2 + b2) / 4

10. • Submitted by-Aurobinda Sthitapragna(B320013)