MOMENT
OF
INERTIA
BSc – I (UNIT IV)
Page |2
Contents
1) Moment of Inertia
Radius of Gyration
2) Theorem of Perpendicular Axes
Theorem of Parallel Axes
3
3
4
4
3) Moment of Inertia of Geometrical Bodies
Circular Disc
Solid Cylinder
Annular Ring
Hollow Cylinder
Solid Sphere
Thin Spherical Shell
Thick Spherical Shell
Rectangular Lamina
Rectangular Bar
Solid Cone
5
6
7
7
8
9
10
11
11
12
4) Kinetic Energy of a Body Rolling on a Horizontal Plane
13
5) Acceleration of a Body Rolling Down on an Inclined Plane
14
6) Products of Moment of Inertia
16
7) Principal Moments
17
öi¡¢no Q¾cÊ
Page |3
For translator motion, value of inertia depends only on the mass of the body. The kinetic energy in such
motion depends on the mass and linear velocity of the body.
When a body rotates about an axis, the kinetic energy of rotation is determined not only by its mass and
angular velocity but also upon the position of the axis about which it rotates and distribution of mass
about this axis.
Let us consider a body of mass
rotating about an axis
with angular velocity . All its particles
have the same angular velocity but as they are at different distances from the axis of rotation, their linear
velocities are different. Hence, we get
where, is the moment of inertia of the body.
Radius of Gyration
If the entire mass of the body is supposed to be concentrated at a point such that the kinetic energy of
rotation is the same as that of the body itself, then the distance of that point from the axis of rotation is
called the radius of gyration of the body about that axis.
where,
is the number of particles each of mass
into which the given mass
is divided.
The dimension of moment of inertia is
and its unit is
.
The dimension of radius of gyration is
and its unit is .
Moment of inertia is scalar because its value about a given axis remains unchanged by reversing its
direction of rotation about that axis.
Greater the moment of inertia of a body, greater is the couple required to produce a given angular
acceleration.
öi¡¢no Q¾cÊ
Page |4
The moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to
the sum of the moment of inertias of the lamina about the two axes at right angles to each other in its own
plane intersecting each other at the point where the perpendicular axis passes through it.
Let
and
be two perpendicular axes in the plane of the lamina. Let
be the mass of a particle at
point at distance from an axis through origin perpendicular to plane
.
Moment of inertia of the particle about - axis =
Moment of inertia of the particle about - axis =
If we divide the whole lamina into a number of particles of masses
at distances
from the axis. Hence, the moment of inertia about - axis and - axis are given as,
Moment of inertia of the lamina about a perpendicular axis through
origin is given as,
=
=
The moment of inertia of a body about any axis is equal to
the sum of its moment of inertias about a parallel axis
through its centre of gravity and the product of its mass and
the square of the distance between the two axes.
Let
be an axis in the plane of paper and
a parallel axis
through , the centre of mass of the body. The perpendicular
distance between the two axes is . Let be the mass of the
body and
, the mass of the element at at a distance
from
.
Moment of inertia of the element having mass
about axis
Moment of inertia of the body about axis
Let be the moment of inertia of the body about
Hence, moment of inertia of the body is given as
, an axis through . Hence,
.
is sum of the moments of all the particles about
passing through , the centre of gravity.
Since the body is balanced about the centre of mass , algebraic sum of all the moments about is zero.
Therefore,
. Hence,
öi¡¢no Q¾cÊ
Page |5
Circular Disc
a) About an axis through the centre of the disc perpendicular to its plane
Let us consider a circular disc of radius . Let us consider an elementary ring of radius and
width
. The area of the disc is given as
. Mass per unit area of the circular disc is
. Hence, mass of the elementary ring is given as,
Moment of inertia of the element about an axis through
its centre perpendicular to its plane is given as,
Hence, the moment of inertia is given as,
b) About the diameter of the circular disc
Let us consider
to be the diameter. Let
of the diameter. Therefore,
By the symmetry of the figure,
. Hence,
c) About the tangent of the circular disc
Let
be a tangent at .
By the theorem of parallel axes,
öi¡¢no Q¾cÊ
and
be the moment of inertia about the end points
Page |6
Solid Cylinder
a) About its own axis of symmetry
Let us consider a solid cylinder of mass
given as,
, length and radius . The volume of the cylinder is
. Hence, the mass density of the cylinder is
cylinder of width
coaxial cylinder is
. Now let us consider a coaxial
at distance from the axis of symmetry
. Hence, the volume of the
. Hence, the mass of the coaxial cylinder is given as,
Moment of inertia of the coaxial cylinder is,
Hence, the moment of inertia is given as,
b) About the axis passing through the centre and perpendicular to its own axis of symmetry
Let
be the axis of symmetry and
be the axis perpendicular to
. Let us consider a
circular disc of width
at a distance from
axis. Mass per unit length of the cylinder is
. Hence the mass of the disc is
. Moment of inertia of this disc about the diameter of the
rod is,
Moment of inertia of the disc about
parallel axes theorem is,
axis is given by
Hence, the moment of inertia of the cylinder is given as,
For a thin rod,
öi¡¢no Q¾cÊ
. Hence moment of inertia is given as,
Page |7
Annular Ring
a) About an axis passing through the origin and perpendicular to its plane
Let us consider a ring having inner radius and outer radius having mass . Area of the face
of the ring is
. Mass per unit area of the ring is given as,
consider a ring having radius
and
. Face area of this ring is
. Let us now
. Mass of this ring is,
Moment of inertia of this ring is given as,
Hence, the moment of inertia is given as,
b) About its diameter
Let us consider
to be the diameter. Let
of the diameter. Therefore,
By the symmetry of the figure,
and
be the moment of inertia about the end points
. Hence,
Hollow Cylinder
a) About its own axis of symmetry
Let us consider a hollow cylinder of mass
volume of the cylinder is given as,
, length and inner radius and outer radius . The
. Hence, the mass density of the cylinder is
. Now let us consider a coaxial cylinder of width
symmetry. Hence, the volume of the coaxial cylinder is
cylinder is given as,
öi¡¢no Q¾cÊ
at distance
from the axis of
. Hence, the mass of the coaxial
Page |8
Moment of inertia of the coaxial cylinder is,
Hence, the moment of inertia is given as,
b) About the axis passing through the centre and perpendicular to its own axis of symmetry
Let
be the axis of symmetry and
be the axis perpendicular to
. Let us consider a
circular disc of width
at a distance from
axis. Mass per unit length of the cylinder is
. Hence the mass of the disc is
. Moment of inertia of this disc about the diameter of the
rod is,
Moment of inertia of the disc about
by parallel axes theorem is,
axis is given
Hence, the moment of inertia of the cylinder is given as,
Solid Sphere
a) About its diameter
Let us consider a solid sphere of radius
and mass
. The volume of the slice is
the sphere. Hence, mass of the slice is given as,
about a diameter
is given as,
öi¡¢no Q¾cÊ
. Consider a thin circular slice of radius,
. Let
be the mass per unit volume of
. Moment of inertia of this slice
Page |9
Hence, the moment of inertia of the disc about its diameter is,
But mass of the sphere is
. Hence, the above
expression can be written as,
b) About a tangent
Let
be a tangent at .
By the theorem of parallel axes,
Thin Spherical Shell
About its diameter
Let us consider a thin spherical shell of radius
is given as,
and mass
. The mass per unit area of the shell
. Let us consider a thin element of shell bounded by two parallel planes
&
at and
. Let the radius of the shell be
and its thickness be
.
The area of the thin element
By the geometry of the figure,
Differentiating we get,
And,
Area of the thin element
Mass of the thin element
öi¡¢no Q¾cÊ
and
.
P a g e | 10
Hence, the moment of inertia about the diameter,
The moment of inertia about the diameter of the spherical shell is given as,
Thick Spherical Shell
About its diameter
Let us consider a thick spherical shell of inner radius , outer radius
volume of the shell is
. The total
. Hence, the mass per unit volume of the shell is
. Let us consider a concentric shell of thickness
centre. Hence, the mass of this concentric shell is,
Moment of inertia of this thin shell about the diameter
is given as,
Moment of inertia of the hollow sphere about its diameter is,
öi¡¢no Q¾cÊ
and mass
at a distance
from the
P a g e | 11
Rectangular Lamina
About an axis perpendicular to its plane and passing through its centre of gravity
Let us consider a rectangular lamina of mass
having length and breadth . The surface area
of the lamina is
strip of width
. Hence, the mass per unit area of the lamina is
at a distance
. Let us consider a thin
from the centre of gravity . The area of the strip is ,
The mass of the strip is given by,
The moment of inertia of the strip about an
axis
parallel to side is given by,
The moment of inertia of the lamina is given as,
The moment of inertia of the lamina about an axis parallel to side
is given by,
Applying perpendicular axes theorem, we find the moment of inertia of the lamina about an axis
perpendicular to its plane and passing through its centre of gravity as,
Rectangular Bar
a) About an axis perpendicular to its plane and passing through its centre of gravity
Let us consider a bar having length and breadth having mass . Let us consider the bar to be
made up of laminas placed one above the other. Hence the moment of inertia of the bar about an
axis perpendicular to its plane and passing through its centre of gravity as,
öi¡¢no Q¾cÊ
P a g e | 12
b) About an axis perpendicular to its length and passing through one of its edges
The moment of inertia of the bar through its centre of gravity is . Let us find the moment of
inertia at the edge . By the geometry of the figure,
By using parallel axes theorem we can find the
moment of inertia of the bar about an axis
perpendicular to its length and passing through
one of its edges as,
For a square bar,
. Hence, the moment of inertia is given as,
Solid Cone
a) About its vertical axis
Let us consider a solid cone having mass
volume of the cone is given as,
us consider a small circular disc of radius
thickness . Hence, volume of the disc is,
, base radius
. Let
be the semi vertical angle of the cone. Let
at a distance
According to the geometry of the figure we have,
Hence, mass of the disc is given by,
Hence, moment of inertia of the disc about the vertical axis
perpendicular to its plane is given as,
öi¡¢no Q¾cÊ
and height . The mass per unit
from the vertex. Let the disc have a
and
P a g e | 13
Hence, the moment of inertia of the solid cone about its vertical axis is given by,
b) About an axis through its vertex and parallel to the base
Let us consider an axis
through the vertex parallel to the base. Let us consider a small circular
disc of radius at a distance from the vertex. Let the disc have a thickness
. Hence, the
moment of inertia of this disc about its diameter is given as,
By parallel axis theorem, the moment of inertia of the disc about the axis parallel through the
vertex and parallel to the diameter of the disc is given as,
Hence, the moment of inertia of the solid cone about the axis parallel through the vertex and
parallel to the base is given as,
Let us consider a body of mass
having radius rolling on a horizontal plane. Let be the linear
velocity and be the angular velocity of the body. The linear and rotational kinetic energy are given as,
Linear kinetic energy
Rotational kinetic energy
Hence, the total kinetic energy is given as,
Total energy
öi¡¢no Q¾cÊ
P a g e | 14
Special Cases
a) Spherical Ball
For a spherical ball moment of inertia about its diameter is
. Hence,
Therefore, the total kinetic energy of the rolling spherical ball is given by,
Total Kinetic Energy
Total Kinetic Energy
b) Circular Disc
For a circular disc moment of inertia about an axis through its centre and perpendicular to its
plane is
. Hence,
Therefore, the total kinetic energy of the rolling circular disc is given by,
Total Kinetic Energy
Total Kinetic Energy
Let us consider a body of mass having radius rolling on a inclined plane. Let be the linear velocity
and be the angular velocity of the body. In one revolution, it will move a distance of
. The total
kinetic energy of the moving body is,
. In moving a distance
travelled is
. Hence the change in potential energy is
Since, there is no slipping, no energy is dissipated. Hence,
öi¡¢no Q¾cÊ
.
, the vertical distance
P a g e | 15
Differentiating the above equation with respect to , we get
Now,
is the velocity
of the body and
is the acceleration
of the body. Hence,
Special Cases
a) Solid Cylinder
For a solid cylinder, moment of inertia about an axis through its centre is
Therefore, the total acceleration of the rolling solid cylinder is given by,
Acceleration,
Acceleration,
b) Solid Sphere
For a solid sphere, moment of inertia about its diameter is
. Hence,
Therefore, the total acceleration of the rolling solid sphere is given by,
Acceleration,
Acceleration,
c) Hollow Sphere
For a hollow sphere, moment of inertia about its diameter is
. Hence,
Therefore, the total acceleration of the rolling hollow sphere is given by,
Acceleration,
Acceleration,
öi¡¢no Q¾cÊ
. Hence,
P a g e | 16
Let us consider a rigid body of mass
momentum is given by,
But,
composing of
particles. For any
. Hence,
Hence, total angular momentum of the rigid body is given by,
But,
The
. Hence, the above equation becomes,
- component of the above expression is,
Similarly equations for
and
can be written as,
Since,
, we get
Hence, the equations for angular momentum can be written as,
This can be written as,
öi¡¢no Q¾cÊ
particle, angular
P a g e | 17
where,
The diagonal elements of the matrix are called the Coefficient of Moment of Inertia.
The other elements of the matrix are called the Products of Moment of Inertia.
Principal moments can be solved by finding the eigen values of the above matrix,
One can find a set of cartesian axes for which the inertia tensor will be a diagonal tensor. The axes are
called as the Principal axes and the corresponding diagonal components are called Principal moments
of inertia.
The angular momentum is given by,
We know that, torque applied to a body is the rate change of angular momentum. Hence,
Now we know that,
Hence,
can be treated as an operator. Hence, we get
If we select the axes of rotation in the body where all the products of moment of inertia vanish, i.e.
These axes where the products of moment of inertia vanish are called the Principal Axes. Hence,
where,
1)
2)
3)
4)
,
and
are called the principal moments of inertia.
Simplified Course in Mechanics and Properties of Matter – C. L. Arora
Physics For Degree Students – C. L. Arora, P. S. Hemne
Concepts of Physics Part 1 – H. C. Verma
Fundamentals of Physics – Halliday, Resnick, Walker
öi¡¢no Q¾cÊ